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hichkok12 [17]
3 years ago
9

The distance between -18 and 15 is equal to _____. 18 - 15 15 - 18 -18 - 15 15 - (-18)

Mathematics
1 answer:
NemiM [27]3 years ago
5 0

Answer: Choice D.  15 - (-18)

This results in 15 - (-18) = 15 + 18 = 33.

So from -18 to 15 on a number line is 33 spaces.

We can also say

|-18 - 15| = |-33| = 33

The use of absolute value is to ensure the distance is never negative.

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In order for the parallelogram to be a square x = ?
yanalaym [24]

Answer: x = 7

Step-by-step explanation: In order for a parallelogram to be a square, the angles at the corners must be 90 degrees, and the diagonals that bisect the corner angles must be 45 degrees.

4x + 17 = 45  Subtract 17 from both sides.

4x = 28   Divide both sides by 4

x = 7

8 0
3 years ago
Simplify this please help <br>the answer choice are in the picture​
pashok25 [27]

Answer:

6^ (1/12)

Step-by-step explanation:

6 ^ 1/3

-------------

6 ^ 1/4

x^a / x^b = x^ (a-b)

6^(1/3-1/4)

Getting a common denominator)

6^(4/12-3/12)

6^1/12

3 0
3 years ago
Read 2 more answers
What is 5c + 12d - 6 ? <br> (simplifying algebraic expression)
g100num [7]
The answer should be the same as the question. 5c + 12d -6 is the answer because to simplify,you collect like tersm yet none of these terms are alike to each other ( different variable and exponent) therefore the answer is 5c+12d-6
5 0
4 years ago
6. Evaluate x³ = -8 *
katrin2010 [14]
-2
-2 x -2 x -2 = -8
3 0
2 years ago
Can someone solve this for me please.
BabaBlast [244]

Answer:

  • The value of PR is 21 ft.

Step-by-step explanation:

<u>We know that:</u>

  • 3(PQ + QT + TP) = PR + RS + SP
  • PQ = 7 ft.
  • QT = 4 ft
  • RS = 12 ft.

<u>Work:</u>

  • 3(PQ + QT + TP) = PR + RS + SP
  • => 3(7 + 4 + TP) = PR + 12 + SP
  • => 21 + 12 + 3TP = PR + 12 + SP
  • => 21 + 3TP = PR + SP
  • => 21 + SP = PR + SP                                                                       [3TP = SP]
  • => 21 = PR

Hence, <u>the value of PR is 21 ft.</u>

Hoped this helped.

BrainiacUser1357

6 0
3 years ago
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