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1 year ago

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A cabinet is the shape of a triangular prism. The triangular base has a base length of 14 inches and a base height of 22 inches.

The cabinet is 67.5 inches tall. What is the volume of the cabinet?

1 answer:

Gekata [30.6K]1 year ago

3 0

Answer:

gat

Step-by-step explanation:

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2 years ago

The question of two numbers is eight the sum of the same two numbers is 72 what are the two numbers

Otrada [13]

Answer:

The two numbers are 64 and 8.

Step-by-step explanation:

You meant "quotient" of two numbers, right?

Represent the numbers by x and y.

Then the quotient is x/y = 8, and beyond that we know that x + y = 72.

Solve this for x and y.

If x/y = 8, then x = 8y. Substituting this into x + y = 72, we get:

(8y) + y = 72, or 9y = 72. Dividing both sides by 9 yields y = 8.

Since x/y = 8, letting y = 8 results in x = 64.

The two numbers are 64 and 8.

5 0

3 years ago

A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t

solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by

V = $V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x$

First, we differentiate V(x) with respect to x, to get,

$\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450$

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0$

Solving, with the help of quadratic formula, we get,

$x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}$ ,

Again differentiation V(x), with respect to x, we get,

$\frac{d^2(V(x))}{dx^2} = 24x - 180$

At x =

$\displaystyle\frac{5(3-\sqrt{3})}{2}$ ,

$\frac{d^2(V(x))}{dx^2} < 0$

Thus, by double derivative test, the maxima occurs at

x = $\displaystyle\frac{5(3-\sqrt{3})}{2}$ for V(x).

Thus, largest volume the box can have occurs when $x = \displaystyle\frac{5(3-\sqrt{3})}{2}}$ .

Maximum volume =

$V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}$

8 0

3 years ago