F(0)=2, f(2)=4, linear equation?

2 answers:

7 0

If you separate those into points (0,2) and (2,4) then figure out the slope by using the equation y2-y1 divided by x2-x1, you'll get 4-2 over 2-0 which gives you 2/2 or 1. You found the slope so you can plug that into the slope-intercept equation y-y1=m(x-x1). For you, it'd be y-2=1(x-0), you end up with the equation y=x+2.

Gelneren [198K]3 years ago

5 0

Could be, but not necessarily.

You have two points on the graph: (0, 2) and (2, 4) .

Yes, there's a straight line that could go through those two points. Y = x + 2 .

But there are a million other graphs that could go through the same two points.
You could draw ellipses, parabolas, circles, exponentials etc. that all go through
them. So when you only know two points on the graph, you can't really tell what
the equation has to be, or whether or not it's linear.

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I don't need you to work it out, just theorem(s) i need to reach the answer :)

VikaD [51]

Not sure why such an old question is showing up on my feed...

Anyway, let $x=\tan^{-1}\dfrac43$ and $y=\sin^{-1}\dfrac35$ . Then we want to find the exact value of $\cos(x-y)$ .

Use the angle difference identity:

$\cos(x-y)=\cos x\cos y+\sin x\sin y$

and right away we find $\sin y=\dfrac35$ . By the Pythagorean theorem, we also find $\cos y=\dfrac45$ . (Actually, this could potentially be negative, but let's assume all angles are in the first quadrant for convenience.)

Meanwhile, if $\tan x=\dfrac43$ , then (by Pythagorean theorem) $\sec x=\dfrac53$ , so $\cos x=\dfrac35$ . And from this, $\sin x=\dfrac45$ .

So,

$\cos\left(\tan^{-1}\dfrac43-\sin^{-1}\dfrac35\right)=\dfrac35\cdot\dfrac45+\dfrac45\cdot\dfrac35=\dfrac{24}{25}$

7 0

4 years ago

PLEASE HELP ME WITH THIS PROBLEM! PLEASE

Vladimir [108]

Answer:

when I find out the two others ill edit this,sorry i'm kinda slow lol

Step-by-step explanation:

<u>Team Wins Losses</u>

Tigers 20

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Pirates 35