The distance in miles if he runs 9 miles per hours is 4.5 miles.
Given, Man’s running speed = 9 miles/hour
Man’s walking speed = $\frac{9}{3}$ = 3 miles/hour
Given, Total time taken by the man to complete the journey is 2 hours.
Let, total distance covered by the man is x miles.
According to question,
$\frac{x}{9}+\frac{x}{3}$ = 2
=> $\frac{x+3x}{9}$ = 2
=> 4x = 18
=> x = $\frac{18}{4}$
x=4.5
Distance covered by the man is 4.5 miles.
Total interest is 466.75 Tk.
As the borrower pays 6.5 percent interest per year on the first Tk. 600 he borrowed
For first Tk. 600, interest, I$_{1}$= $\frac{600\times 6.5\times 1}{100}$ = 39 Tk.
As he apys 7.25% in excess of Tk. 600
Amount in excess of Tk. 600 = Tk. (6,500 – 600) = Tk. 5,900 We have
Interest, I$_{1}$ = $\frac{5900\times 7.25\times 1}{100}$ = 427.75 Tk.
Total interest I$_{1}$ = I$_{1}$ +I$_{2}$ = 39 + 427.75 = 466.75 Tk.
The area of the field 8,800 Sq. feet.
Let, the width = x feet
the length = x + 30 feet
We know, perimeter = 2(Length + breadth)
Perimeter = 2(x + 30 + x)
According to the question
2(x + 30 +x) = 380
=> 4x + 60 = 380
=> 4x = 380 – 60
x = 80 feet.
The length is = 80 + 30 = 110 feet
Area =110 x 80 Sq. feet
= 8,800 Sq. feet
The value is a$^{4}$- 4a$^{2}$+ 2.
Given $\sqrt{x}+\frac{1}{\sqrt{x}}$
=> ( $\sqrt{x}+\frac{1}{\sqrt{x}}$ )$^{2}$ = a$^{2}$
=> $(\sqrt{x})^{2}$+2$\times$$\sqrt{x}$$\times$$\frac{1}{\sqrt{x}}$+$(\frac{1}{\sqrt{x}})$$^{2}$ = a$^{2}$ [(a+b)$^{2}$ = a$^{2}$+2ab+b$^{2}$]
=> x+2+$\frac{1}{x}$ = a$^{2}$
=> x+$\frac{1}{x}$ = a$^{2}$-2
=>(x+$\frac{1}{x}$)$^{2}$ = (a$^{2}$-2)$^{2}$
=> x$^{2}$+2$\times$x$\times$+$\frac{1}{x}$+($\frac{1}{x}$)$^{2}$ = (a$^{2}$)$^{2}$-2$\times$a$^{2}$$\times$2+2$^{2}$
=> x$^{2}$+2+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+4
=> x$^{2}$+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+4-2
x$^{2}$+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+2
