1. A disease named Covid-19 starts with 50 virus and increases exponentially. The relationship between number of virus V and the elapsed time in days 'd' is found to be in the form of an equation i.e. — V=50 x 100/2
i. In how many days will the number of virus increase to reach 5,00,000 in number?
Express the above answer in the exponential form as am.
Can you give the number of days in hours?
Answers
Answer:
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answer
Topic :- This is your punishment for spamming in my question
Differentiation
To Differentiate :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
Solution :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}
dx
d(f(x))
=
dx
d(cotx⋅lnsecx)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅
dx
d(cotx)
+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)(∵
dx
d(fg)
=g⋅
dx
d(f)
+f⋅
dx
d(g)
)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅(−csc
2
x)+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)(∵
dx
d(cotx)
=−csc
2
x)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅
dx
d(secx)
\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)(∵
dx
d(lnt)
=
t
1
⋅
dx
dt
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅secx⋅tanx
\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)(∵
dx
d(secx)
=secx⋅tanx)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)
dx
d(f(x))
=−csc
2
x⋅lnsecx+(cotx⋅tanx)⋅(
secx
1
⋅
secx
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1
dx
d(f(x))
=−csc
2
x⋅lnsecx+1
(\because \cot x \cdot \tan x = 1)(∵cotx⋅tanx=1)
Answer :-
\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}
dx
d(f(x))
=1−csc
2
x⋅lnsecx
Note : csc x = cosec x
\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}
ANSWER
Correct Expression :-
\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}
6x
2
+8
x
3
+12x
=
9y
2
+27
y
3
+27y
To Find :-
\mathtt{x:y\:\:by\:using\:Componendo\:and\:Dividendo.}x:ybyusingComponendoandDividendo.
Concept Used :-
\mathtt{If\:\dfrac{a}{b}=\dfrac{c}{d},then\:using\:Componendo\:and\:Dividendo}If
b
a
=
d
c
,thenusingComponendoandDividendo
\mathtt{\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}
a−b
a+b
=
c−d
c+d
Solution :-
\mathtt{\dfrac{x^3+12x}{6x^2+8}=\dfrac{y^3+27y}{9y^2+27}}
6x
2
+8
x
3
+12x
=
9y
2
+27
y
3
+27y
Using Componendo and Dividendo,
\mathtt{\dfrac{x^3+12x+(6x^2+8)}{x^3+12x-(6x^2+8)}=\dfrac{y^3+27y+(9y^2+27)}{y^3+27y-(9y^2+27)}}
x
3
+12x−(6x
2
+8)
x
3
+12x+(6x
2
+8)
=
y
3
+27y−(9y
2
+27)
y
3
+27y+(9y
2
+27)
Opening brackets,
\mathtt{\dfrac{x^3+12x+6x^2+8}{x^3+12x-6x^2-8}=\dfrac{y^3+27y+9y^2+27}{y^3+27y-9y^2-27}}
x
3
+12x−6x
2
−8
x
3
+12x+6x
2
+8
=
y
3
+27y−9y
2
−27
y
3
+27y+9y
2
+27
Rewriting it,
\mathtt{\dfrac{x^3+3(2)^2(x)+3(2)(x^2)+2^3}{x^3+3(2^2)x-3(2)(x^2)-2^3}=\dfrac{y^3+3(3^2)(y)+3(3)(y^2)+3^3}{y^3+3(3^2)(y)-3(3)(y^2)-3^3}}
x
3
+3(2
2
)x−3(2)(x
2
)−2
3
x
3
+3(2)
2
(x)+3(2)(x
2
)+2
3
=
y
3
+3(3
2
)(y)−3(3)(y
2
)−3
3
y
3
+3(3
2
)(y)+3(3)(y
2
)+3
3
We know that,
\mathtt{(a+b)^3=a^3+3ab^2+3a^2b+b^3}(a+b)
3
=a
3
+3ab
2
+3a
2
b+b
3
\mathtt{(a-b)^3=a^3+3ab^2-3a^2b-b^3}(a−b)
3
=a
3
+3ab
2
−3a
2
b−b
3
Using Identities,
\mathtt{\dfrac{(x+2)^3}{(x-2)^3}=\dfrac{(y+3)^3}{(y-3)^3}}
(x−2)
3
(x+2)
3
=
(y−3)
3
(y+3)
3
\mathtt{\left(\dfrac{x+2}{x-2}\right)^3=\left(\dfrac{y+3}{y-3}\right)^3}(
x−2
x+2
)
3
=(
y−3
y+3
)
3
Taking Cube Root on both sides,
\mathtt{\sqrt[3]{\left(\dfrac{x+2}{x-2}\right)^3}=\sqrt[3]{\left(\dfrac{y+3}{y-3}\right)^3}}
3
(
x−2
x+2
)
3
=
3
(
y−3
y+3
)
3
\mathtt{\dfrac{x+2}{x-2}=\dfrac{y+3}{y-3}}
x−2
x+2
=
y−3
y+3
Using Componendo and Dividendo again,
\mathtt{\dfrac{x}{2}=\dfrac{y}{3}}
2
x
=
3
y
We can write it as,
\mathtt{\dfrac{x}{y}=\dfrac{2}{3}}
y
x
=
3
2
Answer :-
Hence, x : y is equivalent to 2 : 3.
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Answer:
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