I just want help and be done with this :C

The term "quadratic" refers to something that is related to the mathematical concept of a quadratic function or equation.

In mathematics, a quadratic function is a polynomial function of the form f(x) = ax^2 + bx + c, where a, b, and c are constants and x is the variable. The graph of a quadratic function is a parabola, which has a symmetric U-shape.

Quadratic equations can be solved using a variety of methods, including factoring, completing the square, and using the quadratic formula. They have many applications in physics, engineering, economics, and other fields.

In everyday language, the term "quadratic" is sometimes used to describe something that has a parabolic or U-shaped curve, or to refer to something that is complex or difficult to understand.

Brainliest?

Answer:

a function

Step-by-step explanation:

By answering the presented question, we may conclude that As a result, the rectangle has a surface area of 6 square metres (m).

A rectangle is a quadrilateral with four right angles in Euclidean plane geometry. It is also known as an equiangular quadrilateral since each of its angles is equal. A straight angle is another option for the parallelogram. A square has four sides that are all the same length. A rectangle-shaped quadrilateral has four 90-degree vertices and equal parallel sides. As a result, the phrase "equirectangular rectangle" is occasionally used to describe it. Due to the equal and parallel lengths of its opposite sides, a rectangle is frequently referred to as a parallelogram.

To calculate the area of a rectangle, multiply its length by its width using the formula: area = length x width.

Considering the rectangle's dimensions as follows: length = 3 m width = 2 m

To calculate the area, we may plug these numbers into the formula:

3 m x 2 m in size

Multiplying the data results in: area = 6 m2.

As a result, the rectangle has a surface area of 6 square metres (m).

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The first five terms of the sequence are: a1 = 6, a2 = 2a1 - 6 = 2(6) - 6 = 6, a3 = 2a2 - 6 = 2(6) - 6 = 6, a4 = 2a3 - 6 = 2(6) - 6 = 6 and a5 = 2a4 - 6 = 2(6) - 6 = 6

The given sequence is defined recursively, meaning that each term depends on the previous term(s) in the sequence. We are given the initial term a1 = 6, and the recurrence relation an+1 = 2an - 6.

This means that to find any term in the sequence, we can double the previous term and then subtract 6.

Using this recurrence relation, we can find the value of the second term, a2, which is equal to 2a1 - 6. Since a1 = 6, we get a2 = 2(6) - 6 = 6. Similarly, we can find the value of the third term, a3, by substituting a2 in the recurrence relation.

This gives a3 = 2a2 - 6 = 2(6) - 6 = 6. We can continue in this way to find the values of a4 and a5, which are also equal to 6.

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The mixed number equivalent of two and four-sixths is 2 2/3.

To transform the improper fraction 4/6 into a blended variety, we want to divide the numerator (4) with the aid of the denominator (6). The quotient is zero with a remainder of 4. Because of this 4/6 is equal to 0 and 4/6 or in reality 2/3.

Now we will add this fraction to the entire variety 2 to get the blended range equal to 2 and 4/6. To do this, we sincerely write the entire number (2) accompanied by using the fraction (2/3).

Consequently, the blended variety equal of two and 4/6 is 2 2/3. We also can take a look at this answer by converting the blended wide variety lower back to a flawed fraction.

To do this, we multiply the complete number (2) with the aid of the denominator (3) and add the numerator (2) to get 8/3. That is equal to the improper fractions 2 and 4/6, confirming that our answer of 2 2/3 is accurate.

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Answer:

y = (1/4)x + 7/4

Step-by-step explanation:

To find the linearization of f(x) = √(x+3) at a=1, we need to follow these steps:

Find the first derivative of f(x) with respect to x:

f'(x) = 1 / (2√(x+3))

Evaluate f(1) to find the y-coordinate of the point where we want to find the linearization:

f(1) = √4 = 2

Evaluate f'(1) to find the slope of the tangent line at the point (1, f(1)):

f'(1) = 1 / (2√4) = 1/4

Use the point-slope form of the equation of a line to write the equation of the tangent line at (1, 2):

y - 2 = (1/4)(x - 1)

Simplify the equation of the tangent line:

y = (1/4)x + 7/4

This is the linearization of f(x) = √(x+3) at a=1.

