5. Find the product - 2x² (3x² - 4x+7)
2 answers:
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>
You might be interested in
Answer:
A) 3%
B) Product A
Step-by-step explanation:
<u>Exponential Function</u>
General form of an exponential function:
where:
- a is the initial value (y-intercept)
- b is the base (growth/decay factor) in decimal form
- x is the independent variable
- y is the dependent variable
If b > 1 then it is an increasing function
If 0 < b < 1 then it is a decreasing function
<u>Part A</u>
<u>Product A</u>
Assuming the function for Product A is <u>exponential</u>:
The base (b) is 1.03. As b > 1 then it is an <u>increasing function</u>.
To calculate the percentage increase/decrease, subtract 1 from the base:
⇒ 1.03 - 1 = 0.03 = 3%
Therefore, <u>product A is increasing by 3% each year.</u>
<u>Part B</u>
To calculate the percentage change in Product B, use the percentage change formula with two consecutive values of f(t) from the given table:
Check using different two consecutive values of f(t):
Therefore, as 3% > 1%, <u>Product A recorded a greater percentage change</u> in price over the previous year.
Although the question has not asked, we can use the given information to easily create an exponential function for Product B.
Given:
- a = 10,100
- b = 1.01
- n = t - 1 (as the initial value is for t = 1 not t = 0)
To check this, substitute the values of t for 1 through 4 into the found function:
As these values match the values in the given table, this confirms that the found function for Product B is correct and that <u>Product B increases by 1% per year.</u>
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
__
(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
__
(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
__
(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
