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coldgirl [10]
3 years ago
11

563 percent as a fraction

Mathematics
2 answers:
Bogdan [553]3 years ago
6 0

Answer: 5 63/100

Since the numerator is greater than the denominator, we got an IMPROPER fraction. So we can also express it as a MIXED NUMBER, although we know that 563/100 is also equal to 5 63/100 when written as a mixed number.

jekas [21]3 years ago
3 0

Answer:

563/100

or 5 63/100          

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<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
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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
4 years ago
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