Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
![\int_a^b f(r) |r'| dt](https://tex.z-dn.net/?f=%5Cint_a%5Eb%20f%28r%29%20%7Cr%27%7C%20dt)
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
![\sqrt{(x')^2 + (y')^2 + (z')^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x%27%29%5E2%20%2B%20%28y%27%29%5E2%20%2B%20%28z%27%29%5E2%7D)
![\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt](https://tex.z-dn.net/?f=%5Cint_0%5E1%20%282t%2B9t%5E3%29%20%5Csqrt%7B1%2B4t%5E2%20%2B9t%5E4%7D%20dt%20)
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
![\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du](https://tex.z-dn.net/?f=%5Cint_0%5E1%20%5Cfrac%7B2t%2B9t%5E3%7D%7B8t%2B36t%5E3%7D%20%5Csqrt%7Bu%7D%20%20du%20%5C%5C%20%20%5C%5C%20%5Cint_0%5E1%20%5Cfrac%7B2t%2B9t%5E3%7D%7B4%282t%2B9t%5E3%29%7D%20%5Csqrt%7Bu%7D%20%20du%20%5C%5C%20%20%5C%5C%20%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E1%20%5Csqrt%7Bu%7D%20%20du)
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
14:1
?:2 1/2
14:1
x x
2 2
1/2 1/2
35 : 2/12 35 miles Put me as BRAINLIEST!!!!
The answer is B to multiply by 4
Explanation
1 x 4 is 4
4 x 4 is 16
16 x 4 is 64
And 64 x 64 is 256
Hope this helps
Answer: The correct answer would be $17,109.
Step-by-step explanation:
The reasoning behind this answer is quite simple and easy to approach.
The suggested retail price is the base price for the vehicle and any other prices presented in the problem would be in addition to the base price.
So, with this information, we know that the base price is $15,250 and that the additional options would be $359 + $800 + $210 + $490.
Therefore, after completing the math for the additional options and adding them to the base price, you will receive the final cost of $17,109.
Answer:
Volume of slice is 40 in³
Volume of the remaining cake is 197.014 in³
HOPE THIS HELPS!