Find the value of 26/45 times 5/53 times 28/13 please show fingures and work

1 answer:

hjlf3 years ago

3 0

$\bf \cfrac{\stackrel{2}{~~\begin{matrix} 26 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{\underset{9}{~~\begin{matrix} 45 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{53}\cdot \cfrac{28}{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \cfrac{2\cdot 28}{9\cdot 53}\implies \cfrac{56}{477}$

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Required equation to determine two consecutive even number is $x^{2}-16=0$ and required numbers are 4 and 6.

Solution:

Given that product of two consecutive positive integer is 14 more that their sum. Need to create the equation and solve it to get two numbers.

Let’s assume first even number be represented by x.

So second consecutive even number will be $x + 2$

Sum of two consecutive number = $(x) + (x + 2)$

$\Rightarrow$ Sum of two consecutive number = $2x + 2$

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$\begin{array}{l}{\Rightarrow\left(x^{2}+2 x\right)-14=2 x+2} \\\\ {=>x^{2}+2 x-2 x-14-2=0} \\\\ {\Rightarrow x^{2}-16=0} \\\\ {\Rightarrow x^{2}-4^{2}=0} \\\\ {\Rightarrow x^{2}=4^{2}} \\\\ {\Rightarrow x=4}\end{array}$