CPhill

You don't know???

Using a trig identity, we have

1 - (5/7)^2  = (cosx)^2

1 - 25/49 = (cosx)^2

24/49 = (cosx)^2        take the square root of both sides, we have

±√(24)/7 =  ±[2√(6)]/7   = cosx

This makes sense...the sine is positive in the first two quadrants, so the cos is positive in the first, but negative ni the second.......

Hey, sally1, look at this one I just did for you.....

This is certainly an unusual one, huh??    To find the x intercept, let's just let y=0

So we have x^4 = 16   →   x^4 - 16 = 0   ... which factors as

(x^2 - 4) (x^2 + 4) = 0    .... the second thing has a non-real solution but we can set the frist factor = 0  and find two real solutions

(x^2 - 4) = 0 .. factor as ..  (x+2) (x-2) = 0   so the x intercepts occur at (2,0) and (-2,0)

To find the y intercepts, let x = 0....so we have

y^2 = 16   →  y^2 - 16 = 0   factor this as ....  (y+4) (y-4) =0 and setting each factor to 0 produces y intercepts at  (0,4) and (0, -4)

I know that the "xy" term produces the graph of x^4 + y^2 = 16 with a "rotational" element added...heck, I'd even like to look at it myself !!!  (If I could add a stem at the top, it might resemble a strange "apple.')

Wow!!!.....that's certainly an odd duck!!...note that the intercepts are right where we said they would be, but as for the rest?? I couldn't begin to guess the shape of this one!!

x^2+y^2+4x+y+4=0

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Rewrite this as .....   x^2 + 4x +y^2 + y = -4

Now, we need to "complete the square" on x and y......(review how to do this if you don't remember)....

x^2 + 4x + 4   +  y^2 + y + 1/4 =  - 4 + 4 + 1/4

(x+2)^2 + (y+ 1/2)^2 =  1/4

The center is ( -2, -1/2)  and the radius is 1/2

cos^-1(.7439) =

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.743\: \!9}}\right)} = {\mathtt{41.935\: \!295\: \!930\: \!845^{\circ}}}$$

Please realize that this may also be a reference angle with the same measure lying in the 4th quadrant, i.e., 318.064704069155°

Find the equation of the line containing the perpendicular bisector of (2,18) (-2,-4)

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OK...we need to find TWO things, here, before even attempting to write an equation - the midpoint of this segment and the slope between the two points.

The midpoint is [(2 -2)/2, (18-4)/2] = [0, 7]

The slope is      (-4-18 ) / (-2-2) =  -22/-4 = 11/2.....but since we're writing an equation for a perpendicular bisector, we need to take the negative reciprocal of this = -2/11

So....using point-slope from, we have...

y - 7 = (-2/11)(x-0)

y-7 = (-2/11)x

y = (-2/11)x + 7

They're rentals, rosala...I have to give them back to the site if I ever leave.....!!!

Actually...it didn't seem too reasonable to me, either.....just proves the old adage...."The math doesn't lie".......37 bucks for a can of cola???.....must be REALLY good  !!!

Let's find the radius....this is given by ...

√[(-2-9)^2 + (10-7)^2] = √[(-11)^2 + 3^2' = √(121+9) = √130

So the "formula" for an equation of a circle is given by

(x-h)^2 + (y-k)^2 = r^2     where, h,k is the center and r is the radius  ....    so we have...

(x - (-2))^2 + (y-10)^2 = 130

(x +2)^2 + (y-10)^2 = 130

And that's it !!!.....