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fomenos
3 years ago
9

During last year’s volleyball season, the coach concluded that the number of points you scored in each game could be given by th

e equation |x - 7| = 2. How many points did you score in each game ?
Mathematics
2 answers:
Sauron [17]3 years ago
7 0

The two solutions for x are x = 5 or x = 9.

We have some number x in which we subtract off 7. The result of this subtraction (x-7) is the value 2. If the result is negative, we flip it to positive. If the result is positive, we leave it alone. Absolute values are never negative because distance is never negative. Note how x = 5 is 2 units away from 7; similarly, x = 9 is also 2 units away from 7. Here's how you can solve for the two solutions

|x-7| = 2

x-7 = 2 or x-7  = -2

x-7+7 = 2+7 or x-7+7 = -2+7 .... add 7 to both sides

x = 9 or x = 5

x = 5 or x = 9

Therefore you could have scored 5 points or 9 points

kirill115 [55]3 years ago
4 0
I would say 5 but I don’t understand the question much
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Answer:

-35

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27

Step-by-step explanation:

Start by setting up your pairs (x,y)

-5 & 7 are x

3 & 9 are y

X * X

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Y * Y

(-5, 7)

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(3, 9)

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3 0
3 years ago
In the diagram, DC is tangent to the circle, AC = 12 cm, and BD=2134 cm. Find AD.
olasank [31]
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4 0
3 years ago
Please Help! Look at the picture attached.
-BARSIC- [3]

Answer:

23-x=5

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that's the answer

3 0
3 years ago
Read 2 more answers
Use the Pythagorean theorem to find the unknown side of the right triangle.
Nesterboy [21]

The length of hypotenuse is 39.

Step-by-step explanation:

Given,

Vertical side = a = 15

Horizontal side = b = 36

Hypotenuse = c

Using pythagorean theorem

a^2+b^2=c^2\\(15)^2+(36)^2=c^2\\225+1296=c^2\\1521=c^2\\c^2=1521

Taking square root on both sides

\sqrt{c^2}=\sqrt{1521}\\c=39

The length of hypotenuse is 39.

Keywords: square root, addition

Learn more about square roots at:

  • brainly.com/question/9196410
  • brainly.com/question/9178881

#LearnwithBrainly

8 0
3 years ago
Can someone help me out with number 2?
Norma-Jean [14]
\sin255^o=\sin(300^o-45^o)=\sin300^o\cos45^o-\cos300^o\sin45^o= \\\\=- \frac{ \sqrt{3} }{2}\cdot  \frac{ \sqrt{2} }{2}- \frac{1}{2} \cdot  \frac{ \sqrt{2} }{2}=-      \frac{ \sqrt{6} }{4} -  \frac{ \sqrt{2} }{4}=  -\frac{  \sqrt{6}+\sqrt{2} }{4}

5 0
4 years ago
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