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3 years ago

What is the inequality x+5<2

Mathematics

2 answers:

Vlad [161]3 years ago

8 0

Answer:

x < - 3

Step-by-step explanation:

Given

x + 5 < 2 ( subtract 5 from both sides )

x < - 3

Natali [406]3 years ago

4 0

Yo sup??

x+5<2

adding -5 on both the sides

x+5-5<2-5

x<-3

Therefore x can take any value from -3 (-3 not included) to negative infinity.

Hope this helps.

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3 years ago

Intergal of 8sin^4(x)dx

Romashka-Z-Leto [24]

$\int\limits8sin^4(x)dx =$

Take the constant out $\int\limits a*f(x)dx=a* \int\limitsf(x)dx$

$= 8 \int\limits sin^4(x)dx$

$\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx$

$= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)
$

Use the following identify : $sin^2(x) = \frac{1-cos(2x)}{2}$

$= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)$

$\int\limits 1dx = x$

$\int\limits cos(2x)dx = \frac{1}{2} sin(2x)$

Apply the sum rule :

$=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))$

Simplify :

$8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))$

Therefore add a constant to the solution:

$= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C$

hope this helps!

3 0

3 years ago

Simplify: <img src="https://tex.z-dn.net/?f=%288%5E%7B%5Cfrac%7B2%7D%7B3%7D%29%20%5E%7B4%7D%20%7D" id="TexFormula1" title="(8^{\

mrs_skeptik [129]

${ \large{ \sf{ \implies{(8 {}^{ \frac{2}{3} }) {}^{4} }}}}$