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3 years ago

15

A radio station requires DJs to play to commercials for every 10 songs they play. What is the unit rate of this song to commerci

als

1 answer:

deff fn [24]3 years ago

5 0

There will be 10 commercials if one hundred songs

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You and 3 friends divide the proceeds of a garage sale equally. The garage sale earned $412. How much money did you receive?

iren2701 [21]

Answer:

$ 103 for each person

Step-by-step explanation:

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3 years ago

!!URGENT!!WILL GIVE BRAINLIEST!! GEOMETRY DILATIONS!!! PLEASE INCLUDE EXPLANATION

Sergeu [11.5K]

Unless otherwise specified we usually assume the dilation is about the origin. That is you can tell the scale factor by the way the distance to the origin changes.

Point C, for example, is 6 units from the origin. Its image, C' is 4 units from the origin. Then the scale factor image/original is 4/6 = 2/3.

Your green question mark gets filled with 3 to make the fraction 2/3.

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To find the translation, pick any point on A'B'C' and find the corresponding point A'', B'', or C''. Count the squares you have to go left to get from A' to A'', for example. That will be 5 to the left. Similarly, count the squares you have to go down to get from A' to A''. That will be 4 down.

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2 years ago

Richard claims that the value of the expression 23x2+173x+12 is sometimes negative. Which of these values of x supports Richard'

Talja [164]

The correct answer is B.

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2 years ago

Read 2 more answers

Least to greatest 0.23 2.30% 1/5

olga55 [171]

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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)

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3 years ago