Can someone help me with this? Please

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

I'm kind of confused what the question is asking. If they are asking for an expression to represent JL then the answer is 14x -32. If they want to know the distance between J and L then the answer is 66

Step-by-step explanation:

First I set the 2 equations equal to each other (5x-2 = 9x -30) and solved this to get x =7. (ANSWER ONE)

Next I plugged 7 into one of the equations (in this case either one will work bc JK and KL are the same distance. so I plugged 7 into the first one ( 5(7) -2 ) and I got 33. This means the distance of JK (and KL since they are equal) is 33.

Next I multiplied 33 by 2 since JL is JK and KL combined ( JK and KL are both 33 and therefore adding them together (multiplying 33×2) will give you the length of JL) and I got 66 (ANSWER 2)

Sorry if this is super confusing

You might be interested in

Simplify: (8^2/3)^4

tia_tia [17]

Hello!

The answer is 207,126 10/81!

Hopefully this helps! :D

7 0

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]

$\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}$

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.

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

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

b)

A = 10,000, t = 3

$\bf \begin{cases}
A=10000\\
t=3
\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I
\\\\\\
6597.53955 \approx I$

3 0

3 years ago

Which ordered pair is a solution to the inequality? Help please!!

Murrr4er [49]

Choice A is the answer which is the point (1,-1)

=========================================

How I got this answer:

Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.

-------------

checking choice A

plug in (x,y) = (1,-1)

$y - 2x \le -3$