Ask question

Ask question

All categories

3 years ago

9

Are the lines given parallel, perpendicular, or neither? y=3x-7 and y=3x+1. Please help!

1 answer:

kolezko [41]3 years ago

8 0

Answer:

These lines are parallel to each other

Step-by-step explanation:

It has the same slope of 3 and the only thing that is different is the y-intercept

You might be interested in

What is the equation of the line showed in this graph?

Irina-Kira [14]

Answer:

y=1

Step-by-step explanation

3 0

3 years ago

(Help ASAP)(Will Mark Brainliest+5 Star rating+Thanks)

navik [9.2K]

<span> (17.) f(x)=x+3; g(x)=1/x^2

</span><span>If you take B f(x)=x+3 g(x)=1/x^2 plug the f(x) into the g(x) formula (in other words, f(x) becomes the x for g) g(x)=1/x^2 g(x)=1/(x+3)^2

</span>

7 0

3 years ago

If a girl says she likes you but then throws a moldy apple at ur head then laughs abt it and then say she was messing does she r

marysya [2.9K]

13523273748164828482747264

4 0

3 years ago

Read 2 more answers

Find the variable of x. <br> 8x-2=-9+7x

TiliK225 [7]

Answer:

8x-2=-9+7x equals X=-7

7 0

3 years ago

Read 2 more answers

A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer

Anna11 [10]

Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

$L_A \to$ length of play area A

$W_A \to$ width of play area A

$L_B \to$ length of play area B

$W_B \to$ width of play area B

$x \to$ Area of A

$y \to$ Area of B

From the question, we have the following:

$L_A = 1 + 4W_A$

$W_B = 2 + W_A$

$L_B = 2 + 3W_B$

$x = y$

The area of A is:

$x = L_A * W_A$

This gives:

$x = (1 + 4W_A) * W_A$

Open bracket

$x = W_A + 4W_A^2$

The area of B is:

$y = L_B * W_B$

$y = (2 + 3W_B) * ( 2 + W_A)$

Substitute: $W_B = 2 + W_A$

$y = (2 + 3(2 + W_A)) * ( 2 + W_A)$

Open brackets

$y = (2 + 6 + 3W_A) * ( 2 + W_A)$

$y = (8 + 3W_A) * ( 2 + W_A)$

Expand

$y = 16 + 8W_A + 6W_A + 3W_A^2$

$y = 16 + 14W_A + 3W_A^2$

We have that:

$x = y$

This gives:

$W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2$

Collect like terms

$4W_A^2 - 3W_A^2 + W_A -14W_A - 16 =0$

$W_A^2 -13W_A - 16 =0$

Using quadratic calculator, we have:

$W_A = -14.1$ or $W = 1.13$ --- approximated

But the width can not be negative; So:

$W = 1.19$

7 0

3 years ago