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ElenaW [278]
3 years ago
12

What is a vertical line with a rise over run relationship that does not exist.

Mathematics
1 answer:
Valentin [98]3 years ago
6 0
The line would be called undefined because it has no slope.
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MO bisects angle LMN, angel LMO = 6x-20.angle NMO= 2x+32 find angle LMN?
ICE Princess25 [194]
If MO bisects angle LMN:
6 x - 20 = 2 x + 36
6 x - 2 x = 36 + 20
4 x = 56
x = 56 : 4
x = 14
∠ LMN = 2 · ( 2 · 14 + 36 ) = 2 · ( 28 + 36 ) = 2 · 64 = 128
Answer : C )  x = 14, ∠ LMN = 128
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√18+4√2<br><br><br> Can you solve this for me please??
Rudik [331]

Answer:

√18+4√2 = 9.89949494

Step-by-step explanation:

4 0
4 years ago
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How do I solve the square root of 30?
Gre4nikov [31]
There is a way to work it out with a pencil and paper, but that's too complicated
to try and explain with only text, like we have here.

I don't usually recommend this for things that can be done easily on paper ...
but for square roots, your best bet is to use a calculator.

The square root of 30 does not come out even.  It starts out  5.477225...
and it keeps going and never ends.

The negative of  5.477225... is also a square root of 30.  Every number has
two square roots ... the positive and negative of the same number. 
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The regular price of a child's entry ticket to a water park is $6 less than that for an adult's. The park offers half off all en
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The price is just 50 dollars
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the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line? ​
rosijanka [135]

Answer:

A

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2,  5 )

If the slope is - \frac{3}{4} then the point is on the line

Calculate slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = \frac{-1-5}{6+2} = \frac{-6}{8} = - \frac{3}{4} ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = \frac{8-5}{2+2} = \frac{3}{4} ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = \frac{1-5}{-5+2} = \frac{-4}{-3} = \frac{4}{3} ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = \frac{1-5}{1+2} = - \frac{4}{3} ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

4 0
3 years ago
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