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Zigmanuir [339]
2 years ago
11

Write the equation of the line in point slope form for a horizontal line that passes through (5,-6)

Mathematics
1 answer:
denis-greek [22]2 years ago
7 0

Answer:

y=-6

Step-by-step explanation:

y+6=0(x-5)

Solve

y=-6

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Mike wants to save $1,750 in the next 12 months. He plans to save $150 each month for the first 11 months.
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Answer:

$1650

Step-by-step explanation:

Based on the question, we don't need the information that Mike wants to save 1750 in 12 months.

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150*11=1650

He will save $1650 if he saves $150 every month for 11 months.

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

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3 years ago
What are corresponding angles<br>​
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Corresponding angles are equal. They're a F shape and it could be upside down.

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