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Georgia [21]
3 years ago
5

Will give brainliest for correct answer

Mathematics
1 answer:
enyata [817]3 years ago
8 0

Perimeter is 30 units

Semi-perimeter is 15 units

The Area is 39.7 square units

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Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +
slava [35]
\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx

Taking x=7\tan\theta gives \mathrm dx=7\sec^2\theta\,\mathrm d\theta, so that the integral becomes

\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta
=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta
=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta

When \sec\theta>0, we have

=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta
=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta

and from here we can substitute u=\tan\theta to proceed from here.

Quick note: When we set x=7\tan\theta, we are implicitly enforcing -\dfrac\pi2 just so that the substitution can be undone later via \theta=\tan^{-1}\dfrac x7. But note that over this domain, we automatically guarantee that \sec\theta>0, so the absolute value bars can be dropped immediately.
6 0
3 years ago
Number them by order please
Igoryamba
1. D
2. B
3. A
Hope this help :)
5 0
2 years ago
Read 2 more answers
A movie rental store chain membership fee of $10 plus $1.50 per movie rented what is the password I mean it shows the cost of a
djyliett [7]
“The domain is all integers 0 or greater”
4 0
3 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
Ronald spent $133.55 on school clothes. he counted his money and discovered that he had $34.45 left. how much money did he origi
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He had $168 at the begging
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