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Karo-lina-s [1.5K]

3 years ago

15

During a dry year, the reservoir level dropped 512 feet. The next year the level rose 314 feet. Which expression shows the total

change in reservoir height, in feet, during these two periods?

1 answer:

tia_tia [17]3 years ago

5 0

Answer:

You will be hot as shet

Step-by-step explanation:

e

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Solve the system of equations below by graphing both equations with a

Marat540 [252]

Graph using slope and the y intercept

So y=x+5
You would plot the y intercept (0,5) and then go up 1 and to the right 1

The next graph plot y intercept (0,-1) and then go down 2 and to the right 1

And should look like the image

And your solution should be (-2,3) which is option c

4 0

3 years ago

Can someone tell me the answer to this?

lord [1]

Answer:

c is the answer i hope this helps

8 0

3 years ago

Read 2 more answers

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

2 years ago

Jose has -75 points. He wins 200 points;then he loses 25 points. How many points does he have now?

Sav [38]

Answer:

100 points

Step-by-step explanation:

-75 + 200 = 125

125 - 25 = 100

3 0

2 years ago

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HELP IVE FALLEN AND I CANT GET UP

Pani-rosa [81]

Answer:

jjjjjjjjjjjjjjjjjjjjjiiiiiiiiiippppppppppppp

Step-by-step explanation:kk

jjjjjjjjjjjiiiiiiiiiiiiiilllllllllllllllll

3 0

3 years ago

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