qual numero inteiro representa uma divida de 58 reais
A) 58
B) 0
C) -58
D) nenhum
Answers
Answer:
jana hai na to ➡➡➡➡➡➡➡
Step-by-step explanation:
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.
\begin{gathered}\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}} \\ \end{gathered}