Simply you would do
45+10 time the number of days
So
45+10 x 2 and then you would put the answer in the next Collin
Since
, we can rewrite the integral as

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

Both integrals are quite immediate: you only need to use the power rule

to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
3/4 (x-12)=12
3/4x - 9= 12
+9 +9
3/4x=21
*3/4 *3/4
x= 15 3/4
and
3/4y-12=12
+12 +12
3/4y = 24
*3/4 *3/4
y= 18
Answer:
91.42 crates
Step-by-step explanation:
Let
Greatest Number of 140kg crates = x
weight of crate = 140-kilogram
Weight of other shipment = 13200 kilograms
greatest weight of container = 26000 kilograms
Weight of other shipment + (weight of crate * number of crates) ≤ greatest weight of container
13,200 + 140x ≤ 26,000
140x ≤ 26,000 - 13,200
140x ≤ 12,800
x ≤ 12,800/140
x ≤ 91.42 kilograms
Greatest Number of 140kg crates = 91.42 crates
It would be the seconf answer