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2 years ago

11

The drama club is showing a video of the recent play the first showing began at 2:30 PM the second showing was scheduled to star

t at 5:25 PM with in half hour break between the showings how long is the video in hours and minutes

1 answer:

BabaBlast [244]2 years ago

6 0

The video is two hours and 25 minutes

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Justin has $500 in a savings account at the beginning of the summer. He wants to have at least $300 in the account by the end of

SSSSS [86.1K]

Answer:

10 weeks

Step-by-step explanation:

1 because 20 times 10 is 200 and if he starts with 500 then he can withdraw 20 dollars a week for 10 weeks to have 300 left by the end of the summer, hope this helps :)

3 0

2 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

Theres a pic need asap!!!

miskamm [114]

Answer:

Domain: [-1,3)

Range: (-5,3]

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve. <br> 24= 7/8g<br> A. 3/7<br> B. 21<br> C. 23 1/8<br> D. 27 3/7

Mrac [35]

I hope this helps you

3 0

2 years ago

Read 2 more answers

Solve for a: b − 2a = c

vlada-n [284]

Answer:

$a=-\frac{c-b}{2}$

Step-by-step explanation:

Solving for the variable $a$ , given that :

$b-2a=c$

We can first start start by subtracting $b$ from both sides of the equation:

$-2a=c-b$

Now, we can divide both sides of the equation by the coefficient of $a$ , which would be $-2$ :

$a=\frac{c}{-2}-\frac{b}{-2}$

Applying the fraction rule $\frac{a}{c}\pm\frac{b}{c} =\frac{a\pm b}{c}$ :

$a=\frac{c-b}{-2}$

Then, apply the fraction rule $\frac{a}{-b} =-\frac{a}{b}$

$a=-\frac{c-b}{2}$

5 0

2 years ago