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3 years ago

14

Can some one help me answer this

1 answer:

Andrews [41]3 years ago

7 0

Yes, notice for every 10 degrees in Celsius you have 18 degrees Fahrenheit.

32 is the freezing point of water in F, and 0 is the freezing point of water in C.

So it is starting at freezing point, every time you add 10 in C you also add 18 to F.

They show the same temperature just on different scales.

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1. Find the domain and range of the relation and determine whether it is a function.

Katyanochek1 [597]

A. the relation is a function

8 0

3 years ago

A cylinder has a volume of 198 and base has an area of 22 what is the height o the cylinder

Mumz [18]

Answer:

Step-by-step explanation:

The base of the cylinder is in the shape of a circle:

Area of the base= $\pi *r^2$

For Base area=22 square units

$22=\pi *r^2$

$22=3.14*r^2\\r^2=22/3.14\\r^2=7.01\\r=\sqrt{7.01}$

Volume of the cylinder= $\pi *r^2*h$

For volume '198 cubic units' the height(h) is:

$198=\pi *r^2*h$

$198=3.14*(\sqrt{7.01})^2*h$

$198=3.14*7.01*h\\\\198=22.01*h\\\\198/22.01=h\\\\h=9 units\\$

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2 years ago

What is the image of (-5, 6) after a reflection over the line y = -2

prohojiy [21]

Answer:

I'm pretty sure it would be (5,6)

Step-by-step explanation:

Reflecting a point over the Y-axis would change the x-coordinate, but not the y-coordinate

6 0

2 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

5 0

3 years ago

This person has been spamming me so I know they’ll read this. Your a really annoying person. All you do is spam useless links an

zmey [24]

Answer:

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Step-by-step explanation:

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2 years ago