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Gekata [30.6K]
3 years ago
7

Q. # 17 Graph the inequality on a coordinate plane.. - y < 3x - 5

Mathematics
2 answers:
damaskus [11]3 years ago
5 0
-y<3x-5

Multiply each term by -1 to make Y positive: ( when multiplying an inequality by a negative value you also reverse the inequality sign).

multiplying the original equation by -1 you get:
y>-3x +5

The slope is the number in front of the x, so the slope = -3
The Y intercept is the last number in the equation, so y-intercept is 5

When the line crosses x at 0, it would be on Y = 5 and since the slope is negative, the line would  slant downwards from left to right.

Because the inequality sign becomes > (greater than) the shaded portion would be to the right of the line.


 The first graph in your picture is the correct answer.


ad-work [718]3 years ago
4 0
The inequality -y<3x-5 by multiplying by -1 becomes y>-3x+5.
1. Draw line y=-3x+5. When x=0, y=5, when y=0, -3x+5=0, x=5/3.
2. Choose the correct semiplane, substituting (0,0) into the inequality. Since 0 is not greater than 5, you can conclude that point (0,0) isn't a solution and doesn't belong to the shaded area.
3. The correct answer is first choice

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Step-by-step explanation:

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Using Euler's relation, derive the following relationships:a. Cosθ=½(e^jθ+e^−jθ)b. Sinθ=½(e^jθ−e(^−jθ)
Montano1993 [528]

Answer:

a. cosθ = ¹/₂[e^jθ + e^(-jθ)] b. sinθ = ¹/₂[e^jθ - e^(-jθ)]

Step-by-step explanation:

a.We know that

e^jθ = cosθ + jsinθ and

e^(-jθ) = cosθ - jsinθ

Adding both equations, we have

e^jθ = cosθ + jsinθ

+

e^(-jθ) = cosθ - jsinθ

e^jθ + e^(-jθ) = cosθ + cosθ + jsinθ - jsinθ

Simplifying, we have

e^jθ + e^(-jθ) = 2cosθ

dividing through by 2 we have

cosθ = ¹/₂[e^jθ + e^(-jθ)]

b. We know that

e^jθ = cosθ + jsinθ and

e^(-jθ) = cosθ - jsinθ

Subtracting both equations, we have

e^jθ = cosθ + jsinθ

-

e^(-jθ) = cosθ - jsinθ

e^jθ + e^(-jθ) = cosθ - cosθ + jsinθ - (-jsinθ)

Simplifying, we have

e^jθ - e^(-jθ) = 2jsinθ

dividing through by 2 we have

sinθ = ¹/₂[e^jθ - e^(-jθ)]

4 0
3 years ago
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