35-{15-(19+5)÷3 simplify​

Answer:

  96 ft³

Step-by-step explanation:

The front face can be considered as a 5×10 rectangle with a 3×6 rectangle cut from it. Then its area is ...

  A = (5×10) -(3×6) = 32 . . . square feet

The volume of a prism of uniform cross section is ...

  V = Bh . . . . . where B is the area of the base, and h is the height

Our figure has a base area of 32 square feet and a height of 3 feet, so its volume is ...

  V = (32 ft²)(3 ft) = 96 ft³

The expression [tex]\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }[/tex] is equal to [tex]\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}[/tex].

In this question we must use properties of real algebra to simplify a given expression as most as possible. We proceed to show each step with respective reason.

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Answer:

mn(m-4n)/3

Step-by-step explanation:

sorry, I don’t have enough time to write the explanation!

Answer:

Answer:

You see the answer just character limit

Answer:

19.87 ft

Step-by-step explanation:

cross multiply

5/7.8*x/31

31*5=155

7.8*x=7.8x

155/7.8=19.87

x=19.87 ft

Answer:

Step-by-step explanation:

y + 2 = 14x

     y = 14x - 2

When x = 0 , y = 14*0 - 2 = 0-2 = -2

When x= 4  , y = 14*4 - 2 = 56 - 2 = 54

When x = 8 , y = 14*8 - 2 = 112-2 = 110

x             y

0            -2

4            54

8            110

Answer:

[tex]x=\frac{12+\sqrt{138}}{6}\:\:or\:\:x=\frac{12-\sqrt{138}}{6}[/tex]

Step-by-step explanation:

[tex]f(x)=6x^2-24x+1\\\\0=6x^2-24x+1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-24)\pm\sqrt{(-24)^2-4(6)(1)}}{2(6)}\\ \\x=\frac{24\pm\sqrt{576-24}}{12}\\\\x=\frac{24\pm2\sqrt{138}}{12}\\\\x=\frac{12\pm\sqrt{138}}{6}[/tex]

Answer: D

Step-by-step explanation: On edge 2020

Answer:

It would take five months for the total cost of the gyms to be equal.

Step-by-step explanation:

If a set of exam scores forms a symmetrical distribution, then the student mean scores is the median and the mode too.

A symmetrical distribution by definition is the situation which occurs when the values of variables appear at regular frequencies and is characterized by the mean, median, and mode all occuring at the same point in most cases.

Graphically, symmetrical distributions may appear as a normal distribution (i.e., bell curve).

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Answer:

PLSS YOUR IN SCHOOL RN

Step-by-step explanation:

Answer:

[tex]23 L= 23 000ml\\750ml\\50L = 50 000ml\\\\50000=x+23000+750\\x=50000-23000-750\\x=26250ml\\x=26.250L[/tex]

Step-by-step explanation:

Change all the values given to one unit. (Litres to millilitres to avoid working with decimals)

Let x represent the amount of water left to be poured.

Since all will be equal to the volume of the tank, equate all values to the total capacity of the tank.

Change the subject and solve.

Hope you understood.

Have a nice day. :)

Answer:

Angle DCE is a central angle so it's equal to the arc facing it, which is DE, which is given as 116. So angle DCE= 116°

Angle EDF is an angle between a tangent line, DF, and a secant line, DE. So its measure is equal to half of the arc it surrounds, which is arc DE. 116/2= 58°. Angle EDF=58°.

Answer:

C. is the correct answer.

Step-by-step explanation:

l = -0.275t + 81

or, 70 = -0.275×40+81

or, 70 = -11+81

so, 70 = 70 (which is true)

Answer:

i dont know the answer off the top of my head but im "answering" this because the more people answer a question it will pop up at the top for people to see

Step-by-step explanation:

As an example, let's invert the matrix

[tex]\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}[/tex]

We construct the augmented matrix,

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right][/tex]

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right][/tex]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right][/tex]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right][/tex]

• Multiply row 3 by 1/3 :

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]

• Multiply row 2 by 1/7 :

[tex]\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

[tex]\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]

• Multiply row 1 by -1/3 :

[tex]\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right][/tex]

So, the inverse of our matrix is

[tex]\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}[/tex]

Answer:

y = -3

Step-by-step explanation:

Desmos

The "step 2" that Malik should take in order to solve the given system of equations is to add both equations.

