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A) 

After Bob spent \(\frac{1}{3}\)then spent 2/3 of 2/3, which is \(\frac{2}{3}*\frac{2}{3}=\frac{4}{9}\). \(1-\frac{4}{9}-\frac{1}{3}=\frac{2}{9}\). He had 2/9 left.

B)

Set the price of a cap as c, and so each pair of shorts is c+30.

Set the amount of money Bob has at the beginning as x.

We get the equation \(\frac{x}{3}=4c+5(c+30)\), after bob buys 4 caps and 5 pairs of shorts

Bob has already spend 1/3 of his money, so he only has 2/3 left, so spending 2/3 of 2/3 is \(\frac{2}{3}*\frac{2}{3}=\frac{4}{9}\).

So, \(\frac{4}{9}x=16c\). 

We get the system \(\begin{cases}\frac{x}{3}=4x+5(c+30)\\\frac{4}{9}x=16c\end{cases}\).

\(\begin{cases}x=12c+15(c+30)\\x=36c\end{cases}\)

\(36c=27c+450\)

\(9c=450\)

\(c=50\)

\(x=50*36=1800\).

So, Bob had 1800 dollars at the beginning.

Amazing solution thankyou! Mass points was a very nice way of going about solving this. Personally i just used simple triangle properties like ratio bases with same altitudes. I think mass points might be easier though. 

The altitude will have the negative inverse of the slope of \(\overline{BC}\) becasue it is perpedicular. Now with having a point\(A\) on the line and its slope, we know everything about the line therefore:

\(\frac{2-(-8)}{-3-1}=-\frac{5}{2} \)
negative inverse=\(\frac{2}{5}\)
using point-slope form:
\(y-y_1=x(x-x_1)\\ y-3=x(\frac{2}{5}-2) \text{ <----lets plug in (0,y)}\\ y-3=0\\ \boxed{y=3}\)

4) 35pi/4 = 35pi/4 - 8pi = 35pi/4 - 32pi/4 = 3pi/4. 3pi/4 = 135 degrees, which is located in Quadrant II.

3) 5pi/6 is > pi/2 or 90 degrees, but < pi or 180 degrees, so 5pi/6 lies in quadrant two.

-5pi/9 = -5pi/9 + 2pi (coterminal) = 13pi/9. 13pi/9 is greater than pi or 180 degrees, but is less than 3pi/2 = 13.5pi/9 or 270 degrees, so this terminal ray also lies in QUadrant III.

1) -21pi/8 = -21pi/8 + 4pi = 11pi/8. 11pi/8 is > than pi or 180 degrees, but is < 3pi/2 or 270 degrees, so the terminal ray lies in Quadrant III.

Coterminal angles means the same angle, just you rotated the terminal-line around the unit circle a multitude of times. Thus, all coterminal angles of 2pi/3 are in the form 2pi/3 + 2(k)pi, k is in the realm of integers: 2pi/3 - 4pi = -10pi/3, 2pi/3 - 2pi = -4pi/3, 2pi/3 + 2pi = 8pi/3, thus the FIRST, SECOND, and LAST are the only correct answers.

Let S = 0.296296296296...

1000S = 296.296296296... = S + 296

999S = 296

S = 296/999 = (37)(8)/(37)(27) = 8/27 

If \({a}^{b}={a}^{c}\) then \(b=c\).

Following this logic then, for A) \(2y = 5\),  \(y=\frac{5}{2}\).

For B, \({9}^{3u+5}={3}^{6u+10}\) ,\(6u+10=8\), \(u=-\frac{1}{3}\).

For C, \({64}^{5x}={8}^{10x}\).\(4x-1=10x\), \(x=-\frac{1}{6}\).

A general good way to determine if a function is exponential is if it can be written as \(f(x)=a*{b}^{x}+k\), (Note x is not a constant!!) the function is an exponential function. This accounts for all stretches and shifts. Note: \(f(x)=a*{b}^{dx+e}+k\) is not needed, because this can always be transformed into the form I gave.

\(y={2}^{x}\) satisfies this.

\(y={x}^{2}\), does not satisfy this. (x must be in exponent).

\(y=\sqrt{x}\) or \(y={x}^{\frac{1}{2}}\) does not satisfy this.

\(y=\sqrt[3]{x}\) or \(y={x}^{\frac{1}{3}}\) does not satisfy this.

\(y=\frac{1}{x}\) or \(y={x}^{-1}\) does not satisfy this.

