All categories

3 years ago

Determine if the lines are parallel, perpendicular, or neither y=5 and y =-1

Mathematics

2 answers:

valkas [14]3 years ago

6 0

The are parallel to each other because the are both straight horizontal lines

Anna35 [415]3 years ago

4 0

<span> y = 5 and y = -1 is a parallel line .
They both line on the y - axis .
( 0,5 ) and ( 0,-1 ) </span>

You might be interested in

I need help asap someone please help me i dont understand this question so can someone help me

eduard

Answer:

we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

Step-by-step explanation:

Given the expression

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}$

$=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}$

cancel the common factor: 2

$=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}$

cancel the common factor: p

$=\frac{2p+1}{p^2\left(4p^2-1\right)}$

$=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}$

cancel the common factor: 2p+1

$=\frac{1}{p^2\left(2p-1\right)}$

Expanding

$=\frac{1}{2p^3-p^2}$

Thus, we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

7 0

3 years ago

Why did the SDS build a case against the university?

Readme [11.4K]

This isn’t a math equation

7 0

3 years ago

An umbrella costs $6. If Iris purchases 2 umbrellas, how much will these umbrellas cost? Solve an equation to find the answer.

Wewaii [24]

Answer: $12

Step-by-step explanation: $6 x 2 = $12

Hope this helps!

7 0

3 years ago

Read 2 more answers

Solve -9m-4+2m = -9-7m+5, and interpret the results.

Drupady [299]

Answer:

-9m - 4 + 2m = -9 - 7m + 5

-7m - 4 = -7m - 4

-7m + 7m = -4 + 4

0 = 0

This is an identity solution because the same number equals the same number. That means there's an infinite set of solutions.

5 0

3 years ago

Verify sin^4 x - sin^2 x = cos^4 x - cos^2 x is an identity

Citrus2011 [14]

Answer:

(identity has been verified)

Step-by-step explanation:

Verify the following identity:

sin(x)^4 - sin(x)^2 = cos(x)^4 - cos(x)^2

sin(x)^2 = 1 - cos(x)^2:

sin(x)^4 - 1 - cos(x)^2 = ^?cos(x)^4 - cos(x)^2

-(1 - cos(x)^2) = cos(x)^2 - 1:

cos(x)^2 - 1 + sin(x)^4 = ^?cos(x)^4 - cos(x)^2

sin(x)^4 = (sin(x)^2)^2 = (1 - cos(x)^2)^2:

-1 + cos(x)^2 + (1 - cos(x)^2)^2 = ^?cos(x)^4 - cos(x)^2

(1 - cos(x)^2)^2 = 1 - 2 cos(x)^2 + cos(x)^4:

-1 + cos(x)^2 + 1 - 2 cos(x)^2 + cos(x)^4 = ^?cos(x)^4 - cos(x)^2

-1 + cos(x)^2 + 1 - 2 cos(x)^2 + cos(x)^4 = cos(x)^4 - cos(x)^2:

cos(x)^4 - cos(x)^2 = ^?cos(x)^4 - cos(x)^2

The left hand side and right hand side are identical:

Answer: (identity has been verified)

3 0

3 years ago