Are is 16/14 and 64/60
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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:
y = x -- 2
Step-by-step explanation
We need to fill out the formula y = mx + b.
First, we must find out what each part of that formula means.
mx = slope
b = y-intercept
y = solution (Always unknown)
We already know the slope of the formula, so we can fill that out.
y = x + b
If it has a slope of -3/2, then that means it is moving down 3 points and moving to the right 2 points to find the next line. because rise over run.
If it passes through the point (-4, 4) then from their subtract 3 to 4, and 2 to -4.
This means the next point it will reach is (-2, 1). Once again, subtract 3 from 1, and add 2 to -2.
The next point it will reach is (0, -2) This gives us the y-intercept of the equation.
Now we can fill out the rest of the formula!
y = x -- 2