The number of times that a flip comes up heads followed immediately by a flip that is tailsis  10/4.

To solve this problem, we can use the linearity of expectation. Let's consider the first two flips. The probability that they are HT is 1/4, and if they are, then X = 1. If they are not, then we ignore them and consider the next two flips, and so on. So, the expected value of X can be calculated as:

E[X] = (probability of getting HT on the first two flips) * 1 +

(probability of not getting HT on the first two flips) * (expected value of X for the remaining flips)

The probability of getting HT on the first two flips is 1/4, as mentioned earlier. The probability of not getting HT on the first two flips is 3/4. For the remaining eight flips, the situation is exactly the same as the original problem, except that we now have eight flips instead of ten. So, the expected value of X for the remaining flips is E[X] * 8/10.

Putting it all together, we get:

E[X] = (1/4) * 1 + (3/4) * (E[X] * 8/10)

= 1/4 + 6/10 * E[X]

Solving for E[X], we get:

E[X] = 10/4

Therefore, the answer is A) 10/4.

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The line's equation can be expressed as “y = 5x + b".

An array of integer coordinates is evident as the line passes through certain points, characterized by ordered pairs such as (0, 0), (1, 5) and so on until reaching (16, 80).

The slope can be determined by using the formula "m = (y2 - y1) / (x2 - x1)", with the points (x1, y1) = (0, 0), given that no lawns were mowed without spending time on them, and (x2, y2) = (16, 80), as specified in the problem. The resulting m value is found to be equal to 5 upon computation through the equation “m = (80 - 0) / (16 - 0)”. Therefore, the line's equation can be expressed as “y = 5x + b".

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To find the transitive closure of the relation R = {(1,2), (1,4), (2,3), (3,1), (4,2)} on the set (1, 2, 3, 4)
The transitive closure of R is:
{(1,2), (1,4), (2,3), (3,1), (4,2), (1,3), (2,1), (3,2), (3,4), (4,3)}

1. Begin with the original relation R.
2. For each pair of ordered pairs in R, if the second element of the first pair matches the first element of the second pair, add a new ordered pair with the first element of the first pair and the second element of the second pair.
3. Continue until no more new ordered pairs can be added.

Applying these steps, we get:

- From (1,2) and (2,3), we add (1,3).
- From (1,4) and (4,2), we add (1,2) (which is already in R).
- From (2,3) and (3,1),we add (2,1).
- From (3,1) and (1,2), we add (3,2).
- From (3,1) and (1,4), we add (3,4).
- From (4,2) and (2,3), we add (4,3).

The transitive closure of R is:
{(1,2), (1,4), (2,3), (3,1), (4,2), (1,3), (2,1), (3,2), (3,4), (4,3)}

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48 kids only passed mathematics in the first year of the school, according to our calculations.

A mathematical statement known as an equation combines two expressions with the same value. A symbol for it is typically an equals sign (=). Variables, integers, procedures, and functions can all be included in an equation. Given that it makes it easier to relate many factors, it is a crucial tool for solving problems in the actual world. Equations can be used to calculate unknown values or forecast results in the future.
Let's start by abbreviating the number of pupils who passed science and mathematics separately as S and M, respectively. We can claim that 2S = M because twice as many pupils passed science as maths.
In order to determine the overall number of students who passed mathematics, sum the number of students who passed both mathematics and science as well as the number of students who passed just mathematics. M+34 provides for this.
The total number of students who passed science can then be calculated by adding the number of students who passed both mathematics and science as well as the number of students who passed science exclusively. S+34 provides this.
The number of students who failed neither subject can be calculated by deducting the total of M+34 and S+34 from the 116 students who passed at least one subject. This is given by 116 - (M+34 + S+34) = 116 - (2S+68).
Finally, substituting 2S = M, we can calculate that the number of students who passed Mathematics only is M = 116 - 68 = 48.
In conclusion, our calculations show that 48 kids only achieved success in mathematics during the first year of school.

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Answer:

The growth factor for the rabbit population can be calculated using the formula:

Growth factor = 1 + (percent increase as a decimal)

In this case, the rabbit population increases 2.3% each year, so the growth factor can be calculated as:

Growth factor = 1 + (0.023) = 1.023

Therefore, the growth factor for the rabbit population is 1.023, meaning that the population will increase by a factor of 1.023 each year.

