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3 years ago

is it possible to store a 20 foot long pipe in a rectangular room that is 10 feet by 12 feet by 8 feet?

2 answers:

7 0

First we'll consider the floor size.
In a 10 by 12 room the longest pipe that will fit lying on the floor equals the hypotenuse of a 10 by 12 triangle:
hypotenuse² = 10² + 12²
hypotenuse² = 100 + 144
hypotenuse² = 244
hypotenuse = 15.62 feet

Now we have to consider the third dimension (height = 8 feet).
The two sides would be 15.62 and 8
hypotenuse² = 15.62² + 8²
hypotenuse² = 244 + 64 = 308
hypotenuse = 17.55
So, the longest pipe that would fit in that room would be 17.55 feet long.

EleoNora [17]3 years ago

6 0

No, if the pipe is 20 foot long that means the room would have be 20 foot long, tall, or wide which its not.

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A kitchen drawer has a Volume 1,125 cubic inches. The drawer is 15 inches long and 5 inches deep. What is the width of the drawe

Temka [501]

Answer:

The width of the drawer is $15\ in$

Step-by-step explanation:

we know that

The volume of the drawer is equal to

$V=LWH$

we have

$L=15\ in$

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$V=1,125\ in^{3}$

substitute the given values in the formula and solve for W

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3 years ago

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12345 [234]

Answer: 362,880

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8 0

2 years ago

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kramer

Answer:

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Step-by-step explanation:

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3 years ago

Find the inverse function f -1(x) for the given function f(x). f(x) = x + 1/8

velikii [3]

Answer:

f⁻¹(x) = 8x - 1

Step-by-step explanation:

Swap the positions of y (f(x)) and x and solve for y.

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3 years ago

Help thankss answer is <img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="

lyudmila [28]

We have point $P=(x_0,y_0)=(2,-4)$ , so first calculate $f'(x_0)$ . There will be:

$y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}$

Now, we can write the equation of the normal line as:

$y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}$

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

$\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\
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3 years ago