Answer:
Option B. 5 hours
Step-by-step explanation:
The tennis club charges monthly fees = $50.00
and per hour charges of court time = $10.00
Lets assume you play number of hours in a month = x hr
So, per hour charges would be = $10x
Total payment = monthly charges + charges of played hours
$100.00 = $50.00 + 10x
10x = 100 - 50
10x = 50
x = 5
Option B. You used 5 hours court time.
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:
The amount of Polonium-210 left in his body after 72 days is 6.937 μg.
Step-by-step explanation:
The decay rate of Polonium-210 is the following:
(1)
Where:
N(t) is the quantity of Po-210 at time t =?
N₀ is the initial quantity of Po-210 = 10 μg
λ is the decay constant
t is the time = 72 d
The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.
First, we need to find the decay constant:
(2)
Where t(1/2) is the half-life of Po-210 = 138.376 days
By entering equation (2) into (1) we have:
Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.
I hope it helps you!
Answer:
r = 0.5 or 1/2
Step-by-step explanation:
Simple Interest Rate Formula: A = P(1 + r)^t
Simply plug in our known variables:
2700 = 800(1 + r)³
Now we solve for <em>r</em>:
<em>Divide both sides by 800</em>
27/8 = (1 + r)³
<em>Take the cube root on both sides</em>
∛27/8 = ∛(1 + r)³
<em>Simplify</em>
3/2 = 1 + r
<em>Subtract 1 on both sides</em>
r = 1/2
r = 0.5
Answer:
0.95? I hope it's correct
