What are the coordinates of the point on the directed line segment from (-4, 2) to
(-1,-4) that partitions the segment into a ratio of 1 to 5?

The amount of X and Y that can give Adam the most utility is (αI/ Px) units of good X and (βI/ Py) units of good Y.

The Cobb-Douglas utility function represents a consumer's preferences towards two goods, X and Y. The function for Adam's preferences is given by:

U(x,y)=x α * y β, where α + β = 1. Given the price of good X is Px, the price of good Y is Py, and Adam's disposable income is I.

The total expenditure (E) for two goods will be:

E= PxX + PyY, Where X is the quantity of good X and Y is the quantity of good Y.

Adam's income constraint can be represented as:

I = PxX + PyY

We can rewrite the above expression as:

X = (I/ Px) - ((Py/Px)Y)

Thus, Adam's utility function can be written as:

U = X α * Y β

Substituting X with the expression we derived above, we get:

U = [(I/ Px) - ((Py/Px)Y)] α * Y β

To get the optimal consumption bundle, we need to maximize the utility function, which is given by:

MUx/ Px = α(Y/X)β

Muy/ Py = β(X/Y)α

Multiplying the two equations, we get:

MUx * Muy = αβ

Now, substituting the value of α + β = 1 in the above equation, we get:

MUx * Muy = α(1 - α)

Similarly, dividing the two equations, we get:

MUx / Muy = α/β

Now, we have two equations and two unknowns. We can solve them to get the values of X and Y, which maximize Adam's utility.

After solving, we get:

X = (αI / Px)

Y = (βI / Py)

Thus, the amount of X and Y that can give Adam the most utility is (αI/ Px) units of good X and (βI/ Py) units of good Y.

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The behavior of the solution x depends on the sign of the constant C. If C is positive, the solution tends to infinity. If C is negative, the solution tends to negative infinity. If C is zero, the solution remains zero.

To find the solution of the initial value problem, we will solve the differential equation and apply the given initial condition.

(a) The given initial value problem is:

6x' = 4x

To solve this, we can separate variables:

6dx/x = 4dt

Integrating both sides:

6ln|x| = 4t + C

Simplifying:

ln|x| = (2/3)t + C/6

Applying the exponential function to both sides:

|x| = e^((2/3)t + C/6)

Since |x| is the absolute value of x, we can write two cases:

x = e^((2/3)t + C/6) if x > 0

x = -e^((2/3)t + C/6) if x < 0

Combining the two cases, we have the general solution:

x = Ce^((2/3)t)

(b) To describe the behavior of the solution as t approaches infinity, let's consider the value of the exponential term e^((2/3)t). As t becomes larger and tends to infinity, the exponential term also becomes larger and tends to infinity. Therefore, the behavior of the solution is dependent on the value of the constant C.

If C is positive, the solution x = Ce^((2/3)t) will also tend to infinity as t approaches infinity. The function will grow without bound.

If C is negative, the solution x = Ce^((2/3)t) will tend to negative infinity as t approaches infinity. The function will decrease without bound.

If C is zero, the solution x = Ce^((2/3)t) will be identically zero, regardless of the value of t.

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The base of the triangle is changing at a rate of 0.2 cm/min when the altitude is 10 cm and the area is 100 cm². The equation of the tangent line to the curve ysin²x = xcos²y at the point (π/2, π/4) is y = x - π/2.

To find the rate at which the base of the triangle is changing, we can use the relationship between the base, altitude, and area of a triangle. The area of a triangle is given by the formula A = (1/2) * base * altitude. We are given that the altitude is increasing at a rate of 1 cm/min and the area is increasing at a rate of 2 cm²/min.
Using implicit differentiation, we can differentiate the formula A = (1/2) * base * altitude with respect to time t:
dA/dt = (1/2) * (d(base)/dt) * altitude + (1/2) * base * (d(altitude)/dt)
We are given that dA/dt = 2 cm²/min and d(altitude)/dt = 1 cm/min. Substituting these values and the given altitude of 10 cm and area of 100 cm², we can solve for d(base)/dt:
2 = (1/2) * (d(base)/dt) * 10 + (1/2) * base * 1
Simplifying the equation:
2 = 5 * (d(base)/dt) + 0.5 * base
Rearranging and solving for d(base)/dt:
(d(base)/dt) = (2 - 0.5 * base) / 5
Substituting the values base = 10 cm, we get:
(d(base)/dt) = (2 - 0.5 * 10) / 5 = 0.2 cm/m
Therefore, the base of the triangle is changing at a rate of 0.2 cm/min when the altitude is 10 cm and the area is 100 cm².
Now, let's find the equation of the tangent line to the curve ysin²x = xcos²y at the point (π/2, π/4). We can use implicit differentiation to find the derivative of y with respect to x:
d/dx (ysin²x) = d/dx (xcos²y)
Using the chain rule and differentiating each term, we have:
(sin²x)(dy/dx) + 2ysinxcosx = cos²y - x(2cosy)(dy/dx)
Rearranging the equation and solving for dy/dx, we get:
(dy/dx) = (cos²y - 2ysinxcosx) / (sin²x + 2xcosy)
Substituting the point (π/2, π/4), we have y = π/4 and x = π/2:
(dy/dx) = (cos²(π/4) - 2(π/4)sin(π/2)cos(π/2)) / (sin²(π/2) + 2(π/2)cos(π/4))
Simplifying the expression:
(dy/dx) = (1/2 - (π/4)) / (1 + (π/2))
(dy/dx) = (2 - π) / (2 + π)
Therefore, the equation of the tangent line to the curve ysin²x = xcos²y at the point (π/2, π/4) is y =x - π/2.

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The tangent plane to the surface at the point `(2, 2, 1)` is given by: `33(x - 2) - 31(y - 2) + (z - 1) = 0`

Given the equation of the surface as:

`xyz+z²+3x²+6y=11` and the indicated point is `(1, 1, 1)`.

