MATHEMATICS
MIDDLE SCHOOL

Answer: 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)

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)

MIDDLE SCHOOL

There is a drought and the oak tree population is decreasing at the rate of 6% per year. If the population continues to decrease at the same rate, how long will it take for the population to be half of what it is? If necessary, round your answer to the nearest tenth.

Answer:

in approximately 10 years the oak tree population will be 50% of what it is. because 5 x 10 = 50. hope this helps :)

Step-by-step explanation:

MIDDLE SCHOOL

What is the definition of rotation

Rotation is the action of turning/spinning around an axis or center.

MIDDLE SCHOOL

Clara estimates that the length Of a hiking trail is 5.75 kilometers. She later learnsthat its Actual length is 6.25 km. What is the percent error in Clara's estimate, to the nearest tenth of a percent? A. 7.7%

B. 8.0%

C. 11.0%

D. 11.5%

When you calculate results that are aiming for known values, the percent error formula is useful tool for determining the precision of your calculations. The formula is given by: The experimental value is your calculated value, and the theoretical value is your known value.

% error = | ( experimental – theoretical )/ theoretical| * 100

% error = | ( 5.75 – 6.25)/ 6.25| * 100

% error = 8.0 %

8.0

Hope this helped it probbaly didnt just have a nice day

HIGH SCHOOL

Find the annual interest rate. I=$54, P=$900, t=18 months

Well, there are 12 months in a year, so 18 months will then be 18/12 of a year

Answer:

4%

Step-by-step explanation:

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