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worty [1.4K]
3 years ago
7

Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe

nt variable
A. f(q) = 1/2q - 3/2


B. f(q) = 2s + 3


C. f(s) = 1/2s - 3/2


D. f(q) = 2q + 3
Mathematics
2 answers:
Ugo [173]3 years ago
7 0

Answer:

  D.  f(q) = 2q + 3

Step-by-step explanation:

Choices B and C are eliminated right away because there is no "q" in the definition, just "s". If the function is to be a function of "q", then "q" is expected to appear in the definition.

__

The two variables in the given equation are "s" and "q". If "q" is designated as the independent variable, then the dependent variable is "s". The equation must be solved for "s":

  6q = 3s - 9

We observe that the coefficient of "s" is 3, and that all numbers are multiples of 3, so we can divide by 3 to simplify this a bit:

  2q = s - 3

Since we want an expression for s alone, we can add 3 to get ...

  2q +3 = s

Now, we can write ...

  s = f(q) = 2q +3

IRINA_888 [86]3 years ago
5 0

Answer:D

Step-by-step explanation:

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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

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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.

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

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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}

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

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