The unit normal vector to the surface at the indicated point `P(x, y, z)` is given by the formula: `grad f(P)`

The gradient of the function `f(x, y, z) = xyz+z²+3x²+6y` is:

grad f(x, y, z) = (fx(x, y, z), fy(x, y, z), fz(x, y, z))

where fx(x, y, z), fy(x, y, z), and fz(x, y, z) are the partial derivatives of f with respect to x, y, and z, respectively.

fx(x, y, z) = `yz + 6x`fy(x, y, z)

= `xz + 6`fz(x, y, z)

= `xy + 2z`

Now, put the values `x = 1, y = 1, z = 1` in grad f(x, y, z) to get the gradient at point `(1, 1, 1)` as follows:

grad f(1, 1, 1) = (fx(1, 1, 1), fy(1, 1, 1), fz(1, 1, 1))

= (7, 7, 3)

The unit normal vector at point `(1, 1, 1)` is: `n = (7, 7, 3)/√83`

Let's write the given equation of the surface as: `2e^(x²−y²)+xy=5`The indicated point is `(2, 2, 1)`.

To find a normal vector at point `(2, 2, 1)`, we need to find the gradient of the function `f(x, y) = 2e^(x²−y²)+xy` and evaluate it at point `(2, 2)`.

The gradient of the function `f(x, y) = 2e^(x²−y²)+xy` is:

grad f(x, y) = (fx(x, y), fy(x, y))where fx(x, y) and fy(x, y) are the partial derivatives of f with respect to x and y, respectively.

fx(x, y) = `4xe^(x²−y²) + y`fy(x, y)

= `-2ye^(x²−y²) + x`

Now, put the values `x = 2, y = 2` in grad f(x, y) to get the gradient at point `(2, 2)` as follows:

grad f(2, 2) = (fx(2, 2), fy(2, 2))

= (33, -31)

Thus, the normal vector at point `(2, 2, 1)` is:

`n = (33, -31, 1)/√2141`

The tangent vector is obtained by taking the partial derivatives of the function with respect to x and y.

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(a) To find the average velocity of the ball for different time intervals, we need to calculate the change in height (∆y) over each interval and divide it by the duration of the interval (∆t).

(i) For ∆t = 0.5 seconds:
∆y = y(t + ∆t) - y(t) = (48(t + 0.5) - 16(t + 0.5)^2) - (48t - 16t^2)
Average velocity = ∆y / ∆t = ((48(t + 0.5) - 16(t + 0.5)^2) - (48t - 16t^2)) / 0.5 ft/s

(ii) For ∆t = 0.1 seconds:
Average velocity = ∆y / ∆t = ((48(t + 0.1) - 16(t + 0.1)^2) - (48t - 16t^2)) / 0.1 ft/s

(iii) For ∆t = 0.05 seconds:
Average velocity = ∆y / ∆t = ((48(t + 0.05) - 16(t + 0.05)^2) - (48t - 16t^2)) / 0.05 ft/s

(iv) For ∆t = 0.01 seconds:
Average velocity = ∆y / ∆t = ((48(t + 0.01) - 16(t + 0.01)^2) - (48t - 16t^2)) / 0.01 ft/s

(b) To estimate the instantaneous velocity when t = 2 seconds, we can use the results from part (a) and take the limit as ∆t approaches 0. In other words, we need to find the average velocity as ∆t approaches 0.

Using the answers from part (a), we substitute t = 2 and evaluate the expressions as ∆t approaches 0 to estimate the instantaneous velocity.

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Suppose y= 3x. We are to find dy and Δy when x=27 and Δx=0.1, From part A to approximate 3/27.1A. We are to find dy and Δy when x=27 and Δx=0.1.

We know that y = 3xTherefore dy/dx = 3Since x = 27 and Δx = 0.1Therefore, dy = 3(0.1) + 3(27) - 3(27)dy = 0.3Therefore, Δy = dyΔx= 0.3/0.1= 3Now, we are to approximate 3/27.1We know that Δy = 3 and x = 27Therefore, 3/27.1 = 3/27 + 3/270=0.11.

Therefore,  Δy=3 and 3/27.1=0.11 when x=27 and Δx=0.1B.From part A to approximate 3/27.1Let x = 27.1Then y = 3(27.1) = 81.3Therefore, 3/27.1 = Δy/y= 0.3/81.3= 0.00369∴ 3/27.1 ≈ 0.00369

Therefore, 3/27.1 ≈ 0.00369. Suppose y= x1.

We are to find dy and Δy when x=4 and Δx=0.1Since y = x1, we know that dy/dx = 1Therefore, dy = 1(0.1) + 1(4) - 1(4)dy = 0.1Therefore, Δy = dyΔx= 0.1/0.1= 1Now, we are to approximate 4.1We know that Δy = 1 and x = 4Therefore, 4.1 = x + Δx = 4 + 0.1 = 4.1∴ 4.1 = x + Δx = 4 + 0.1 = 4.1.

Therefore,  Δy=1 and 4.1=4.1 when x=4 and Δx=0.1B.

From part A to approximate 4.11Let x = 4.1Then y = 4.1  = 1.478296938×101.

Therefore, 4.1  = Δy/y= 0.1/1.478296938×101= 6.758277317×10-10

∴ 4.1  ≈ 6.758277317×10-10Therefore, 4.1  ≈ 6.758277317×10-10.

Therefore ,We have found that Δy=3 and 3/27.1=0.11 when x=27 and Δx=0.1.

We have found that Δy=1 and 4.1=4.1 when x=4 and Δx=0.1. The approximate are 3/27.1 ≈

0.00369 and 4.1  ≈ 6.758277317×10-10 respectively.

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The elements a and b of the vector are given as follows:

The coordinates of the vector are given as follows:

(x,y) = (2, -4).

The rule for a reflection over the line y = x is given as follows:

(x,y) -> (y,x).

Hence:

(-4,2).

The rule for a clockwise rotation of 3π/2 radians is given as follows:

(x,y) -> (-y,x).

