∫ e^(3x)*(cosh(2x)dx
= ∫ [e^(3x)*(e^(2x)+e^(-2x))/2]dx
= ∫ [(e^(5x)+e^x)/2]dx
=e^(5x)/10+e^x/2+C
=(1/10)(e^(5x)+5e^x)+C
The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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Answer:
1.7
Step-by-step explanation:
1/2 (2 x anything) = anything
1/2 cancels out the 2
Answer:
The solved given expression is
Step-by-step explanation:
Given expression is 
To solve the given expression :
Therefore
Therefore the solved given expression is
Factor out 18.
The factors for 18 are as listed:
1, 2, 3, 6, 9, 18
Now factor out 30.
The factors for 30 are as listed:
1, 2, 3, 5, 6, 10, 15, 30
Now look at the factors that are both factors for 18 and 30.
So, the common factors are 2, 3, and 6.
