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Mashcka [7]
3 years ago
6

30 POINTS! HELP! URGENT! THANK YOU!

Mathematics
1 answer:
N76 [4]3 years ago
6 0
The 1st option is correct
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Write the equation of the line that passes through (–2, 1) and is perpendicular to the line 3x – 2y = 5.
Eddi Din [679]
3x - 2y = 5
2y = 3x - 5
y = (3/2) x - 5/2

so the gradient (m)  of the required line = -1 / (3/2) = -2/3

using the general form 

  y - y1 = m(x - x1)  ,                m = -2/3 , x1 = -2 and y1 = 1

 y - 1 = -(2/3) ( x - (-2))


y = -(2/3)x - 4/3 + 1

y = -(2/3)x  - 1/3 

multiply through by 3

3y = - 2x  - 1 

in standard form its

2x + 3y = -1
3 0
3 years ago
Please WILL GIVE BRAINLIEST!<br> What is this expression in radical form?<br><br> (4x^3y^2)^3/10
svetoff [14.1K]
It should be 10√(4x3y2)3
5 0
3 years ago
Solve the equation. 5y - 10 = -25
algol [13]

Answer: 5y-10= -25

5y -10+10= -25 +10

Step-by-step explanation:

8 0
3 years ago
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Solve n3 + 2n2 - 15n = 0 by factoring.
azamat
N³ - 2n² - 15n = 0

"n" 

n (n² + 2n - 15) = 0

n = 0

n² + 2n - 15 = 0

Δ = (2)² - 4(1)(-15)
Δ = 4 + 60 = 64

n' = (-2+8) / 2 = 6/2 =3
n'' = (-2-8) / 2 = -10/2 = -5

Solution:

S {-5 , 0 , 3 }
5 0
3 years ago
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Please help due today i beg of u .......A compute technician keeps track of his earning through tout each months. The technician
Mnenie [13.5K]

Answer:

y=35x+200

The slope is 35 and it means the technician earns $35 per hour.

Step-by-step explanation:

The technician finds that

  • when he works 55 hours during the month, he earns $2,125;
  • when he works 30 hours, he earns $1,250.

Let

x = number of hours worked

y = amount of money earned

Find the slope of the linear equation:

\dfrac{2,125-1,250}{55-30}=\dfrac{875}{25}=35

Equation of the linear function:

y-1,250=35(x-30)\\ \\y=35x-1,050+1,250\\ \\y=35x+200

The slope is 35 and it means the technician earns $35 per hour.

6 0
3 years ago
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