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3 years ago

7

Howie wants to get a car loan with 4% simple interest rate. If the car costs $25,000, how much would Howie need to pay back at t

he end of one year, including the interest?

1 answer:

likoan [24]3 years ago

7 0

The amount Howie will pay back at the end of one year is $26000.00

The given parameters are:

$P =25000$ -- the principal amount

$R = 4\%$ -- the interest rate

$T=1$ --- the duration

The amount to pay back in this duration is then calculated using:

$A = P(1 + RT)$

So, we have:

$A = 25000(1 + 4\% \times 1)$

$A = 25000(1 + 4\% )$

Express percentage as decimal

$A = 25000(1 + 0.04)$

$A = 25000(1 .04)\\$

Multiply

$A = 26000$

Hence, the amount to pay back is $26000.00

Read more about simple interest at:

brainly.com/question/1115815

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What second degree polynomial function has a leading coefficient of -2 and root 4 with a multiplicity of 2

OverLord2011 [107]

The second degree polynomial with leading coefficient of -2 and root 4 with multiplicity of 2 is:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

<h3>How to write the polynomial?</h3>

A polynomial of degree N, with the N roots {x₁, ..., xₙ} and a leading coefficient a is written as:

$p(x) = a*(x - x_1)*(x - x_2)*...*(x - x_n)$

Here we know that the degree is 2, the only root is 4 (with a multiplicity of 2, this is equivalent to say that we have two roots at x = 4) and a leading coefficient equal to -2.

Then this polynomial is equal to:

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2 years ago

A bag contains 5 red marbles, 8 blue marbles and 6 green marbles. If three marbles are drawn out of the bag, what is the probabi

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3 years ago

A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6

katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is $49\ ft$

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

$h(t)=-16t^{2} +63t+4$

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

$h(3)=-16(3)^{2} +63(3)+4=49\ ft$

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

so

Convert the function into a vertex form

$h(t)=-16t^{2} +63t+4$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$h(t)-4=-16t^{2} +63t$

Factor the leading coefficient

$h(t)-4=-16(t^{2} -(63/16)t)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})$

$h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})$

Rewrite as perfect squares

$h(t)-(67,600/1,024)=-16(t-(63/32))^{2}$

$h(t)=-16(t-(63/32))^{2}+(67,600/1,024)$

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

so

$h(t)=-16t^{2} +63t+4$

For h(t)=0

$0=-16t^{2} +63t+4$

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0

3 years ago