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3 years ago

Anusha needs to find a rectangular container large enough to hold a volume of 130 in.. Which container should sh

Mathematics

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

"a container with the dimensions 3 by 5 by 10 inches"

Step-by-step explanation:

To find volume of rectangular box, we multiply the length, width, and height. Or if given in 3 dimensions form, we multiply all 3.

Lets find volume of each of the answer choices:

a container with the dimensions 2 by 5 by 10 inches = 2 * 5 * 10 = 100

a container with the dimensions 3 by 4 by 10 inches = 3 * 4 * 10 = 120

a container with the dimensions 3 by 5 by 10 inches = 3 * 5 * 10 = 150

a container with the dimensions 4 by 2 by 10 inches = 4 * 2 * 10 = 80

We want a container LARGE ENOUGH to contain a volume of 130 cubic inches. Which of the containers has a volume GREATER than 130?

Yes, only the 3rd option. It is 150 cubic inches in volume.

So, this is the answer:

a container with the dimensions 3 by 5 by 10 inches

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Find the total area the regular pyramid. T.A. =

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Answer:

20 +4√2

Step-by-step explanation:

Find the total area the regular pyramid = A + 1/2ps

A is the Base area

p is the perimeter of the base

s is slant height

A = 1/2 * 4 * 4

A = 16/2

A = 8

p = 4 + 4 + (√4^2+4^2)

p = 8 + √32

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Substitute

TSA = 8 + 8 + 4√2 + 4

TSA = 20 +4√2

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Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and i

Minchanka [31]

Answer:

$3x - 4y + z = 1$

Step-by-step explanation:

Given

$Point\ 1 = (4,3,1)$

$Point\ 2 = (4,1,-7)$

Perpendicular to $8x + 7y + 4z = 18$

Required

Determine the plane equation

The general equation of a plane is:

$a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

For $n =$

$(x_1,y_1,z_1) = (4,3,1)$

$(x_2,y_2,z_2) = (4,1,-7)$

First, we need to determine parallel vector $V_1$

$V_1 =$

$V_1$ is parallel to the required plane

From the question, the required plane is perpendicular to $8x + 7y + 4z = 18$

Next, we determine vector $V_2$

$V_2 =$

This implies that the required plane is parallel to $V_2$

Hence: $V_1$ and $V_2$ are parallel.

So, we can calculate the cross product $V_1 * V_2$

$V_1 =$

$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$ $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$

Calculate the product

$V_1 * V_2 = i(2*4- 8*7) - j(0*4- 8*8) + k(0*7 - 2 * 8)$

$V_1 * V_2 = i(8- 56) - j(0- 64) + k(0 - 16)$

$V_1 * V_2 = i(-48) - j(- 64) + k(- 16)$

$V_1 * V_2 = -48i +64j - 16k$

So, the resulting vector, n is:

$n =$

Recall that:

$n =$

By comparison:

$a = -48$ $b = 64$ $c = -16$

Substitute these values in $a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

$-48(x-x_1) + 64(y - y_1) -16(z-z_1) =0$

Recall that: $(x_1,y_1,z_1) = (4,3,1)$

So, we have:

$-48(x-4) + 64(y - 3) -16(z-1) =0$

$-48x + 192 + 64y -192 - 16z +16 = 0$

Collect Like Terms

$-48x + 64y - 16z = 0 - 192 + 192 - 16$

$-48x + 64y - 16z = -16$

Divide through by -16

$3x - 4y + z = 1$

Hence, the equation of the plane is $3x - 4y + z = 1$

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