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Assuming the question is "What is the maximum value of  c  such that. . ."

\(y\ =\ \frac13x^2\\~\\ y\ =\ x+c\)

Equate both expressions of  y

\(\frac13x^2\ =\ x+c\\~\\ \frac13x^2-x-c\ =\ 0\)

This quadratic equation will have only one solution when the discriminant equals zero.

\((-1)^2-4(\frac13)(-c)\ =\ 0\\~\\ 1+\frac43c\ =\ 0\\~\\ \frac43c\ =\ -1\\~\\ c\ =\ -\frac34\)

Yah....still not working for me either.

LCM, so they are all finishing a donut hole at the same time...ie synchronized.

Ooops....see Hecticar's answer below....I used the wrong numbers...it should be

6^2    8^2   10^2

36     64     100     LCM 14400

14400/36 = 400

I use the same math thing as the user who posted this, and I came here to check my answer, and i put in sqrt33.

it was the wrong answer

I thought it was sqrt(32+1)

just like CPhill...

Note that  angle CAD  = 30°

Since this an inscribed angle.....then the measure of  minor arc CD is twice this  = 60°

And angle ACB  = 40°

Then......by simiar reasoning.....minor arc AB  = 80°

Then the sum of  of the measures of minor arcs  AD  and BC  must be  360 - 60 - 80  =  360 - 140  = 220°

But these arcs are intercepted by angles ACD  and CAB

So.....the sum of their measures must = (1/2) of the sum of these arcs  = (1/2)(220°) = 110°

x^2 + 2y^2  = 16

2y^2  =  16 - x^2

y^2  =  (16 - x^2)/2      take the positive root on both sides

y  =  [ (1/2)(16 - x^2)]^(1/2)    =   [ 8 - x^2/2]^(1/2)

xy  =  x [ 8 - x^2/2]^(1/2)       take the derivative of this with respect to x   and set to 0

y '  = [ 8 - x^2/2]^(1/2)  - (x)(1/2) [ 8 - x^2/2] ^(-1/2) (x)  =

[8 - x^2/2]^(1/2) - (x^2/2) [ 8 - x^2/2]^(-1/2)  = 0      factor

[8 -x^2/2]^(-1/2)  [ [8-x^2/2  -x^2/2]  = 0     

[8 -x^2/2]^(-1/2)  [  8 - x^2]  = 0       divide both sides by the first factor and we have that

8 - x^2  = 0

x^2  =  8       take the positive root

x = √8  = 2√2

And y  =  √ [ 8 - (√8)^2 / 2 ]  =  √[ 8 -8/2] = √ 8 -4]  = √4  = 2

So.....the max value of xy on the graph is  (2√2) (2)  =  4√2

y = x^2 + a

y = ax

Set the y's equal

x^2 + a  =  ax          rearrange as

x^2 - ax + a = 0 

For this to have real solutions.......the discriminant must  be ≥  0

So

a^2  -  4a ≥  0

a ( a - 4) ≥  0

Setting each factor to 0  and solving for a  produces the following solutions  a = 0   and a = 4

So  we have the following possible intervals that  will produce solutions

(-inf, 0 ]  or  ( 0 , 4 )   or  [ 4, inf )

If a  is in the first interval, then a (a - 4) ≥ 0  so this interval produces a solution

If  a is in the second interval, then a(a - 4)  < 0.....so  no solutions are found here

If a  is in the third interval, then a (a - 4) ≥ 0 .....so this interval produces a solution

So....the solution intervals are  

(-inf, 0 ]  U [ 4, inf )

© from cphill 

\(\angle\)

from latex enter \ angle

oh okay thank you Melody much appreciated

What is the radius, in inches, of a right circular cylinder if its lateral surface area is 3.5 square inches and its volume is 3.5 cubic inches?

The lateral surface area  is given by  2pi *r * height

The volume is given by  pi * r^2 * h

So  we have that

2pi * r * h  =  3.5   ⇒    h  =   [ 3.5] / [ 2 pi *r]    (1)

pi * r^2 * h  = 3.5  (2)

Sub ( 1) into (2) for h     and we have that

pi * r^2  *  [ 3.5] / [ 2 pi * r ]  =  3.5     simplify

r * 3.5/2  =  3.5     multiply both sides by  [ 2] / 3.5

r  = 2 inches

It should be  4 pi 6^2    then later    14400/36  =  400

And can you explain why do you get the LCM of those three numbers?

LCM{4, 36} = 36

GCD{4, 36} = 4

Ratio =36:4 = 9:1

Sum =36 + 4 = 40

surface area of sphere = 4 pi r^2

4 pi 3^2       4 pi 8^2    4 pi 10^2     divide all by 4pi

3^2                    8^2            10^2

9                       64               100           LCM is 14400

14400/9 = 1600 holes    by Niraek

Help help I need help quick

anybody help!!!

hectictar cphill or EP?