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2 years ago

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CORINA INDRAGESTE FOARTE MULT ANIMALELE IN DAR DE ZIU EI A PRIMIT O CARTE DESPRE LUMEA LOR MINUNATA IN PRIMA ZI EA A CITIT 36 DE

PAGINI IAR IN A DOUA ZI A TREIA ZI DIN REST CATE PAGIN MAI ARE DE CITIT STIIND CA ALBUMUL ARE IN TOTAL 159DE PAGINI?

1 answer:

aliya0001 [1]2 years ago

7 0

Paging that’s what I got for the answer

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The total operating cost to run the organic strawberry farm varies directly with its number of acres. What is the constant of va

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Answer:

$16\ acres$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the number of acres

y ---> total operating costs

we have the ordered pair (20,551,520)

For x=20 acres, y=$551,520

<em>Find the value of the constant of proportionality k</em>

$k=\frac{y}{x}$

substitute the values of x and y

$k=\frac{551,520}{20}=\$27,576\ per\ acre$

The linear equation is equal to

$y=27,576x$

Find out how many acres are on a farm with a total operating cost of $441,216

so

For y=$441,216

substitute in the linear equation the value of y and solve for x

$441,216=27,576x$

$x=441,216/27,576$

$x=16\ acres$

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2 years ago

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Joe receives an average of 780 emails in his personal account and 760 emails in his work account each month. After changing his

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Answer:

702 emails

Step-by-step explanation:

<h2>This problem bothers on depreciation of value, in this context it is Joe's email that has depreciated by 10%.</h2>

Given data

Average personal emails received monthly = $780 emails$

Average work emails received monthly $= 760 emails$

We are required to solve for the new amount of emails Joe will be receiving after changing his address, to find this value we need to solve for the depreciation of his personal mails.

After solving for the depreciation , we then need to subtract the depreciation from the initial number of mails to get the new number of mails.

let us solve for 10% depreciation.

$depreciation= \frac{10}{100} *780\\depreciation=0.1*780= 78 emails$

The new number of mails

$= initial number of mail- depreciation\\ =780-78= 702 emails$

Joe will be receiving an average of 702 emails in his personal account monthly

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2 years ago

A school district has a student population of $6,734 students. If the maximum class size is 25 students. How many more teacher m

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Answer:

53 teachers

Step-by-step explanation:

Basically, what we need to do here is to find how many teachers there need to be, first. If there are 6,734 students in the school district and if maximum class size is 25, then the number of teachers needed is:

6,734 / 25 = 269.36

Of course, it's obvious that we can't have a decimal number of teachers, so we need to find integer (269 or 270).

If we take 269 teachers and 25 students per class, we get:

269 • 25 = 6,725 students, which is not enough, since there are 6,734 students.

That means that the number of teachers needed is 270.

It is given that there are already 217 teachers, meaning that 270-217=53 teachers have to be supplemented.

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The parent function f(x) = x3 is represented by graph A. Graph A is transformed to get graph B and graph C. Write the functions

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2 years ago

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It takes 32 hours for a motorboat moving downriver to get from pier A to pier B. The return journey takes 48 hours. How long doe

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Answer:

Time taken by the un powered raft to cover this distance is T = 192.12 hr

Step-by-step explanation:

Let speed of boat = u $\frac{km}{hr}$

Speed of current = v $\frac{km}{hr}$

Let distance between A & B = 100 km

Time taken in downstream = 32 hours

$32 = \frac{100}{u +v}$

$u + v = \frac{100}{32}$

u + v = 3.125 ------ (1)

Time taken in upstream = 48 hours

$48 = \frac{100}{u -v}$

$u -v = \frac{100}{48}$

u - v = 2.084 ------- (2)

By solving equation (1) & (2)

u = 2.6045 $\frac{km}{hr}$

v = 0.5205 $\frac{km}{hr}$

Now the time taken by the un powered raft to cover this distance

$T = \frac{100}{v}$

Because un powered raft travel with the speed of the current.

$T = \frac{100}{0.5205}$

T = 192.12 hr

Therefore the time taken by the un powered raft to cover this distance is

T = 192.12 hr

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