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3 years ago

Can somebody explain how to do this question

Mathematics

1 answer:

evablogger [386]3 years ago

7 0

Answer:

The answer is D

Step-by-step explanation:

Remember:

Contrapositive means that if you reverse the hypothesis and the conclusion the statement will mean the same thing

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a diver begins a sea level and dives down 200 feet. he ascends at a steady rate of 12 and 1 thirds feet per minute for 4.5 minut

Komok [63]

12 1/3 x 4.5 = 55 1/2 feet

-200 + 55 1/2 = -144 1/2

the diver is at -144 1/2 feet

4 0

3 years ago

Read 2 more answers

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:

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

Notice that:

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

5 0

3 years ago

Which best describes the relationship between the line that passes through the points (–6, –1) and (–11, 1) and the line that pa

yaroslaw [1]

The two are perpendicular to each other because the two slopes are negative reciprocals

(Y2 - Y1) / (X2 - X1)

First slope:
( 1 - (-1)) / (-11 - (-6))
2/-5

Second slope:
(-13 - (-8)) / (-5 - (-3))
-5/-2 or 5/2

You know when two aliens are perpendicular when you multiply the two slopes and get -1 as the product

-2/5 X 5/2 = -1

Thus the two lines are perpendicular to each other.

3 0

3 years ago

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If it’s right i’ll give brainliest :)

fomenos

Answer:

x=126

Step-by-step explanation:

6 0

3 years ago

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