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3 years ago

What is the midpoint of the line segment graphed below?(-12-3)(5-10)

Mathematics

2 answers:

wel3 years ago

6 0

Good evening ,

Answer:

the midpoint C(-3.5 , -6.5)

Step-by-step explanation:

C(x,y)

then x=(-12+5)/2=-3.5

y=(-3-10)/2=-6.5

Look at the photo below for more Details.

kodGreya [7K]3 years ago

5 0

Answer:

(-7/2, -13/2)

Step-by-step explanation:

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C=2πr<br> Solve for r<br><br> R=2π/C<br> R=C/2π<br> R=2C/π<br> R=π/2C

Evgen [1.6K]

I hope this helps you

2pi.r=C

r=C/2pi

7 0

3 years ago

Hello please help with my question, I will give brainlyest!

lilavasa [31]

No, John is incorrect.

<h3>Correct work shown:</h3>

$\sf x^2 - 6x - 7 = 0$

$\sf x^2 - 6x = 7$

$\sf x^2 - 6x + 9= 7 + 9$

$\sf (x-3)^2= 16$

$\sf x-3=\pm \sqrt{16}$

$\sf x-3= \pm 4$

$\sf x= + 4 + 3 \quad or \quad x = -4 + 3$

$\sf x= 7 \quad or \quad x = -1$

The correct answer should be x = 7 or x = -1

3 0

1 year ago

Which represents the solution(s) of the graphed system of equations, y = x2 – 2x and y = –2x – 1? (1, –1) (0, 0) and (0, –1) (0,

olganol [36]

ANSWER

No solution

EXPLANATION

The first equation is

$y = {x}^{2} - 2x$

and the second equation is

$y = - 2x - 1$

We equate the two equations to obtain;

${x}^{2} - 2x = - 2x - 1$

This implies that

${x}^{2} = - 2x + 2x - 1$

${x}^{2} = - 1$

There is no real number whose square is -1.

Therefore, the equation has no solution.

6 0

3 years ago

Read 2 more answers

What is the scale factor that takes the large triangle to the small triangle?

Lerok [7]

Step-by-step explanation:

such a scale factor applies to all sides and lines.

we see that DE is 12 units, and AB is 4 units.

so,

12×f = 4

f = 4/12 = 1/3 is the scale factor

control : check the other side

FE = 6 units

6 × 1/3 = 2

so, the corresponding side in the small triangle must be 2 units long.

and indeed, CB is 2 units long. all correct.

5 0

2 years ago

Find the square root by factor <br>21904

mihalych1998 [28]

$21904|2\\10952|2\\.\ 5476|2\\.\ 2738|2\\.\ 1369|37\\.\ \ \ \ 37|37\\.\ \ \ \ \ \ 1|\\\\21904=2\times2\times2\times2\times37\times37=2^4\times37^2\\\\\sqrt{21904}=\sqrt{2^4\times37^2}=\sqrt{2^4}\times\sqrt{37^2}=2^2\times37=4\times37=148$

$\sqrt{21904}=\sqrt{4\times5476}=2\sqrt{4\times1369}=2\times2\sqrt37\times37}=4\times37=148$

8 0

3 years ago

Read 2 more answers

What is the midpoint of the line segment graphed below?(-12-3)(5-10)​

What is the midpoint of the line segment graphed below?(-12-3)(5-10)