F (x) =x-2 whats the answer Complete the table below for the function f

Which of the following formulas would find the surface area of a right cylinder where h is the height, r is the radius, LA is th

Sedbober [7]

I'd say it is A but I think you have made a tpyo
It should be 2*PI*radius^2 + 2*PI*radius*height

8 0

4 years ago

Please help! provide thorough explanation and use elimination method!!

ICE Princess25 [194]

Multiply both sides of the second equation by 100 to get rid of the decimals:

0.05n + 0.10d = 1.50

==> 5n + 10d = 150

Multiply both sides of the first equation by -5:

n + d = 21

==> -5n - 5d = -105

Add the two equations together:

(5n + 10d) + (-5n - 5d) = 150 + (-105)

Notice that the terms containing n get eliminated and we can solve for d :

(5n - 5n) + (10d - 5d) = 150 - 105

5d = 45

d = 45/5 = 9

Plug this into either original equation to solve for n. Doing this with the first equation is easiest:

n + 9 = 21

n = 21 - 9 = 12

So Donna used 12 nickels and 9 dimes.

8 0

3 years ago

For every $20 that Danika saves, her grandmother gives her $30. If Danika saves $150, how much money will her grandmother give h

Colt1911 [192]

The answer is 225 dollars because you divide 150/20=7.5 then multiply 7.5(30) and you get 225

7 0

2 years ago

Read 2 more answers

What is the result of adding these two equations? 4x – 4y = -2 –9x – 4y = -3

kondaur [170]

Answer: -5x-8y=-5

Step-by-step explanation:

add right sides of equation and even them with result of adding right sides of equation

4x-4y-9x-4y=-2-3

-5x-8y=-5

5 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:

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

Notice that:

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

5 0

3 years ago