All categories

3 years ago

Help what is the function

Mathematics

1 answer:

RideAnS [48]3 years ago

7 0

The function is: F(x)= x+1

You might be interested in

The proof, with a missing reason, proves that the measure of angle ECB is 54°.

aniked [119]

Answer:

Midsegment of a Triangle Theorem

Step-by-step explanation:

$In\: \triangle ADE\\m\angle A + m\angle D + m\angle AED = 180°\\\therefore 90°+36°+m\angle AED= 180°\\\therefore 126° +m\angle AED = 180°\\\therefore m\angle AED = 180°-126°\\\therefore m\angle AED = 54°\\$

$In\: \triangle ADE,$ D and E are mid points of sides AB and AC respectively.

Therefore, DE || BC

Hence, by Midsegment of a Triangle Theorem

$m\angle ACB = m\angle AED =54°(corresponding\:\angle 's) \\\therefore m\angle ECB = m\angle AED =54°(\because A - E - C)$

3 0

3 years ago

The sum of twice the number x and 25 more than three times the number y

inn [45]

Answer:

Correct expression:

$2x + 3y + 25$

6 0

2 years ago

Helppppppppppppppppppppppppppp

Nata [24]

Answer:

14 cm

Step-by-step explanation:

One side of the composite has a length of 6 and the other side has a length of 8.

If we add these two numbers, we'll get the missing side length of the rectangle

6 + 8 = 14 cm

6 0

3 years ago

there are 2 dozen calories in a candy bar. Jason ate 8 candy bars.How many calories are there in 21 candy bars

kifflom [539]

504
............................................................

8 0

3 years ago

Read 2 more answers

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