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Mashutka [201]
3 years ago
9

Can someone help me out pls

Mathematics
1 answer:
lozanna [386]3 years ago
6 0

Answer:

a

Step-by-step explanation:

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If RT is a diameter of Q, Find m A.88°<br> B.89°<br> C.92°<br> D.101°
kumpel [21]

Answer:

A.88°

Step-by-step explanation:

Given,

RT is the diameter of the circle with center Q.

We have to find the measure of ∠RQS.

Solution,

m\angle RQS=7x+18\ \ \ and\\m\angle TQS=9x+2

Since RT is the diameter. That is it makes a straight angle at the center Q.

And measure of straight angle is 180°.

So we can say that  ∠RQS and ∠SQT makes a straight angle.

\therefore \angle RQS+\angle SQT=180\°

Now substituting the given values, we get;

(7x+18)+(9x+2)=180\°\\\\7x+9x+20=180\°\\\\16x=180-20=160\°\\\\x=\frac{160}{16}=10\°

We get the value of 'x'.

By substituting the value of x, we can find the measure of ∠RQS.

m\angle RQS=7x+18=7\times10+18=70+18=88\°

m\angle TQS=9x+2=9\times10+2=90+2=92\°

Hence the measure of ∠RQS is 88°.

7 0
4 years ago
Solve for x. 9(x + 1) = 25 + x
Anit [1.1K]

Answer:

Given the equation: 9(x+1) = 25+x

Distributive property of multiplication over addition states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately i,e,

a\cdot (b+c)=a\cdot b+a\cdot c

Then, by using distributive property on LHS in the given equation we have;

9x+9 =25 +x

Subtract 9 from both the sides;

9x+9-9=25+x-9

Simplify:

9x = 16+x

Subtract x from both the sides we get;

9x-x=16+x-x

Simplify:

8x=16

Divide both sides by 8 we have;

\frac{8x}{8}= \frac{16}{8}

On Simplify:

x =2

Therefore, the value of x for the equation 9(x+1) = 25+x is, 2


8 0
3 years ago
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18) Find the opposite of the opposite.
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The answer is number 3
3 0
3 years ago
Give me the right answer and get 30 points ASAP
BabaBlast [244]

Answer:

THE CORRECT ANSWER IS... A

Step-by-step explanation:

7 0
3 years ago
Name the property of equation 8.2 + (-8.2)=0​
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Answer:

Inverse property of addition.

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