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2 years ago

X-2y=4 going through (1,-6)

Mathematics

2 answers:

kakasveta [241]2 years ago

4 0

Answer:

−

Explanation:

First, rewrite the equation in slope-intercept form:

−

⇒

So the slope of the given line is

Lines that are perpendicular have slopes that are negative reciprocals of each other. Meaning, if a line has slope

, then a line perpendicular to this has slope

−

. That means the slope of our perpendicular line is

−

Knowing the new slope and a point, we can find an equation for our perpendicular line by using the slope-point equation

−

(

−

)

, or by plugging in the given

(

)

point (and the new slope

) into

to find

for the new line.

−

(

−

)

⇒

−

[

−

(

−

)

]

⇒

−

[

]

⇒

−

⇒

−

(

−

)

⇒

−

∴

becomes

−

Step-by-step explanation:

attashe74 [19]2 years ago

4 0

Answer:

false

Step-by-step explanation:

Okay so we plug in the values for x and y to make the equation:

1 - 2 x -6 = 4

And then we just solve the left side:

2 x -6 = -12

1 - -12 = 1 + 12 = 13

So this is false

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3 years ago

The equation y−5=−2(x−3) is in point-slope form. Which shows this equation in slope-intercept form?

MAXImum [283]

y - 5 = -2(x - 3)

Distribute -2.

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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:

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