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zavuch27 [327]
2 years ago
5

determine the length of a chord whose central angle is 75° in the circle with the radius of 12 inches. ​

Mathematics
1 answer:
Black_prince [1.1K]2 years ago
6 0

Answer:

Step-by-step explanation:

other two sides joining center to the ends of chord=radius of circle=12 in

using cosine formula

cos A=\frac{b^2+c^2-a^2}{2bc} \\2bc cos~A=b^2+c^2-a^2\\\angle A=75^\circ\\b=c=r=12~in\\2*12*12~cos~75=12^2+12^2-a^2\\a^2=2*12^2-2*12^2 cos ~75\\a^2=2*12^2(1-cos 75)\\a=12\sqrt{2(1-cos~75)} \approx 14.6~inches

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saul85 [17]
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6c = 30   Divide both sides by 6
c = 5   
7 0
2 years ago
Explain for both problems and tell me the work that comes with it please and thank you
Anton [14]

Problem 33

Use the distributive property on the side containing parentheses of each expression and compare it to the other side.

a) 3(5a + 3) = 15a + 9 not equal to 15a + 6

b) 2(7b - 2) = 14b - 4 not equal to 14b + 4

c) 5(2c + 3) = 10c + 15 not equal to 7c + 8

d) 3(d + 5/3) = 3d + 5 which is equal to 3d + 5

Answer for problem 33: d)

Problem 34

Use a proportion. Let the unknown number of bowls be x. The proportion is made up of two ratios that are set equal to each other. Set each ratio as a ratio of the number of avocados per bowls of guacamole. 3 avocados per 1 bowl (3/1) equals 17 avocados per x bowls (17/x).

\dfrac{3}{1} = \dfrac{17}{x}

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x = \dfrac{17}{3}

x = 5 \dfrac{2}{3}

Answer: 17/3 full bowls which is the same as 5 2/3 full bowls.

3 0
3 years ago
What two numbers product is 72 and sum is 17
iVinArrow [24]
That would be 8 and 9.

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8 0
3 years ago
Read 2 more answers
What is the following product?
VMariaS [17]

For this case we must find the product of the following expression:

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\sqrt [n] {a} * \sqrt [n] {b} = \sqrt [n] {ab}

So, we have:

\sqrt {12 * 18} =\\\sqrt {216} =

We rewrite the 216 as6 * 6 * 6 = 6 ^ 3 = 6 ^ 2 * 6

\sqrt {6 ^ 2 * 6} =

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Option D

6 0
2 years ago
Read 2 more answers
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0
2 years ago
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