One possible value of ∠U is 80° (to the nearest degree).

A triangle is a three-sided polygon with three vertices. The triangle's internal angle, which is 180 degrees, is constructed.

To find the possible values of ∠U, we can use the Law of Cosines:

c² = a² + b² - 2ab cos(C)

Where c is the side opposite the angle we want to find (∠U), a and b are the other two sides, and C is the angle opposite side c.

In this case, we want to find ∠U, so we'll use side u as c and sides t and s (which we don't know yet) as a and b, respectively:

u² = t² + s² - 2ts cos(U)

Substituting the given values, we get:

340² = 620² + s² - 2(620)(s)cos(U)

Simplifying:

115600 = 384400 + s² - 1240s cos(U)

Subtracting 384400 and rearranging:

s² - 1240s cos(U) + 268800 = 0

Now we can use the quadratic formula to solve for s:

s = [1240 cos(U) ± √(1240² cos²(U) - 4(1)(268800))]/(2)

Simplifying under the square root:

s = [1240 cos(U) ± √(1537600 cos²(U) - 1075200)]/(2)

s = [1240 cos(U) ± √(409600 cos²(U) + 1742400)]/(2)

s = [620 cos(U) ± √(102400 cos²(U) + 435600)]

Since s must be positive, we can discard the negative solution, and we have:

s = 620 cos(U) + √(102400 cos²(U) + 435600)

Now we can use the fact that the sum of angles in a triangle is 180° to find ∠U:

∠U = 180° - ∠T - ∠S

Since we know ∠T = 110°, we just need to find ∠S. We can use the Law of Sines to do this:

sin(S)/s = sin(T)/t

sin(S) = (s/t)sin(T)

Substituting the values we know:

sin(S) = (620 cos(U) + √(102400 cos²(U) + 435600))/620 * sin(110°)

sin(S) ≈ (1.481 cos(U) + 2.225)/6.959

Now we can use a calculator to find the arcsin of both sides to get ∠S:

∠S ≈ arcsin((1.481 cos(U) + 2.225)/6.959)

Finally, we can substitute the values we found for ∠S and ∠T into the equation we found earlier for ∠U:

∠U = 180° - 110° - arcsin((1.481 cos(U) + 2.225)/6.959)

Simplifying:

∠U = 70° - arcsin((1.481 cos(U) + 2.225)/6.959)

Now we can use trial and error or a graphing calculator to find the values of ∠U that satisfy this equation. One possible solution is:

∠U ≈ 80°

Therefore, one possible value of ∠U is 80° (to the nearest degree).

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The length of the hypotenuse of the right triangle is approximately 350.7 inches.

What is meant by hypotenuse?

The hypotenuse is the longest side of a right triangle, which is opposite the right angle and is also the side that connects the two legs.

What is meant by a right triangle?

A right triangle is a triangle where one angle measures 90 degrees (a right angle). The other two angles are acute, meaning they measure less than 90 degrees. The side opposite the right angle is the hypotenuse.

According to the given information

The area of a right triangle can be calculated as:

Area = 1/2 x base x height

We are given the area of the triangle as 10,710 square inches and the height as 612 inches. Therefore, we can solve for the base of the triangle as:

10,710 = 1/2 x base x 612

10,710 = 306 x base

base = 10,710 / 306

base = 35

Now that we know the height and base of the triangle, we can use the Pythagorean theorem to find the length of the hypotenuse (c):

c² = a² + b²

c² = 35² + 612²

c² = 122,881

c =√(122,881)

c ≈ 350.7

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Let an be a sequence, and bn, cn be test sequences and If an = o(bn) and bn = o(cn), then an = o(cn).

Let an be a sequence, and bn, cn be test sequences. Given that an = o(bn) and bn = o(cn), we can infer the following:

1. an is bounded by some constant times bn as n approaches infinity.
2. bn is bounded by some constant times cn as n approaches infinity.

Since an is bounded by bn and bn is bounded by cn, we can conclude that an must also be bounded by cn, meaning an = o(cn). In other words, an grows no faster than cn as n approaches infinity.

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The percentage of games in which at least 55 points were scored is 32.14%.

We have,

We know that the third quartile (Q3) is 55, which means that 75% of the data is below 55.

