hectictar

1.   D  is correct.

2.   (P U R) ∩ Q

First let's find  P U R .

P U R   =   {red, blue, green, yellow, black, purple}

Now find the intersection of   P U R  and   Q

The intersection of  {red, blue, green, yellow, black, purple}  and  {red, green, black}

is   {red, green, black} .  So...

(P U R) ∩ Q   =   {red, green, black}

3.   C  is correct.   

The four points  A(-4,0) ,  B(0,-4) ,  X(0,8) ,  and  Y(14,k)  are grouped on the Cartesian plane.

If segment AB is parallel to segment XY, then...

the slope of AB   =   the slope of XY

\(\frac{-4-0}{0--4}\,=\,\frac{k-8}{14-0}\\~\\ \frac{-4}{4}\,=\,\frac{k-8}{14}\\~\\ -1\,=\,\frac{k-8}{14}\\~\\ -1\cdot14\,=\,k-8\\~\\ -14\,=\,k-8\\~\\ -14+8\,=\,k\\~\\ -6\,=\,k\)

4.   Let  f  be a function such that   f(x + y)  =  x + f(y)   for any two real numbers  x  and  y .

      If  f(0)  =  2  ,  then what is  f(2012)  ?

f(x + y)  =  x + f(y)

f( -2012 + 2012 )   =   -2012 + f(2012)

f(0)   =   -2012 + f(2012)

2   =   -2012 + f(2012)

2 + 2012   =   f(2012)

2014   =   f(2012)

Ah...thanks..but....my final answer happened to be right but my working out was all messed up!!  I had to fix it!!

2.   Let   f(x)  =  3x² - 4x  .  Find the constant  k  such that   f(x)  =  f(k - x)   for all real numbers  x .

f(x)   =   f(k - x)

                                                 And   f(x)  =  3x² - 4x

3x² - 4x   =   f(k - x)

                                                 And   f(k - x)  =  3(k - x)² - 4(k - x)

3x² - 4x   =   3(k - x)² - 4(k - x)

3x² - 3(k - x)² - 4x + 4(k - x)   =   0

3[x² - (k - x)²] - 4[x - (k - x)]   =   0

3[x + (k - x)][x - (k - x)] - 4[x - (k - x)]   =   0

3[x + k - x][x - k + x] - 4[x - k + x]   =   0

3[ k ][ 2x - k ] - 4[ 2x - k ]   =   0

( 2x - k )( 3k - 4 )   =   0

2x - k  =  0      or      3k - 4  =  0

k  =  2x                      k  =  4/3

The constant value that works is   k  =  4/3

1.   The function  \(f\)  satisfies   \(f(\sqrt{x+1})=\frac1x\)   for all   \(x\geq-1\,,\quad x\neq0\) .   Find  \(f(2)\) .

We want to find an  x  value such that...

\(\sqrt{x+1}=2\\~\\ x+1=4\\~\\ x=3\)

So......

\(f(\sqrt{x+1})=\frac1x\\~\\ f(\sqrt{3+1})=\frac13\\~\\ f(\sqrt{4})=\frac13\\~\\ f(2)=\frac13\)

2.

m∠A  +  m∠C   =   180°

(2x + 9)°  +  (3x + 1)°   =   180°

2x + 9 + 3x + 1   =   180

5x + 10   =   180

5x   =   170

x   =   34

m∠A   =   (2x + 9)°

m∠A   =   (2(34) + 9)°

m∠A   =   (68 + 9)°

m∠A   =   77°

3.

m∠B  +  m∠D   =   180°

(4x - 20)°  +  x°   =   180°

4x - 20 + x   =   180

5x - 20   =   180

5x   =   200

x   =   40

m∠B +  m∠D   =   180°

m∠B  +  40°   =   180°

m∠B   =   140°

4.

m∠D  +  m∠B   =   180°

x°  +  148°   =   180°

x°   =   180°  -  148°

x°   =   32°

x   =   32

m∠A   =   (2x + 1)°

m∠A   =   (2(32) + 1)°

m∠A   =   65°

5.

m∠B + m∠D   =   180°

(x + 10)° + (x + 24)°   =   180°

x + 10 + x + 24   =   180

2x + 34   =   180

2x   =   146

x   =   73

m∠A  +  m∠C   =   180°

(x + 15)° + m∠C   =   180°

(73 + 15)° + m∠C   =   180°

88° + m∠C   =   180°

m∠C   =   180° - 88°

m∠C   =   92°

Given that   \(\frac{8-\sqrt{18}}{\sqrt2}=a+b\sqrt2\)   where  \(a\)  and  \(b\)  are integers, find the values of  \(a\)  and  \(b\) .

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\(\quad\frac{8-\sqrt{18}}{\sqrt2}\\ =\\ \quad\frac{8-\sqrt{18}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18}\sqrt2}{\sqrt2\sqrt2}\\ =\\ \quad\frac{8\sqrt2-\sqrt{18\cdot2}}{\sqrt{2\cdot2}}\\ =\\ \quad\frac{8\sqrt2-\sqrt{36}}{\sqrt{4}}\\ =\\ \quad\frac{8\sqrt2-6}{2}\\ =\\ \quad4\sqrt2-3\\ =\\ \quad-3+4\sqrt2\)

So     \(a=-3\)     and     \(b=4\)  .

Write   \(\frac{12}{\sqrt2}+\sqrt{18}\)   in the form   \(b\sqrt2\)   ,  where   \(b\)   is an integer.

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\(\ \quad\frac{12}{\sqrt2}+\sqrt{18}\\ =\\ \quad\frac{12}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}+\sqrt{18}\\ =\\ \quad\frac{12\sqrt2}{2}+\sqrt{18}\\ =\\ \quad6\sqrt2+\sqrt{18}\\ =\\ \quad6\sqrt2+\sqrt{2\cdot3\cdot3}\\ =\\ \quad6\sqrt2+3\sqrt2\\ =\\ \quad9\sqrt2\)