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3 years ago

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Which situation can be represented by the equation 114×6=712?

1 answer:

maria [59]3 years ago

8 0

Answer:

There are 114 students in Westville middle school. Each of them would like to bring 6 cookies to have a party. How many total cookies will be in the school if all the kids brought cookies?

712 total cookies!

114x6=712

Glad I could help!!

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How do I get X for this question?

exis [7]

Answer:

The value of x is 12

Step-by-step explanation:

To find the value of x, we need to note that the interior angles are equal to 180. We also know that angle R is equal to 180 - (8 + 6x). So we can add all of this together and set equal to 180.

180 - (8 + 6x) + 4x + 2 + 30 = 180

180 - 8 - 6x + 4x + 2 + 30 = 180

-2x + 24 = 0

-2x = -24

x = 12

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3 years ago

What is an equation of the line?

Harman [31]

Your answer is going to be B.y-2=-3/8(x+4)

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3 years ago

Find the product of the given polynomials. <br> (5x + 8 – 6r) (4 + 2r – 7)<br><br><br>

Inessa05 [86]

Answer:

-12r² + 10xr - 15x + 34r - 24

Step-by-step explanation:

1. Organize it, variables first - as well as adding constants such as 4 and -7

(5x - 6r + 8) (2r - 3)

2. Start by multiplying the 5x by (2r - 3), then -6r, followed by 8

(10xr - 15x) + (-12r^2 + 18r) + (16r - 24)

3. Simplify

10xr - 15x - 12r² + 18r + 16r - 24

4. Order by greatest to smallest factorial

-12r² + 10xr - 15x + 18r + 16r - 24

5. Combine like variables

-12r² + 10xr - 15x + 34r - 24

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3 years ago

A rectangle is removed from a right triangle to create the shaded region shown below. Find the area of the shaded region. Be sur

julsineya [31]

I hope this helps you

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3 years ago

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

3 years ago