Question

Can someone help me with this? I dont know whats its trying to ask. It decreases by 1/2 every year. URGENT

Answer 1

this is just a quick addition to "jsimpson11000" good reply above

so we know the decrease is exponential, that means we have an equation about V = abᵗ, now, hmmm who knows what "ab" is, now, once we know that, then we can get the "slope" from t=2 to t=4, so let's use the table to get it.

`{\Large \begin{array}{llll} V=ab^t \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} t=3\\ V=100 \end{cases}\implies 100=ab^3 \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} t=5\\ V=25 \end{cases}\implies 25=ab^5\implies 25=ab^3b^2\implies \stackrel{\textit{substituting from above}}{25=(100)b^2} \\\\\\ \cfrac{25}{100}=b^2\implies \cfrac{1}{4}=b^2\implies \sqrt{\cfrac{1}{4}}=b\implies \boxed{\cfrac{1}{2}=b} \\\\[-0.35em] ~\dotfill`

`100=ab^3\implies 100=a\left( \cfrac{1}{2} \right)^3\implies 100=\cfrac{a}{8}\implies \boxed{800=a} \\\\\\ ~\hfill {\Large \begin{array}{llll} V=800\left( (1)/(2) \right)^t \end{array}}~\hfill`

now let's find the slope

`\begin{array}{llll} f(x)~from\\\\ x_1 ~~ to ~~ x_2 \end{array}~\hfill slope = m \implies \cfrac{ \stackrel{rise}{f(x_2) - f(x_1)}}{ \underset{run}{x_2 - x_1}}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array} \\\\[-0.35em] ~\dotfill\\\\ V(t)= 800\left( (1)/(2) \right)^t\qquad \begin{cases} t_1=2\\ t_2=4 \end{cases}\implies \cfrac{V(4)-V(2)}{4 - 2} \\\\\\ \cfrac{(50)~~ - ~~(200)}{2}\implies \text{\LARGE -75}`

so is a negative slope, because is Decay or decrement, however you're expected to enter it as positive, so in essence just the absolute value change.

Answer 2

Below

Explanation:

Year 0 = 800

Year 1 = 400

Year 2 = 200

Year 3 = 100

Year 4 = 50

Year 5 = 25

From year 2 to year 4 = 150 dollars decrease in 2 years = 75 dollars per year