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AleksAgata [21]
2 years ago
9

Can pls someone help me :(

Mathematics
2 answers:
Bumek [7]2 years ago
4 0

Answer:

tanF = \frac{77}{36}

Step-by-step explanation:

tanF = \frac{opposite}{adjacent} = \frac{EG}{FG} = \frac{77}{36}

Over [174]2 years ago
3 0

Answer:

ok multiply the two first the 85 and 77 and divide it by 36

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Bruno bought 96 pounds of sugar for his bakery. Every day he used the same amount of sugar to make bakery items. After 10 days,
Paladinen [302]
96 - 15.2 = 80.8
80.8 / 10 = 8.08

He was using up 8.08 pounds of sugar daily for his baking, so the answer is the second option.
6 0
3 years ago
I NEED HELP ON THIS QUESTION PLS.
STatiana [176]
Answer: F
Explanation: you already know you have a 90° and the straight lines means it’s 180° so you need to find x and z which are the same so 2x + 90=180 which you would then subtract 90 from 180 which would be 90 then divide 90 by two which would be 45
5 0
3 years ago
What is the length of side BC of the triangle? Make sure to show all of your work!
Mrrafil [7]
Isosceles triangle
so
x + 8 = 2x -1
x = 9

BC = 3x - 3
BC = 3(9) - 3
BC = 27 - 3
BC = 24
5 0
3 years ago
The price of an item has been reduced by 60%. The original price was $84
IceJOKER [234]

Answer:

$33.60

Step-by-step explanation:

First, we get 60 percent of 84. That is 50.40. Next, we do 84 minus 50.40, which is 33.60. Therefore, your answer is $33.60. Hope this helped!

4 0
2 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
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