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maw [93]
3 years ago
12

Please help thank you.

Mathematics
1 answer:
Cloud [144]3 years ago
4 0
Question 1:
All sides of an equilateral triangle are equal.
7x - 5 = 3x + 15
Add both sides by 5 and subtract both sides by 3x
4x = 20
x = 5
(Whoa! Watch out for the first choice! We're looking for the length of a side, not the value of x)
Plug x = 5 into 3x + 15.
You get 30. All the sides have a length of 30 in.
The last choice is your answer.

Question 2:
Draw it (or just look at the image).
I'm a terrible artist.
Angle CAB = Angle ACB
Angle CAB + Angle ACB + Angle ABC = 180
2 * Angle CAB + Angle ABC = 180
2x - 6 + 4x - 3 = 180
6x = 189
x = 31.5
(Again! Watch out! We're not looking for the value of x!)
Angle CAB = Angle ACB = x -3 = 31.5 - 3 = 28.5
The first choice is your answer.

Question 3:
The second choice is your answer.
There's not much explanation... That's what it's called.

Have an awesome day! :)

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Find the value of $B - A$ if the graph of $Ax + By = 3$ passes through the point $(-7,2),$ and is parallel to the graph of $x +
Salsk061 [2.6K]

Answer:

-6

Step-by-step explanation:

We know that since Ax + By = 3 passes through (-7, 2), then if we plug -7 in for x and 2 in for y, the equation is satisfied. So, let's do that:

Ax + By = 3

A * (-7) + B * 2 = 3

-7A + 2B = 3

We also know that this line is parallel to x + 3y = -5, which means their slopes are the same. Let's solve for y in the second equation:

x + 3y = -5

3y = -x - 5

y = (-1/3)x - (5/3)

So, the slope of this line is -1/3, which means the slope of Ax + By = 3 is also -1/3. Let's solve for y in the first equation:

Ax + By = 3

By = -Ax + 3

y = (-A/B)x + 3/B

This means that -A/B = -1/3. So, we have a relationship between A and B:

-A/B = -1/3

A/B = 1/3

B = 3A

Plug 3A in for B into the equation we had where -7A + 2B = 3:

-7A + 2B = 3

-7A + 2 * 3A = 3

-7A + 6A = 3

-A = 3

A = -3

Use this to solve for B:

B = 3A

B = 3 * (-3) = -9

So, B = -9 and A = -3. Then B - A is:

B - A = -9 - (-3) = -9 + 3 = -6

The answer is -6.

<em>~ an aesthetics lover</em>

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