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elena-14-01-66 [18.8K]

3 years ago

8

How to find the roots of a graph?

1 answer:

ExtremeBDS [4]3 years ago

8 0

Find the roots of a graph by looking at where the parabola hits the x-axis on both sides. In the calculator you work put 2nd+trace+2 and go below the point, press enter and above the point and press enter 2 times for your answer.

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Each side of WRD is 3 inches long what id the measure of each angle in WRD

ira [324]

Answer:

60 degrees

Step-by-step explanation:

If each side of the triangle WRD is 3 inches long, then WRD is an equilateral triangle, meaning all angles are the same size. Since there are 180 degrees in a triangle, 180/3 = 60 degrees.

6 0

3 years ago

Which expression is equivalent to x + y + x + y + 3(y + 5)?

elixir [45]

Answer:

C

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

PLEASE HELP I GIVE BRAINLIEST

Galina-37 [17]

I Think it would either be D or C

(I’m not sure, let me know if I’m wrong)

4 0

3 years ago

If sally had 3 bapples and billy has 2 bapples how want out of sally and billy would have the most apples

pychu [463]

Sally has the most apples than the rest which is by one more and her total is -(5)

7 0

3 years ago

17. A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find

Pavlova-9 [17]

Answer:

$r = 3.74cm$

Step-by-step explanation:

Given

Hemisphere

$Radius = 7cm$

Cone

$Height = 49m$

Required

Determine the radius of the cone

First, we calculate the volume (V1) of the hemisphere.

$V_1 = \frac{2}{3}\pi r^3$

Substitute 7 for r

$V_1 = \frac{2}{3}\pi * 7^3$

$V_1 = \frac{2}{3}\pi * 343$

$V_1 = \frac{686}{3}\pi$

Volume (V2) of a cone is:

$V_2 = \frac{1}{3}\pi r^2h$

Because the lead is cast into a cone, then they have the same volume.

So, we have:

$\frac{686}{3}\pi = \frac{1}{3}\pi r^2h$

$686 = r^2h$

Substitute 49 for h

$686 = r^2 * 49$

Divide both sides by 49

$r^2 = \frac{686}{49}$

$r^2 = 14$

$r = \sqrt{14$

$r = 3.74cm$

<em>Hence, the radius of the cone is 3.74cm</em>

5 0

3 years ago