\(y=8x-4\) does not satisfy this.

Only A is correct.

Great problem! I had to think some time about this one.

We first notice these are all cevian lines, so mass points work perfectly!

Becuase [BFC] = [FCE] then \(\overline{BF} = \overline{FE}\).

Similarly, \(\frac{\overline{DF}}{\overline{FC}}=\frac{3}{7}\).

Label the mass of D as 7, the mass of C as 3, so the mass of F is 3+7 = 10.

Because \(\overline{BF} = \overline{FE}\), then the mass is split evenly between B and E, so they are both 5.

The mass of D is 7, and the mass of B is 5, so the mass of A is 7-5 = 2.

Becuase mass of A * length of AD = mass of B * length of BD, we get that the ratio of the lengths of AD : BD is \(\frac{AD}{BD}=\frac{5}{2}\).

Similarly, \(\frac{AE}{EC}=\frac{3}{2}\). With these ratios, we also know the ratios of the area of some triangles, which we can use to our advantage.

Set the area of the shaded quadrilateral as x, and the area of the triangle ABC as Y.

We get two systems, by finding the area of triangle ADC and AEB, each two ways.

\(\begin{cases} \frac{3}{5}y=x+3 \\ \frac{5}{7}y=x+7 \end{cases}\). We solve and get x = 18. So the area of the shaded quadrilateral is 18.

Any permutation starting 1, 5 , ...    4! of them 

Any permuttion starting 1, 6, ...       4! of them

Any permutation ending ..., 5, 1      4! of them

Any permutation ending ..., 6, 1      4! of them

Any permutation starting 5, 1, 6, ...  3! of them

Any permutation starting 6, 1, 5, ...  3! of them

etc. ...

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}} \)

Working from  the bottom-up

4 + 1/2 = 9/2

1/ (9/2) =  2/9

4+ 2/9  = 38/9

1/ (38/9)  =9/38

1 + 9/38 = 47/38

1 / (47/38)  = 38 / 47

\((a+b)^2=a^2+b^2+2ab\\ 16=6+4ab\\ ab=\cfrac{5}{2}\)
Then 
\((a+b)^3=a^3+b^3+3ab(a+b)\\ 64=a^3+b^3+3(\frac{5}{2}){4}\\ \boxed{a^3+b^3=34}\)

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}}.=\cfrac{38}{47}\)

x^2 + y^2  =  1

(x - 1)^2 +y^2  =  1        subtract the second from  the first

x^2 - ( x -1)^2  =  0

x^2 - x^2 + 2x  - 1   =0

2x - 1  = 0

x = 1/2

y^2  =  1  - (1/2)^2

y^2  = 3/4

y = sqrt (3) / 2          and  y =   -sqrt (3) / 2

Pts of intersection  are   ( 1/2 , sqrt (3) / 2)    and ( 1/2 , -sqrt (3) / 2)

Since the x values are the same, the slope is undefined  

Construct perpendicullar line  to the given one passing through (7.1)

The equation of this line  is

y = (x - 7) + 1

y = x  -6

The x intersection of  these lines is

-x + 5 = x - 6

2x =11

x = 11/2  =  5.5

And the y coordinate of the intersection is

5.5 - 6  =   -.5

The intersection point is ( 5.5 , -.5)

The reflected point  is  given by  [ 5.5 - (7 - 5.5) , -.5 - ( 1 -  -.5) ]  = [( 5.5 - 1.5) ,( -.5 - 1.5) ] =

( 4 , -2)

Construct a circle with a radius of  (2)  at the origin

The  line l will be tangent to this  circle

The equation of this  circle is

x^2 + y^2  = 4     (1)

The slope of a tangent line to this  circle at any  point will be -x /y

So

-x/ y =  1

-x = y  (2)

Sub (2) into (1)

x^2 + (-x)^2 = 4

2x^2  = 4

x^2  = 2

x =  sqrt2

y =-sqrt 2

The equation of the line is

y = 1 (x - sqrt 2)  - sqrt 2

y = x - 2sqrt 2

y=  x -sqrt 8

When x = 0, y= - sqrt 8

When y = 0 x = sqrt 8

So points (0.-sqrt 8)  and (sqrt 8,0)  are on the line

The base of  the triangle is  sqrt [ (sqrt 8)^2 + (-sqrt 8)^2 ] ]=  sqrt (16)  = 4

Its area is  (1/2) base * height  =  ( 1/2)(4) (2)  =  4