We also know that the median (Q2) is 42, which means that 50% of the data is below 42.

The interquartile range (IQR) is:

IQR = Q3 - Q1

= 55 - 39

= 16

To find the lower limit for "at least 55 points", we add 0.5 times the IQR to Q3:

Lower limit = Q3 + 0.5 × IQR = 55 + 0.5 × 16 = 63

This means that any score of 55 or higher falls into the category of "at least 55 points".

We also know that the smallest value is 36 and the largest value is 64.

So, the range of scores that fall into the category of "at least 55 points" is 55 to 64.

This is a range of 9 points out of a total range of 28 points (64 - 36).

The percentage of games in which at least 55 points were scored is:

= (9/28) × 100%

= 32.14% (rounded to two decimal places)

Thus,

The percentage of games in which at least 55 points were scored is 32.14%.

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Assuming the function has the form r(t) = (t^2 + 3, 2t - 1, t + 1), the curvature can be found as follows:

To find the curvature, κ(t), of a vector-valued function r(t), you can use the formula:

κ(t) = || r'(t) × r''(t) || / || r'(t) ||^3

First, compute the first and second derivatives of r(t):

r'(t) = (2t, 2, 1) and r''(t) = (2, 0, 0)

Next, compute the cross product r'(t) × r''(t):

r'(t) × r''(t) = (0, -2, 4)

Now, find the magnitudes:

|| r'(t) × r''(t) || = sqrt(0^2 + (-2)^2 + 4^2) = sqrt(20)
|| r'(t) || = sqrt((2t)^2 + 2^2 + 1^2) = sqrt(4t^2 + 5)

Then, compute the curvature κ(t):

κ(t) = sqrt(20) / (sqrt(4t^2 + 5))^3

This is the curvature of the given vector-valued function for any value of t.

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Answer:

The correct answer is 3/5.

Step-by-step explanation:

Let's assume that the total number of passengers on the train initially are '100x'.

Now, it is given that 1/5 of them get off at station A. So, the number of passengers who get off at station A are -

1/5*100x = 20x

Hence, the passengers left on the train will be - 100x - 20x = 80x

Now, it is given that a quarter of the remaining passengers are dropped off at station B. So, the passengers being dropped off at station B are -

1/4*80x = 20x

So, after the drop off at station B the remaining passengers in the train will be - 80x - 20x = 60x

Therefore, the fraction of passengers who remained on the train after station B will be - 60x/100x = 3/5

f(x) - g(x) ≥ 0 for all real numbers x. Let r be a rational number and let f(r)-g(r)=a. Since f and g are continuous functions, for any ε>0, there exist δ1 and δ2 such that if |x-r|<δ1, then |f(x)-f(r)|<ε/2, and if |x-r|<δ2, then |g(x)-g(r)|<ε/2.

Choose ε=|a|/2. Then we have δ1 and δ2 such that if |x-r|<δ1, then |f(x)-f(r)|<|a|/2 and if |x-r|<δ2, then |g(x)-g(r)|<|a|/2.

Now let δ=min(δ1, δ2). Then if |x-r|<δ, we have:

|f(x)-g(x)-(f(r)-g(r))| = |(f(x)-f(r)) - (g(x)-g(r))| ≤ |f(x)-f(r)| + |g(x)-g(r)| < |a|/2 + |a|/2 = |a|

Therefore, we have shown that for any rational number r, and any a=f(r)-g(r), there exists a δ such that for all x with |x-r|<δ, we have |f(x)-g(x)-(f(r)-g(r))|<|a|.

Since the set of rational numbers is dense in the set of real numbers, this implies that for any real number r and any a=f(r)-g(r), we have the same result. Therefore, we can conclude that f(x)-g(x) is continuous for all x.

Given that f: R → R and g: R → R are continuous functions, and for all rational numbers r, f(r) - g(r) ≥ 0. We need to show that f(x) - g(x) ≥ 0 for all x.

Since f and g are continuous functions, the difference function h(x) = f(x) - g(x) is also continuous. By the given condition, we know that h(r) ≥ 0 for all rational numbers r.

Now, let x be any real number. We can find a sequence of rational numbers {r_n} converging to x. Since h is continuous, we have that the limit as n approaches infinity of h(r_n) equals h(x). Since h(r_n) ≥ 0 for all rational numbers, we can conclude that h(x) ≥ 0.

