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forsale [732]
3 years ago
12

A 10 foot pole casts a 15 ft shadow. At the same time, a tree casts a 24 foot shadow. How tall is the tree

Mathematics
2 answers:
Ludmilka [50]3 years ago
7 0

1. Set up a ratio for the pole comparing height to shadow length:

   10 feet tall / 15 feet shadow

2. Set this ratio equal to a similar ratio for the tree:

    \dfrac{10 \text{ feet tall}}{15 \text{ feet shadow}}  =\dfrac{x \text{ feet tall}}{24 \text{ feet shadow}}

 

3. Solve the equation by multiplying by 24 on both sides of the equation:

    24 \cdot \dfrac{10}{15}  = x

   

      x = 16 feet

hichkok12 [17]3 years ago
4 0

Answer:

41

Step-by-step explanation:

10=15

24=x

cross multiply

24×15=410

410÷10=41

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3 years ago
A model for the population in a small community after t years is given by P(t)=P0e^kt.
LUCKY_DIMON [66]
\bf \textit{Amount of Population Growth}\\\\&#10;A=Ie^{rt}\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;I=\textit{initial amount}\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\\&#10;\end{cases}

a)

so, if the population doubled in 5 years, that means t = 5.  So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

\bf \textit{Amount of Population Growth}\\\\&#10;A=Ie^{rt}\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &2\\&#10;I=\textit{initial amount}\to &1\\&#10;r=rate\\&#10;t=\textit{elapsed time}\to &5\\&#10;\end{cases}&#10;\\\\\\&#10;2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r&#10;\\\\\\&#10;\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\&#10;\textit{how long to tripling?}\quad &#10;\begin{cases}&#10;A=3\\&#10;I=1&#10;\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}

\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t&#10;\\\\\\&#10;\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t

b)

A = 10,000, t = 3

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3 years ago
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