A wire was bent into the shape of a rectangle with width 5 and length 7 . If this wire

Mathematics

2 answers:

kvasek [131]3 years ago

7 0

Answer:

i think the answer it 6

Step-by-step explanation:

because squares have the same sides so a 6 by 6 square would work.

Flauer [41]3 years ago

3 0

Answer:

Step-by-step explanation:

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erik [133]

Answer:

<u>Exponential model</u>

$y=Ae^{rt}$

where:

y = value at "t" time
A = initial value
r = rate of growth/decay
t = time (in years)

Given:

y = 100 g
t = 0 years

Substituting given values into the formula and solving for A:

$\begin{aligned}y & =Ae^{rt}\\\implies 100 & = Ae^{r \times 0}\\100 & = Ae^0\\100 & = A(1)\\A & = 100\end{aligned}$

Given:

A = 100 g
y = 50 g when t = 30.17

Substituting the given values into the equation and solving for r:

$\begin{aligned}y& =Ae^{rt}\\\\\implies 50 & =100e^{30.17r}\\\\\dfrac{1}{2} & = e^{30.17r}\\\\ln \dfrac{1}{2} & = \ln e^{30.17r}\\\\\ln 1-\ln2 & =30.17r \ln e\\\\0-\ln 2 & =30.17r(1)\\\\-\ln 2 & =30.17r\\\\r & = \dfrac{-\ln 2}{30.17}\end{aligned}$

Therefore, the final equation is:

$y=100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}$

<h3><u>Question 1</u></h3>

Q: From 100g how much remains in 80 years?

$\begin{aligned}t=80 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)80}\\& = 15.91389949 \: \sf g\end{aligned}$

Q: How long will it take to have 10% remaining?

10% of 100 g = 10 g

$\begin{aligned}y=10 \implies 10 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{10} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{10} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 10}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 100.2225706\: \sf years\end{aligned}$

<h3><u>Question 2</u></h3>

Q: How much remains after 50 years (time)?

<u></u>

$\begin{aligned}t=50 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)50}\\& = 31.70373153 \: \sf g\end{aligned}$

Q: How long to reach 20 g (amount remaining)?

<u></u> $\begin{aligned}y=20 \implies 20 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{5} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{5} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 5}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 70.05257062\: \sf years\end{aligned}$