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ozzi
3 years ago
12

How many solutions does the equation have |r-6|=0

Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

Only one solution

Step-by-step explanation:

|r - 6|  = 0 \\   r - 6 =  \pm \: 0 \\ r - 6 = 0 \\ r = 6

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Please simplify the equation
ss7ja [257]
Simplify
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6 0
3 years ago
What is the value of x? (will give brainliest if you show your work) PLEASE HELP
andrew-mc [135]
Check the picture below.

\bf \cfrac{8}{12}=\cfrac{x+4}{2x+1}\implies 16x+8=12x+48\implies 4x=40
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8 0
3 years ago
For the given line segment, write the equation of the perpendicular bisector.
Jlenok [28]

Answer:

D) y = -4/5x – 47/10

Step-by-step explanation:

Step 1. Find the <em>midpoint of the segmen</em>t.

The two end points are (-6, -4) and (-2, 1).

The midpoint is at the average of the coordinates.

(xₚ, yₚ) = ((x₁ + x₂)/2, (y₁ + y₂)/2)

(xₚ, yₚ) = ((-6 - 2)/2, (-4 + 1)/2)

(xₚ, yₚ) = (-8/2, -3/2)

(xₚ, yₚ) = (-4, -3/2)

===============

Step 2. Find the <em>slope (m₁) of the segment</em>

m₁ = (y₂ - y₁)/(x₂ - x₁)

m₁ = (1 - (-4))/(-2 - (-6))

m₁ = (1 + 4)/(-2 + 6)

m₁ = 5/4

===============

Step 3. Find the <em>slope (m₂) of the perpendicular bisector </em>

m₂ = -1/m₁

m₂ = -4/5

====================

Step 4. Find the <em>intercept of the perpendicular bisector</em>

y = mx + b

y = -(4/5)x + b

The line passes through (-4, -3/2).

-3/2 = -(4/5)(-4) + b

-3/2 = 16/5 + b      Multiply each side by 10

 -15 = 32 + 10 b    Subtract 32 from each side

 -47 = 10b             Divide each side by 10

    b = -47/10

===============

Step 5. Write the <em>equation for the perpendicular bisector</em>

y = -4/5x – 47/10

The graph shows the midpoint of your segment at (-4, -3/2) and the perpendicular bisector passing through the midpoint and (0, -47/10).

4 0
3 years ago
1. -8(-2x+6)+(-10)<br> 2.9(x-8)-17x<br> show work please
Damm [24]
1)-8(-2x+6)+(-10)
16x-48-10
16x-50
2) 9(x-8)-17x
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3 0
2 years ago
Read 2 more answers
75 + s = -2 . what is s
babunello [35]

Answer:

s = -77

Step-by-step explanation:

3 0
3 years ago
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