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2 years ago

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Jesus Christ is coming soon, Believe in him and get saved, Muhammad is dead, Buddha is dead, Krishna is dead, But Jesus rose fro

m the dead and I don't say this in a hateful way to anyone, but Jesus can change your life like he changed mine, Jesus is coming back soon, Repent and Get right with God

2 answers:

WINSTONCH [101]2 years ago

8 0

Answer:

PREACH IT I've witnessed his miracles first hand

yulyashka [42]2 years ago

5 0

Amen, I agree with you. I've been in the deep with myself for way too long now... but now I gave myself to God and he changed my life forever.

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TRUE or FALSE: Your boat travels 480 miles in 20 hours. That is the same rate as 860 miles in 40 hours.

Morgarella [4.7K]

Answer:

false

Step-by-step explanation:

860 divided by 2 is 430 NOT 480

5 0

3 years ago

Read 2 more answers

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$ , we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$ .

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

7 0

3 years ago

I would really appreciate some help.

VLD [36.1K]

2 to the power of 61 is = 2305843009213694000

then just minus that by 1 and divide it by 3. if it’s divisible then it’s a multiple of 3.

7 0

2 years ago

Please solve the following problem

kirza4 [7]

Answer:

x = 27

Step-by-step explanation:

Given

$\frac{x}{9}$ = 3

Multiply both sides by 9 to clear the fraction

x = 9 × 3 = 27

5 0

3 years ago

30 POINTS FOR BRAINLIEST ANSWER!!!!!!!

skad [1K]

I think b not sure it’s the only one that makes sense

6 0

3 years ago