V(t)=27,500(0.90)t. Find the value after 11 years

Mathematics

1 answer:

Alika [10]4 years ago

4 0

Answer:

8,630

Step-by-step explanation:

$V(t) = 27,500(0.90)^t$

$V(11) = 27,500(0.90)^{11}$

$V(11) = 27,500(0.3138)$

$V(11) = 8,630$

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Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?

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$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}$

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$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2$