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frez [133]
3 years ago
5

The phone company charges 10 cents to connect a call for one minute and 8 cents per minute after that. How long could you talk o

n the phone for 1 dollar?
Mathematics
1 answer:
Lunna [17]3 years ago
6 0
<span>The phone company charges 10 cents to connect a call for one minute and 8 cents per minute after that. 
We call x minute to talk after the first minute.
We have 10+8x</span>≤100 (cents)
and 8x≤100-10
or 8x≤90 and x≤90:8
and x≤11.25
So: <span>How long could you talk on the phone for 1 dollar: it is 11 minute and 15 seconds</span>
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Peter spent half the money on his gift card and coffee and he loaded another $10 on the gift card how much was on the gift card
cestrela7 [59]

Answer:

$80 was on the gift card to begin with

Step-by-step explanation:

Let the initial money be x

Peter spent (1/2)*x

Peter added 10

<em>Before adding more money on the gift card Peter had x -(1/2)x left on the gift card which is (1/2)x</em>

<em>After loading more money Peter had (1/2)x + 10</em>

(1/2)x = 40 <em>(money spent by Peter)</em>

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3 years ago
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

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2 years ago
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Step2247 [10]

Answer: 2

Step-by-step explanation:

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