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3 years ago

I need help with this... 4 more than some number d. and 3 is less than X. and the others :(

Mathematics

1 answer:

Greeley [361]3 years ago

3 0

2) C: 4+d

3) A: x-3

4) C: 5(1-f)

I know that this is a few weeks late...

but hope this helps!

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The price of products may increase due to inflation and decrease due to depreciation. Marco is studying the change in the price

likoan [24]

Answer:

A) 3%

B) Product A

Step-by-step explanation:

Exponential Function

General form of an exponential function: $y=ab^x$

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

Part A

Product A

Assuming the function for Product A is exponential:

$f(x) = 0.69(1.03)^x$

The base (b) is 1.03. As b > 1 then it is an increasing function.

To calculate the percentage increase/decrease, subtract 1 from the base:

⇒ 1.03 - 1 = 0.03 = 3%

Therefore, product A is increasing by 3% each year.

Part B

$\sf percentage\:change=\dfrac{final\:value-initial\:value}{initial\:value} \times 100$

To calculate the percentage change in Product B, use the percentage change formula with two consecutive values of f(t) from the given table:

$\implies \sf percentage\:change=\dfrac{10201-10100}{10100}\times 100=1\%$

Check using different two consecutive values of f(t):

$\implies \sf percentage\:change=\dfrac{10303.01-10201}{10201}\times 100=1\%$

Therefore, as 3% > 1%, Product A recorded a greater percentage change 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)

$\implies f(t) = 10100(1.01)^{t-1}$

To check this, substitute the values of t for 1 through 4 into the found function:

$\implies f(1) = 10100(1.01)^{1-1}=10100$

$\implies f(2) = 10100(1.01)^{2-1}=10201$

$\implies f(3) = 10100(1.01)^{3-1}=10303.01$

$\implies f(4) = 10100(1.01)^{4-1}=10406.04$

As these values match the values in the given table, this confirms that the found function for Product B is correct and that Product B increases by 1% per year.

4 0

1 year ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20-%20%20%5Cfrac%7B2%7D%7B9%7D%20%29%28x%20%2B%20%20%5Cfrac%7B1%7D%7B2%7

Aleonysh [2.5K]

$f(x) = (x - \dfrac{2}{9} )(x + \dfrac{1}{2} )$

$\text {When } f(x) = 0 :$

$(x - \dfrac{2}{9} )(x + \dfrac{1}{2} ) = 0$

$(x - \dfrac{2}{9} ) = 0 \text { or } (x + \dfrac{1}{2} ) = 0$

$x = \dfrac{2}{9} \text { or } x = -\dfrac{1}{2}$

3 0

3 years ago

?? So lost with this

labwork [276]

Answer:

$A)\ 1,\ -1,\ -3\\\\B)\ d=-2\\\\C)\ \left\{\begin{array}{ccc}a_1=11\\a_n=a_{n-1}-2\end{array}\right\\\\D)\ f(n)=-2n+13$