You will need $4270 more
(9500+1500+1870)-(6100+2500)
12870-8600
4270
Answer:
we would need to know what the experiment was or we can't awnser
Answer:
-19.68
Step-by-step explanation:
(0.2 + (-5) × (4.1)
-4.8 × (4.1)
-19.68
0.08 equals 2/25
0.4 equals 2/5
Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector


Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt

After integrating using power rule, replace 'u' with function for 't' and evaluate limits: