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3 years ago

Whats The quotient of 1332 divided by 9

Mathematics

2 answers:

Flura [38]3 years ago

8 0

Answer:

148

Step-by-step explanation:

kvv77 [185]3 years ago

3 0

Answer:

148

Step-by-step explanation:

=1332÷9

=148

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lesya692 [45]

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4 0

3 years ago

Point b has coordinates (-8,15) and lies on the circle whose equation is x^2+y^2=289. If an angles is drawn in standard position

Alenkinab [10]

Answer:

$\cos \theta=-\dfrac{8}{17}$

Step-by-step explanation:

Coordinates of Point b $=(-8,15)$

b lies on the circle whose equation is $x^2+y^2=289$

$x^2+y^2=17^2$

Comparing with the general form a circle with center at the origin: $x^2+y^2=r^2$

The radius of the circle =17 which is the length of the hypotenuse of the terminal ray through point b.

For an angle drawn in standard position through point b,

x=-8 which is negative

y=15 which is positive

Therefore, the angle is in Quadrant II.

$\cos \theta=\dfrac{Adjacent}{Hypotenuse} \\$Adjacent=-8\\Hypotenuse=17\\\cos \theta=\dfrac{-8}{17} \\\cos \theta=-\dfrac{8}{17}$

8 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:

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