Hence:

(-4,2) -> (-2, -4).

The values of a and b are given as follows:

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To find the derivative of the function f(x) = ([tex]x^{2}+4)^{279} + (\sqrt[]{x} +x)^{401}[/tex], we need to apply the chain rule. f'(x) = [tex]2x(x^{2}+4)^278(x^{2}+4) + 401(\sqrt[]{x} +x)^{(400)}(1/2\sqrt[]{x} + 1)[/tex]

Let's differentiate each term separately:

The derivative of (x²+4)^279:

[279[tex](x^{2}+4)^(279-1)] * (2x) = 2x * (x^{2}+4)^278 * (x^{2}+4)[/tex]

The derivative of [tex](\sqrt[]{x} +x)^{401}[/tex]:

[[tex]401(\sqrt[]{x} +x)^{(401-1)}] * (1/2\sqrt[]{x} + 1)[/tex]= 401 * [tex](\sqrt[]{x} +x)^{(400)}[/tex] * (1/2√x + 1)

Now, we can add the derivatives of both terms to get the derivative of f(x):

f'(x) = 2x * (x²+4)^278 * (x²+4) + 401 * [tex](\sqrt[]{x} +x)^{(400)}[/tex] * (1/2√x + 1)[tex](x²+4)^279 + (\sqrt[]{x} + x)^401[/tex]

Therefore, the derivative of f(x) is:

f'(x) = [tex]2x(x^{2}+4)^{278}(x^{2}+4) + 401(\sqrt[]{x} +x)^{(400)}(1/2\sqrt[]{x} + 1)[/tex]

Note that we used the variable x in the derivative since it was not specified to differentiate with respect to z.

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Therefore, the interest on the notes would be:(a) $70 (b) $45 (c) $92 (d) $70.

To determine the interest on the given notes, we can use the simple interest formula:

Interest = Principal * Rate * Time

(a) For $5,600 at 5% for 90 days:

Principal = $5,600, Rate = 5% (or 0.05), Time = 90 days/360 (converted to years)

Interest = $5,600 * 0.05 * (90/360) = $70

(b) For $1,280 at 9% for 5 months:

Principal = $1,280, Rate = 9% (or 0.09), Time = 5 months/12 (converted to years)

Interest = $1,280 * 0.09 * (5/12) = $45

(c) For $6,900 at 8% for 60 days:

Principal = $6,900, Rate = 8% (or 0.08), Time = 60 days/360 (converted to years)

Interest = $6,900 * 0.08 * (60/360) = $92

(d) For $2,000 at 7% for 6 months:

Principal = $2,000, Rate = 7% (or 0.07), Time = 6 months/12 (converted to years)

Interest = $2,000 * 0.07 * (6/12) = $70

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A) The slope of the secant line of a function, f., B) The slope of the tangent line of a function, f., and C) The instantaneous rate of change of a function. The derivative represents the slope of the tangent line and the secant line at a point, as well as the instantaneous rate of change of the function. Option A, B , C

The derivative represents several important concepts in calculus. Let's explore each option to determine which ones are correct.

A) The slope of the secant line of a function, f: This is true. The derivative of a function at a specific point represents the slope of the tangent line to the function at that point. By taking two points on the function and finding the slope of the line connecting them (secant line), the derivative provides the instantaneous rate of change of the function at a specific point.

B) The slope of the tangent line of a function, f: This is true. As mentioned earlier, the derivative represents the slope of the tangent line to a function at a particular point. The tangent line touches the curve at that point, and its slope is given by the derivative.

C) The instantaneous rate of change of a function: This is true. The derivative measures the rate at which the function is changing at a specific point, thus representing the instantaneous rate of change. It provides information about how the function behaves locally.

D) The average rate of change of a function: This is not true. The derivative does not represent the average rate of change of a function. Instead, it focuses on the instantaneous rate of change at a specific point.

E) The slope of a function between two points: This is not entirely accurate. While the derivative gives the slope of the function at a specific point, it does not directly represent the slope between two arbitrary points on the function.

F) The slope of a function at a single point: This is true. The derivative provides the slope of the function at a particular point, which indicates how the function behaves near that point.

Option A, B and C are correct.

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The rectangular form of the given polar equation is (x, y) = (-5 cos θ, -5 sin θ)

The given polar equation is r = -5. To convert the given polar equation to rectangular form, we will use the conversion formula, x = r cos θ and y = r sin θ.

Here, r = -5. We can substitute r = -5 in the formulae and obtain:

x = -5 cos θ

y = -5 sin θ

Thus, the rectangular form of the given polar equation is (x, y) = (-5 cos θ, -5 sin θ).

To sketch the graph of the given polar equation, we need to plot points for different values of θ. We can take some values of θ such as 0, π/4, π/2, 3π/4, π, etc, and substitute them in the above equation to obtain the corresponding values of x and y.

We can then plot these points on the graph and join them to get the required graph. Here is the graph of the given polar equation

Thus, we have converted the polar equation r = -5 to rectangular form and sketched its graph. We used the conversion formula x = r cos θ and y = r sin θ to convert the polar equation to rectangular form and plotted points for different values of θ to sketch the graph of the equation.

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The function f(x) = Ax² + 4, we can use the power rule of differentiation. Therefore, the correct answer for f'(x) is A. 2A·x.

To find the derivative of the function f(x) = Ax² + 4, we can use the power rule of differentiation, which states that the derivative of x^n with respect to x is nx^(n-1).

Applying the power rule to each term of the function f(x), we get:

f'(x) = 2Ax + 0 = 2Ax

f'(x) = d/dx(A·x² + 4)

= A·d/dx(x²) + d/dx(4)

= A·2x + 0

= 2A·x

Therefore, the correct answer for f'(x) is A. 2A·x.'

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The volume of solid obtained by rotating the region bounded by the graphs y=1/x,y=0,x=1 and x=8 about y=7 is ∞.

