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3 years ago

Heeeeellllpppppppppppp!!!!

Mathematics

1 answer:

DochEvi [55]3 years ago

8 0

$f(x)=8x^2-4x\\\\\text{substitute}\ x=\dfrac{m}{4}\\\\f\left(\dfrac{m}{4}\right)=8\cdot\left(\dfrac{m}{4}\right)^2-4\cdot\dfrac{m}{4}=8\cdot\dfrac{m^2}{16}-m=\dfrac{m^2}{2}-m$

$\text{Answer:}\ D.\ \dfrac{m^2}{2}-m$

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Answer:

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3 years ago

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3 years ago

A zoo is building a new large-cat exhibit. Part of the space will be used for lions and part for leopards. The exhibit will have

Fittoniya [83]

The equations for the expenses for maintaining 4 lions and 4 leopards in the new large-cat exhibit are as follows:

Lions Leopard

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<h3>What is an equation?</h3>

An equation is a mathematical statement that shows the values of two mathematical expressions are equal.

An equation is made up of two mathematical expressions connected by an equal sign.

<h3>Data and Calculations:</h3>

Annual expenses for lion per unit, x = $8,000

Annual expenses for leopard per unit, y = $6,000

Total number of larget-cats required = 8

Let the cost of maintaining 4 lions = 4x

Let the cost of maintaining 4 leopards = 4y

Total cost for 4 lions = $32,000 (8,000 x 4)

Total cost for 4 leopards = $24,000 (6,000 x 4)

Total cost for eight large cats per year = $56,000 ($32,000 + $24,000)

Equation for lions, 4x = 32,000

Equation for leopards, 4y = 24,000

The total costs for maintaining eight large cats in all will be <u>$56,000</u> per year.

Thus, the equations for the expenses for maintaining 4 lions and 4 leopards in the new large-cat exhibit are <u>4x = 32,000</u> and <u>4y = 24,000</u>, respectively.

Learn more about equations at brainly.com/question/13763238

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2 years ago

Help please

ddd [48]

Length of the box = 25 - 2x
Width of the box = 14 - 2x
Height of the box = x

a.) Volume of the box = x(25 - 2x)(14 - 2x) = x(350 - 50x - 28x + 4x^2) = 4x^3 - 78x^2 + 350x
Therefore, the function that models the volume of the box is V(x) = 4x^3 - 78x^2 + 350x

b.) 4x^3 - 78x^2 + 350x ≥ 240
4x^3 - 78x + 350x - 240 ≥ 0
x = 0.834, 5.438
The values of x for which the volume is greater than 240 in^3 is 0.834 ≤ x ≤ 5.438

c.) For maximum volume, dV/dx = 0
dV/dx = 12x^2 - 156x + 350 = 0
x = 2.882910931

Therefore, maximum volume = 4(2.882910931)^3 - 78(2.882910931)^2 +350(2.882910931) = 453.798 in^3

8 0

4 years ago