Answer:
D. -x2 + 2x -3
Step-by-step explanation:
the first step is we open all the brackets then it will be :
x2 - 3x - 2x2 + 5x - 3
= x2 - 2x2 - 3x + 5x - 3
= -x2 + 2x -3
Answer:
hiiii
Step-by-step explanation:
but u have posted all the pictures upside down so can't understand anything
Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx
Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx
Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx
For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds
Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds
Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp
Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5
Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)
Which is equal to:
Answer: = log(18)
Answer:
Option C.
Step-by-step explanation:
It is given that radius of circle X is 6 units and radius of circle Y is 2 units.
Here, circle Y is smaller and circle X is larger.
If we translate the circles so they share a common center point, and after that we dilate circle Y by a scale factor of 3, then we get circle which is equal to circle X.
We know that, after dilation the figure and its image are similar.
Hence, we can say that both circles are similar.
Therefore, the correct option is C.