Therefore, f(x) - g(x) ≥ 0 for all real numbers x.

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- Asset r1 with price q1 = 1
- Asset r2 with price q2 = 1.5
- Asset r3 with price q3 = 2
- Asset r4 is a linear combination of assets r1, r2, and r3: r4 = 4r3 - 2r2 + r1
- Our goal is to compute the no-arbitrage price of asset r4, denoted as q4.

Step 1: Determine the price of r4 using the prices of r1, r2, and r3.
Since r4 is a linear combination of r1, r2, and r3, we can calculate the price of r4 (q4) using the given prices of the other assets:
q4 = q1 + (-2)q2 + 4q3

Step 2: Substitute the given prices and calculate q4.
Now, we'll substitute the given prices for q1, q2, and q3 into the equation:
q4 = 1 + (-2)(1.5) + 4(2)

Step 3: Solve for q4.
q4 = 1 - 3 + 8
q4 = -2 + 8
q4 = 6

The no-arbitrage price of asset r4 is 6.

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Answer: There are 6 ways to form the classes.

Step-by-step explanation: We need to distribute 80 indistinguishable students into 3 distinct classrooms such that each classroom has at most 30 students. Let the number of students in the three classrooms be x, y, and z, respectively. Then we have:

x + y + z = 80 (since the students are indistinguishable)

We want to find the number of non-negative integer solutions to this equation, subject to the condition that each variable is at most 30.

We can represent this problem using generating functions as follows:

The generating function for each variable is:

(1 + x + x^2 + ... + x^30) (since each variable can take on values from 0 to 30)

The generating function for the number of solutions is the product of the three generating functions:

(1 + x + x^2 + ... + x^30)^3

We need to find the coefficient of x^80 in this generating function, which will give us the number of ways to form the classes.

Using the binomial theorem, we can expand the generating function as follows:

(1 + x + x^2 + ... + x^30)^3 = (1 - x^31)^-3

Expanding the above using the binomial theorem, we get:

(1 - x^31)^-3 = ∑(n+2)C(2)x^(31n)

where ∑(n+2)C(2) represents the sum of the binomial coefficients (n+2)C(2) for n ranging from 0 to infinity.

To find the coefficient of x^80, we need to set n = 2, since 31n must be less than or equal to 80.

Thus, the coefficient of x^80 is (2+2)C(2) = 6.

Therefore, there are 6 ways to form the classes.

For a standard normal distribution, the probability of having a z-score less than -0.58 is approximately 0.2807. This can be found by looking up the area under the standard normal curve to the left of -0.58 using a z-table or a calculator. Rounding this to 4 decimal places gives the answer of 0.2807.

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There are 52,360 integer solutions in part (a), 27,405 integer solutions in part (b), and 4,845 integer solutions in part (c).

(a) This problem can be solved using stars and bars. We need to distribute 31 identical stars among 5 bins (corresponding to the variables x1, x2, x3, x4, and x5) with no restrictions on the number of stars in each bin. Using the stars and bars formula, the number of solutions is:

[tex]C(31+5-1, 5-1) = C(35, 4) = 52,360[/tex]

(b) To ensure that xi > 0, we can subtract 1 from each variable and then use the same method as in part (a).

Let [tex]y1 = x1 - 1, y2 = x2 - 1, y3 = x3 - 1, y4 = x4 - 1, and y5 = x5 - 1[/tex].

Then we have:

[tex]y1 + y2 + y3 + y4 + y5 = 26[/tex]

Using the stars and bars formula, the number of solutions is:

[tex]C(26+5-1, 5-1) = C(30, 4) = 27,405[/tex]

(c) To ensure that xi ≥ i, we can subtract i-1 from xi and then use the same method as in part (a).

Let [tex]z1 = x1 - 1, z2 = x2 - 2, z3 = x3 - 3, z4 = x4 - 4, and\ z5 = x5 - 5.[/tex]

Then we have:

[tex]z1 + z2 + z3 + z4 + z5 = 16[/tex]

Using the stars and bars formula, the number of solutions is:

[tex]C(16+5-1, 5-1) = C(20, 4) = 4,845[/tex]

Therefore, there are 52,360 integer solutions in part (a), 27,405 integer solutions in part (b), and 4,845 integer solutions in part (c).

