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Fofino [41]
3 years ago
9

Maria took out an unsubsidized Stafford loan of $6,925 to pay for college. She plans to graduate in 4 years. The loan had a dura

tion of ten years and an interest rate of 5.0%, compounded monthly. By the time Maria graduates, how much greater will the amount of interest capitalized be than the minimum amount that she could pay to prevent interest capitalization? Round all dollar values to the nearest cent.
Mathematics
1 answer:
Finger [1]3 years ago
4 0

Answer:

$144.70

Step-by-step explanation:

Calculation to determine how much greater will the amount of interest capitalized be than the minimum amount that she could pay to prevent interest capitalization

First step is to determine the Interest only monthly repayments

Using this formula

I=Prt

where,

P=$6925

r=0.05/1

t=1

Let plug in the formula

I=6925*0.05/12

I= $28.854166666

Second step is to determine the amount she will owe after 4 years

Using this formula

S=P(1+r)n

Let plug in the formula

S=6925*(1+0.05/12)4*12

S=6925*(1+0.05/12)48

S=$8454.70

Third step is to determine the Interest part

Interest =8454.70 - 6925

Interest = $1529.70

Now let determine the how much greater will the amount of interest capitalized be

Interest capitalized=1529.70 - 1385.00

Interest capitalized =$144.70

Therefore how much greater will the amount of interest capitalized be than the minimum amount that she could pay to prevent interest capitalization is $144.70

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3 years ago
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sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
3 years ago
If cos 0=2/3, what are the values of sin 0 and tan 0?
morpeh [17]

Answer:

Below

Step-by-step explanation:

● cos O = 2/3

We khow that:

● cos^2(O) + sin^2(O) =1

So : sin^2 (O)= 1-cos^2(O)

● sin^2(O) = 1 -(2/3)^2 = 1-4/9 = 9/9-4/9 = 5/9

● sin O = √(5)/3 or sin O = -√(5)/3

So we deduce that tan O will have two values since we don't khow the size of O.

■■■■■■■■■■■■■■■■■■■■■■■■■

●Tan (O) = sin(O)/cos(O)

● tan (O) = (√(5)/3)÷(2/3) or tan(O) = (-√(5)/3)÷(2/3)

● tan (O) = √(5)/2 or tan(O) = -√(5)/2

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