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3 years ago

SLOPE AND Y INTERCEPT IM GONNA BREAKDOWN?????????

Mathematics

2 answers:

VikaD [51]3 years ago

6 0

Answer:

The Slope is -1/2

The Y intercept is 2

ollegr [7]3 years ago

5 0

Answer:

slope: -1/2

y-intercept: 2

Step-by-step explanation:

y=mx+b is the equation to a linear function

m=slope

b=y-intercept

In this case, -1/2 is the m, and 2 is the y-intercept.

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1 year ago

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3 years ago

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What is the exact value of sin(theta + beta)?

NISA [10]

Answer: $\frac{-3\sqrt{13}+4\sqrt{3}}{20}$

This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.

sqrt = square root

=======================================================

Explanation:

Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative

Use the pythagorean trig identity to get the following:

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\$

Angle beta is in Q1 where sine and cosine are positive.

Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.

Since $\tan \beta = \frac{3}{4}$ , this means $\sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}$

Use these ideas:

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.

-------------------------------------

To recap:

$\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\$

They lead to this

$\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\$

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3 years ago