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:
21
Step-by-step explanation:
He will guess correctly 25% of the time.
He has 84 questions
25% of 84
.25 * 84
21
We should expect that he will get 21 questions correct
2ways you can solve that is by multiplying 9 by 1 then by 5 or 9by 5 then by 1
Answer:
246 ft is the maximum height
Step-by-step explanation:
The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t² + bt + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex in the question s(t) = 124 + 64t − 16t², is at t
t = -64/2(-16) = 64/32 = 2 seconds
Thus, 2 seconds after the object was thrown, it reaches its highest point (maximum value of h) which is given by
h = -16(2)² + 64 (2) + 124 = 246eet
Answer:
you can be my bestie
Step-by-step explanation:
hello
