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3 years ago

There are 9 numbers written, beginning with: 8, 5, 4, 9, 1, ... Finish the sequence.

Mathematics

1 answer:

alexdok [17]3 years ago

4 0

Answer:

14 -50 -29 -541

Step-by-step explanation:

8-3=5-1=4+5=9-8=1+13=14-64=-50+21=-29-512=-541

i got this by looking up sequence pattern finder in google and clicking on the second option then inserting the numbers you gave hope this helps

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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

55. (a) If alpha and beta are the roots of the equation xsquare+ px+q=0 and beta>alpha find the square of the

densk [106]

Answer:

√(p²-4q)

Step-by-step explanation:

Using the Quadratic Formula, we can say that

x = ( -p ± √(p²-4(1)(q))) / 2(1) with the 1 representing the coefficient of x². Simplifying, we get

x = ( -p ± √(p²-4q)) / 2

The roots of the function are therefore at

x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus

( -p + √(p²-4q)) / 2 - ( ( -p - √(p²-4q)) / 2)

= 0 + 2 √(p²-4q)/2

= √(p²-4q)

7 0

2 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$