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It intersects one time if they have ONE solution.

We substitute Y.

\(x^2+a=ax\)

Then we set it equal 0 so it matches the standard quadratic form.

\(x^2-ax+a=0\)

In order to find when it has ONE solution, the discriminant of the equation ABOVE has to be equal to 0.

The discriminant is:

\(b^2-4ac\)

We plug in values.

\((-a)^2-4(a)=0\)

We have:

\(a^2-4a=0\)

We factor:

\(a(a-4)=0\)

a = 0 or a = 4

So the sum of the values is 0 + 4 = \(\boxed{4}\)

Just find TWO x inputs that are 3 units apart....like -5 and -2   and calculate the difference of f(x) between those two

x = -5    f(x) = -11

x = -2     f(x) = ?  (what do you see on your table?)    then calculate the difference....usually the absolute value of the difference....

                                   I think you can do that......can ya? 

Can you post the image please, its always blocked on my computer.

First find the total number of ways four cords can be chosen out of 21 cords.

Can you do that?

Then count the number of bow tie cords there are.

We do this by listing the TYPES of bow tie cords.

Then counting for each cord type, after this, we add them up all together.

We take the result and divide it by the total number of ways four cords can be chosen out of 21 cords.

NOTICE HOW the value of x² is the golden ratio!!!!

I know its been a week since Halloween, but I couldn't resist it. Just don't let your imagination go too far or else my post is rated R.

Do not scroll down if you love cats and have a wild imagination

RECIPE FOR CAT FOOD

cats are yummy.

their flesh is tangy, semi-sweet, and a touch of sourness _{(probably because I left it in the basement for too long)}. 

Its best when fungus is GROWN on it, not ADDED to it. After adding a mixture of honey and milk and pouring it on the fungus, microwave for 99 minutes.

The result is a crisp, teriyaki-styled delicacy!

I recommend lighting the cat on fire in order to remove the fur. Make sure the cat is deceased when you light the fire, or else you will make a lot of noise and/or the cat will make a MESS on your floor! (Based on my experience!)

A good side snack is to remove the eyeballs. Then heating it on a stove top in order to melt it. Try sunnyside style, and add a pinch of salt. This will be like a sunny-side up egg, except its cat eyeballs.

Another good side is to remove the tail, and burn it (of course to remove fur). After burning, I recommend pouring gasoline over it and lighting in a controlled fire. Add ONE GALLON of HAND OIL and A POUND OF SALT. What you get is a giant tail fry.

again?.....

You need to add a link to the original question.

Too bad, you might have to do some homework on your own.

c = 3*pi

that is exactly what  i wanna hear :D good job learning and understanding the material :) 

Replace y with (4t-3) and then make x the subject.

y  =   x

       _____

      √[x - 3]

This function will not exist   on  ( -infinity , 3 ]

When x  approaches  3  from the right, y approaches infinity

Using Calculus, we can find the x value where the minimum occurs

y '  = (x - 3)^(-1/2)  + (-1/2)x(x - 3)^(-3/2)

Set this to 0

(x - 3)^(-1/2)  + (-1/2) x (x - 3)^(-3/2)  = 0       factor

(x - 3)^(-3/2)  [ (x - 3) - (1/2)x)  ]  = 0

(x - 3) - (1/2)x  =  0

(1/2)x -  3  =  0

(1/2)x  =  3

x  = 6

So....the minimum occurs at 

6 / √[6 - 3 ]   =   6 / √3

So.....the range  is    [ 6 / √3, infinity   )

If (x + 1)(y + 1) = 40 and xy(x + y) = 380, then find all possible values of x^2 + y^2.

I'm only assuming integer values for  x , y

(x + 1) ( y + 1)  =  40             xy ( x + y)  =  380     (2)

xy  + x + y  +  1  = 40

xy + x + y  =  39

x + y  =  39  - xy    (1)

Sub (1)  into (2)  and we have that

xy ( 39 - xy)  =  380

-(xy)^2 + 39(xy)  = 380

-(xy)^2  + 39(xy) - 380  =  0   multiply through by  -1

(xy)^2  - 39xy  + 380   =  0       factor

(xy  - 20) ( xy - 19)  =  0

Set both factors to  0   and we have that

xy  = 20         xy  = 19

The possible values for  x, y  are

x         y

1       19

19      1

1       20

2       10

4        5

5        4

10      2

20      1

Because  ( x + 1)  ( y  + 1)  =  40

Then  only

(1, 19)   and (19, 1)  will work

So

x^2 + y^2  =   362

Anything to the zero power = 1            x⁰ = 1             (4a)⁰ = 1

k = -8/3

both of those answers didn't work

I have a solution to this that is very involved. But I will try to outline it even though it may not be the most elegant solution. Here it goes:

