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Sati [7]
3 years ago
8

Which pair of triangles can be proven similar?

Mathematics
1 answer:
sladkih [1.3K]3 years ago
8 0

Answer:l and ll

Step-by-step explanation:

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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.

The substitute will therefore be   x = 7 sin\theta

2.) Given x = 7 sin\theta

\frac{dx}{d \theta} = 7cos \theta

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

8 0
3 years ago
Find the value of y.
True [87]

Answer:

6

Step-by-step explanation:

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2 years ago
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