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
<span>Answer:
a million) d/154 = 14/40 4 bypass multiply: 44d = 2156 d = 40 9 2) i will apply ^ to signify exponents, because of the fact superscript numbers do no longer artwork nicely at right here. c is continuous and represents the ratio of v to d^3 the dating is, subsequently: v = cd^3 3) enable w = wages for 5 adult adult males w/5 = $fifty two.08/3 bypass multiply: 3w = $260.40 w = $86.80 4) enable x = value of a 9-minute call x/9 = $0.ninety/6 bypass multiply: 6x = $8.10 x = $a million.35 5) i'm no longer completely confident what's being asked for in this one, yet i'm vulnerable to circulate with: P/a million = V/ok or P:a million = V:ok</span>
Answer:
dfgsdfgdsfgsd
Step-by-step explanation:
1/2 cup = 3/6 cup
1/3 cup =2/6 cup
Just dump a much from the 1/2 cup into the 1/3 cup.
The 1/2 cup will have 1/6 left.
Answer:
-8
Step-by-step explanation:
1. x + 23 = 15
2. x + 23 = 15
-23 -23
3. x = -8
