please help its math!!! will give brainly and more points if correct please I need help

The equations of four lines are given. Identify which lines are perpendicular.


Line 1: y+8x=−5


Line 2: x+16y=−5


Line 3: y=−6x−7


Line 4: y+3=18(x−1)


Lines 2 and 3 are perpendicular.

Lines 1 and 2 are perpendicular.

All four lines are perpendicular.

Lines 1 and 4 are perpendicular.

Answers

Answer 1

Answer:

None of the options match

Step-by-step explanation:

First let's check option 1: lines 2 and 3.

line 2 in mx+b form: y=-1/16x -5

line 3, already in mx+b form: y=-6x-7

if it was perpendicular, the slopes should have been -1/16 and 16 or -6 and 1/6.

now for the 2nd option: lines 1 and 2.

line 1 in mx+b form: y=-8x-5

line 2 in mx+b form: y=-1/16x-5

Again, the slopes do not match, they should have been -8 and 1/8 or -1/16 and 16.

For the 3rd option: all lines

this is obviously wrong since lines 1, 2 and 2, 3 are not perpendicular themselves. No need to bother checking this option.

Lastly, 1 and 4.

line 1 in mx+b form: y=-8x-5

line 4 --> y+3=18x-18 --> y=18x-21

This is also wrong since the slopes should have been -8 and 1/8 or 18 and -1/18.

To conclude, I think the answer is none. None of the combinations on here are perpendicular anyways...

Answer 2

Answer:

D

Step-by-step explanation:

We have the four lines:

[tex]\displaystyle\text{Line 1: }y+8x&=-5 \\\\ \text{Line 2: } x+\frac{1}{6}y&=-5\\\\\text{Line 3: } y&=-6x-7 \\\\\text{Line 4: } y+3&=\frac{1}{8}(x-1)[/tex]

Remember that for two lines to be perpendicular, their slopes are negative reciprocals of each other.

So, let’s determine the slope of each line.

For Line 1, we can rewrite our equation as:

[tex]y=-8x-5[/tex]

So, the slope of Line 1 is -8.

We can rewrite Line 2 as:

[tex]\displaystyle \begin{aligned} \frac{1}{6}y&=-x-5\\y&=-6x-30\end{aligned}[/tex]

So, the slope of Line 2 is -6.

The slope of Line 3 is also -6.

And the slope of Line 4 is 1/8.

So, the slopes of perpendicular lines must be negative reciprocals.

The negative reciprocal of -8 (slope of Line 1) is 1/8. We multiply by a negative and then take the reciprocal.

Since the slope of Line 4 is 1/8, this means that Lines 1 and 4 are perpendicular.

For Lines 2 and 3, they have the same slope, not negative reciprocals of each other.

So, they will instead of parallel instead of perpendicular.

And Line 1 is perpendicular only to Line 4, and no other line.

Therefore, our answer is D.


Related Questions

What is the largest possible integral value in the domain of the real-valued function

f(x)=[tex]\frac{1}{\sqrt{800-2x} }[/tex]

Answers

Answer:

Max Value: x = 400

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

AntiderivativesIntegral Property: [tex]\int {cf(x)} \, dx = c\int {f(x)} \, dx[/tex]Integration Method: U-Substitution[Integration] Reverse Power Rule: [tex]\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C[/tex]

Step-by-step explanation:

Step 1: Define

[tex]f(x) = \frac{1}{\sqrt{800-2x} }[/tex]

Step 2: Identify Variables

Using U-Substitution, we set variables in order to integrate.

[tex]u = 800-2x\\du = -2dx[/tex]

Step 3: Integrate

Define:                                                                                                            [tex]\int {f(x)} \, dx[/tex]Substitute:                                                                                         [tex]\int {\frac{1}{\sqrt{800-2x} } } \, dx[/tex][Integral] Int Property:                                                                                     [tex]-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx[/tex][Integral] U-Sub:                                                                                           [tex]-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du[/tex][Integral] Rewrite:                                                                                          [tex]-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du[/tex][Integral - Evaluate] Reverse Power Rule:                                                 [tex]-\frac{1}{2}(2\sqrt{u}) + C[/tex]Simplify:                                                                                                         [tex]-\sqrt{u} + C[/tex]Back-Substitute:                                                                                            [tex]-\sqrt{800-2x} + C[/tex]Factor:                                                                                                           [tex]-\sqrt{-2(x - 400)} + C[/tex]

Step 4: Identify Domain

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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Answers

Answer:

B

Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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or

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Step-by-step explanation:

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Answers

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Answers

Answer:

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Step-by-step explanation:

if you look at the equation it says

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you could then divide 30 by 3 in order to get M by itself and you would get M=10.

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E 2.5
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Answers

Answer:

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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25(5) + 4=129

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Answers

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Step-by-step explanation:

Answer:

step one

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Answers

Answer:

Yes this statement is correct.

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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