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Zielflug [23.3K]

2 years ago

7

A machine at a bakery poured out flour for 4 minutes. At the end of that time, the machine was empty. The graph shows the amount

of flour (In grams) remaining in the machine versus time (in minutes). Find the range and the domain of the function shown. 1000 900 GO 600+ Amount of flour (grams) 500- 300- 200 100 8 10 12 11 16 Time (minutes) Write your answers as inequalities, using xory as appropriate. Or, you may Instead click on "Empty set" or "All reals" as the answer.

1 answer:

kykrilka [37]2 years ago

5 0

Answer:

Domain: 0<x<4

Range:0<y<400

The signs should be less than or equal too

Step-by-step explanation:

The signs are less than or equal to

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Sam works as a baggage carrier at an airport facility. He owes a certain amount of money, in dollars, to his friend Daniel. The

Ivenika [448]

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2 years ago

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

3 0

2 years ago

When Val checks her cable bill, she is shocked by her movie-rental bill for the month. Renting movies costs $2.99 for each movie

astraxan [27]

Answer:

She should buy the monthly plan for the unlimited movies rather than pay $2.99 per movie. This is because, the more she pay that amount for each movie, the higher her expenses would become at the end of each month.

For example, let assume, in a month, she 8 free days (Saturday and Sunday). She paying for each movie each of those days would supersede the amount she could have spent assuming she did the unlimited monthly plan of $7.99.

That notwithstanding other days which will feel like watching movies or the public holidays which she would be free to relax.

Step-by-step explanation:

3 0

2 years ago

Eudora ran from her home to her secret laboratory at an average speed of 12\text{ km/h}12 km/h12, start text, space, k, m, slash

OleMash [197]

Answer:

Eudora spent 0.5 hours from home to library and 1.5 hours from library to school

Step-by-step explanation:

Given

Home to Library:

$S_1 = 12km/h$ -- Average Speed

Library to School

$S_2 = 76km/h$ -- Average Speed

Total

$Distance = 120km$

$Time = 2\ hr$

Required

Determine the time taken from home to library and from library to school

Let the time spent from home to library be x. So, from library to school will be: 2 - x

So, we have:

Home to Library:

$S_1 = 12km/h$ -- Average Speed

$T_1 =x$ -- Time

Library to School

$S_2 = 76km/h$ -- Average Speed

$T_2 =2 - x$ -- Time

Distance is calculated as:

$Distance = Speed * Time$

Home to Library:

$D_1 = S_1 * T_1$

$D_1 = 12 * x$

$D_1 = 12x$

Library to School:

$D_2 = S_2 * T_2$

$D_2 = 76 * (2- x)$

$D_2 = 152- 76x$

The total distance is 120km. So, we have:

$120km = D_1 + D_2$

$120 = 12x + 152 - 76x$

Collect Like Terms

$120 - 152 = 12x - 76x$

$-32 = -64x$

$-64x=-32$

Solve for x

$x = \frac{-32}{-64}$

$x = 0.5$

Recall that:

$T_1 =x$

$T_2 = 2 - x$

So, we have:

$T_1 = 0.5$

$T_2 = 2 -0.5 = 1.5$

4 0

1 year ago