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3 years ago

Please answer this correctly

Mathematics

1 answer:

kupik [55]3 years ago

5 0

Answer:

Look at the money bags below!!! (but I'll give you the answer)

Step-by-step explanation:

John F: 7 full bags - 1 half

Juan A: 9 full bags

Jason A: 3 full bags

Nick J: 3 full bags- 1 half

Alfonso S: 8 full bags

Hope this helped and wasn't confusing!!! xx - Asia

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Explain how to divide a over b divided by c over d

kow [346]

Flip c over d to d over c and multiply it so:
$\frac{a}{b} \times \frac{d}{c}$

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3 years ago

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Find the purchase price of a car if after 3 years it was worth $15000 and after 7 years it was worth $8000

navik [9.2K]

$\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{3 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&15000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &3\\ \end{cases} \\\\\\ 15000=P(1-r)^3 \\\\[-0.35em] ~\dotfill$

$\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{7 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&8000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &7\\ \end{cases} \\\\\\ 8000=P(1-r)^7 \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf 15000=P(1-r)^3\implies \cfrac{15000}{(1-r)^3}=P\qquad \qquad \stackrel{\textit{doing some substitution}}{8000=\left( \cfrac{15000}{(1-r)^3} \right)(1-r)^7} \\\\\\ \cfrac{8000}{15000}=\cfrac{(1-r)^7}{(1-r)^3}\implies \cfrac{8}{15}=(1-r)^{7-3}\implies \cfrac{8}{15}=(1-r)^{4}$

$\bf \sqrt[4]{\cfrac{8}{15}}=1-r\implies r=1-\sqrt[4]{\cfrac{8}{15}}\implies r\approx 0.15 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 15000=P(1-r)^3\implies 15000=P(1-0.15)^3\implies 15000=P(0.85)^3 \\\\\\ \cfrac{15000}{0.85^3}=P\implies 24424.995\approx P$

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3 years ago

1797 is 31% of a number. What number?

abruzzese [7]

You should
1797/g = 31/100
g= 1797 x 100 ÷ 31
g= 5796.77

8 0

3 years ago

Can u guys answer this and fast (3.6*10^2)*(2.8*10^2)

VashaNatasha [74]

so basically we only use the associative property of multiplication and one of the laws of exponents

associative=a(bc)=(ab)c
also law of exponent=a^x times a^y=a^(x+y)

(3.6 times 10^2) times (2.8 times 10^2)
associative
3.6 times 10^3 times 2.8 times 10^2
3.6 times 2.8 times 10^2 times 10^2
law of exponents
10^2 times 10^2=10^(2+2)=10^4
3.6 times 2.8=10.08
the answer is 10.08 times 10^4 or (10.08*10^4) or (1.001*10^5)

the second one


(3.6 times 10^6) times (2.2 times 10^-4)
associative
3.6 times 10^6 times 2.2 times 10^-4
3.6 times 2.2 times 10^6 times 10^-4
law of exponents
10^6 times 10^-4=10^(6-4)=10^2
3.6 times 2.2=7.92
the answer is 7.92 times 10^2 or (7.92*10^2)


A is (1.001*10^5)
 B is (7.92*10^2)

7 0

4 years ago

Evaluate the expression. × , when =13 and =12

densk [106]

Answer:

The answer you would be looking for is 156

Step-by-step explanation:

If M=13 and N=12 then it would just be 12x13 I hoped this helped

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3 years ago

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