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4 years ago

3/5 times 2&5/6=? Please answer me quickly

Mathematics

2 answers:

RSB [31]4 years ago

6 0

I got 54/30 which simplified would be... 1 12/15

Rus_ich [418]4 years ago

3 0

I believe your answer is 1 if you meant 3/5 multiplied by 2 and /6

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F(x)=2x+3 find the f(4)<br><br> a.11 b.8<br><br> c.9 d.5

Ivan

The answer is a because you fill in 4 where 2(x) is for the x

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3 years ago

Write an expression for the sequence of operations described below. triple u, then multiply t by the result

nikklg [1K]

Answer:

t(3u)

Step-by-step explanation:

First we have to triple (x3) u, which is what gets us 3u. Then we multiply that result by t.

This gives us: t(3u)

Remember, this problem is asking for you to write an expression, not and equation. That is why there is not an equal sign included in the answer.

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3 years ago

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Allushta [10]

What???... um yes? 2

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3 years ago

What is v in 55 -7v 6?

Oxana [17]

The v is the variable

7 0

3 years ago

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