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3 years ago

8

Reflection (in Y)

1 answer:

Tanzania [10]3 years ago

7 0

(3;4) percent of the piano 4

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At the beginning of an experiment, the number of bacteria in a colony was counted at time t=O. The number of bacteria in the col

yan [13]

Answer:

1092

Step-by-step explanation:

We have been given that the number of bacteria in the colony t minutes after the initial count modeled by the function $B(t)=9(3)^t$ . We are asked to find the average rate of change in the number of bacteria over the first 6 minutes of the experiment.

We will use average rate of change formula to solve our given problem.

$\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}$

Upon substituting our given values, we will get:

$\text{Average rate of change}=\frac{b(6)-b(0)}{6-0}$

$\text{Average rate of change}=\frac{9(3)^6-9(3)^0}{6}$

$\text{Average rate of change}=\frac{9(729)-9(1)}{6}$

$\text{Average rate of change}=\frac{6561-9}{6}$

$\text{Average rate of change}=\frac{6552}{6}$

$\text{Average rate of change}=1092$

Therefore, the average rate of change in the number of bacteria is 1092 bacteria per minute.

8 0

2 years ago

A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's hal

mars1129 [50]

Answer:

$t=5.0days$

Step-by-step explanation:

Using the formula for the exponential decay that is $N=N_{0}e^{-kt}$ , we have N= $\frac{1}{2}{\times}46=23$ , $N_{0}=46$ and k=0.1374.

Thus, $N=N_{0}e^{-kt}$ becomes

$23=46{\times}e^{-0.1374t}$

$\frac{23}{46}=e^{-0.1374t}$

$\frac{1}{2}=e^{-0.1374t}$

Taking log on both sides, we get

$ln(\frac{1}{2})={-0.1374t}$

$t=\frac{ln\frac{1}{2}}{-0.1}$

$t=\frac{-0.6931}{-0.1374}$

$t=5.0days$

4 0

3 years ago

Which statement is true regarding the graphed functions?

Novosadov [1.4K]

I’m not exactly sure but maybe f(0) = 4 and g(–2) = 4

4 0

3 years ago

You invested a total of $60,000 in 2 funds earning 8.5% and 10% simple interest. During 1 year, the 2 funds earned a total of $5

maksim [4K]

The amount invested in the fund earning 8.5% is $40,000 whereas $20,000 was invested in the fund earning 10%

What is the return on each fund?

The return on each fund is determined as the amount invested multiplied by the interest rate on the fund

;Let X be the amount invested at 8.5%

Interest=8.5%*X=0.085X

The interest of the fund invested at 10%=($60,000-X)*10%

The interest of the fund invested at 10%=$6000-0.10X

Total interest=0.085X+6000-0.10X

Total interest earned during year 1=$5,400

5400=0.085X+6000-0.10X

5400=6000-0.015X

0.015X=6000-5400

0.015X=600

X=600/0.015

X=$40,000(amount invested at 8.5%)

amount invested at 10%=$60,000-$40,000

amount invested at 10%=$20,000

Find out more about simple interest on:https://brainly.in/question/37840965

#SPJ1

7 0

2 years ago

Solve the inequality -3x<54

kifflom [539]

Answer:

x<-18

Step-by-step explanation:

you divide -3 by both sides.

6 0

2 years ago