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3 years ago

15

What is the quotient of 15÷3?

2 answers:

Ludmilka [50]3 years ago

6 0

Answer:

5

Step-by-step explanation:

Marrrta [24]3 years ago

6 0

5

Hope this helps you :D

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2. Graph each inequality on a number line:<br><br> 5x + 15 < 0

Georgia [21]

Answer:

x < -3

Step-by-step explanation:

1. To put this inequality on a number line, we first must solve it.

Step 1: Subtract 15 from both sides.

$5x + 15 - 15 < 0 - 15$

$5x < -15$

Step 2: Divide both sides by 5.

$\frac{5x}{5}$
$x < -3$

2. Now that we know x < -3, let's graph it!

Since it's less than, we make the number line point to the left, and we make the circle open.

3 0

2 years ago

PLEASE ANSWER ASAP!!!

adell [148]

Answer:

A

Step-by-step explanation:

100fps=68.182mph

68.182/3.8=17.94

8 0

3 years ago

Convert 32% to a decimal <br><br> And convert 129% into a decimal

Aliun [14]

the decimals are 0.32 and 1.29

4 0

3 years ago

Anne, Mary, and Lynn team up to pick apples. Anne can pick 4 baskets of apples per hour, and Mary can pick 5 baskets per hour. A

krek1111 [17]

Answer:

Lynn can pick nine baskets in three hours

Step-by-step explanation:

Anne and Mary pick 9 Baskets an hour

so in a half an hour they pick 4 and a half.

But it says they pick 6 together so lynn picked one and a half baskets in a half an hour

so over an hour she picks three

and over three hours she picks nine.

4 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

3 years ago