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Answer:

(a) Box model for keeping track of net gain:

Win $3 with probability 2/10 (represented by +3)

Lose $2 with probability 3/10 (represented by -2)

Lose $1 with probability 5/10 (represented by -1)

Box model for keeping track of the number of winning plays:

Win with probability 2/10

Lose with probability 8/10

(b) The expected value for the number of winning plays can be calculated as:

E(X) = np = 5 * 2/10 = 1

The variance can be calculated as:

Var(X) = np(1-p) = 5 * 2/10 * 8/10 = 0.8

The standard error can be calculated as:

SE = sqrt(Var(X)/n) = sqrt(0.8/5) = 0.4

(c) Yes, it would be appropriate to use the normal approximation for the number of winning plays since the number of trials is large enough (n=5) and the probability of success (p=2/10) is not too close to 0 or 1. We can assume that the number of winning plays follows a normal distribution with mean 1 and standard deviation 0.4.

Step-by-step explanation:

The probability that more than 20 new registrations occur in 5 seconds is 0.9794

We know that the conditions for the Poisson distribution.

1) The occurrence of one event does not affect the probability another event will occur. i.e., the events are independent events.

2) The average rate i.e., the ratio events per time period is constant.

3) Two events cannot occur at the same time.

and the formula for the Poisson distribution is:

[tex]P(X = k)=\frac{e^{-\lambda}\times {\lambda}^k}{k!}[/tex]

Here, the average number of new registrations is 2.66 per second.

We need to find the probability that more than 20 new registrations occur in 5 seconds.

λ = 5 × 2.66

λ = 13.3

P(X  > 20)

= 1 - P(X = 20)

= [tex]1-\frac{e^{-13.3}\times {13.3}^{20}}{20!}[/tex]

= [tex]1-\frac{0.00000167449\times 2.9993892e+22}{2.432902e+18}[/tex]

= 1 - 0.0206

= 0.9794

Therefore, the required probability is 0.9794

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Answer:

[tex]145 \text{ cm}^2[/tex]

Step-by-step explanation:

We can represent the surface area of the composite figure as:

[tex]SA = 4(\text{area of triangle side}) + 4(\text{area of rectangle side}) + (\text{area of pyramid base})[/tex]

First, we can solve for the area of one of the triangle sides.

[tex]A_\triangle = \frac{1}{2}bh[/tex]

[tex]A_\triangle = \frac{1}{2} \cdot 5 \cdot 4[/tex]

[tex]A_\triangle = 10 \text{ cm}^2[/tex]

Next, we can solve for the area of one of the rectangle sides.

[tex]A_\square = lw[/tex]

[tex]A_\square = 5 \cdot 4[/tex]

[tex]A_\square = 20 \text{ cm}^2[/tex]

Next, we can solve for the area of the pyramid base.

[tex]A_\text{base} = lw[/tex]

[tex]A_\text{base} = 5 \cdot 5[/tex]

[tex]A_\text{base} = 25 \text{ cm}^2[/tex]

Finally, we can solve for the total surface area of the composite figure by plugging the values we just solved for into the uppermost equation.

[tex]SA = (4 \cdot A_\triangle) + (4 \cdot A_\square) + A_\text{base}[/tex]

[tex]SA = 4(10 \text{ cm}^2) + 4(20\text{ cm}^2) + 25\text{ cm}^2[/tex]

[tex]SA = 40 \text{ cm}^2 + 80\text{ cm}^2 + 25\text{ cm}^2[/tex]

[tex]\bold{SA = 145 \, \textb{ cm}^2}[/tex]

Answer:

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5 is larger than -7.4.

Hope this helps :)

Answer:

My answer is -7.4 < 5 ( less than).

We use 0 (zero) is a number to compare with those numbers .

Because you know -7.4 is less than 0 (negative always less than 0)

and 0 is less than 5 so in this conclusion -7.4 must less than 5.

Answer:

9x

Step-by-step explanation:

All we need to do is add like terms wich would be the 2s.

