All Answers 358095 Answers

Thanks for the further explanation....I was trying to see how you determined the radius and that made it clear !!!

I'm going to look at his one a little more......

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\ x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\ \underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+ (\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{ (x+m)^2} + \underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+( \frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{ (y+n)^2} =-q +(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 +(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\ (x+m)^2+(y+n)^2=m^2+n^2-q\\ \boxed{ \Rightarrow x_0 = -m \qquad y_0=-n \qquad R = \sqrt{m^2+n^2-q}}$$

You put one digit before the decimal point and all the rest after it.  Then multiply the number by 10 to the power of however many places you moved the decimal point to the left. So for this problem, since you moved the decimal point 5 places to the left:

$${\mathtt{425\,950}} = {\mathtt{4.2595\times 10^{5}}}$$

I haven't seen that one before, heureka....but I like it !!!

I'll have to remember that technique !!

Thumbs Up and Points from me!!

Anynomous!

OK...after 4 years, the computer loses $3000 of its original value

So...the slope of our equation repersents the change in value over the change in time, i.e.,

-3000/4    .....  so we have...an the original value is just expressed as (0, 3200) in terms of (t, v) this can be seen as the "y intercept" on the "t-v" axis  .... so we have

v = -(3000/4)t + 3200

Note that...the machine will become fully depreciated in the 5th year.......

Hi sally1,

$$\boxed{x^2+y^2-6x+6y+9=0}$$

$$compare: ax^2+2bxy+cy^2+2dx+2ey+f=0$$

$$b=0 \quad and \quad a=c \quad
\Rightarrow \textcolor[rgb]{1,0,0}{\mbox{this is a circle}}$$

$$\mbox{compare for a circle}: x^2+y^2+2mx+2ny+q=0$$

$$2m=-6 \quad and \quad 2n=6 \\
\Rightarrow \\
m=-3 \\
n=3\\
q=9$$

$$\\radius: R = \sqrt{m^2+n^2-q}=\sqrt{(-3)^2+3^2-9}=\sqrt{9}=3\\
Center: x_0=-m=3 \quad and \quad y_0=-n=-3$$

radius = 3

center (3, -3)

$$\boxed{x^2+y^2+2mx+2ny+q=0}$$

Find the standard form:

$$\\x^2+2mx+y^2+2ny=-q\\
x^2+\textcolor[rgb]{1,0,0}{2m}x+y^2+\textcolor[rgb]{0,0,1}{2n}y=-q\\
\underbrace{x^2+\textcolor[rgb]{1,0,0}{2m}x+
(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2 }_{
(x+m)^2}
+
\underbrace{y^2+\textcolor[rgb]{0,0,1}{2n}y+(
\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2}_{
(y+n)^2} =-q
+(\frac{ \textcolor[rgb]{1,0,0}{2m} }{2})^2
+(\frac{ \textcolor[rgb]{0,0,1}{2n} }{2})^2\\
(x+m)^2+(y+n)^2=m^2+n^2-q\\
\boxed{
\Rightarrow x_0 = -m \qquad y_0=-n \qquad
R = \sqrt{m^2+n^2-q}}$$

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.....

12/6/14

This is Chris's favourite for the day.  It's just a little different from the norm.

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

---------------------------------------------------------------------------------------------------

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)

-----------------------------------------------------------------------------------------------------

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

It is the giant, economical VAT !

It’s over 38 liters of soda. You could bathe in it! LOL

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 !!!.....

I like the abbreviation, CP!

I was actually kind of stumped on this one. I wrote it all out, but my answer really didn't seem that rational, becuause the party can costed so much.

After you posted, I saw we had the same answer so I posted mine quick. Thanks for the verification! :)

Hello sally1,

The equation for a circle is: (x-h)² + (y-k)² = r² 

(h, k) is the center.

So the center of the circle will be halfway between the endpoints of the diameter (the midpoint). Use the Midpoint Formula: [(x₁+x₂)/2 ,  (y₁+y₂)/2]

= [(-1 + 5)/2 , (1 + 9)/2]

= (2, 5)

So the equation so far is: (x-2)² + (y-5)² = r²

Now find the radius, which is the diameter divided in half. 

Use the Distance Formula to find the diameter:

Plug in the endpoints of the diameter: d = sqrt[(5 - -1)² + (9- 1)²]

Simplify.

d = sqrt( 36 + 64)

d = sqrt(100)

d = 10

Half of that = radius = 5

Radius squared = 25

So the final equation: (x-2)² + (y-5)² = 25

Nice job ND !!....I don't see any labels on those cans...must be "generic" cola.....

Thumbs Up + Points......

It looks like we have two cans (cylinders) here. One that is smaller (regular size) and one that is larger (party size) We also know the price of the smaller can. If they want to base the price of the party-sized can proportional to the smaller can, (in other words, you don't get a discount for buying the bigger can) the following should help you find the price of the party-sized can.

These cans probably look something like this:

Let's find the volume of these cans, and then find out how much bigger the party-sized can is, to see how much more it should cost.

Our equation for the volume of a cylinder is V = (pi)(r²)(h)        ---Volume = (pi)*(radius²)*(height)

The regular can has radius of 3 cm and a height of 12.5 cm, so let's put this in and find it's volume.

V = (pi)(r²)(h)

V = (3.14)(3²)(12.5)

V = 353.25

Now let's find the volume of the party-sized can:

V = (pi)(r²)(h)

V = (3.14)(18²)(37.5)

V = 38,151

Now, we can divide the volume of the regular-sized can by the party-sized can to see how much bigger the party-sized can is compared to the regular sized can. 

38,151 ÷ 353.25

108

The party-sized can is 108 times as big, so it should be priced 108 times bigger than the regular-sized can.

.35*108 = 37.80

$37.80 

Assuming that we want the price to be proportional to the volume...

The volume of a cylinder is given by  pi*r^2 * h    where r is the radius and h is the height

So the volume of the smaller can is given by

pi*(3cm)^2 *(12.5cm)= 353.42 cm^3

And the volume of the larger can is given by

pi*(18cm)^2 *(37.5cm) = 38170.35 cm^3

So...using a ratio...

38170.35/353.42 = x/.35      where x is the price of the larger can = $37.80

Wow!!...that's a pretty pricey can of cola....but...it IS a party, after all!!

Complemetary angles are angles that add up to 90 degrees.

So you would set up the equation:

2x + 6x = 90

8x = 90

x = 11.25