I think these are correct:
1) C
2) D
3) C
4) D
5) A
6) C
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
Answer:
w = 32
Step-by-step explanation:
Remark
The value of the exterior angle (marked w + 33) = the sum of the two angles not connected to it. w and w + 1
Equation
w + w + 1 = w + 33
Solution
2w + 1 = w + 33 Subtract w from both sides.
2w - w + 1 = 33 Combine
w + 1 = 33 Subtract 1 from both sides.
w = 33 - 1
w= 32
15.4 seconds = 0.00428 hours and 100 meters = 0.062 miles soo the student ran approximately 14.5 mph