To calculate the volume of the solid obtained by rotating the region bounded by the graphs y=1/x,y=0,x=1 and x=8 about y=7, the following steps are needed:

Step 1: First, sketch the region bounded by the graphs y=1/x,y=0,x=1 and x=8 about y=7.

Step 2: Since we are rotating the region about y=7, we need to express the radius of rotation as a function of y by subtracting the value of y from the distance between y=7

The top function and between y=7 and the bottom function.

r(y) = (7 - 1/y) - 7 = -1/y

Step 3: Next, find the limits of integration, which will be the y-values where the functions intersect.

Setting 1/x = 0, we have x = 1 and setting 1/x = y/7, we have

x = 7/y.

So the limits of integration are 0 ≤ y ≤ 7.

Step 4: Using the formula for the volume of a solid of revolution, the volume of the solid is given by:

V = π∫[a, b]r(y)²dy

V = π∫[0,7] (-1/y)²dy

V = π∫[0,7] 1/y²dy

V = π(-1/y)|[0,7]

V = π(-1/7) + π(-1/0)

(division by zero is undefined)

V = ∞

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Part A  - The total cost of the   adjustable-rate mortgage is $483,582.34.

Part B  - The total cost   of the fixed-rate mortgage is  $ 561,486.27.

Part C  -  Advantages and disadvantages are given belwo.

Part A  -  I used a spreadsheet  to calculate the total cost of the adjustable-rate mortgage. The spreadsheet included the following columns  -

I calculated   the total payment   for each year by multiplying the monthly payment by 12.  I then totaled thepayments for each  year to get the total cost of the mortgage.

The total cost of the adjustable  rate mortgage is $483,582.34.

Part B  -  I also utilized a spreadsheet to compute the total cost of the fixed-rate mortgage. The spreadsheet included the following columns  -

Note that I calculated the  monthly payment for the fixed-rate mortgage by using the following formula  -

monthly payment = principal x  interest rate/ (1 - (1 + interest rate) ^ -term)

Thetotal cost of the fixed-rate mortgage is   $561,486.27.

Part C  -  The adjustable-rate mortgage has a lower initial monthly payment than the fixed-rate mortgage.

However, the adjustable  rate mortgage has the potential to become more expensive over time if the interest rates go up.

The advantages of the adjustable-rate mortgage are  -

Note that the disadvantages of the adjustable-rate mortgage are  -

The advantages of the fixed-rate mortgage are  -

The disadvantages of the fixed-rate mortgage are  -

Conclusievely, the best loan type for you will depend on your individual circumstances and risk tolerance.

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The lower limit of integration, a, is 0.

The upper limit of integration, b, is 10.

The volume of the solid is 1250π/3 cubic units.

Since the diameter 2r is determined by the distance x from the base, the lower limit of integration, a, is 0, and the upper limit of integration, b, is 10, representing the full range of x values along the altitude of the triangle.

Finally, the cross-sectional area A(x) of each semicircular cross-section is given by:

A(x) = (π/2) r² = (π/2) (x + 5)².

Now we can calculate the volume of the solid using the formula:

V = [tex]\int\limits^{10}_0[/tex]A(x) dx

= [tex]\int\limits^{10}_0[/tex] (π/2)  (x + 5)² dx.

= (π/2) [((1/3)(1000) + 5(100) + 250) - ((1/3)(0) + 5(0) + 25(0))]

= (π/2) [((1000/3) + 500 + 250) - 0]

= (π/2) (1000/3 + 500 + 250)

= (π/2)  (1000/3 + 1500)

= (π/2)  (2500/3)

= π (2500/6)

= 1250π/3.

Therefore, the volume of the solid is 1250π/3 cubic units.

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The complete question is attached at the end.

Therefore, the point of intersection of the two lines r1 and r2 is P(⟨-3, -10, 2⟩).

To find the vector parametrization of the line L that passes through the points (1, 3, 1) and (3, 8, 4), we can use the two-point form.

Let's denote the parameter t and find the direction vector of the line:

Direction vector d = (3, 8, 4) - (1, 3, 1) = (2, 5, 3)

Now, we can write the vector parametrization of the line L:

r(t) = (1, 3, 1) + t(2, 5, 3)

Simplifying:

r(t) = (1 + 2t, 3 + 5t, 1 + 3t)

Therefore, the vector parametrization of the line L is r(t) = ⟨1 + 2t, 3 + 5t, 1 + 3t⟩.

Now, let's find the point of intersection of the two lines r1(t) and r2(s).

Given:

r1(t) = ⟨-3, -14, 10⟩ + t⟨0, -1, 2⟩

r2(s) = ⟨-15, -10, -1⟩ + s⟨-4, 0, -1⟩

To find the point of intersection, we need to equate the x, y, and z components of the two parametric equations:

x1 + t * 0 = x2 + s * (-4)

y1 + t * (-1) = y2 + s * 0

z1 + t * 2 = z2 + s * (-1)

Solving these equations will give us the values of t and s at the point of intersection.

From the first equation:

-3 = -15 - 4s

Simplifying:

4s = 12

s = 3

Substituting s = 3 into the second equation:

-14 - t = -10

Simplifying:

t = -4

Now we can substitute the values of t and s into either of the parametric equations to find the point of intersection.

Substituting t = -4 into r1(t):

r1(-4) = ⟨-3, -14, 10⟩ + (-4)⟨0, -1, 2⟩

= ⟨-3, -14, 10⟩ + ⟨0, 4, -8⟩

= ⟨-3, -10, 2⟩

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According to the question The center of mass of the homogeneous lamina in the first quadrant, bounded by [tex]\(y = 0\), \(x = 0\)[/tex], and [tex]\(f(y) = -y^2 + 9\)[/tex], is approximately [tex]\((\bar{x}, \bar{y}) = (4.8, 0.75)\).[/tex]

To find the center of mass of a homogeneous lamina in the first quadrant, we need to calculate the coordinates of its center of mass, denoted as [tex]\((\bar{x}, \bar{y})\).[/tex]

The center of mass coordinates are given by the formulas:

[tex]\(\bar{x} = \frac{1}{A} \int\int x \, dA\)[/tex]

[tex]\(\bar{y} = \frac{1}{A} \int\int y \, dA\)[/tex]

where [tex]\(A\)[/tex] represents the area of the lamina.

