The simplified expression of 2√8x³(3√10x⁴ - x√5x²) is 24x³√5x - 4x³√10x
<h3>How to determine the simplified product?</h3>
The complete question is added as an attachment
From the attached figure, the product expression is:
2√8x³(3√10x⁴ - x√5x²)
Evaluate the exponents
2√8x³(3√10x⁴ - x√5x²) = 2 *2x√2x(3x²√10 - x²√5)
Evaluate the products
2√8x³(3√10x⁴ - x√5x²) = 4x√2x(3x²√10 - x²√5)
Open the bracket
2√8x³(3√10x⁴ - x√5x²) = 12x³√20x - 4x³√10x
Evaluate the exponents
2√8x³(3√10x⁴ - x√5x²) = 24x³√5x - 4x³√10x
Hence, the simplified expression of 2√8x³(3√10x⁴ - x√5x²) is 24x³√5x - 4x³√10x
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The total cost when 881 minutes is used is $477.50.
<h3>What are the equation that model the question?</h3>
a + 480b = 277 equation 1
a + 990b = 532 equation 2
Where:
- a = flat fee
- b = variable fee
<h3>What is the flat fee and the variable fee?</h3>
Subtract equation 1 from equation 2
510b = 255
b = 255 / 510
b = $0.50
In order to determine the flat fee, substitute for b in equation 1
a + 480(0.5) = 277
a + 240 = 277
a = 277 - 240
a = $37
<h3>What is the total cost when 881 minutes is used?</h3>
Total cost = flat fee + (variable cost x number of minutes spoken)
$37 + (881 x 0.5) = $477.50
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50/80, then divide it by ten so the answer is 5/8. Once the number can divide both numbers your good.
If 32 is divided by 6, the answer is 5.3. However because she cannot hit part of a ball, the number would be 5. Therefore, she would hit 5 out of 32 balls.
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)
