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timurjin [86]
3 years ago
8

Write the equation of the line, in slope-intercept form, that passes through the point (-8,-1) and that is parallel to x-3=0

Mathematics
1 answer:
nataly862011 [7]3 years ago
5 0
Y=-5X which is the equation you use to figure out the answer
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#MissionExam001
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Evaluate the equation.
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Answer:

C

Step-by-step explanation:

193b=212.3

193b/193=212.3/193

(divide by 193 to get the value of b)

193 cancels 193

b= 1.098445596

(round off to one decimal place)

b=1.1

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2 years ago
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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
21/7 rewrite as a whole number
saw5 [17]
Do the division :-  21 / 7  = 3

Answer is 3
3 0
3 years ago
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