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Marysya12 [62]
2 years ago
5

A local print shop has a big order of pamphlets to print, so they decide to use two of their printers for the one job. The newes

t printer can print the pamphlets four times as fast as the old printer Working together the printers can complete the job in 4 hours. How many hours would it take cach printer to print all of the pamphlets by itself a . Use the table to set up a rational equation to describe the situation. Use the variable x to represent the time it takes the newest printer to complete the job .
Mathematics
1 answer:
nlexa [21]2 years ago
6 0

If both printers work together. Then the number of the pamphlets will be 20x in four hours.

<h3>What is Algebra?</h3>

Algebra is the study of mathematical symbols, and the rule is the manipulation of those symbols.

A local print shop has a big order of pamphlets to print, so they decide to use two of their printers for the one job.

The newest printer can print the pamphlets four times as fast as the old printer Working together the printers can complete the job in 4 hours.

Let x be the number of the pamphlets are printed by the old printer in one hour.

Then the number of the pamphlets are printed by the new printer in one hour will be 4x.

Then if they work together, then the  number of the pamphlets in one hour will be 5x.

Then the number of the pamphlets in 4 hour will be

→ 5x × 4

→ 20x

More about the Algebra link is given below.

brainly.com/question/953809

#SPJ1

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An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click
Sergeeva-Olga [200]

Answer:

A.    \mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

B.    \mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

A = Pe^{rt}

where;

A = (8000) \ e ^{0.7t}

The instantaneous rate of change = \dfrac{dA}{dt}

\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )

= 8000 \dfrac{d}{dt}e^{0.07 \ t}

\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 2 years; the instantaneous rate of change is:

\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}

\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

12000 = (8000)e^{0.07 \ t}

\dfrac{12000 }{8000}= e^{0.07 \ t}

1.5= e^{0.07 \ t}

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 5.79

\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}

\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

7 0
3 years ago
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