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loris [4]
3 years ago
15

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. . . A = 50°

, b = 30 ft, c = 14 ft CHOICES: A) 160.87 ft^2 B) 420 ft^2 C) 134.99 ft^2 D) 321.74 ft^2
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0
Area of the triangle:
A = 1/2 · c · b · sin A =
=  1/2 · 30 · 14 · 0.766 ≈ 160.87 ft²
Answer: 160.87 ft²
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kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

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\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

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RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
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= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
(05.05)
Bas_tet [7]

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Step-by-step explanation:

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