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Elza [17]
2 years ago
8

Write an equation for the line that passes through (−8.5,11) and (5,−2.5)

Mathematics
2 answers:
enot [183]2 years ago
5 0

y = -x + 2.5

slope formula:

\sf \dfrac{y_2-y_1}{x_2-x_1}

find slope:

\rightarrow \sf \dfrac{-2.5-11}{5-(-8.5)}

\rightarrow \sf \dfrac{-13.5}{13.5}

\rightarrow \sf -1

equation formula:

\sf y_1-y = m(x_1-x)

find equation:

\hookrightarrow \sf y-11=-1(x-(-8.5)

\hookrightarrow \sf y-11=-x-8.5

\hookrightarrow \sf y=-x-8.5+11

\hookrightarrow \sf y=-x+2.5

Mkey [24]2 years ago
5 0
  • (-8.5,11)
  • (5,-2.5)

Slope:-

  • m=-2.5-11/5+8.5
  • m=-13.5/13.5
  • m=-1

Equation in point slope form

  • y-11=-1(x+8.5)
  • y-11=-x-8.5
  • y=-x+2.5
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(2x + y)2 – (3x – 2y)2 + (x – 4y)(x + 4y<br>11​
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Step-by-step explanation:

(a+ b)² = a² +  b² + 2ab

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(a- b)² = a² + b² - 2ab

(3x - 2y)² = (3x)² + (2y)² - 2*3x *2y

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(x - 4y(x + 4y) = x² - (4y)²

                     = x² - 16y²

(2x + y)² - (3x - 2y)² + (x - 4y)(x +4y)

  = 4x²  + y² + 4xy - (9x² + 4y² - 12xy) + x² - 16y²

  = 4x²  + y² + 4xy - 9x² - 4y² + 12xy + x² - 16y²

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8 0
2 years ago
Find the equation of the line through (8,−6) which is perpendicular to the line y=x3−7.
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Answer:

y = (-1/3)(x + 10)

Step-by-step explanation:

The slope of the new (perpendicular) line is the negative reciprocal of the slope of the given line, which appears to be 3.  Thus, the perpendicular line has the slope -1/3.

Using the slope-intercept form y = mx + b, and substituting the givens, we obtain:

y = mx + b    =>    -6 = (-1/3)(8) + b, or

-6 = -8/3 + b.  We must solve for the y-intercept, b:

Multiplying all three terms by 3 removes the fraction:

-18 = -8 + 3b.  Thus, -10 = 3b, and so b must be -10/3.

The desired equation is    

y = (-1/3)x - 10/3, or

y = (-1/3)(x + 10)

8 0
3 years ago
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