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Zielflug [23.3K]
3 years ago
13

What is the area of the following triangle? A. 300 in B. 60 in C. 150 in D. 72 in

Mathematics
2 answers:
babymother [125]3 years ago
6 0
Answer: C. 150
Hopes this helps
NemiM [27]3 years ago
4 0

Answer:

C, 150 in

Step-by-step explanation:

12 in x 25 = 300

300/2

=150 in

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Evaluate the following integral using trigonometric substitution.
wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function \int\limits {\sqrt{a^{2} -x^{2} } } \, dx, using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as asin \theta i.e x = a sin\theta.

All integrals in the form \int\limits {\sqrt{a^{2} -x^{2} } } \, dx are always evaluated using the substitute given where 'a' is any constant.

From the given integral, \int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx where a = 7 in this case.

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2.) Given x = 7 sin\theta

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cross multiplying

dx = 7cos\theta d\theta

3.) Rewriting the given integral using the substiution will result into;

\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta  } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)}   } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)}   }}} \, 7cos\theta d\theta\\

= \int\limits343 cos^{2}  \theta \, d\theta

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i hope this helps!
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The perpendicular equation that passes through the point would be:

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