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4 years ago

Evaluate: (-2/3*3/4) -1

Mathematics

1 answer:

slamgirl [31]4 years ago

5 0

Answer:

$= ( - \frac{2}{3} \times \frac{3}{4}) \times ( - 1)$

$= \frac{( - 2) \times ( - 1)}{4}$

$= \frac{ + 2}{4}$

$= + \frac{1}{2}$

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Practice solving quadratic equations.

Dennis_Churaev [7]

In order for the equation to be true, one of the factors on the left has to equal zero

So, 2x-3=0 or x-4=0 (the 3 on the left for multiplication doesn’t affect the possible answers for x)

Once you solve them, you conclude that x has to be 4 or 3/2.

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3 years ago

Writing the quation for a line that goes through the following points (0,3) (1,1

Tomtit [17]

$\huge\boxed{y=-2x+3}$

Start off by finding the slope of the line with the formula $\frac{y_2-y_1}{x_2-x_1}$ .

Substitute the values. $\frac{1-3}{1-0}$

Subtract. $\frac{-2}{1}$

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Substitute the values. $y-3=-2(x-0)$

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3 years ago

Precalculus. Please look at the picture

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