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2 years ago

If R is the midpoint of TS, TR = 5x - 12, and RS = 3x + 8, find TS.

Mathematics

2 answers:

Mademuasel [1]2 years ago

8 0

Answer:

Step-by-step explanation:

GenaCL600 [577]2 years ago

6 0

Answer:

5x -12 = 3x +8 (set the two = each other because they are the same length)

2x- 12= 8 (subtract 3x from both sides)

2x = 20 (add 12 to both sides)

x=10 (what x= for both expressions)

5(10) -12 (plug it into the first one to see what the length is and to see they're =

50 - 12 ( I already multiplied, now subtract)

38 (what the length of TR is)

3(10) +8 (plug it in again but into the other expression)

30+8 (multiply and add)

38 (the two have the same answer, so the x-value is correct.)

38+38= 76 (add the lengths of RS and TR and you get the length of TS)

Step-by-step explanation:

I hope this helps :)

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3 years ago

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scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-(-1)}{(-4)-(-4)}$

m= $\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-6}{(-4)-(-1)}$

m= $\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-4)-6}{2-2}$

m= $\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-(-4)}{(-4)-(2)}$

m= $\frac{3}{-6}$

m= $\frac{-1}{2}$

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{2-3}{3-(-2)}$

m= $\frac{-1}{5}$

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-2}{2-3}$

m= $\frac{-3}{-1}$

m=3

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{0-(-1)}{(-3)-2}$

m= $\frac{-1}{5}$

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{3-0}{(-2)-(-3)}$

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

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3 years ago

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schepotkina [342]

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40/10=4

Width=4, Length=10

Perimeter=L+L+W+W

4+4+10+10=28

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MrRa [10]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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