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3 years ago

Find the midpoint rounded to the nearest hundreth a (-2,1) b (-1,1)

Mathematics

1 answer:

Veronika [31]3 years ago

8 0

Answer:

$(-1.5, 1)$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Midpoint Formula: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Step-by-step explanation:

Step 1: Define

Endpoint A (-2, 1)

Endpoint B (-1, 1)

Step 2: Find Midpoint

Simply plug in your coordinates into the midpoint formula to find midpoint

Substitute [MF]: $(\frac{-2-1}{2},\frac{1+1}{2})$
Subtract/Add: $(\frac{-3}{2},\frac{2}{2})$
Divide: $(-1.5, 1)$

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Nastasia [14]

Answer:

Step-by-step explanation:

let the plane intersects the join of points in the ratio k:1

let (x,y,z) be the point of intersection.

$x=\frac{3k+2}{k+1} \\y=\frac{5k+4}{k+1} \\z=\frac{-4k+16}{k+1} \\\because ~(x,y,z)~lies~on~the~plane.\\2(\frac{3k+2}{k+1} )-3(\frac{5k+4}{k+1} )+\frac{-4k+16}{k+1} +6=0\\multiply~by~k+1\\2(3k+2)-3(5k+4)+(-4k+16)+6(k+1)=0\\6k+4-15k-12-4k+16+6k+6=0\\-7k+14=0\\k=2\\x=\frac{3*2+2}{2+1} =\frac{8}{3} \\y=\frac{5*2+4}{2+1}= \frac{14}{3} \\z=\frac{-4*2+16}{2+1} =\frac{8}{3}$

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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.

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)

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3 years ago