In this case, the lamina is bounded by [tex]\(y = 0\), \(x = 0\), and \(f(y) = -y^2 + 9\).[/tex]

Let's calculate the coordinates of the center of mass by substituting the given values into the formulas.

The lamina is bounded by [tex]\(y = 0\)[/tex] and [tex]\(f(y) = -y^2 + 9\).[/tex]

To find the limits of [tex]\(y\)[/tex], we set [tex]\(f(y) = 0\)[/tex] and solve for [tex]\(y\)[/tex]:

[tex]\(-y^2 + 9 = 0\)\(y^2 = 9\)\(y = \pm 3\)[/tex]

Since the lamina is in the first quadrant, we consider the positive value [tex]\(y = 3\)[/tex] as the upper limit.

For \(x\), the lower limit is [tex]\(x = 0\)[/tex] and the upper limit is given by [tex]\(x = f(y) = -y^2 + 9\).[/tex]

Now, let's calculate the area of the lamina, [tex]\(A\):[/tex]

[tex]\[A = \int_{0}^{3} (-y^2 + 9) \, dx\][/tex]

[tex]\[A = \int_{0}^{3} (-y^2 + 9) \, dy\][/tex]

[tex]\[A = \left[-\frac{1}{3} y^3 + 9y\right]_{0}^{3}\][/tex]

[tex]\[A = \left(-\frac{1}{3} (3)^3 + 9(3)\right) - \left(-\frac{1}{3} (0)^3 + 9(0)\right)\][/tex]

[tex]\[A = (0 + 27) - (0 + 0)\][/tex]

[tex]\[A = 27\][/tex]

Now, let's calculate the integrals to find the coordinates of the center of mass.

[tex]\[\bar{x} = \frac{1}{A} \int_{0}^{3} \int_{0}^{-y^2 + 9} x \, dx \, dy\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \int_{0}^{3} \left[\frac{1}{2} x^2\right]_{0}^{-y^2 + 9} \, dy\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \int_{0}^{3} \frac{1}{2} ((-y^2 + 9)^2 - 0) \, dy\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \int_{0}^{3} \frac{1}{2} (y^4 - 18y^2 + 81) \, dy\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \left[\frac{1}{5} y^5 - 6y^3 + 81y\right]_{0}^{3}\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \left(\frac{1}{5} (3)^5 - 6(3)^3 + 81(3) - 0\right)\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \left(\frac{1}{5} (243) - 6(27) + 81(3)\right)\][/tex]

[tex]\[\bar{x} = \frac{1}{A} \left(48.6 - 162 + 243\right)\][/tex]

[tex]\[\bar{x} = \frac{1}{A} (129.6)\][/tex]

Similarly,

[tex]\[\bar{y} = \frac{1}{A} \int_{0}^{3} \int_{0}^{-y^2 + 9} y \, dx \, dy\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \int_{0}^{3} \left[yx\right]_{0}^{-y^2 + 9} \, dy\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \int_{0}^{3} y((-y^2 + 9) - 0) \, dy\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \int_{0}^{3} (-y^3 + 9y) \, dy\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \left[-\frac{1}{4} y^4 + \frac{9}{2} y^2\right]_{0}^{3}\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \left(-\frac{1}{4} (3)^4 + \frac{9}{2} (3)^2 - 0\right)\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \left(-\frac{1}{4} (81) + \frac{9}{2} (9)\right)\][/tex]

[tex]\[\bar{y} = \frac{1}{A} \left(-20.25 + 40.5\right)\][/tex]

[tex]\[\bar{y} = \frac{1}{A} (20.25)\][/tex]

Now, let's substitute the value of [tex]\(A = 27\)[/tex] and calculate the coordinates of the center of mass.

[tex]\[\bar{x} = \frac{1}{27} (129.6) = 4.8\][/tex]

[tex]\[\bar{y} = \frac{1}{27} (20.25) = 0.75\][/tex]

Therefore, the center of mass of the homogeneous lamina in the first quadrant, bounded by [tex]\(y = 0\), \(x = 0\)[/tex], and [tex]\(f(y) = -y^2 + 9\)[/tex], is approximately [tex]\((\bar{x}, \bar{y}) = (4.8, 0.75)\).[/tex]

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The solution using Empirical rule are:

1) μ - 3σ =  100

2) μ - 2σ = 150

3) μ - 1σ = 200

4) μ + 1σ = 300

5) μ + 2σ = 350

6) μ + 3σ = 400

The empirical rule is also referred to as the "68-95-99.7 rule" and it serves as a guideline for how data is distributed in a normal distribution.

The rule states that (approximately):

- 68% of the data points will fall within one standard deviation of the mean.

- 95% of the data points will fall within two standard deviations of the mean.

- 99.7% of the data points will fall within three standard deviations of the mean.

We are given:

Mean: μ = 250

Standard deviation: σ = 50

Thus:

1) μ - 3σ = 250 - 3(50)

= 100

2) μ - 2σ = 250 - 2(50)

= 150

3) μ - 1σ = 250 - 1(50)

= 200

4) μ + 1σ = 250 + 1(50)

= 300

5) μ + 2σ = 250 + 2(50)

= 350

6) μ + 3σ = 250 + 3(50)

= 400

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The value of the integral ∫∫∫_W f(x, y, z) dV is 1944.

The bounds of integration must first be established based on the provided area W before we can evaluate the integral _W f(x, y, z) dV.

x2 y 36, 0 x 6, x - y z, and x + y define the region W.

As a result, the integral is: _W f(x, y, z) dV = _W 18z dV.

Each variable is taken into account separately in order to establish the boundaries of integration.

The range of values for x is 0 to x 6.

