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vaieri [72.5K]
3 years ago
12

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

ct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.
Complete the following proof. (Note: Your teacher has additional instructions on how to complete this problem, if needed.)

Prove: In an equilateral triangle the three medians are equal.

Mathematics
1 answer:
Juliette [100K]3 years ago
3 0
Hello,
Please, see the attached files.
Thanks.

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Carlos dropped the phone he had received for his birthday and broke the screen. He knows he has to pay for the replacement himse
NeX [460]

Answer:

you would want B

Step-by-step explanation:

why because if you get a discount you would be able to have to not pay as much as if you dont get a discount.

7 0
3 years ago
Help this is due tomorrow!!!
r-ruslan [8.4K]

Answer:

V=4/3*3.14*r^3

V=4/3*3.14*3^3

V=113.1

Volume of Himiphere= Volume of sphere/2

113.1/2

V=56.6

Step-by-step explanation:

The answer is 56.6

4 0
2 years ago
Read 2 more answers
3. Find x. Round to the nearest tenth. *
Ad libitum [116K]
13.3 I’m very positive
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3 years ago
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Find the value of y, rounded to the nearest tenth. Please help me, I'd appreciate it!!
Blababa [14]
If a secant<span> and a </span><span>tangent of a circle </span><span>are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.
</span><span>
y</span>² = 7(15+7)
<span>y</span>² = 7*22
<span>y</span>² = 154
<span>y = </span>√154
<span>y = 12.4  </span>← to the nearest tenth<span>
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8 0
3 years ago
What is the speed (approximately) of a 2.5-kilogram mass after it has fallen freely from rest through a distance of 12 meters?
Norma-Jean [14]

Answer:

15.3 m/s

Step-by-step explanation:1) Find the gravitational potential energy using the equation :

     PE_g=mgh

    PE_g=2.5(12)(9.8)\\PE_g=294

2) Then use the equation for kinetic energy to solve for the velocity:

     KE=\frac{1}{2} mv^{2}

     294=\frac{1}{2}(2.5)(v^2)\\235.2=v^2\\\sqrt{235.2} =v\\15.3 m/s=v

6 0
3 years ago
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