Which real number property is illustrated here? *

Given the function f(x) = 5^x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3. Part A: Find the average r

Masteriza [31]

Part A)

f(x) = 5^x
f(0) = 5^0
f(0) = 1
----------
f(x) = 5^x
f(1) = 5^1
f(1) = 5
----------
f(x) = 5^x
f(2) = 5^2
f(2) = 5*5
f(2) = 25
----------
f(x) = 5^x
f(3) = 5^3
f(3) = 5*5*5
f(3) = 125

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Rate of change for section A = (f(1) - f(0))/(1 - 0)
Rate of change for section A = (5 - 1)/(1 - 0)
Rate of change for section A = 4/1
Rate of change for section A = 4

Rate of change for section B = (f(3) - f(2))/(3 - 2)
Rate of change for section B = (125 - 25)/(3 - 2)
Rate of change for section B = 100/1
Rate of change for section B = 100

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Part B)

From part A) above, we found,

Rate of change for section A = 4
Rate of change for section B = 100

Which means that section B's rate of change is 25 times greater (since 100/4 = 25, or 25*4 = 100)

Answer for part B: 25

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Extra: Explain why one rate of change is greater than the other.

The rate of change for section B is larger because the exponential function is growing faster as x increases. This is shown visually by the sharper and steeper incline as the function curve goes upward. The function starts off with relatively slower growth but it accelerates in speed.

6 0

3 years ago

Question 1 [5 points]

RideAnS [48]

If I’m not wrong, it should be 1000-5+9.10% and add it to the tune and divide with the 9.90% and -3 and then you can get your answer

6 0

3 years ago

The sum of the first 10 terms of an arithmetic sequence is 235 and the

Degger [83]

Answer:

$\bold{a_1 = -42.65}\\\\\bold{d=14.7}$

Step-by-step explanation:

The sum of the first 10 terms of an arithmetic sequence is:

$S_{10}=a_1+a_2+...+a_{10}=\dfrac{a_1+a_{10}}{2}\cdot10=\dfrac{2a_1+(10-1)d}{2}\cdot10$

$\dfrac{2a_1+(10-1)d}{2}\cdot10=235\\\\(2a_1+9d)\cdot5=235\\\\2a_2+9d=47$

the sum of the second 10 terms is: a₁₁ + a₁₂+...+ a₂₀

And the sum of the first 20 terms of an arithmetic sequence is:

$S_{20}=a_1+a_2+...+a_{10}+a_{11}+...+a_{20}=\dfrac{2a_1+(20-1)d}{2}\cdot10$

so the sum of the second 10 terms is:

$a_{11}+a_{12}+...+a_{20}=S_{20}-S_{10}$

Therefore we have:

$\dfrac{2a_1+(20-1)d}{2}\cdot10-\dfrac{2a_1+(10-1)d}{2}\cdot10=735\\\\(2a_1+19d)\cdot5-(2a_1+9d)\cdot5=735\\\\2a_1+19d-(2a_1+9d)=147\\\\10d=147\\\\d=14.7$