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Harrizon [31]
2 years ago
13

What is the measure of ZC?

Mathematics
1 answer:
Studentka2010 [4]2 years ago
8 0

Answer:

55°

Step-by-step explanation:

m\angle C =  \frac{1}{2} (184 \degree - 74 \degree) \\  \\  \implies \: m\angle C =  \frac{1}{2} (110 \degree) \\  \\  \implies \: m\angle C =  55 \degree

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Someone please help ASAP
kondor19780726 [428]

I haven't done this in a while, but I'm pretty sure the answer is D, (-3, -1.)

If you were to substitute x for -3 and y for -1 in the first inequalities and simplify, it would look like this:

-1 > -3-2

-1 > - 5

And that inequality is true, because -1 is bigger than -5. Let's also substitute in the other inequality, which would be:

-1 > 2(-3) + 2

-1 > -6 + 2

-1 > -4

And -1 is bigger than -4. So, I think the answer would be D because substituting for those values of x and y would stil. make the inequalities true.

5 0
3 years ago
Hi, can someone help me pretty please?
Tju [1.3M]
It’s 174 square feet
3 0
3 years ago
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Equation<br><img src="https://tex.z-dn.net/?f=%28z%20%2B%20%20%5Cfrac%7B6%7D%7B7%7D%20%29%20%2B%20%20%5Cfrac%7B2%7D%7B14%7D%20%2
ddd [48]

Given:

The equation is

\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1

To find:

The solution of the given equation.

Solution:

We have,

\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1

It can be written as

\left(z+\dfrac{6}{7}\right)+\dfrac{1}{7}=-1

\left(z+\dfrac{6}{7}\right)=-1-\dfrac{1}{7}

Multiply both sides by 7.

7z+6=-7-1

7z+6=-8

7z=-8-6

7z=-14

Divide both sides by 7.

z=\dfrac{-14}{7}

z=-2

Therefore, the value of z is -2.

5 0
3 years ago
In the end it says solve for X <br> show work
elena55 [62]

Answer:

x = 53

Step-by-step explanation:

angle 3 + angle 8 = 180 because of linear pair

66 + 2x + 8 = 180

74 + 2x = 180

2x = 180 - 74

2x = 106

x = 106/2

x = 53

3 0
2 years ago
Read 2 more answers
The coordinate plane shows the floor plan for a swimming pool. What is the area of the pool’s border? A. 50 square meters B. 65
podryga [215]

Answer: Choice B

Step-by-step explanation:

7 0
3 years ago
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