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OLEGan [10]
3 years ago
13

Find the equation of the line that is perpendicular to y = 2x + 8

Mathematics
1 answer:
Natasha2012 [34]3 years ago
6 0

Answer:

\boxed {\boxed {\sf B. \ y= - \frac{1}{2} x-4}}

Step-by-step explanation:

We are given the line y=2x+8. It is in slope-intercept form or y= mx+b, where m is the slope and b is the y-intercept.

The slope of the line is 2.

We want to find the equation that is perpendicular. Perpendicular lines have negative reciprocal slopes.

1. Negate

  • Negate the slope, or switch the sign
  • (+) 2 --> -2

2. Reciprocal

  • Flip the numerator and denominator of the slope.
  • -2/1 --> -1/2

The perpendicular slope is -1/2, so the correct equation must be B.

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Bob can paint a fence in 12 hours, and working with Jen can paint a fence in 4 hours. How long would it have taken Jen to paint
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Answer:

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Step-by-step explanation:

For Bob

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Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

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Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

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HELP
anyanavicka [17]
The axis of symmetry is x=2.
6 0
3 years ago
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