R(t) = integral of r'(t) = integral of ti + e^tj + te^tk = 1/2t^2i + e^tj + (te^t - e^t)k + c
r(0) = j - k + c = i + j + k
c = i + 2k
Therefore, r(t) = (1/2t^2 + 1)i + e^tj + (te^t - e^t + 2)k
Differentiating the function
... g(x) = 5^(1+x)
we get
... g'(x) = ln(5)·5^(1+x)
Then the linear approximation near x=0 is
... y = g'(0)(x - 0) + g(0)
... y = 5·ln(5)·x + 5
With numbers filled in, this is
... y ≈ 8.047x + 5 . . . . . linear approximation to g(x)
Using this to find approximate values for 5^0.95 and 5^1.1, we can fill in x=-0.05 and x=0.1 to get
... 5^0.95 ≈ 8.047·(-0.05) +5 ≈ 4.598 . . . . approximation to 5^0.95
... 5^1.1 ≈ 8.047·0.1 +5 ≈ 5.805 . . . . approximation to 5^1.1
[(7**13)**3]*[7**0]
[7**39]*[1]..........> 9.0954 E 32 Strawberries in the field
Answer:ABD is equal to CDB
Step-by-step explanation:
The reason you can tell is if you was to fold them both they would match up