For y:

The limits for y are determined by the inequality [tex]x^2[/tex] ≤ y ≤ 36.

Since we have 0 ≤ x ≤ 6, the lower limit for y is x^2, and the upper limit is 36.

For z: The inequality x - y z x + y determines the bounds for z.

We can now construct the integral:

_W f(x, y, and z) dV = [0,6] [x, 36] 18z dz dy dx [x-y, x+y].

The intended outcome will be obtained by evaluating this triple integral.

We may carry out the integration to determine the integral _W f(x, y, z) dV, where f(x, y, z) = 18z and the region W is defined by x2 y 36, 0 x 6, and x - y z x + y.

Integrating with regard to z comes first:

∫∫∫_W 18z dV = ∫[0,6] ∫[[tex]x^2[/tex],36] ∫[x-y, x+y] 18z dz dy dx.

Evaluating the innermost integral with respect to z:

∫[x-y, x+y] 18z dz = 9z² ∣[x-y, x+y]

= 9(x+y)² - 9(x-y)².

Finally, we integrate with respect to x:

∫[0,6] 9[(x+36)³/3 - (x+x^2)³/3] dx = 9[(6+36)³/3 - (6+6^2)³/3 - (0+36)³/3 + (0+0)³/3].

Evaluating this expression gives the value of the integral ∫∫∫_W f(x, y, z) dV.

∫∫∫_W f(x, y, z) dV = ∫[0, 6] ∫[[tex]x^2[/tex], 36] ∫[x-y, x+y] 18z dz dy dx

Integrating with respect to z:

∫[x-y, x+y] 18z dz = 9[tex]z^2[/tex]∣[x-y, x+y]

[tex]= 9(x+y)^2 - 9(x-y)^2[/tex]

= 18xy

Now we have:

∫∫∫_W f(x, y, z) dV = ∫[0, 6] ∫[[tex]x^2, 36] 18xy dy dx[/tex]

Integrating with respect to y:

∫[[tex]x^2, 36] 18xy dy = 9xy^2[/tex]∣[[tex]x^2[/tex], 36]

[tex]= 9x(36)^2 - 9x(x^2)^2[/tex]

= 324x - 9x^5

Finally, integrating with respect to x:

∫[0, 6] 324x - [tex]9x^5 dx = (162x^2 - (9/6)x^6)[/tex]∣[0, 6]

[tex]= (162(6)^2 - (9/6)(6)^6) - (162(0)^2 - (9/6)(0)^6)[/tex]

= 1944 - 0

= 1944

Therefore, the value of the integral ∫∫∫_W f(x, y, z) dV is 1944.

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The exact value of sin G is

A.  √10/10

The bearing from B to A is worked using SOH CAH TOA

Sin = opposite / hypotenuse - SOH

Cos = adjacent / hypotenuse - CAH

Tan = opposite / adjacent - TOA

Sin G = 4 / hypotenuse

We find the hypotenuse using Pythagoras

hypotenuse = √4² + 12²

hypotenuse = 4√10

Sin G = 4 / 4√10

Sin G = 1 / √10

Sin G = √10 / 10

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To find the points on the surface of the given closest and farthest from the origin, we can use the method of Lagrange multipliers.  The closest points to the origin on the surface are indeed the points (1, √2, 1).

Let's define the function f(x, y, z) = x^2 + y^2 + z^2, which represents the distance squared from any point (x, y, z) to the origin. We need to find the extreme values of f subject to the constraint g(x, y, z) = (x - 1)^2 + (y - √2)^2 + (z - 1)^2 - 1 = 0, which represents the given surface equation.

Using Lagrange multipliers, we introduce a Lagrange multiplier λ and set up the following system of equations:

∇f = λ∇g,

g(x, y, z) = 0.

Taking the partial derivatives of f and g with respect to x, y, and z, we have:

∂f/∂x = 2x,

∂f/∂y = 2y,

∂f/∂z = 2z,

∂g/∂x = 2(x - 1),

∂g/∂y = 2(y - √2),

∂g/∂z = 2(z - 1).

Setting up the equations:

2x = λ(2(x - 1)),

2y = λ(2(y - √2)),

2z = λ(2(z - 1)),

(x - 1)^2 + (y - √2)^2 + (z - 1)^2 - 1 = 0.

From the first three equations, we can deduce:

x = λ(x - 1),

y = λ(y - √2),

z = λ(z - 1).

Considering the cases separately:

Case 1: λ = 0

From the equations x = λ(x - 1), y = λ(y - √2), and z = λ(z - 1), we find that x = 0, y = 0, and z = 0. However, these values do not satisfy the constraint equation.

Case 2: λ ≠ 0

From the equations x = λ(x - 1), y = λ(y - √2), and z = λ(z - 1), we can solve for x, y, and z:

x/(x - 1) = y/(y - √2) = z/(z - 1) = λ.

Simplifying the expressions, we find x = 1, y = √2, and z = 1. These values satisfy the constraint equation and are the closest points to the origin on the surface.

To confirm the answer, we can solve the constraint equation using an alternate method. By rearranging the constraint equation, we have:

(x - 1)^2 + (y - √2)^2 + (z - 1)^2 = 1.

Expanding the terms, we get:

x^2 - 2x + 1 + y^2 - 2√2y + 2 + z^2 - 2z + 1 = 1.

Simplifying further, we have:

x^2 + y^2 + z^2 - 2x - 2√2y - 2z + 3 = 1.

Rearranging the terms, we obtain:

x^2 + y^2 + z^2 - 2x - 2√2y - 2z + 2=0

This equation represents the equation of a sphere with center (1, √2, 1) and radius √2. The origin lies outside this sphere, so the closest points to the origin on the surface are indeed the points (1, √2, 1).

Therefore, using both the Lagrange multipliers method and an alternate approach, we have confirmed that the points (1, √2, 1) on the surface are the closest points to the origin.

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To make the function f(x) continuous, the value of k should be 10.

