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True [87]
3 years ago
13

HELP...The volume of a rectangular prism is 36x^6y^8. The length of the prism is 4x^3y^5, and the width of the prism is 3xy^2. W

hat is the height of the prism? Write a justification of your result. v=lwh​
Mathematics
1 answer:
Margaret [11]3 years ago
4 0

Answer:

H = 3x^2y

Step-by-step explanation:

Given

V = 36x&^6y^8

L = 4x^3y^5

W = 3xy^2

Required

Find the height of the prism

Volume (V) is calculated as:

V =LWH

Substitute values for V, L and W

36x^6y^8 =4x^3y^5 * 3xy^2 * H

Make H the subject

H = \frac{36x^6y^8}{4x^3y^5 * 3xy^2}

H = \frac{36x^6y^8}{4*3x^3*x*y^5y^2}

H = \frac{36x^6y^8}{12x^4*y^7}

Divide:

H = 3x^{6-4}y^{8-7}

H = 3x^2y

The height of the prism is: 3x^2y

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Find the derivative of the function f(x) = (x3 - 2x + 1)(x – 3) using the product rule.
julsineya [31]

Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}

Now, we distribute the expression of f(x) and find the derivative afterwards.

f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

6 0
3 years ago
A rectangular trunk has a volume of 26,880 cubic inches. The trunk is 48 inches long by 28 inches wide. What is the trunk’s heig
Law Incorporation [45]
Answer: 240

Exclamation:

26880=4(28)(h)

26880= 112. (h)
_____. ____
112. 112

Answer: 240=h

I hoped this helped!
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Find the AOS for y = x^2 – 2x – 15
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Answer:

Step-by-step explanation:

x=1

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3 years ago
80% of 65 please help
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The answer is 52.

65 x 0.8 = 52.
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3 years ago
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A pile of coins, consisting of quarters and half dollars, is worth $11.75. If there are 2 more quarters than half dollars, how m
Black_prince [1.1K]

The pile contains 17 quarters and 15 half-dollars.

Let <em>x</em> = the number of quarters and <em>y</em> = the number of half-dollars.

We have two equations:

(1) $0.25<em>x</em> + $0.50<em>y</em>  = $11.75

(2) <em>x</em> = <em>y</em> +2

Substitute the value of <em>x</em> from Equation (2) into Equation (1).

0.25(<em>y</em>+2) + 0.50<em>y</em> = 11.75

0.25<em>y</em> + 0.50 + 0.50<em>y</em> = 11.75

0.75<em>y</em> = 11.75 – 0.50 = 11.25

<em>y</em> = 11.25/0.75 = 15

Substitute the value of <em>y</em> in Equation (2).

<em>x</em> = 15 + 2 = 17

The pile contains 17 quarters and 15 half-dollars.

<em>Check</em>: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.

4 0
3 years ago
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