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2 years ago

15

You invested a total of $12,000 at 4.5 percent and 5 percent simple interest. During one year, the two accounts earned $570. How

much did you invest in each account?

1 answer:

Luda [366]2 years ago

6 0

Answer: You invested $6000 in both accounts.

Step-by-step explanation:

Let x represent the amount invested in the account earning 4.5% interest.

Let y represent the amount invested in the account earning 5% interest.

You invested a total of $12,000 at 4.5 percent and 5 percent simple interest. This means that

x + y = 12000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the account earning 4.5%

I = (x × 4.5 × 1)/100 = 0.045x

Considering the account earning 5%

I = (y × 5 × 1)/100 = 0.05y

During one year, the two accounts earned $570. . This means that

0.045x + 0.05y = 570 - - - - - - - - - - 1

Substituting x = 12000 - y into equation 1, it becomes

0.045(12000 - y) + 0.05y = 570

540 - 0.045y + 0.05y = 570

- 0.045y + 0.05y = 570 - 540

0.005y = 30

y = 30/0.005 = 6000

x = 12000 - y = 12000 - 6000

x = $6000

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Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

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$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

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