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:
See Below
Step-by-step explanation:
The surface area of cylinder is given by the formula:
Where
r is radius ( diameter is 4, so radius is 4/2 = 2)
h is height ( h = 9)
Lets find original surface are:
<u>Halving diameter:</u>
diameter would be 4/2 = 2, so radius would be 2/2 = 1
So, SA would be:
<u>Halving height:</u>
Height is 9, halving would make it 9/2 = 4.5
Now, calculating new SA:
Original SA is ,
Halving diameter makes it
Halving height makes it
So, halving diameter does not have same effect as halving height.