In order for a function to be continuous, it needs to be defined at every point in its domain, and the left-hand and right-hand limits must be equal at any point of potential discontinuity. In this case, the function is defined as -2x² + 16 for x < 3, and it is defined as 5x - k for x ≥ 3.

To make the function continuous at x = 3, we need the left-hand limit and the right-hand limit to be equal. The left-hand limit can be found by evaluating the function -2x² + 16 as x approaches 3 from the left side, giving us -2(3)² + 16 = 10. The right-hand limit can be found by evaluating the function 5x - k as x approaches 3 from the right side, giving us 5(3) - k = 15 - k.

Setting the left-hand limit equal to the right-hand limit, we have 10 = 15 - k. Solving for k, we find k = 5.

Therefore, the value of k that makes the function f(x) continuous is k = 10. With this value of k, the function will be defined as -2x² + 16 for x < 3 and 5x - 10 for x ≥ 3, ensuring continuity at x = 3.

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(b) By solving the dual problem graphically, we find that the optimal solution occurs at the vertex (0, 2) with a minimum value of W = 10. The shadow price for resource 1 (x1 + x3 ≤ 3) in the primal problem is 10. This means that an additional unit of resource 1 would increase the objective function value by 10. The shadow price for resource 2 (x2 + 2x3 ≤ 5) in the primal problem is 0. This indicates that the optimal solution is not sensitive to changes in resource 2.

(a) To construct the dual problem, we will assign dual variables y1 and y2 to the constraints of the primal problem and convert the objective function to the dual form. The dual problem has two constraints corresponding to the resources in the primal problem. The objective function in the dual problem represents the costs or prices associated with the resources.

(b) To solve the dual problem graphically, we will plot the feasible region and find the optimal solution.

Constraint 1: y1 + y2 ≥ 2

Rearranging the inequality, we get:

y2 ≥ 2 - y1

Constraint 2: y1 + 2y2 ≥ 6

Rearranging the inequality, we get:

y2 ≥ (6 - y1) / 2

The feasible region is the intersection of the two constraint inequalities, taking into account the non-negativity constraints y1 ≥ 0 and y2 ≥ 0.

Now, let's plot the feasible region on a graph:

Starting with the y-axis, plot points (0, 2) and (0, 3) to represent the non-negativity constraint y2 ≥ 0.

Next, draw a line with a slope of -1 passing through the point (0, 2).

Draw another line with a slope of -0.5 passing through the point (0, 3).

The feasible region is the area bounded by these two lines and the y-axis.

To find the optimal solution, we need to evaluate the objective function at the vertices of the feasible region.

Vertices of the feasible region:

(0, 2)

(0, 3)

Now, substitute these vertices into the objective function:

For (0, 2):

W = 3(0) + 5(2) = 10

For (0, 3):

W = 3(0) + 5(3) = 15

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First, we need to know the expression for the motion of a simple harmonic oscillator, which is the following.  

[tex]$$x(t) = A cos(ωt + φ)$$[/tex]  Where,A= Amplitude, ω = Angular frequency, t = Time, φ = Phase angle.

Now we have the equation of the simple harmonic oscillator which is given below.  

[tex]$$x=0.5^\circ cos (0.7t)$$[/tex]

Comparing with the above expression, we can see that,

[tex]$$A=0.5^\circ$$[/tex] and ω=0.7.

Therefore, The period, T is given by the formula below.

[tex]$$T = \frac{2\pi }{\omega}$$[/tex]

Substituting the given value of ω=0.7 in the above formula, we get

[tex]$$T = \frac{2π}{0.7}$$ $$T = 8.979 \approx 9\ s$$.[/tex]

Therefore, the period of the oscillator is approximately 9 seconds.

Thus, we can conclude that the period of the simple harmonic oscillator is approximately 9 seconds.

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The gradient field of f(x, y) = (1/6)(e^(-6y))^2 is (1/3)(ex-6y)e^(-6y)i + (2/3)(ex-6y)(6)e^(-6y)j. The gradient field represents the vector field formed by taking partial derivatives of the function with respect to its  variable.

The function f(x, y) = (1/6)(e^(-6y))^2 represents a surface in three-dimensional space. The gradient field of this function describes the direction and magnitude of the steepest ascent at any point on the surface.

To find the gradient field, we calculate the partial derivatives of the function with respect to x and y. The partial derivative with respect to x, denoted as ∂f/∂x, gives us the rate of change of f in the x-direction. In this case, the partial derivative is (1/3)(ex-6y)e^(-6y)i.

Similarly, the partial derivative with respect to y, denoted as ∂f/∂y, gives us the rate of change of f in the y-direction. The partial derivative in this case is (2/3)(ex-6y)(6)e^(-6y)j.

Combining these partial derivatives, we obtain the gradient field of f(x, y) as a vector field with the x-component (1/3)(ex-6y)e^(-6y)i and the y-component (2/3)(ex-6y)(6)e^(-6y)j.

The gradient field provides valuable information about the direction and intensity of the function's change at each point, which can be useful in various applications, such as optimization problems or understanding the flow of a physical quantity.

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the estimated error involved in this approximation is approximately 25,502,498.79794309684.

To approximate the sum of the series using the sum of the first 100 terms, we can rewrite the series as follows:

Σ(1/[tex](n^3[/tex]+ 1)) ≈ Σ(1/

We can use the formula for the sum of the first n cubes, which is given by:

Σ([tex]k^3[/tex]) =[tex](n(n+1)/2)^2[/tex]

In this case, we want to find the sum of the first 100 cubes:

Σ[tex](n^3) = (100(100+1)/2)^2[/tex]

= [tex](5050)^2[/tex]

= 25,502,500

Now, let's calculate the approximation using the sum of the first 100 terms:

Approximation = Σ[tex](1/n^3)[/tex]

≈ [tex]1/1^3 + 1/2^3 + 1/3^3 + ... + 1/100^3[/tex]

To calculate this sum, we can use a calculator or a programming language:

Approximation ≈ 1.20205690316

To estimate the error involved in this approximation, we can calculate the difference between the exact sum and the approximation:

Error = Exact Sum - Approximation

Error = 25,502,500 - 1.20205690316

Error ≈ 25,502,498.79794309684

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The graph of y = x^2 + x - x can be described as a simple parabola that opens upward. The domain is all real numbers (-∞, ∞), there are no intercepts, there are no vertical or horizontal asymptotes, it increases for x > -1/2 and decreases for x < -1/2, there is no local maximum or minimum, the concavity is upward throughout the entire graph, there is no point of inflection, and it can be sketched as a U-shaped curve.

