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faltersainse [42]
3 years ago
6

P = q/4 -r solve for q

Mathematics
1 answer:
Ivan3 years ago
4 0

Answer:

4(p+r) = q

Step-by-step explanation:

p = q/4 -r

Add r to each side

p+r = q/4 -r+r

p+r = q/4

Multiply each side by 4

4(p+r) = q/4*4

4(p+r) = q

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Answer:

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5 0
3 years ago
If mABC= 116’ what is m ABC
Vikentia [17]

 

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4 0
3 years ago
Find the value of (2^−2 − 4^0) × 4^−2​
Dafna1 [17]

Answer:

-0,046875

Step-by-step explanation:

( {2}^{ - 2}  -  {4}^{0} ) \times  {4}^{ - 2}  \\  = (0.25 - 1) \times 0.0625 \\  =  - 0.75 \times  0.0625 \\  =  - 0.046875

8 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!! explain too please
ale4655 [162]

Answer:

8. Not similar

9. Similar

Step-by-step explanation:

8. You could try to us the AA theorem which means that two angles in the triangle are equal, which would mean that the triangles are similar. But 46+58=104 and 180-104=76 so there are not two similar angled between the triangles.

9. You can use the same theorem for this problem (AA). So 106+31=137 and then 180-137=43 which is one of the angles in the second triangle. Since at least two of the angles are equal, you can use AA theorem yo prove they are similar.

Hope this helps!

4 0
3 years ago
Read 2 more answers
Which point is a solution to the inequality shown in this graph?
kykrilka [37]

Answer:

The only solution can be (0,-3) point.

Step-by-step explanation:

We have to judge whether the points in options are the solution to the graphed inequality or not.

The first point is (5,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The second point is (6,0) which not included in the shaded region of the graph. Hence, it can not be a solution.

The third point is (0,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The fourth point is (0,-3). It is on the firm red line which is included in the shaded region of the graph. Hence, it is a solution.

Therefore, the only solution can be (0,-3) point. (Answer)

5 0
3 years ago
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