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balu736 [363]
3 years ago
11

Solve this inequality: j/4 – 8 < 4. A. j < –48 B. j < 48 C. j > –12 D. j < 12

Mathematics
2 answers:
postnew [5]3 years ago
4 0
J/4 - 8 < 4

j/4 < 4 + 8
j4 < 12
j < 12 X 4
j < 48
Kruka [31]3 years ago
3 0
\frac{j}{4} - 8 \ \textless \  4 \\ \\  \frac{j}{4} \ \textless \  4 + 8 \\ \\  \frac{j}{4} \ \textless \  12 \\ \\ j \ \textless \  12 \times 4 \\ \\ j \ \textless \  48 \\ \\

The final result is: j < 48.
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Answer:

Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

                           f'(x) = -\frac{1}{1 + x^{2}} ---(1)

We know that:

                  \frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}} ---(2)

Comparing (1) and (2)

                           f'(x)=-(tan^{-1}x) ---- (3)

Using power series expansion for tan^{-1}x

f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx

= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]

=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}

=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

as

                 tan^{-1}(0)=0 \implies C=0

Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

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Step-by-step explanation:

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