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myrzilka [38]
3 years ago
6

Solve the inequality (-3,-3) (3,-1) on a graph

Mathematics
1 answer:
Firdavs [7]3 years ago
5 0
The first number is x axis while the second is y

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When you solve a problem involving money, what can a negative answer represent?
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That you owe someone some money
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its A. moderate and positive

Step-by-step explanation:

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2 years ago
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What is the equation of the line through (1, 6) and (0, 2)?
ASHA 777 [7]

Answer:

y = 4x + 2

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y = mx + b

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3 years ago
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.

6 0
3 years ago
(-3*10^9)(1.27*10^-12) in scientific notation
katen-ka-za [31]
It would be = -3.81 * 10^-3
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3 years ago
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