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2 years ago

13

Which equation has no solution?

2 answers:

Firdavs [7]2 years ago

3 0

A, because the absolute value is always non negative.<span />

son4ous [18]2 years ago

3 0

|4x - 2| = -6 is your answer.

Solve each equation:
|4x - 2| = -6
The equation has no solution because absolute value equations are always positive.

|-2 - x| = 9
-2 - x = 9 <em>or</em> -2 - x = -9

-2 - x = 9
-x = 11
x = -11

-2 - x = -9
-x = -7
x = 7

x = -11 or x = 7.

|3x + 6| = 6
3x + 6 = 6 <em>or </em>3x + 6 = -6

3x + 6 = 6
3x = 0
x = 0

3x + 6 = -6
3x = -12
x = -4

x = 0 or x = -4.

|-2x + 8| = 0
-2x + 8 = 0
-2x = -8
x = -4.

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Find the length of midsegment YZ in trapezoid ABCD when A(-5,-6) B(-6,-2) C(-4,0) and D(3,2)

Luda [366]

Answer:

Option C

Step-by-step explanation:

Since, Y and Z are the midpoints of sides AB and CD of the given trapezoid.

Segment YZ will the midsegment of trapezoid ABCD.

By the theorem of midsegment,

m(YZ) = $\frac{1}{2}(AD+BC)$

By using expression for the length of a segment between two points,

Length of a segment = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between two points A(-5, -6) and D(3, 2),

AD = $\sqrt{(3+5)^2+(2+6)^2}$

AD = $\sqrt{64+64}$

AD = $\sqrt{128}$

AD = $8\sqrt{2}$

Distance between B(-6, -2) and C(-4, 0)

BC = $\sqrt{(-6+4)^2+(-2-0)^2}$

BC = $\sqrt{8}$

BC = $2\sqrt{2}$

Therefore, m(YZ) = $\frac{1}{2}(8\sqrt{2}+2\sqrt{2})$

= $\frac{1}{2} (10\sqrt{2})$

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Option C will be the answer.

7 0

2 years ago