? persons can finish work in=35 days
$= \dfrac{100\times7}{35}$
$=20persons$
1 hour=$60 \times 60=3600seconds$
therefore,Percentage of hour in 36 seconds= $\dfrac{36}{3600}\times100$=1%
$\dfrac{x^{m^{2}}\times x^{mn}}{x^{nm}\times x^{n^{2}}}\times \dfrac{x^{n^{2}\times x^{ln}}}{x^{ln}\times x^{l^{2}}}\times \dfrac{x^{l^{2}}\times x^{lm} }{x^{lm}\times x^{m^{2}} }=1$
$Given question becomes,\dfrac{1}{5}\times2\div\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}=?$
$=[\dfrac{1}{5}\times2\times\dfrac{5}{1}]+[\dfrac{1}{5}-\dfrac{1}{10}]$
$=2+[\frac{2-1}{10}]$
$=2+\frac{1}{10}$
$=2+0.1$
$=2.1$
$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\div \dfrac{x^{2}+x+1}{3(x+3)}$
For multiply the fraction will get reciprocal
$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\times \dfrac{3(x+3)}{x^{2}+x+1}$
=3(x-1)
=3x-3$Geometric progression series$
$r=\dfrac{x^{\dfrac{3}{2}}}{x}=x^{\dfrac{1}{2}}$
$t_{n}=a\times r^{n-1}=l$
$x \times [x^{\dfrac{1}{2}}]^{85-1}=x\times x^{\dfrac{1}{2}\times84} $
$=x\times x^{42}$
$=x^{43}$
$l^{2}=r^{2}+h^{2}$
$h^{2}=l^{2}-r^{2}$
$h=\sqrt{l^{2}-r^{2}}$
$V=\dfrac{1}{3}\pi r^{2}\sqrt{l^{2}-r^{2}}$
Arithmetic Progresson series
$t_{n}=a+(n-1)d=l$
a=100, n=20, d=-5
$ t_{n}=(100)\times(20)\times(-5)$
=100-100
=0$=2x^4-162:\quad 2\left(x^4-81\right)$
$=\mathrm{Factor}\:2x-6:\quad 2\left(x-3\right)$
$=\frac{2\left(x^4-81\right)}{2\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1$
$\frac{x^4-81}{\left(x^2+9\right)\left(x-3\right)}$
$\:x^4-81:\quad \left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$x^4-81$
$x^4-9^2$
$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^4-9^2=\left(x^2+9\right)\left(x^2-9\right)$
$=\left(x^2+9\right)\left(x^2-9\right)$
$\mathrm{Factor}\:x^2-9:\quad \left(x+3\right)\left(x-3\right)$
$=\left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$=\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:}\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}:\quad x+3$
$\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:the\:common\:factor:}\:x-3$
$\frac{\left(x+3\right)\left(x^2+9\right)}{x^2+9}$
$\mathrm{Cancel\:the\:common\:factor:}\:x^2+9$
$=x+3$
LCM*HCF=Product of two numbers
$Z \times HCF=X \times Y$
$HCF=\dfrac{X \times Y}{2}$
l. Senkuttuvan
2. Emayavaramban
3. Kanaikal Irumporai
4. Kuttuvan Cheral
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