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3 years ago

Question in picture attached

Mathematics

1 answer:

mars1129 [50]3 years ago

4 0

Answer:

Go to the search bar and type up photos. Press on the photo/screenshot you want and it should give you a button you can use to crop.

Step-by-step explanation:

You might be interested in

Write an equation that you can use to solve for x. Enter your answer in the box.

Temka [501]

Answer:

60 + x =100

Step-by-step explanation:

Since the angle with 100 degrees is opposite the angle combining 60 and x, they will have congruent measures.

5 0

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

3 0

2 years ago

A graph titled Calories in Strawberries. The x-axis shows the number of strawberries, numbered 1 to 9, and the y-axis shows the

seraphim [82]

From the equation y = 4x, each strawberry contains 4 calories.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let y represent the number of calories in x strawberries. Hence:

y - 0 = [(27 - 0)/(6.8 - 0)](x - 0)

y = 4x

From the equation y = 4x, each strawberry contains 4 calories.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

4 0

2 years ago

What is the solution-2|2.2x-3.3|=-6.6

mylen [45]

$Solution, solve\:for\:x,\:-2\left|2.2x-3.3\right|=-6.6\quad :\quad x=0\quad \mathrm{or}\quad \:x=3$

$Steps:$

$-2\left|2.2x-3.3\right|=-6.6$

$\mathrm{Divide\:both\:sides\:by\:}-2,\\\frac{-2\left|2.2x-3.3\right|}{-2}=\frac{-6.6}{-2}$

$\mathrm{Simplify},\\\left|2.2x-3.3\right|=3.3$

$|f\left(x\right)|=a\quad \Rightarrow \:f\left(x\right)=-a\quad \mathrm{or}\quad \:f\left(x\right)=a,\\2.2x-3.3=-3.3\quad \quad \mathrm{or}\quad \:\quad \:2.2x-3.3=3.3$

$2.2x-3.3=-3.3,\\\mathrm{Add\:}3.3\mathrm{\:to\:both\:sides},\\2.2x-3.3+3.3=-3.3+3.3,\\\mathrm{Refine},\\2.2x=0,\\\mathrm{Multiply\:both\:sides\:by\:}10,\\.2x\cdot \:10=0\cdot \:10,\\\mathrm{Refine},\\22x=0,\\\mathrm{Divide\:both\:sides\:by\:}22,\\\frac{22x}{22}=\frac{0}{22},\\\mathrm{Simplify},\\x=0$

$2.2x-3.3=3.3,\\\mathrm{Multiply\:both\:sides\:by\:}10,\\2.2x\cdot \:10-3.3\cdot \:10=3.3\cdot \:10,\\\mathrm{Refine},\\22x-33=33,\\\mathrm{Add\:}33\mathrm{\:to\:both\:sides},\\22x-33+33=33+33,\\\mathrm{Simplify},\\22x=66,\\\mathrm{Divide\:both\:sides\:by\:}22,\\\frac{22x}{22}=\frac{66}{22},\\\mathrm{Simplify},\\x=3$