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dusya [7]
2 years ago
11

On Melissa's 6th birthday, she gets a $2000 CD that earns 7% interest compounded quarterly. If the CD matures on her 13th birthd

ay, how much money will be available?​
Mathematics
1 answer:
Alexandra [31]2 years ago
8 0

Answer:

A\simeq3250.83

Step-by-step explanation:

The amount formula in compound interest is:

A=P(1+\frac{r}{n} )^{nt}

where:

P = principal amount

r = annual interest

n = number of compounding periods

t = number of years

We already know that:

P = $2000

r = 7\% = \frac{7\%}{100\%}=0.07

t = 7 (number of years from 6th to 13th bday)

n = 4 (quarterly in a year)

Then,

A=2000(1+\frac{0.07}{4} )^{(4)(7)}\\\\A=2000(1+\frac{0.07}{4} )^{28}\\\\A=3250.825792\\\\A\simeq3250.83

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A bus is going to a museum 135 miles away. The bus will travel at 30 miles per hour. How long will it take the bus to get
TEA [102]

Answer:

4.5 hours

Step-by-step explanation:

distance equals rate times time, a relationship that can be solved for time:

time = distance / rate

Here the time is (135 miles) / (30 mph) = 4.5 hours

The bus requires 4.5 hours to get to the museum.

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What is the domain and range of this graph
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Answer:

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I hope it will help =)

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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.

8 0
2 years ago
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