Label x-intercepts, vertex and axis of symmetry for f(x) = (x+3)(x-5)

You wish to borrow $8000 for debt consolidation. The bank's interest rate for a personal loan is 7.2%. What is your monthly paym

Lisa [10]

<h3>The monthly payment is $242.44 for loan to paid off in 3 years.</h3>

Step-by-step explanation:

The amount borrowed = Principal = $8000

The rate of interest = 7.2%

Time (T) = 3 years

Now, Simple Interest = $\frac{P\times R \times T}{100}$

$\implies SI = \frac{8000 \times 7.2 \times 3}{100} = 1,728$

So, the total interest = $1728

Now, <u>Amount to be paid = Principal + Interest</u>

⇒ A = $8000 + $1728 = $8728

Also, 1 year = 12 months

⇒ 3 years = 3 x 12 months = 36 months

So, total amount to be paid in 36 months is $8728.

⇒The amount to be paid in 1 month is $(\frac{8728}{36} ) = 242.44$

Hence, the monthly payment is $242.44 if loan is to paid off in 3 years.

3 0

3 years ago

Please help. I’ll mark you as brainliest if correct.

stiks02 [169]

Answer:

$4600

Step-by-step explanation:

$3000 to pay off the car + $1600 to repaint the kitchen = $4600 owed on the home equity line of credit

3 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$