Simplify the problems

Mathematics

1 answer:

antoniya [11.8K]3 years ago

6 0

For most of them you can use a fraction simplifier calculator but so far i have b) 21/2 c) 203/2

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If I have $40 and a cookie is $3. How many cookies can I get and how much money would I have left.

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13 cookies with 33 cents left over

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The chorus at an elementary school is made up on fifth and sixth graders. This year, the chorus had 12 fifth-grade girls, 9 fift

Sveta_85 [38]

Answer:

A. Ratio boys to girls ---> 19 : 26

B. Ratio of 5th grade to 6th grade ---> 21 : 24

C. Ratio of girls to total number of chorus students ---> 26 : 45

Step-by-step explanation:

A. the amount of boys is 9 + 10, the amount of girls is 12 + 26

B. the amount of 5th graders is 12 + 9, the amount of 6th graders is 14 + 10

C. the number of girls is 12 + 14, and the number of chorus students is 12+9+14+10

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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]

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

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

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

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Consider the given function and the given interval. f(x) = (x − 4)2, [3, 6]

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Answer:

f average = 1

smaller value c = 3

larger value c = 5

Step-by-step explanation:

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