Next we would divided but sense -2+2 is 0 we just have x (which is 1) so our answer would be 9x

PLLLSS GIVE BRAINLIST MEAN THE WORLD

Answer:

[tex]5\sqrt{5}[/tex]
ROUNDED: 11.18

Step-by-step explanation:

Use the Pythagorean Theorem to solve for a right triangle: [tex]a^{2} +b^{2} =c^{2}[/tex]

The legs are "a" and "b".

The hypotenuse (diagonal across from the right angle) is "c".

To start, fill in the values for the formula, then simplify.
[tex]a^{2} + 10^{2} = 15^{2}[/tex]
[tex]a^{2} + (100) = (225)[/tex]
Subtract 100 from both sides to leave [tex]a^{2}[/tex] by itself:
[tex]a^{2} = 125[/tex]

Next, take the square root of each side.
[tex]\sqrt{a^{2} } = \sqrt{125}[/tex]
The ^2 and square root cancel out for "a", giving you:
[tex]a=\sqrt{125}[/tex]

Now, depending on if you need your answer in radical form or not, we simplify [tex]\sqrt{125}[/tex] to get your answer.

In radical form, simplify it to get: [tex]5\sqrt{5}[/tex]

In decimal form, take the square root of 125 and round to the needed value (likely tenths or hundredths). You should also round your answer, giving you: 11.18

As we have proved that every natural number N is congruent to the sum of its decimal digits modulo 9, by expressing N in terms of its digits modulo 9 and using the fact that 10 is congruent to 1 modulo 9.

To begin with, let's represent any natural number N in decimal notation as follows:

N = dₙ x 10ⁿ + dₙ₋₁ x 10ⁿ⁻¹ + ... + d₁ x 10 + d₀

Where dᵢ denotes the ith decimal digit from the right and n is the number of digits in N minus 1.

We can also express N in terms of its digits modulo 9 as follows:

N ≡ dₙ x 1ⁿ + dₙ₋₁ x 1ⁿ⁻¹ + ... + d₁ x 1 + d₀ (mod 9)

The reason for this is that 10 is congruent to 1 modulo 9, which means that any power of 10 is also congruent to 1 modulo 9. Therefore, we can replace 10ⁿ by 1ⁿ in the above expression without changing its congruence modulo 9.

Now, notice that each digit dᵢ can be written as a multiple of 9 plus a remainder rᵢ, such that 0 ≤ rᵢ ≤ 8. In other words:

dᵢ = 9 x qᵢ + rᵢ, where qᵢ is the quotient of dᵢ divided by 9.

Substituting this in the previous expression for N, we obtain:

N ≡ (9 x qₙ + rₙ) x 1ⁿ + (9 x qₙ₋₁ + rₙ₋₁) x 1ⁿ⁻¹ + ... + (9 x q₁ + r₁) x 1 + (9 x q₀ + r₀) (mod 9)

Expanding the products and using the fact that 9 is congruent to 0 modulo 9, we get:

N ≡ rₙ x 1ⁿ + rₙ₋₁ x 1ⁿ⁻¹ + ... + r₁ x 1 + r₀ (mod 9)

But the right-hand side of this expression is precisely the sum of the remainders of the digits of N when divided by 9, which is the same as the sum of its decimal digits modulo 9. Therefore, we have shown that N is congruent to the sum of its decimal digits modulo 9, as required.

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Complete Question:

Prove that every natural number N is congruent to the sum of its

decimal digits mod 9.

After calculating and conversion of the given vector to component form we get

for part A

−7(u • v) = 64.209

for part B

u • w = 26.353

In order to convert the vectors to component form,

u = 6(cos 60°i + sin60°j) = 6(0.5i + 0.866j) = (3i + 5.196j)

v = 4(cos 315°i + sin315°j) = 4(-0.707i + 0.707j) = (-2.828i + 2.828j)

−7(u • v) = −7[(3)(-2.828) + (5.196)(2.828)] = −7(-8.484 + 14.697)

= 64.209

In order to convert the vectors to component form,

u • w = (6)(cos60°)(-12)(cos330°) + (6)(sin60°)(-12)(sin330°)

= (-36 + 62.353) = 26.353

Hence, u • w is not equal to zero and u and w are not parallel or orthogonal.

After calculating and conversion of the given vector to component form we get

for part A

−7(u • v) = 64.209

for part B

u • w = 26.353

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