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Ymorist [56]
2 years ago
9

A boat's value over time is given as the function f(x) and graphed below. Use A(x) = 400(b)x + 0 as the parent function. Which g

raph shows the boat's value increasing at a rate of 25% per year?

Mathematics
1 answer:
Lelu [443]2 years ago
3 0

Answer:

Last Graph (4th one)

Step-by-step explanation:

The equation given is  A(x)=400*b^x + 0\\A(x) = 400b^x

Since we want increase of 25%, the value of b would be 1.25, so we have:

A(x)=400(1.25)^x

The graph should be increasing so we can immediately eliminate 2nd and 3rd graphs since they are clearly decreasing. We now have to choose between 1st graph and 4th graph.

<u>1st graph:</u>

Let's take the point (-20,30)

A(x)=400(1.25)^{-20}\\A(x)=4.65

<u>2nd graph:</u>

Let's take the point (-20,4.5)

A(x)=400(1.25)^{-20}\\A(x)=4.65

hence, this makes sense. So the correct graph is 4th one.

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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

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3 years ago
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