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Ann [662]
3 years ago
11

Which unit rates would be appropriate for describing the speed of cars in a race?

Mathematics
1 answer:
meriva3 years ago
5 0

Answer:

Miles per hours and feet per second.

Hope this helps! Jesus loves you! :)

-Seth


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The integral of (5x+8)/(x^2+3x+2) from 0 to 1
Gnom [1K]
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)
5 0
3 years ago
In​ March, a family starts saving for a vacation they are planning for the end of August. The family expects the vacation to cos
Makovka662 [10]
Deposited money:

March: 120
April: 120+(120×20)/100=120+24=144
May: 144+(144×20)/100=144+18,8=162,8
June: 162,8+(162.8×20)/100=162,8+32.56=195.36
July: 195.36+(195.36×20)/100=195.36+23.87=219.23
August: 219.23+(219.23×20)/100=219.23+43.85=263.08

120+144+162.8+195.36+219.23+263.08=1105.19
So the answer is no.
7 0
3 years ago
What’s the correct answer to 9(7+x)=100?
Natalija [7]
Answer: x= 37/9

Decimal Form: 4.1(repeating)

Mixed Number: 4 1/9
6 0
2 years ago
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Jill made chocolate shakes for a party she used 1/4 cup of milk for each shake how many ounces of milk did Jill need to make for
PIT_PIT [208]
Since we know that Jill needs 1/4 of a cup of milk per shake and that she wants to make 36 shakes, you would set up an expression that involves multiplying 1/4 and 36 together: 1/4*36
Then, when you multiply, you get 36/4 which simplifies to 9.
Your answer would be 9 ounces
6 0
3 years ago
A serving of crackers has 1.5 grams of fat.
Viefleur [7K]
D. You multiply 3.75x1.5 :)
4 0
3 years ago
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