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ololo11 [35]
3 years ago
10

Write the slope intercept form of the equation through (2,-4) parallel to y=-1

Mathematics
2 answers:
VLD [36.1K]3 years ago
6 0
-3/2 you take your x value as (-2,4) and Y intercept of (0,1). To make it easier throw that in a graph and do rise over run
Angelina_Jolie [31]3 years ago
3 0

Answer:

  • y = -x - 2

Step-by-step explanation:

Parallel lines have equal slopes.

Given slope, m = - 1 and point (2, -4).

<u>Slope- intercept form:</u>

  • y = mx + b

<u>Th line will be:</u>

  • y = -x + b

<u>Substitute x and y- values of the point into equation to find the value of b:</u>

  • - 4 = - 2 + b
  • b = 2 - 4
  • b = - 2

<u>The line is:</u>

  • y = -x - 2
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3 years ago
What are the zeros of the quadratic function f(x) = 6x ^ 2 - 24x + 1 ?
ZanzabumX [31]

Answer:

x = \frac{12+\sqrt{138} }{6} and x = \frac{12-\sqrt{138} }{6}

Step-by-step explanation:

Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: x=\frac{-b+\sqrt{b^2-4ac} }{2a} or x=\frac{-b-\sqrt{b^2-4ac} }{2a}.

Here, a = 6, b = -24, and c = 1. Plug these in:

x=\frac{-b+\sqrt{b^2-4ac} }{2a}

x=\frac{-(-24)+\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24+\sqrt{552} }{12}=\frac{24+2\sqrt{138} }{12} =\frac{12+\sqrt{138} }{6}

AND

x=\frac{-b-\sqrt{b^2-4ac} }{2a}

x=\frac{-(-24)-\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24-\sqrt{552} }{12}=\frac{24-2\sqrt{138} }{12} =\frac{12-\sqrt{138} }{6}

Thus, the zeroes are: x = \frac{12+\sqrt{138} }{6} and x = \frac{12-\sqrt{138} }{6}.

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Answer:

Step-by-step explanation:

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Snowcat [4.5K]

Answer:

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Step-by-step explanation:

Given

The attached table

(a)\ f^{-1}(-15)

This represents an inverse function.

So, we look into x row for its value.

i.e.

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Just like (a)

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