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:
a=5.25
Step-by-step explanation:
Answer:
See below for answers and explanations
Step-by-step explanation:
Top left: Since y can't be greater than 0 but is equal to 0, then the range is (-∞,0] and the domain is (-∞,∞) since there are no domain restrictions
Top right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)
Bottom left: Since y cannot be less than 2 but equal to it, and x holds no domain restrictions, then the domain is (-∞,∞) and the range is [2,∞)
Bottom right: Since both x and y have no restrictions, then the domain is (-∞,∞) and the range is (-∞,∞)
Answer:
pick a graph from google
Step-by-step explanation:
there is prablly some u can pick from
I think you have to follow someone and ask them to be your tutor.
