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dangina [55]
3 years ago
12

What is the first step in solving the inequality StartFraction m minus 2 Over 6 EndFraction < –1?

Mathematics
1 answer:
nlexa [21]3 years ago
8 0

Answer:

A.Multiply both sides by 6. B.Add 2 to both sides. C.Change the direction of the inequality. D.Change the inequality to ≤.

1

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Simplify: 4x-5y -2y+x
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5x-7y 
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-5y and -2y combine to get -7y.
This is all done using the like terms method.

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3 years ago
The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your
Alenkinab [10]
<span>Triangle PQR is a right triangle. First we have to find the length of each side of the triangle. This can be done using the points provided, along with the Pythagorean theorem, which is a^2+b^2=c^2. PR^2 = (7- -2)^2+(3-5)^2 = 85 => PR = sqrt(85) QR^2 = (7- -1)^2+(3-1)^2 = 68 => QR = sqrt(68) QP^2 = (1-5)^2+(-1 - -2)^2 =17 => QP = sqrt(17) Now that we have the sides of the triangle, we can put them into the Pythagorean theorem again to see that it works out: (Sqrt(17))^2 + (sqrt(68))^2 = (sqrt(85))^2 17 + 68 = 85 85 = 85 Since the Pythagorean theorem works for right triangles, the triangle is indeed a right triangle.</span>
6 0
4 years ago
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HELP!!!! I can’t figure this out
grin007 [14]

Answer:

A.\,\orange{ \bold{ \frac{x - 3}{3x + 1} }}

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\frac{x + 2}{ {x}^{2} + 5x + 6 }  \div  \frac{3x + 1}{ {x}^{2}  - 9}  \\  \\  = \frac{x + 2}{ {x}^{2} + 3x  + 2x+ 6 }  \div  \frac{3x + 1}{ {x}^{2}  -  {(3)}^{2} }  \\  \\  = \frac{\cancel{x + 2}}{ \cancel{{({x}+ 3)}} \cancel{(x+ 2) }}   \times   \frac{\cancel{( {x}  +   3)}( {x} -  {3)}}{3x + 1}  \\  \\  =  \purple{ \bold{ \frac{x - 3}{3x + 1} }}

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3 years ago
PLEASE HELP NO LINKS​
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Answer:

wow I wish I could help. I'm not the best at math

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