Can someone simplify pls

Mathematics

2 answers:

algol [13]2 years ago

8 0

we are given to simplify

$\\ \ (x^6y^5)(5xy^2)(-6x^4)\\ \\$

Open the bracket and write the like terms together

$\\ \ =x^6y^5*5xy^2*(-6)x^4\\ \\ \ =-30x^6*x*x^4*y^5*y^2\\ \\$

when the bases are same then the exponents gets added

$\\ \ =-30x^{6+1+4}y^{2+5}\\ \\ \ =-30x^{11}y^7$

const2013 [10]2 years ago

6 0

(x^6y^5)(5xy^2)(-6x^4)

Remove parentheses and multiply.

-30x^6x^4xy^5y^2

Apply exponent rules.

-30x^11y^7

Your answer is $-30x^{11}y^{7}$

I hope this helps ;)

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Factor the polynomial. 7x2 + 68xy - 20y2 A) (7x - 2y)(x + 10y) B) (7x - 10y)(x - 2y) C) (7x + 10y)(x + 2y) D) (7x + 2y)(x - 10y)

Otrada [13]

For this case we have the following polynomial:
7x2 + 68xy - 20y2
Factoring we have:
(7x-2y) (x + 10y)
We verify the factorization:
7x2 + 70xy - 2xy - 20y2
Rewriting we have:
7x2 + 68xy - 20y2
Therefore, the factorization is correct.
Answer:
A) (7x - 2y) (x + 10y)

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∫(cosx) / (sin²x) dx

kirza4 [7]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2822772

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\
=\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}$

Make the following substitution:

$\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}$

and then, the integral (i) becomes

$=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}$

Integrate it by applying the power rule:

$\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\
\mathsf{=-\,\dfrac{1}{u}+C}$

Now, substitute back for u = sin x, so the result is given in terms of x:

$\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

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user100 [1]

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