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Let's make some careful observations about the problem. 

First, note that Triangle ABC is a 45-45-90 right triangle

From this, we can conclude that

\(AB = BC  =   12/\sqrt 2 =  6\sqrt 2 \)

From the problem, we also know that

\(AD = AD \\ BD = BD \\ AB = BC\)

So triangles  ABD  and CBD  are congruent

Thus, we have that \([ ABD ]   =  (1/2) [ABC]   =  (1/2) ( 1/2) ( 6\sqrt 2)^2  =  (1/4) (72)   = 18\)

So 18 is the answer. 

Thanks! :)

The problem states it must be an ACUTE triangle. The acute triangle states that the sum of the squares of two of the sides must always be greater than the square of the third side. 

Essentially, it states that

\(7^2 +12^2 >x^ 2\\ 7^ 2 + x^ 2 > 1 2^ 2\\ \)

Now, solving each condition for x (gonna leave this up to you), we get that

\(x < \sqrt{193}\\ x > \sqrt{95}\)

Since x has to be an integer, we must round down for the first equation and round up for the second equation to get

\(x \leq 13\\ x \geq 10\)

Thus, we have \(10 \leq x \le 13\)

There are 4 numbers that work. 10, 11, 12,13. 

I hope this explenation was clear enough to understand. Also, I'm not confident in my answer, but it should be right! :)

Thanks! :)

\(\frac{3-z}{z+1} \ge 2(z + 4)\)

simplified:  \(\frac{(2z+1)(z+5)}{z+1} \ge 0\)
 

interval notation: \((-\infty, -5] \cup (-1,- \frac{1}{2}]\)

We don't have the expand everything, as we can just use factoring to our advantage. 

Moving all terms to the right side of the equation, we get

\( (x - 2)(x + 6) - (x - 2)(x + 5) \leq 0\)

Now, since both terms have x- 2, we can factor it out to get

\((x-2)(x+6 - (x+5)) \le 0\\ (x-2)(x+6-x-5) \leq 0 \\ x-2 \leq 0\)

Isolating x from the equation, we get

\(x \le 2\)

In interval notation, this is \((-\infty, 2]\)

Thanks! :)

\(f(3)=4\)
 

the second equation vertically compresses the original function by 1/5, horizontally compressed by 5, and horizontally shifted 5 to the left. 
 

orig. point: \((3,4)\)

new point: \((-0.4,-4.2)\) or \((-2/5,-21/5)\)

First, let's distribute in x on the right hand side of the equation. We get

\(x^2+6x > 16-x+14+x^2\)

Now, let's move all terms to one side and combine all like terms. The x^2 cancels out, which simplifies the problem for us. We get the inequality

\(7x-30>0\)

Now, we isolate x. We find that

\(7x > 30\\ x > 30/7\)

In interval notation, this is

\((\frac{30}{7}, \infty)\)

Thanks! :)

*1600 points!

We need the numerator to be nonzero, so the largest possible value not in the domain is x = 2.

Using brute force , which is something you should NOT do (even though this time, I'm just lazy):

\(11250x^6 + 12750(x^5)(y) - 1800(x^4)(y^2) - 4740(xy)^3\)\( - 1662(x^2)(y^4)-234(x)(y^5) - 12y^6\)

\(\text{The coefficient of} (x^2)(y^4) \text{​ is -1,662.}\)

Hope this helped .

2 = 2

2 + 7 = 9

2 + 7 + 12 = 21

2 + 7 + 12 + 17 = 38

2 + 7 + 12 + 17 + 22 = 60

2 + 7 + 12 + 17 + 22 + 27 = 87

2 + 7 + 12 + 17 + 22 + 27 + 32 = 119

The answer is 32.

Simplify this equation to 

3y = -5x -12 

y = -5/3 x - 4        X intercept occurs when y = 0 

                                    Then x = -12/5 

                               Y intercept occurs when x = 0 

                                      Then y = -4     

• Find distance between  (-12/5 , 0 )     And   (0, -4)       Using distance formula     D = 4.66 units 

The line in y = Mx+b form is then 

       Y = 8x +14           The x-axis intercept occurs when y = 0     Put in y = o 

       0 = 8x + 14 

        X = -14/8  = - 1 3/4      <==== x intercept

ONE real root ( actually a double root) would mean the discrimination (from the Quadratic Formula) would be = 0 

b^2 - 4ac = 0        

X^2 + 4m x + m = 2x-6 

x^2 - (4m - 2) x + (m+6) = 0          A = 1      B = 4m-2     C = m+6 

(4m-2)^2 - 4 (1)(m+6) = 0 

16 m^2 -16m +4  - 4m - 24 = 0 

16 m^2 - 20m - 20 = 0          Now use quadratic formula    To find   M = 1.90587

First, find the slope between the two points:

slope     M = (y1-y2) / (x1-x2)   = (2 - -7 ) /(-8-4) =  - 9 / 12 = - 3/4 

Now the line in y = Mx + b form becomes 

                      Y = -3/4 x + b     Sub in either point to find 'b'  = -4 

then the line equation is    Y = =-3/4 x - 4     

     Re-arrange to 

          3/4 x   + y = -4       Multiply thru by 4 

            3x  + 4y = -16      Done .

Sorry, just realized, for this question, there IS no max value. 

The MIN value is 13 since the parabola open upwards. 

My bad.

What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?          

Combine terms and arrange                                    t² + 4t – 96          

Take the first derivitive and set equal to zero          2t + 4  =  0    

                                                                                      2t  =  –4     

                                                                                        t   =  –2     

_.   

If s is a real number, then what is the smallest possible value of 2s^2 - 8s + 19 - 7s^2 - 8s + 20?   

Combine terms               –5s² – 16s + 39     

If you set this equal to y and graph it, you see that    

the curve is a parabola that opens downward.    

When a parabola opens upward, the smallest y-value is the vertex.   

When a parabola opens downward, there is no smallest value.    

_.   

The maximum possible value of the expression occurs at the vertex of the graph. 

So let's find the y value of the vertex.

First, simplifying and combining all like terms, we get the equation

\(4t^{2}+4t+14\)

Now, the x value of the vertex of the graph is \(\frac{-b}{2a}\), so we have

\(\frac{-4}{8} = -1/2\)

Plugging this value back into equation and subbing out t, we get

\(y = 13\)

So 13 is the our answer. 

Thanks! :)

It's actually more simple than you think! All we have to do is apply difference of squares. 

We have

\(Slope = [ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2 \)

So 2 is our answer. 

Thanks! :)