Ask question

Ask question

All categories

3 years ago

5

Mrs. Lee uses 0.025 of wax to make a candle. On Monday, she made 50 candle. On Tuesday, she made 4 times as many candles as on M

onday. How much wax did she use to make the candles on Tuesday?

1 answer:

kvv77 [185]3 years ago

6 0

I think it is
Monday = 50 candles
Tuesday = 4*50 = 200 candles

Wax used for each candle = 0.025 kg
<span>Total wax used on tuesday = 0.025 × 200 = 5 kg
i Hope i helped u =)!!!!!.....</span>

You might be interested in

HELP ME PLEAEEEE .HOW TO SOLVE IT

Mrac [35]

You need to clarify whether you need the x or the y or both

4 0

3 years ago

Tyrone likes to snack on his big bag of candy. He takes 8 pieces of candy from the bag each time he snacks. After eating 18 snac

Igoryamba

Answer:

c= ur mom

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.

7 0

3 years ago

A) what is average rate of change of f(x) over the interval from x=5 to x=9? (table shows values of f(x). graph shows function o

SVETLANKA909090 [29]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

The average rate of change of a function over the interval $[a,b]$ is equal to $\frac{f(b)-f(a)}{b-a}$ , hence:

$\frac{f(b)-f(a)}{b-a}\\\\\frac{f(9)-f(5)}{9-5}\\\\\frac{14-(-4)}{9-5}\\ \\\frac{14+4}{4}\\ \\\frac{18}{4}\\ \\\frac{9}{2}$

Therefore, the average rate of change of $f(x)$ over the interval $[5,9]$ is $\frac{9}{2}$ .

<u>Part B</u>

Do the same thing as in Part A:

$\frac{f(b)-f(a)}{b-a}\\ \\\frac{f(1)-f(0.25)}{1-0.25}\\ \\\frac{2-5}{0.75}\\ \\\frac{-3}{0.75}\\ \\-4$

Therefore, the average rate of change of $g(x)$ over the interval $[0.25,1]$ is $-4$ .

<u>Part C</u>

To interpret our answer from Part B in terms of the real world it represents, we say that between 0.25 seconds and 1 second, the ball falls at a rate of 4 feet per second (since our average rate of change is negative).

5 0

2 years ago

Please anwser this thank you!!

Kamila [148]

Step-by-step explanation:

v>29 or 29,infinity sign i used an app called and this is what it showed

4 0

3 years ago

Read 2 more answers