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3 years ago

Help!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Stels [109]3 years ago

7 0

(1,4) because when you substitute 1 in for x and 4 for y, you get 5(1) - 7(4) or 5 - 28 which equals -23

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Franco se quedó en un hotel por 5 noches. La habitación del hotel cuesta $ 92 por noche. Tenía un cupón de $ 10 de su factura to

lesya [120]

Answer:

Serian $460 per si tiene un cupon de $10 serian $450

Step-by-step explanation:

Para encontrar el resultado multiplica 92x5= 460 y si tiene un cupon de $10 lo restas y da a un total de $450

3 0

3 years ago

What is 0.371 as a decimal

sweet [91]

Answer:

It is in a decimal.

In a fraction it’s 371/1000

In a percentage 37.1%

7 0

3 years ago

Read 2 more answers

HELP!!!!!!!!!!!!!!!!!!!!!!!!!! EXPLAIN TOO PLZ!

asambeis [7]

Ok so you would just take the two fractions and simply if needed and that the answer!

7 0

3 years ago

Read 2 more answers

Alyssa wants to buy a new car and is planning to take out a loan to pay for the car.

g100num [7]

Answer: the amount of interest that Alyssa will pay on her loan is $1320

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the loan.

P represents the principal or amount taken as loan

R represents interest rate

T represents the duration of the loan in years.

From the information given,

P = $22000

R = 4.5%

T = 4 years

Therefore,

I = (22000 × 1.5 × 4)/ 100

I = 132000/100 = $1320

4 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.

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:

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

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

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

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.

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

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

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

So, lets check this limit:

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

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago