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3 years ago

Help please! Will give brainly to correct answer.

Mathematics

1 answer:

vazorg [7]3 years ago

6 0

Use inverse to solve this problem. Cos-1(2/5)=66.421=66.

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Is 152 points in 8 games 171 points in 9 games equivalent

max2010maxim [7]

Answer:

Yes, they are equivalent.

3 0

3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

6 0

3 years ago

(x^2 + 5x + 2) – (5x^2 – x – 2) =

zlopas [31]

Answer:

${ \tt{( {x}^{2} + 5x + 2) - (5 {x}^{2} - x - 2) }} \\ = { \tt{(1 - 5) {x}^{2} + (5 + 1)x + (2 + 2)}} \\ = { \tt{ - 4 {x}^{2} + 6x + 4 }} \\ = { \tt{ - 2(2 {x}^{2} - 3x - 2) }}$