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2 years ago

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

Mathematics

1 answer:

Natasha2012 [34]2 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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Use complete sentences to describe why √-1 ≠ -√1

tekilochka [14]

Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

$\displaystyle \large{ {y}^{2} = x} \\ \displaystyle \large{ y = \pm \sqrt{x} }$

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

$\displaystyle \large{ {a}^{2} = a \times a = |b| }$

Where b is the result from a×a. Let's see an example.

Examples

$\displaystyle \large{ {2}^{2} = 2 \times 2 = 4} \\ \displaystyle \large{ {( - 2)}^{2} = ( - 2) \times ( - 2) = | - 4| = 4}$

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

$\displaystyle \large{ {y}^{2} = - 1}$

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

$\displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{ - 1} }$

Then where does plus-minus come from? It comes from one of Absolute Value propety.

Absolute Value Property I

$\displaystyle \large{ \sqrt{ {x}^{2} } = |x| }$

Solving absolute value always gives the plus-minus. Therefore...

$\displaystyle \large{ y = \pm \sqrt{ - 1} }$

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

$\displaystyle \large{ {y}^{2} = 1} \\ \displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{1} } \\ \displaystyle \large{ y = \pm \sqrt{1} }$

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

5 0

2 years ago

Read 2 more answers

A high school athletic department bought 40 soccer uniforms at a cost of $3,000. After soccer season they returned some uniforms

quester [9]

Interesting problem.

First - let's figure cost of each uniform at purchase.

3,000/40 = $75 each

When some uniforms were returned at $40 - there was a difference of $35 in what they paid and what they rec'd in return. ($75 - 35 = $40)

8 0

3 years ago

The area of triangle LMN is 45 in². What is the area of the parallelogram LMNO? A.45 in² B. 90 in² C. 135 in² D.180 in²