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Arte-miy333 [17]

2 years ago

8

1 point

1 answer:

romanna [79]2 years ago

3 0

Answer:

Y=50x + 200

Step-by-step explanation:

For every month, which is represented by X, the overall price, represented by y, goes up by 50 so we must multiply 50 by the number of months on the plan, X. We also have to factor in the one-time cost of the phone plan itself, which is 200 and ALSO does not go up with every month, so we add 200 to the equation of 50x, giving us 50x + 200, all we have to do from here is show what this equation is doing, which is giving us the answer to Y (the overall price of the plan), and we finish with Y=50x + 200

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Cost Price, etc.

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9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

SSSSS [86.1K]

Answer:

Part 4) $r=84\ units$

Part 9) $sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) $sin(\theta)=-\frac{9\sqrt{202}}{202}$

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

$C=2\pi r$

using proportion

$\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}$

simplify

$\frac{r}{180^o}=\frac{56}{120^o}$

solve for r

$r=\frac{56}{120^o}(180^o)$

$r=84\ units$

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

$cos^2(\theta)+sin^2(\theta)=1$

we have

$cos(\theta)=-\frac{2}{3}$

substitute the given value

$(-\frac{2}{3})^2+sin^2(\theta)=1$

$\frac{4}{9}+sin^2(\theta)=1$

$sin^2(\theta)=1-\frac{4}{9}$

$sin^2(\theta)=\frac{5}{9}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{5}}{3}$

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

$sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?

see the attached figure to better understand the problem

In the right triangle ABC of the figure

$sin(\theta)=\frac{BC}{AC}$

Find the length side AC applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=11^2+9^2$

$AC^2=202$

$AC=\sqrt{202}\ units$

so

$sin(\theta)=\frac{9}{\sqrt{202}}$

simplify

$sin(\theta)=\frac{9\sqrt{202}}{202}$

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

$sin(\theta)=-\frac{9\sqrt{202}}{202}$

5 0

3 years ago