All categories

4 years ago

Simplify into one fraction

Mathematics

2 answers:

ella [17]4 years ago

8 0

Answer and Explanation :

We have simplify the expressions into one fraction :

1) Expression $\frac{7}{x-3}+ \frac{3}{x-5}$

Taking LCM in the denominator,

$=\frac{7(x-5)+3(x-3)}{(x-3)(x-5)}$

$=\frac{7x-35+3x-9}{x^2-5x-3x+15}$

$=\frac{10x-44}{x^2-8x+15}$

The required form is $\frac{7}{x-3}+ \frac{3}{x-5}=\frac{10x-44}{x^2-8x+15}$

2) Expression $\frac{-5}{x-3}-(-4\cdot(x+2))$

Taking LCM in the denominator,

$\frac{-5}{x-3}+(4x+8)$

$=\frac{-5+(4x+8)(x-3))}{x-3}$

$=\frac{-5+4x^2-12x+8x-24}{x-3}$

$=\frac{4x^2-4x-29}{x-3}$

The required form is $\frac{-5}{x-3}-(-4\cdot(x+2))=\frac{4x^2-4x-29}{x-3}$

3) Expression $\frac{6}{x+7}- \frac{3}{x-2}$

Taking LCM in the denominator,

$=\frac{6(x-2)-3(x+7)}{(x+7)(x-2)}$

$=\frac{6x-12-3x-21}{x^2-2x+7x-14}$

$=\frac{3x-33}{x^2+6x-14}$

The required form is $\frac{6}{x+7}- \frac{3}{x-2}=\frac{3x-33}{x^2+6x-14}$

artcher [175]4 years ago

3 0

1. 10x-44/x(squared)-8x+15
2. X+22/x(squared) -x-6
3.3x-33/x(squared)+5x-14

You might be interested in

Calculating the degrees of freedom, the sample variance, and the estimated standard error for evaluations using the t statistic

Yuliya22 [10]

Answer:

a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

b) $s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

c) For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

Step-by-step explanation:

Part a

For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

Part b

From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

$SS = \sum_{i=1}^n (X_i -\bar X)^2$

And the sample variance for this case can be calculated from this formula:

$s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

Part c

For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

8 0

3 years ago

Sarah borrowed some money from her father at a simple interest rate of 5.5% to pay for her college tuition. At the end of the ye

gavmur [86]

$\bf ~~~~~~ \textit{Simple Interest Earned Amount}
\\\\
A=P(1+rt)\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &\$2555\\
P=\textit{original amount deposited}\\
r=rate\to 5.5\%\to \frac{5.5}{100}\to &0.055\\
t=years\to &1
\end{cases}
\\\\\\
2555=P[1+(0.055)(1)]\implies \cfrac{2555}{1+(0.055)(1)}=P
\\\\\\
\cfrac{2555}{1.055}=P\implies 2421.800947867 \approx P$