Solve 4 1/4 - 2 11/20

2 answers:

8 0

This could be solved as a group of mixed numbers, but I like to work with improper fractions more.
17/4 - 51/20
Next, you need to find a common denominator.
85/20 -51/20
When adding and subtracting fractions, you only work in the numerator, and the denominator stays the same.
34/20
Then, you just simplify.
Improper fraction: 17/10
Mixed number: 1 7/10
Decimal: 1.7

AysviL [449]4 years ago

6 0

$4\frac{1}{4} - 2\frac{11}{20}$
Change them into improper fractions by multiplying the denominator(bottom) by the whole number then adding the numerator
$\frac{17}{4}$ - $\frac{51}{20}$
You must have a common denominator
Multiply $\frac{17}{4}$ by $\frac{5}{5}$ ⇒ $\frac{85}{20}$
$\frac{85}{20}$ - $\frac{51}{20}$ = $\frac{34}{20}$
Simplify by dividing the numerator by the denominator
34÷20= $1\frac{14}{20}$ ⇒ $1 \frac{7}{10}$ reduced
You answer is $1 \frac{7}{10}$

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Integral of x"2+4/x"2+4x+3

dolphi86 [110]

I'm guessing you mean

$\displaystyle \int\frac{x^2+4}{x^2+4x+3}\,\mathrm dx$

First, compute the quotient:

$\displaystyle \frac{x^2+4}{x^2+4x+3} = 1 + \frac{4x-1}{x^2+4x+3}$

Split up the remainder term into partial fractions. Notice that

x ² + 4x + 3 = (x + 3) (x + 1)

Then

$\displaystyle \frac{4x-1}{x^2+4x+3} = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52$

So the integral becomes

$\displaystyle \int \left(1 + \frac{13}{2(x+3)} - \frac{5}{2(x+1)}\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}$

We can simplify the result somewhat:

$\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^{13}\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|\frac{(x+3)^{13}}{(x+1)^5}\right| + C \\\\ = \boxed{x + \ln\sqrt{\left|\frac{(x+3)^{13}}{(x+1)^5}\right|} + C}$