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3 years ago

What is 90 divided by 12???? pls help i need to get my grades up

Mathematics

2 answers:

Delicious77 [7]3 years ago

8 0

Answer:

7.5

Step-by-step explanation:

Andrew [12]3 years ago

5 0

Answer:

7.5

Step-by-step explanation:

cant u just g oogle it?

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Find the pressure if a force of 750 N acts on an area of 12 cm3

Leto [7]

The answer is P=F/A I had a quiz on these

5 0

3 years ago

The approximate height, h, in meters, traveled by golf balls hit with two different clubs over a horizontal distance of d meters

inna [77]

Answer:

a) 100 metres

b) 10 metres

Step-by-step explanation:

Given: the approximate height, h, in meters, travelled by golf balls hit with two different clubs over a horizontal distance of d meters is given by the functions: $h=-0.002d^2+0.3d\,,\,h=-0.004d^2+0.5d$

To find: distances when the ball is at the same height when either of the clubs are used. Also, to find this height

Solution:

(a)

$-0.002d^2+0.3d=-0.004d^2+0.5d\\d(-0.002d+0.3)=d(-0.004d+0.5)\\-0.002d+0.004d=0.5-0.3\\0.002d=0.2\\d=\frac{0.2}{0.002}=100\,\,metres$

(b)

Put d = 100 in $h=-0.002d^2+0.3d$

$h=-0.002(100)^2+0.3(100)\\=-20+30\\=10\,\,metres$

4 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:

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