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cricket20 [7]
3 years ago
9

Simplify the expression:

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0
The answer Is A. 13p hope it helps
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What's the y-intercept of the equation: 5x+2y=20<br>A. (4,0)<br>B. (5,0)<br>C. (0,2)<br>D. (0,10)​
AveGali [126]
Answer: D. (0,10)

Why:
Your given equation is 5x+2y=20, so in order to find the y-intercept, we must put it in slope-intercept form, a.k.a. y=max+b.

To do this, subtract 5x from both sides. This will then give you 2y=-5x+20.

And to fully simply, divide both sides by 2. Finally, you get y=(-5/2)x+10. And since we know b in slope-intercept form is the y-intercept, and 10 is b in our final function/equation. We get the answer of y-intercept=(0,10)
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If x &lt; 5, then which of the following must be true?
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Step-by-step explanation:

8 0
3 years ago
The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to pro
oksian1 [2.3K]

Here is the full question

A snack food company wishes to have a cylinder package for it's almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer:

The radius and height are both dimension in the cylinder; in order to minimize the cost

radius = 2.515 cm

height = 18.93 cm

Step-by-step explanation:

We denote the radius of the cylinder to be = r

and the height of the cylinder = h

The volume of a cylinder is known to be = πr²h

Also, from the question; we are also told that the cylinder contains 120 πcm³

i.e πr²h = 120π

Dividing both sides with π; we have:

r²h = 120

h = \frac{120}{r^2}

The base and sides will cost $.01 per cm² to produce

Total cost of the base and side c_1 = 0.01 ( πr² + 2πrh)

but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

i.e

cost of the top cylinder c_2 = 0.02 ( πr²)

Overall Total cost = c_1 + c_2

= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

= 0.01 πr² + 0.02 πrh + 0.02 πr²

= 0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )

= 0.03 \pi r^2 + 2.4 \frac{\pi}{r}

Taking the differentiation to find the radius dimension to minimize cost; we have:

\frac{dc}{dr} =0 ⇒ 0.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0

0.06 \pi r^2 = \frac{2.4 \pi}{r^2}

r^4 = \frac{2.4 \pi}{0.06 \pi}

r^4 = 40

r = \sqrt[4]{40}

r= 2.515 cm

However, \frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}

\frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0

Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

To determine the height ; we have:

h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}

h = \frac{120}{6.34}

h = 18.93 cm

7 0
4 years ago
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