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mihalych1998 [28]
3 years ago
10

Can someone help me with these? 25 points.

Mathematics
2 answers:
frosja888 [35]3 years ago
6 0

Answer:

the answers are e,f,g,b

Step-by-step explanation:

And whats the question?

natka813 [3]3 years ago
3 0

Step-by-step explanation:

The area of a triangle is 1/2 bh, where b is the base and h is the height.

A. 1/2 * 6 * 9 = 27 sq. ft

B. 1/2 * 6 * 4 = 12 sq. ft

C. 1/2 * 7 * 8 = 28 sq. ft

D. 1/2 * 10 * 3 = 15 sq. ft

E. 1/2 * 14 * 5 = 35 sq. ft

F. 1/2 * 15 * 10 = 75 sq. ft

G. 1/2 * 20 * 13 = 130 sq. ft

H. 1/2 * 4 * 11 = 22 sq. ft

Hope this helps!

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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}

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

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

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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

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

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

square root both sides

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we know that

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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

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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}

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