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3 years ago

One long-distance company offers a plan such that each minute costs $0.10 and each call has a $0.25 service

Mathematics

1 answer:

gladu [14]3 years ago

3 0

Answer:

y=0.10(t)+0.25

27 minutes and 30 seconds

Step-by-step explanation:

We know that you start with 25 cents as a service fee, this is for making the call. For each minute you talk, 10 cents are added. Multiply the number of minutes spent by 10 cents a minute for the cost based on the call. Then add the 25 cents service fee.

If you talked for 5 minutes:

y=0.10(5)+0.25

y=0.50+0.25

y=0.75

75 Cents

For three dollars, you would plug in 3 for y

3.00=0.10(t)+0.25

-0.25 -0.25

2.75=0.10(t)

divide both sides by 0.10 to get

27.5=t

You can talk for 27 minutes and 30 seconds

27 minutes

Hope this helps :-)

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3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

Hence, the amount of oil was decreasing at 69300 barrels, yearly

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2 years ago

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andrezito [222]

Answer:

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2 years ago

How do you solve 5x-2y= 18 and -5x+3y= -22 by solving a system of linear equations by elimination? Please add the steps in which

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Adding them, you get:

y=-3

So now that you know y, you can substitute the value of y into one of the original equations. Let's substitute it in the first one:

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Solving this, we get:

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3 years ago

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Let A and B represent the governor's salaries of states A and B, respectively.

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The governor of state A has a salary of 171,915.
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