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3 years ago

Pls answer this for me

Mathematics

1 answer:

nekit [7.7K]3 years ago

8 0

Answer:

a 166 67/100

b 20 1/4

c 829 77/100

d 664 47/50

e 208 4/25

Step-by-step explanation:

you need to find the common denominator (the number on the bottom) so for the first one when can 10 = 100 if you multiply 10 x 10 you get 100 so multiply the entire fraction by 10 and you will get 10/100 so the equation would be 381 10/100 - 214 43/100 which equals 166 67/100 then just do the same thing for the rest of them

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PLEASE HELP 100 PTS!

Dafna11 [192]

Answer:

infinite solutions

Distributive Property

Step-by-step explanation:

5x + 3(x − 5) = 6x + 2x − 15

Distribute

5x+3x -15 = 6x+2x -15

Combine like terms

8x -15 = 8x-15

Subtract 8x from each side

8x-8x-15 = 8x-8x-15

-15=-15

This is always true so there are infinite solutions

We used the Distributive Property

And the Subtraction Property of Equality

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3 years ago

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Ksivusya [100]

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3 years ago

Ms. Griffith took a cab ride which cost $16. She gave the driver $20 and said, "keep the change." What percent did Ms. Griffith

Sophie [7]

Answer:

25%

Step-by-step explanation:

Given data

Cost of ride= $16

Amount paid= $20

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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.

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

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

b)

A = 10,000, t = 3

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

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Pls answer this for me​

Pls answer this for me