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Anything to the power of 0 is 1, so if x = 0, 9^x would be 1. Lets assume x = 0. Then \((9^x - 2) \cdot (3^x + 1)\) would be equal to \(-1 \cdot 1\) or -1. So, the minimum value of \((9^x - 2) \cdot (3^x + 1)\) is -1.

            Prove that given any set of $17$ integers, there exist nine of them whose sum is divisible by $2.$    

            How about demonstrating an exception?  A set of seventeen 3's.  You did say "any" set.  

            3 3 3   3 3 3   3 3 3    

            3 3 3   3 3 3   3 3    

            Choose and add up any nine of them.  The total is 27.  Not divisible by 2.  

Nevermind, I figured it out myself, the answer is:

3sqrt(10)

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! :)

Alright. Let's first put y in terms of x. 

We get \(y=25-x\). 

Subbing this value back into the second equation, we get 

\(3x+75\)

Since x can't be negative, the smallest possible value of x is 0. 

When x is 0, we have \(3(0)+75=75\)

So 75 is our final answer. 

Thanks! :)

\(m=0, k = 2\)

The answer is 2000 * 1.0015^60.

The answer is (B) 27/16.

The answer is (B) 27/16.

               Real numbers a and b satisfy  
               a + ab = 250  
               a - ab = -240  
               Enter all possible values of a, separated by commas.     

               Really?  Just add them together.....   

               a + ab  =    250    

               a – ab  =  –240    

               2a  =  10    

                 a  =  5    

_.   

                                                  ab⁴ = 48           (eq 1)    

                                                  ab² = 4             (eq 2)    

solve (eq 2) for a                       a = 4 / b²     

sub a back in to (eq 1)               (4 / b²) • b⁴  =  48    

                                                   (b⁴ / b²)  =  48 / 4    

                                                             b²  =  12   

                                                               b  =  2 • sqrt(3)    

The problem doesn't ask for a     

 but sub b² back in to (eq 2)             a • 12  =  4    

                                                                a  =  4 / 12    

                                                                a  =  1 / 3     

_.   

All members of our painting team paint at the same rate. If $10$ members can paint a $1000$ square foot wall in $10$ minutes, then how long would it take for $5$ members to paint a $1000$ square foot wall, in minutes?     

Same size wall, but half as many painters.  It will take twice the time.  20 minutes     

_.   

In quadrilateral $PQRS$, $PQ = QR = RS = SP$, $PR = \sqrt{2}$, and $QS = \sqrt{2}$. Find the perimeter of $PQRS$.   

All four sides are equal; that makes it at least a rhombus.  

Both diagonals are the same; and that makes it a square.   

The diagonal of the square is the square root of two. 

That makes the sides equal one.  Therefore the perimeter is 4 units.    

_.   

The answer is \(18\)

I believe the rhombus is a square for main reasons. 

For one thing, the opposite angles of a rhombus are 90 degrees, so we have two 90 degree angles.  

The adjacent angle also should be 90 degrees, so all 4 angles are 90 degrees. 

Therefore, we just have a square. 

We have \(12^2 = 144\)

So the area is 144. 

Thanks! :)

From the problem, we can write the system

 where t is toast, b is bagel and m is muffin. 

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

From the first two equations, we have

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Subbing this in to the third equation, we have 

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is about 262 cents or 261 cents and 2/3. 

Thanks! :)