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3 years ago

If y represents a student's age, which inequality shows that you must be older than 14 to

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

B. y > 14

y which is students age is greater than 14

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У = 3х + 1 у= 2х Using substitution

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Can't quite figure out the limit problem

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Use the squeeze theorem; if

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2 years ago

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5C%20%28x%5E3-6x%5E2%2B9x%2B3%29%283x%5E2-12x%2B9%29%20dx" id="TexFormula1" title="\int

AlexFokin [52]

If the integral is simply

$\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx$

then notice that

$\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx$

which means you can compute the integral easily with a substitution

$u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx$

Under this transformation, the integral is

$\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}$

On the other hand, in case you're missing a symbol and the integral is actually

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx$

then first carry out the division:

$\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}$

Now, $3x^2-12x+9=3(x-3)(x-1)$ , so to integrate the remainder term you can decompose it into partial fractions:

$-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}$

$9-2x=a(x-1)+b(x-3)$

$x=1\implies7=-2b\implies b=-\dfrac72$

$x=3\implies3=2a\implies a=\dfrac32$

$\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}$

Then the integral would be

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}$

which can be rewritten in several ways, such as

$\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C$

6 0

3 years ago