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Tema [17]
3 years ago
14

Please help.

Mathematics
2 answers:
Tatiana [17]3 years ago
5 0
The answer is d account history

aev [14]3 years ago
4 0
D account history is the answer
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What is the area of this parallelogram?
MariettaO [177]

Answer:

A= 28

Step-by-step explanation:

To find the area of a parallelogram, multiply the base by the height. The formula is: A = B * H where B is the base, H is the height, and * means multiply.

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2 years ago
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Find another point on the line containing the points (-2,-3) and (3,-4)
Orlov [11]

Answer:

(8,-5)

Step-by-step explanation:

The slope of the line is -1/5, therefore one of the points is located at (8,-5)

Hope this helps!

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3 years ago
What is the degree of the power function represented in the table?
allochka39001 [22]

Answer:

3

Step-by-step explanation:

-2+1= -1

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3 years ago
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8) through: (-3, -2), perp. to y = x – 1<br> A) y=-5x – 1 B) y=-4x – 5<br> C) y=-x – 5 D) y=-5x – 4
nexus9112 [7]

<u>Answer:</u>

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

<u>Solution:</u>

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1

We know that, <em>product of slopes of perpendicular lines is -1. </em>

So, 1 \times slope of perpendicular line =  -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

8 0
3 years ago
What is The measure of angle L
bixtya [17]

Answer:

Angle L = 90 degrees

Step-by-step explanation:

5 0
3 years ago
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