Here I assume that x is the number of people stranded on the island, n is the number of coconuts gathered and piled before dark, and k₁, k₂, k₃, k₄, and k₅ are the number of coconuts, respectively, in each of the x piles after the additional coconut is given to the monkey and before the night watches 1 through 5 hide their own portion. We can write a system of 5 equations involving the 7 variables(n, x,  k1, k2, k3, k4, and k5) as follows:

\(k_1x= n -1\)

\(k_2x=n-k_1-2 \)

\(k_3x=n-(k_1+k_2)-3\)

\(k_4x=n-(k_1+k_2+k_3)-4\)

\(k_5x=n-(k_1+k_2+k_3+k_4)-5\)

This system evidently has an infinite number of solution since the number of variables is larger than the number of equations and can be approached in multiple ways, but all methods would involve a reduction of the number of variables. Here we are asked to find the minimum value of n for a given value of x, so we will reduce the number of equations to only one, a quintic polynomial equation in the variable x, the number of stranded people, and containing two other variables, n and k₅. I will only do step one of this process since the rest can be accomplished similarly.

Multiply the 5th equation by x to get:

\(k_5x^2=nx-(k_1+k_2+k_3+k_4)x-5x\)

Replace k₁x, k₂x, k₃x, and k₄x by their values given by the 1st through 4th equation in the system to get :

\(k_5x^2=nx-[(n-1)+(n-k_2-2)+(n-k_1-k_2-3)+(n-k_1-k_2-k_3-4)]-5x\)

\(=nx-4n+10+3k_1+2k_2+k_3-5x\)

Note that k₄ has been eliminated from the equation and by repeating this process 3 more times , we can also eliminate k₁, k₂, and k₃ to get the quintic we were aiming at:

\(k_5x^5=(n-5)x^4+(10-4n)x^3+(6n-10)x^2+(5-4n)x+(n-1)\)

Moving all terms to the left side we get:

\(k_5x^5+(5-n)x^4+(4n-10)x^3+(10-6n)x^2+(4n-5)x+(1-n)=0\)

Well this is the formula we were looking for. The author of this question claims that if the there are only six people stranded on the island, the initial number of coconuts would have to be a minimum of 7,771. Let's see if this holds true. Substituting 6 for x in the above equation we get:

\(0=7,776k_5+1,296(5-n)+216(4n-10)+36(10-6n)+6(4n-5)+(1-n)\)

\(7,776k_5+4,651-625n\), which implies that

\(n=\frac{7,776k_5+4,651} {625}\)

Here we need to find the smallest value of k₅ that results in an integer value for n. I gave the problem to EXCEL to solve. EXCEL did a search for k₅=1, 2, ..., 10,000, we decided to limit the search, and came up with 16 values for k₅ that resulted in integer values for n.The smallest of these values was k₅=624 and the largest was k₅=9,999. Plugging these for k₅ in the formula for n, gave us, what do you know, n= 7,771 and n= 124,411.

1. The circumradius is 2*sqrt(5).

4.  Angle AOC = 122 degrees.

Here is desmos graph

https://www.desmos.com/calculator/h99cfiuvht

Using the formula for a circle   (x-h)^2 + (y-k)^2 = r ^2

for the first point this becomes   (1-h)^2 + (3-k)^2 = r^2

for the second point , this becomes  (3-h)^2 +(-2-k)^2 = r^2     these are both equal to r^2, so equate them

    then expand, collect 'like' terms and simplify to

4h-10k-3=0   (1)

Do all of this to the THIRD set of points and equate to the second set of points (red above)

   expand and simlify to

4h+10k-21 =0     (2)

solve (1) and (2) for h,k    =   3, 9/10

     then use distance formula from the center (3,9/10)  to any of the three points to find r^2 = 8.41  

(x-3)^2 + (y-9/10)^2 = 8.41

maybe show the method with a different set of values

"I NEED HELP PLEASE! CORRECT ANSWERS ONLY PLEASE. PLEASE ANSWERS ASAP THANK YOU."

These words tell me that you are only trying to get answers for your homework, and that you are not actually interested in learning the material or engaging in a productive discussion.  I refuse to help anyone who displays such a poor attitude.

thanks, I like both of these methods, I struggle with this topic so I will change the numbers and use both to see what I like better, I will post later asking if it is right or not.

A little different approch....

same method to find the line    y = -2x +11

Now use the distance formula    d²= (x1-x2)^2 +(y1-y2)^2           d^2 = (6 sqrt5)^2 = 180

Using the point given (3,5) as x2    and y₂ = -2x+11 as shown above

(x-3)^2 + (y- (-2x+11)^2 = 180   

x^2 + 6x + 9 +4x^2 -24x +36     -180  = 0

5x^2-30x -135 = 0          Use quadratic formula to find x = 9      or      -3

Sub these values into the red line equation to find y = -7    and 17                           (9,-7)    and (-3,17)