A. Domain: The equation y = x^2 + x - x is defined for all real numbers, so the domain is (-∞, ∞).
B. Intercepts: To find the x-intercepts, we set y = 0 and solve for x:
0 = x^2 + x - x
0 = x^2
x = 0
There is only one x-intercept at (0, 0). To find the y-intercept, we substitute x = 0 into the equation:
y = 0^2 + 0 - 0
y = 0
Thus, the y-intercept is at (0, 0).
C. Asymptotes: There are no vertical or horizontal asymptotes for this equation.
D. Intervals of Increase and Decrease: The coefficient of the x^2 term is positive, indicating that the parabola opens upward. Therefore, the function increases for x > -1/2 and decreases for x < -1/2.
E. Local Maximum and Minimum Values: Since the parabola opens upward, there is no local maximum or minimum.
F. Interval of Concavity: The concavity of the graph is upward throughout the entire curve.
G. Point of Inflection: There is no point of inflection.
H. Sketch: The graph can be sketched as a U-shaped curve, opening upward. The vertex of the parabola is at the minimum point, which occurs at (-1/2, -1/4). The curve passes through the origin (0, 0) and increases to the right and decreases to the left of the vertex. The graph is symmetric with respect to the vertical line x = -1/2.

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The function y = x³ + 9x² + 15x + 3 has a local minimum at (-2, -9) and no local maximum points.

To find the local maximum and minimum points, we first need to find the critical points of the function. The critical points occur where the first derivative is equal to zero or undefined. Taking the derivative of y = x³ + 9x² + 15x + 3, we have y' = 3x² + 18x + 15. Setting y' equal to zero and solving for x, we find x = -2 and x = -5 as the critical points.

Next, we apply the second derivative test to determine whether these critical points correspond to local maximum or minimum points. Taking the second derivative of y, we have y'' = 6x + 18.

Evaluating y'' at x = -2, we find y''(-2) = 6(-2) + 18 = 6, which is positive. This indicates that at x = -2, the function has a local minimum.

Since there are no other critical points, there are no other local maximum or minimum points.

Therefore, the function y = x³ + 9x² + 15x + 3 has a local minimum at (-2, -9) and no local maximum points. The point (-2, -9) represents the lowest point on the graph of the function.

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The solution of the given differential equation using Laplace transforms and partial fractions is x(t) = 1/6(-1/3 - e^-3t)

Given the differential equation,2 dt dx+6x(t)=t, x(0) = −1

Let's solve the given differential equation using Laplace transforms.

The Laplace transform of the given differential equation is:2L{dx/dt} + 6L{x(t)} = L{t}

Taking the Laplace transform of the first derivative: L{dx/dt} = sL{x} - x(0)

Here, x(0) = −1L{dx/dt} = sL{x} + 1

Substituting these values in the equation, we get:2(sL{x} + 1) + 6L{x} = L{t}2sL{x} + 2 + 6L{x} = L{t}2sL{x} + 6L{x} = L{t} - 2 ... (1)

Now, we need to find the Laplace transform of the function t and for that we use the following property:

L{tn} = n!/(sn+1)where n = 1 in this caseL{t} = 1/(s2)

Substituting this value in equation (1), we get:2sL{x} + 6L{x} = 1/(s2) - 2L{x}(2s + 6)L{x} = (1/(s2)) / (2s + 6)L{x} = (1/(2s(s + 3))) = 1/6((1/s) - (1/(s + 3)))Let's rewrite this using partial fraction decomposition:

(1/s) = A/s, (1/(s + 3)) = B/(s + 3)

Multiplying both sides by s(s + 3), we get:1 = A(s + 3) + B(s)

Solving for A and B, we get: A = -1/3 and B = 1/3

Therefore, L{x} = 1/6(-1/3 * (1/s) + 1/3 * (1/(s + 3)))

Now, let's take the inverse Laplace transform of L{x}:x(t) = 1/6(-1/3 - e^-3t)

Now, we will verify the solution for the given differential equation.

The given differential equation is:2 dx/dt + 6x(t) = t

The first derivative of x(t) is: dx/dt = (e^-3t)/2 - 1/6

Substituting these values in the differential equation, we get:2 ((e^-3t)/2 - 1/6) + 6((1/6) - (e^-3t)/3) = t

This reduces to: t = t

Therefore, the solution satisfies the given differential equation.

Now, we will verify the solution for the initial condition, x(0) = −1.x(0) = 1/6(-1/3 - e^-0) = -1

The initial condition is also satisfied.

Therefore, the solution of the given differential equation using Laplace transforms and partial fractions is x(t) = 1/6(-1/3 - e^-3t).

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The linearization L(x) of the function at [tex]f(x)=\sqrt(x)[/tex],a=25 is 5.1.

The linearization of f at a is L(x) = f(a) + f'(a)(x-a)

Given function is  [tex]f(x)=\sqrt(x)[/tex] ,a=25

Use a=25 and x=26 to approx. √26, √25=5

f '(x)= 1/2√x

f'(25)=1/10

L(x)=√25 + 1/10(26-25)

L(x)=5 + (1/10)= 5.1

Therefore,  The linearization L(x) of the function at [tex]f(x)=\sqrt(x)[/tex],a=25 is 5.1.

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Complete Question:

Find the linearization L(x) of the function at a. f(x)=sqrt(x),a=25 nd use it to find an approximation for √26.