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2 years ago

10

A 10 pound bag of grapes costs $14.00. How many pounds of grapes can you get for $7.00?

1 answer:

GarryVolchara [31]2 years ago

6 0

5pounds cfdthjhgddvnirdvjitdxvj

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The formula for the volume of a cone is V = jarh, where V is the volume, r is the radius, and h Is the heightWhat is the volume

olga nikolaevna [1]

Given:

The radius, r=4x

The height,

$h=21xy^2$

To find the volume of the cone:

The volume of the cone formula is,

$V=\frac{1}{3}\pi r^2h$

Substitute the values of r and h in the above formula we get,

$\begin{gathered} V=\frac{1}{3}\pi\times(4x)^2\times21xy^2 \\ =\frac{1}{3}\pi\times16x^2\times21xy^2 \\ =\pi\times16x^2\times7xy^2 \\ =112\pi x^3y^2 \end{gathered}$

Hence, the volume of the cone is

$V=112\pi x^3y^2$

Thus, the correct option is option D.

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7 months ago

What is the equation of the trend line in the scatter plot ?

ryzh [129]

Answer:

Equation: -3x + 18

Hope this helps!

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2 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:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

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2 years ago

Which of the following is the solution set for -3t + 11 > 20?

Keith_Richards [23]

answer A.

FOLLOW ME PLEASE

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2 years ago

Please help with math!

svet-max [94.6K]

You're going to have to reupload it, I'm sorry! Only have of the question shows!

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2 years ago