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Each of these points is the same radius distance from the center to the point.....using distance formual for the first two points 

(x-22)^2 + (y-15)^2      =   ( x+25)^2 + (y-0)^2     reduces to 

-94x - 30y + 84 = 0 

For the last two points:

(x+25)^2 + y^2 = ( x-22)^2 + ( y+29)^2   reduces to 

94x-58y-700 = 0 

Add the two reduced equations to get 

-88y = 616

y = -7 

Now put this valueof 'y' into either of the reduced equations to find   

x = 3.12765

Center = ( 3.12765, -7) 

I will assume you have this :

(1+i) / ( 2-i) (3+i) )       Multiply the bottom part (the denominator ) to get :

(1+i) / ( 7 -i)                 Now rationalize 

(1+i) (7+i)    /   (7-i)(7+i) 

(6 +8i)   / ( 50) 

1/25 ( 3 + 4i)  

Simplify the real and imaginary parts of the expression. 

2i 

1. Figure out combinations that multiply to 2000. 

2. Then, figure out the number of permutations in each sequence.

3. Add the number of permutations together. 

In order to have an infinite number of solutions, we want two equivalent equations  

Subtracting 3a from both sides gives us 3a+2b=k+mb. 

Because we want both equations to be equal to each other (meaning the left side matches), we can incorporate the equation k+mb=2

We want M to be an integer-- other than zero (except it cannot be other than zero because we want the equations to line up) 

We need m=0 so K would have to be 2

M=0

K=2

The side length is 6.

The answer is 5.

By graphing on Desmos, find   a = –1.79129 and b = 2.79129    

Doing the required addition     –1.79129     

                                                +2.79129     

                                                +3        

                                                +3        

total                                          +7      

_.   

Double root occurs when discrimonant  (b^2 - 4ac)   of the quadratic equation = 0 

First one:   x^2 -2x +1

                               a = 1     b = -2      c = +1   

                             ( -2) ^2 - 4 (1)(1) = 0   ?             yes  =====>    the first one has a double root 

\((a-b)^3=a^3-3a^2b+3ab^2-b^3\)=

\(1-3ab(a-b)\)=

\(1-3ab\)=1

3ab=0,

ab=0

Because ab=0, we know \((a-b)^2\)=\(a^2+b^2+0\)=1

Because of this, \((a+b)^2=1+0=1\), so (a+b) is +/-1

Finally because a+b is 1, and one of either a or b is 0, because ab is 0, the possible pairings of (a,b) are

(0,1), or (1,0)

Hope this helps!

 🤙 🤙 🤙

\(9^{x}-2\cdot 3^{x}+1\):

Set a = \(3^x\)

\(a^2-2a+1=0\)

\((a-1)^2=0 \)

a=1, so \(3^x\)=1

That means x=0 because anything to the 0th power is one

Hope this helps! 

 🤙 🤙 🤙

First, we can write a handy equation to solve for the constant and find the 3 terms. 

Let's let the constant added to every number be x. 

Since the three terms for a geometric series, we have the equation

\(\frac{100+x}{60+x} = \frac{140+x}{100+x}\)

Now, when we crossmultiply and then expand everything out, we get

\((x+100)(100+x)=(x+60)(140+x)\\ 200x+x^{2}+10000=140x+x^{2}+{8400+60x}\\ 200x+x^{2}+10000=200x+x^{2}+8400\)

Now, we bring all terms to one side of the equation. We have

\(200x+x^{2}+10000-x^{2}-200x-8400=0\\200x+1600-200x=0 \\ 1600=0\)

However, this statement is obviously not true, meaning that x is invalid. 

This also means that there are NO solutions to this given problem. 

*Note, I may have made a mistake. Not sure. 

Thanks! :)

First of all, let's focus on the second equation. 

We can factor the right side of the equation to get \((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since x+y is equal to 10 from the first equation, the equation becomes

\(10(x^2+xy+y^2)=162+x^2+y^2\)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now, isolating x in the first equation, we get

\(x=10-y\)

Now plugging this value of x into the second equation, we get

\(9(10-y)^2+9y^2+10(10-y)y=162\\ 9(100-20y+y^2)+9y^2+100y-10y^2=162\\900-180y+9y^2+9y^2+100y-10y^2=162\\ 8y^2-80y+900=162\\8y^2-80y+738=0\)

However, plugging this into the quadratic formula, notice that unofruntately, the descriminant is less than 0, menaing the eventual solutions are NONREAL numbers. 

Thanks! :)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

We could probably use casework to solve this problem.

The first case to look for is only two digit numbers. 

We have 23 and 32, so there are 2 possibilities. 

Next, consider 3 digit numbers. 

We have 311, 131, 113 for 3 more. 

We also have 221, 212, and 122 to add 3 more on. 

Next, consider 4 digit numbers. 

The only numbers that work are numbers with three 1s and one 2. There are 4 ways we could do this. 

Lastly, we have 5 digit numbers. The only one we have to consider is \(111111\), one possibility. 

Now, we add up the numbers. We get \(2+3+3+4+1 = 13.\)

So 13 is our answer!

Thanks! :)