Solve the inequality c+12<16

2.6.4 practice algebra 2 you are helping to design a road for a high mountain pass. there are two routes over the pass, but both

lora16 [44]

Answer:

Step-by-step explanation:

8 0

3 years ago

Assume that it costs Apple approximately E(x) 25,600 + 100x + 0.012 dollars to manufacture x 32GB iPods in a day. (a) The averag

uranmaximum [27]

Answer:

(a) $C'(x)=\dfrac{x^2-2560000}{x^2}$

(b)x=1600, Minimum Average Cost Per iPod=$132

(c) $C''(x)=\dfrac{5120000}{x^3}$

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

Step-by-step explanation:

Given that it costs Apple approximately $ C(x) to manufacture x 32GB iPods in a day, where:

$C(x)=25,600+100x+0.01x^2$

(a)The average cost per iPod when they manufacture x iPods in a day is given by:

$Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}$

The average cost per iPod is therefore:

$C'(x)=\dfrac{x^2-2560000}{x^2}$

(b)To minimize average cost of x iPods per day, we set the average cost per iPod=0 and solve for x.

$C'(x)=\dfrac{x^2-2560000}{x^2}=0\\x^2-2560000=0\\x^2=2560000\\x=\sqrt{2560000}=1600$

The resulting minimum average cost (at x=1600) is given as:

$Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}\\\dfrac{25,600+100(1600)+0.01(1600)^2}{1600}\\=\$132$

<u>Second derivative test</u>

(c)The answer above is a critical point for the average cost function. To show it is a minimum, we calculate the second derivative of the average cost function.

$C''(x)=\dfrac{5120000}{x^3}$

At the critical point, x=1600

$C''(1600)=\dfrac{5120000}{1600^3}=0.00125$

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

3 0

3 years ago

Find all derivatives of y^n of the function y=√x+5

quester [9]

Hello,

As somebody has distroyed my question, (may be he have not understood ?)

i will reply to y=(√x)+5

$\displaystyle \boxed{\dfrac{d^n y}{dx^n} =(-1)^n*n!! *\dfrac{1}{2^n} *x^{-\frac{2n+1}{2} }\\}\\y'=\dfrac{1}{2} *x^{-\frac{1}{2}} \\\\y''=-\dfrac{1}{4} *x^{-\frac{3}{2}} \\...\\$

3 0

3 years ago

Here is an expression: 2 + 3t

Dima020 [189]

2+3t
When t is 1
2+3(1)
2+3
5

When t is 4
2+3(4)
2+12
14

3 0

3 years ago

Read 2 more answers

Y-4=3(x-1) identify the slope

ElenaW [278]

Slope is 1/1 hope this helps

5 0

3 years ago