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3 years ago

Plese someone answer me b asap

Mathematics

2 answers:

Oduvanchick [21]3 years ago

8 0

Answer:

$\huge\boxed{\sf 9y^2 -22y-24}$

Step-by-step explanation:

$\sf (3y-1)^2 -(2y+4)^2+(2y-3)(2y+3)\\\\Using \ formula:\\\\1) (a-b)^2 = a^2-2ab+b^2\\\\2) (a+b)^2= a^2+2ab+b^2\\\\3) (a+b)(a-b) = a^2-b^2\\\\Applying \ Formulas\\\\(3y)^2-2(3y)(1)+(1)^2-[(2y)^2+2(2y)(4)+(4)^2]+(2y)^2-(3)^2\\\\9y^2-6y+1-[4y^2+16y+16]+4y^2-9\\\\9y^2-6y+1 -4y^2-16y-16+4y^2-9\\\\Combining \ like \ terms\\\\9y^2-4y^2+4y^2-6y-16y+1-16-9\\\\9y^2 -22y-24\\\\\rule[225]{225}{2}$

Hope this helped!

Ipatiy [6.2K]3 years ago

3 0

Answer:

a) $\frac{1}{3}$ k² - 9

b) -7y² - 3

Step-by-step explanation:

a) ( $\frac{2}{3}$ k - 6)² - ( $\frac{1}{3}$ k + 3)²

= $\frac{4}{9}$ k² - 36 - $\frac{1}{9}$ k² + 9

= $\frac{4}{9}$ k² - $\frac{1}{9}$ k² + 9 -36

= $\frac{3}{9}$ k² - 27

= $\frac{1}{3}$ k² - 9

____________________________________

b) (3y - 1)² - (2y + 4)² + (2y - 3) (2y + 3)

= 9y² - 1 - 4y² + 16 + (2y - 3) (2y + 3)

= 9y² - 4y² - 1 + 16 + (- 3) (4y² + 6y)

= 5y² + 15 - 12y² - 18

= 5y² - 12y² + 15 - 18

= -7y² - 3

Hope this helps!

(NOTE: Do you mind marking me as Brainliest?)

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Evaluate the following integral using trigonometric substitution

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Answer:

The result of the integral is:

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Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

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3 years ago

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abruzzese [7]

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3 years ago

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nadya68 [22]

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Step-by-step explanation:

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Mekhanik [1.2K]

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Answer:

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