$\frac{7x^{2}+14x-2}{(x-5)(x+4)(x+1)}$
$\frac{5x+2}{x^{2}-x-20}+\frac{2x-1}{x^{2}-4x-5}$
=$\frac{5x+2}{x^{2}-x-20}+\frac{2x-1}{x^{2}-4x-5}$
=$\frac{5x+2}{x^{2}-5x+4x-20}+\frac{2x-1}{x^{2}-5x+x-5}$
=$\frac{5x+2}{x(x-5)+4(x-5)}+\frac{2x-1}{x(x-5)+1(x-5)}$
=$\frac{5x+2}{(x-5)(x+4)}+\frac{2x-1}{(x-5)(x+1)}$
=$\frac{(5x+2)(x+1)+(2x-1)(x+4)}{(x-5)(x+1)(x+4)}$
=$\frac{7x^{2}+14x-2}{(x-5)(x+4)(x+1)}$
137
Let, total number of travelers = n (A U B U C)
Number of traveler who had been to Spain, n(A) = 55
Number of traveler who had been to France, n(B) = 53
Number of traveler who had been to Germany. n((C) = 79
Number of traveler who had been to Spain and France. n($A\bigcap_{}^{}B$) = 18
Number of traveler who had been to Spain aml Germany,n($A\bigcap_{}^{}C$)= 17
Number 01‘ traveler who had been to France and Germany. n ($B\bigcap_{}^{}C$) = 25
Number of traveler who had been to all three countries. n ($A\bigcap_{}^{}B\bigcup_{}^{}C$)= 10
SO, Travelers took part in the survey=$(A\bigcup_{}^{}B\bigcup_{}^{}C)$=n(A)+n(B)+ n(C)-n($A\bigcap_{}^{}B$)-n($B\bigcap_{}^{}C$)-n($A\bigcap_{}^{}C$)+n($A\bigcap_{}^{}B\bigcup_{}^{}C$)
= 55+53+79-18-17-25+10=197-60= 137
1%
Let, cost price of frist shirt = x Tk.
cost price of second shirt = y TK
According to the question
x+x$\times$10%= y-y$\times$10%
=> x+$\frac{10x}{100}$ = y- $\frac{10y}{100}$
=> $\frac{10x+x}{10}$ = $\frac{10y+y}{10}$
=> 11x = 9y
=> $\frac{x}{y} = \frac{9}{11}$
That means cost price of first shirt 9 Tk. & second shirt 11 Tk.
Total cost = 9 + 11 = 20 Tk.
Now at 10% profit on first shirt the selling price will be = 9 + 9 X 10% = 9.9Tk.
At 10% loss on second shirt the selling price will be = 11 – 11 x 10% = 9.9Tk.
Total selling price will be = 9.9 + 9.9 = 19.8 Tk.
Loss amount = 20 – 19.8 = 0.2 Tk.
Percentage of loss = $\frac{0.2}{20}$ x 100 = 1%
Loss 1% (ans)
3 cm
let. the tangent AB = 5cm
Distance at the point A from the centre the circle 0A = 4cm
now drawing a perpendicular from the tangent on the centre of the circel OB. we get a right angled triangle $\Delta$AOB
Appliying Pythagorean theorem
OA$^{2}$+OB$^{2}$= AB$^{2}$
=> 4$^{2}$+OB$^{2}$=5$^{2}$
=> OB$^{2}$= 25-16
=> OB$^{2}$ = 9 =3$^{2}$
So, OB = 3
The radius of the circle is 3cm.
x(x+4)
First portion,
$x^{3}-16x$
=$x(x^{2}-16)$
=$x(x^{2}-4^{2})$
=$x(x+4)(x-4)$
Second portion,
$2x^{3}+9x^{2} +4x$
=$x(2x^{2}+9x+4)$
=$x(2x^{2}+8x+1x+4)$
=x{2x(x+4)+1(x+4)}
=x(x+4)(2x+1)
Third portion
$2x^{3}+x^{2}-28x$
$= x (2x^{2} + x -28)$
$= x(2x^{2} + 8x -7x -28)$
= x{2x(x+4) – 7( x+4)}
=x(x+4)(2x-7)
Now, H.C.F = x(x+4)
