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2 years ago

Solve for f -f +2 + 4f = 8 - 3f

Mathematics

1 answer:

Ulleksa [173]2 years ago

4 0

Answer:

f=1

Step-by-step explanation:

Let start off with doing -f plus 4f which gives you 3f. The equation would now be 2+3f=8-3f. Add 3f so you can cancel out the -3f and add the 3f to the other 3f. Now your equation is 2+6f=8. Cancel out do by doing -2. (Don't forget to to it to the 8 also.) Now your equation is 6f=6. Divide 6 from both sides to get 1. (You divide because you want to get f alone.

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1 year ago

Unit 3 homework 6 Gina Wilson

Sonja [21]

Answer:

5) The equation of the straight line is 2 x - y + 1 =0

6) The equation of the straight line is x + y -5 =0

7) The equation of the straight line is 5 x + 6 y - 24 =0

8) The equation of the straight line is x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

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

Given points are (1,3) , ( -3,-5)

$m = \frac{-5-3 }{-3-1 } = \frac{-8}{-4} = 2$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 3 = 2 ( x - 1 )$

y = 2x - 2 +3

2 x - y + 1 =0

The equation of the straight line is 2 x - y + 1 =0

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

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

Given points are (1,4) , ( 6,-1)

$m = \frac{-1-(4) }{6-1 } = \frac{-5}{5} = -1$

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

$y - 1 = -1 ( x - 4 )$

y - 1 = - x +4

The equation of the straight line is x + y -5 =0

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

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

Given points are (-12 , 14) , ( 6,-1)

$m = \frac{-1-(14) }{6+12 } = \frac{-15}{18} = \frac{-5}{6}$

$y - 14 = \frac{-5}{6} ( x - (-12) )$

6( y - 14 ) = - 5 ( x +12 )

6 y - 84 = - 5x -60

5 x + 6 y -84 + 60 =0

5 x + 6 y - 24 =0

The equation of the straight line is

$y - y_{1} = m ( x - x_{1} )$

Slope of the line

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

Given points are (-4 , -2) , ( 4 , 0)

$m = \frac{0+2}{4 +4} = \frac{2}{8} = \frac{1}{4}$

$y - (-2) =\frac{1}{4} ( x - (-4) )$

4 ( y + 2) = x + 4

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The equation of the line y = 3x + 6 is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒ 3x + y + k =0

⇒ 12 + 7 + k =0

⇒ k = -19

The equation of the parallel line is 3x + y -19 =0

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Answer:

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Step-by-step explanation:

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