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3 years ago

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Help???????????????????

1 answer:

Nataliya [291]3 years ago

4 0

The answer is 69

hope this helps (;

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What is the length of the diameter of a circle if the endpoints of the diameter are at the points (4, 2) and (–6, –1)?

Oksana_A [137]

Answer:

d= 10.44030651

Step-by-step explanation:

The diameter is the length between the endpoints. We can find it using the distance formula.

d= sqrt((x2-x1)^2+(y2-y1)^2 )

d = sqrt((-6-4)^2+ (-1-2)^2)

d = sqrt((-10)^2+(-3)^2)

d= sqrt(100+9)

d = sqrt(109)

d= 10.44030651

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3 years ago

Evelyn goes to a middle school that includes grades 6, 7, and 8. She wants to estimate the percentage of students in the school

morpeh [17]

<span>B. The sample might not be representative of the population because it only includes students who are attending an after-school activity.</span>

7 0

3 years ago

Read 2 more answers

Which expression is equivalent to the one below?

Anarel [89]

Answer:

B

Step-by-step explanation:

Start by breaking down the equation

-1/2 x 10 = -5

-1/2x1/4=-1/4

Then combine your answer

-5=1/8

8 0

3 years ago

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.

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3 years ago

The mass of a vase is 0.3kilgram. A flower has a mass that is 0.03 times the mass of the vase. What is the mass of the flower?

deff fn [24]

Hello!

0.3 × 0.03 = 0.009

The mass of the flower is 0.009 kilogram.

8 0

3 years ago