Class 11 (CBSE/NCERT)
Mathematics
Limits And Derivatives
can anyone please help me to solve these questions, please
detailed answers
(step by step, if possible with formula, if possible all steps, please no final answer)
find the derivative of
★ cos⁶x + sin⁵x
★ tan 3x + sec 3x
★ 3sinx + x cos x
★sec²x + tan x
★cos ^m n + x sin x
★ x cos²x
Answers
Step-by-step explanation:
Explanation:
Diagram:-
\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf D}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4){\bf A}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf B}\put (4,4.2){\sf a\:cm}\put (6.5,2){\sf b\;cm}\end{picture}
Given:-
The ratio of two sides of a parallelogram is 3:5.
perimeter of the parallelogram =64cm
To find:-
Length of sides
Solution:-
Let the sides be 3x and 5x
As we know that in a parallelogram
\boxed {\sf Perimeter =2 (a+b)}
Perimeter=2(a+b)
Where a and b are the sides of the parallelogram
Substitute the values
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\longrightarrow 2 (3x+5x)=64 \end{gathered}
:⟶2(3x+5x)=64
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\longrightarrow 6x+10x=64 \end{gathered}
:⟶6x+10x=64
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\longrightarrow 16x=64 \end{gathered}
:⟶16x=64
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\longrightarrow x=\dfrac {64}{4}\end{gathered}
:⟶x=
4
64
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\longrightarrow x=16 \end{gathered}
:⟶x=16
________________________________
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\rightarrowtail 3x=3\times 4 =12cm \end{gathered}
:↣3x=3×4=12cm
\begin{gathered}\\\qquad\qquad\displaystyle\sf {:}\rightarrowtail 5x=5\times 4=20cm \end{gathered}
:↣5x=5×4=20cm
\begin{gathered}\\\\\therefore\sf Sides\:of\:the\:parallelogram \:are\:12cm\:and\:20cm.\end{gathered}
∴Sidesoftheparallelogramare12cmand20cm.
Step-by-step explanation:
click on above page you will got the solution