Given to us

Step 1: Multiply the first equation with 3, while the second equation with 4,

6x + 12 y = 60

20x - 12 y = 44

Step 2: Add both the equations,

(6x + 12y) + (20x - 12y) = 60+44

Step 3: caluculate the value of x,

x = 4

Step 4: Substitute the value of x in equation 1,

2(4) + 4y = 20

8 + 4y = 20.

4y = 12

Y = 3

Step 5: The solution is (4, 3).

Hence, the "step 2" that Malik should take in order to solve the given system of equations is to add both equations.

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Answer:

C

Step-by-step explanation:

Simple interest is the amount charged on the principal amount with a fixed rate of interest for a time period.

The balance of the last month of Jerome's account is $880

Simple interest is the amount charged on the principal amount with a fixed rate of interest for a time period. Simple interest calculated only on the principal amount.

The formula for the simple interest can be given as,

[tex]I=\dfrac{P\times r\times t}{100}[/tex]

Here, [tex]I[/tex] is the interest rate on the principal amount of [tex]P[/tex] with the rate of [tex]r[/tex] in the time period of [tex]t[/tex].

Given information-

Jerome's average balance checking account pays simple interest of 7.2% annually.

Jerome made $5.28 in interest last month.

Convert the interest rate monthly as,

[tex]r=\dfrac{7.2}{12} \\r=0.6[/tex]

Thus the monthly interest rate is 0.6 percent.

Suppose the balance of the last month is [tex]P[/tex]. Use the above formula to find the principal amount,

[tex]I=\dfrac{P\times r\times t}{100}\\5.28=\dfrac{P\times 0.6\times 1}{100}[/tex]

Solve it for [tex]P[/tex],

[tex]P=\dfrac{5.28\times100}{0.6\times1} \\P=\dfrac{528}{0.6} \\P=880[/tex]

Hence, The balance of the last month of Jerome's account is $880.

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Answer:

10 hr

Step-by-step explanation:

i divided 990 by 90 and got 10 hr

Answer:

Step-by-step explanation:

11hrs.

Source: https://www.quora.com/How-long-does-it-take-a-truck-traveling-at-a-velocity-of-90km-hr-to-travel-990-km

Answer: 14 problems worth 5 points and 15 problems worth 2 points

Step-by-step explanation:

Answer:

Juan

Step-by-step explanation:

3x8=24, but in this case, you are doing exponents. So you have to do this equation

3x3x3x3x3x3x3x3=6561

Answer:

whenever you're multiplying terms that have exponents you multiply the coefficients and add their exponents:

Step-by-step explanation:

I think the answer to your first question should be:

30a²b - 30ab²

Your second answer is good

whenever you're multiplying terms that have exponents you multiply the coefficients and add their exponents:

for example, 6a²b³c x 2abc = 6(2)(a²)(a)(b³)(b)(c)(c)

= 12a³b⁴c²

Answer:

area 12m2 i think thats supposed to be it i did this a couple of weeks ago and i think thats it

Answer:

3

Step-by-step explanation:

The  slope of the line that passes through the points (9, -5) and (-1, -9) is 2/5

The formula for calculating the slope of a line is expressed as;

Slope = y2-y1/x2-x1

Substitute the coordinates

Slope = -9+5/-1-9
Slope = -4/-10

Slope = 2/5

Hence the  slope of the line that passes through the points (9, -5) and (-1, -9) is 2/5

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Answer:

Not possible

It doesn't factorise because we can't pull out any highest common multiple.

Also, to be able to factorise a quadratic, it must be in the general form of ax^2 + bx + c.

Hope this helps!

Answer:

B

Step-by-step explanation:

B is the best choice because there is no y intercept, but C and D also show the same slope

After the store gives 20% off sale, the expression of the item at price p can be written as, p-0.2=0.8p.

Given to us

A store is having a 20% off sale.

The sale price of an item with price p is p – 0. 2p.

We know that for any item the price can be 100%(1.00) of itself, therefore, the price of the item will be p.

We know that for any item the price can be 100%(1), therefore, after giving 20%(0.2) off sale, the price will be 80%(1-0.2 = 0.8) of the original.

Thus, the expression is equivalent to

[tex](100\%)p-(20\%)p = 80\%p\\\\1p-0.2p = 0.8p\\\\p-0.2p=0.8p[/tex]

Hence, after the store gives 20% off sale, the expression of the item at price p can be written as, p-0.2=0.8p.

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