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wow my answer was way off.... thank you :)

Let's see...using the Law of Sines, we have

b/sin40 = c/sin90       (sin90 = 1)

b/sin40 = 20

b =20sin40 = 12.855

sorry correction Y intercept (0,-6)

x-intercept= (3,0)

Y-intercept=(0,6)

To emphasize what Melody did:

The equation:  |3x| + |4y|  =  12

Can be split into 4 separate equations; 2 for the  |3x| = ± 3x  and  2 for the  |4y| = ± 4y :

--->                 3x + 4y  =  12       --->   y  =  (12 - 3x)/4

                      -3x + 4y  =  12       --->   y  =  (12 + 3x)/4

                       3x - 4y  =  12       --->   y  =  (12 - 3x)/-4

                      -3x - 4y  =  12        --->   y  =  (12 + 3x)/-4  

Graphing these individually will give you the parallelogram that Melody got.

Oops I meant  

cos = 40º = .766

Melody and I both cheated...(but she cheated first.....and as chief mod, she should set a BETTER EXAMPLE to the people of Camelot!!!)...LOL!!!!

You always give excellent answers saseflower :)

I 'cheated' a bit because I ran off and graphed it.

Here is the graph.

√[ 4 + √(16 + 16a) ]  +  √[ 1 + √(1 + a) ]  =  6

     √[ 4 + √(16 + 16a) ]      +  √[ 1 + √(1 + a) ]  =  6 

=>  √[ 4 + √[16(1 + a) ] ]     +  √[ 1 + √(1 + a) ]  =  6

=>  √[ 4 + √16·√(1 + a)  ]    +  √[ 1 + √(1 + a) ]  =  6

=>  √[ 4 + 4·√(1 + a)  ]        +  √[ 1 + √(1 + a) ]  =  6

=>  √[ 4(1 + √(1 + a) )  ]      +  √[ 1 + √(1 + a) ]  =  6

=>  √4·√[ 1 + √(1 + a) ]        +  √[ 1 + √(1 + a) ]  =  6

=>  2√[ 1 + √(1 + a) ]          +    √[ 1 + √(1 + a) ]  =  6

=>  3√[ 1 + √(1 + a) ]  =  6

=>  √[ 1 + √(1 + a) ]  =  2       (divide both sides by 3)

=>   1 + √(1 + a)  =  4             (square both sides)

=>    √(1 + a)  =  3                   (subtract 3 from both sides)

=>  1 + a  =  9                          (square both sides)

=>     a  =  8

log_1/39  =  x     --->   (1/3)^x  =  9     --->     (3^-1)^x  =  3²     --->     3^-x  =  3²

     --->     -x  =  2     --->     x  =  -2

In exponential form, this says

(1/3)^x = 9 

Note that 1/3  = 3^-1 and 9 = 3²   so we can write

(3^-1)^x  = 3²

3^-x = 3²    and, equating exponents, we have

-x  = 2     so   x = -2

So

log_1/3 9 = -2

I think this might be  [csc^2x secx] / [sec^2x+csc^2x ].....if so, convert to sines and  cosines

[ 1 / sin^2x * 1 / cosx ] / [ 1 / cos^2x  + 1 / sin^2x ]     simplify the denominator

[ 1 / sin^2x * 1 / cosx ] / [(sin^2x + cos^2x) / (cos^2x sin^2x) ] =

[ 1 /(sin^2x * cosx)] / [1 /( cos^2x * sin^2x) ] =

( cos^2x * sin^2x) ] / (sin^2x * cosx) =

cosx

33% is a tiny bit less than a third

A third of 46 = a tiny bit more than 15

15 is close enough

$${\frac{{\mathtt{4}}}{{\mathtt{410}}}} = {\frac{{\mathtt{2}}}{{\mathtt{205}}}} = {\mathtt{0.009\: \!756\: \!097\: \!560\: \!975\: \!6}}$$

|x - 4| - 10  =  2

Add 10 to both sides:  |x - 4|  =  12

So, either                       x - 4  =  -12         or         x - 4 = 12

                                            x  =  -8           or              x  =  16

Those are the two possible values of x; the product of these two numbers will be your answer. 

$$\\ tan^2(4pi/3)- Sec^2(-11pi/6)\\\\
=tan^2(pi/3)- Sec^2(pi/6)\\\\
=(\sqrt3)^2-(\frac{2}{\sqrt3})^2\\\\
=3-\frac{4}{3}\\\\
=\frac{9-4}{3}\\\\
=\frac{5}{3}\\\\$$

csc²(x) · sec(x) / sec²(x)  +  csc²(x)

=  csc²(x) · [ sec(x) / sec²(x) ]  +  csc²(x)                           

=  csc²(x) · [ 1 /sec(x) ]  +  csc²(x)                        <---   simplify:  sec(x) / sec²(x)

=  csc²(x) · ( cos(x) )  +  csc²(x)                            <---    cos(x)  =  1 / sec(x)

=  csc²(x) · [ cos(x) + 1 ]                                        <---    factor

=  [ 1 / sin²(x) ] · [ cos(x) + 1 ]                              <---    csc²(x)  =  1 / sin²(x)

=  [ cos(x) + 1 ] / sin²(x)

=  [ cos(x) + 1 ] / [ 1 - cos²(x) ]                             <---    sin²(x)  =  1 - cos²(x)

=  [ cos(x) + 1 ] / [ ( 1 - cos(x) )·( 1 - cos(x) ) ]      <---    factor

=  1 / [ 1 - cos(x) ]                 <--- provided that sec²(x) ≠ 0, cos(x) ≠ 0

Nice, saseflower.....!!!

Thanks CPhill!! 

Thank you! Still need some help figuring out where on the graph I need to apply these.

Saseflower has almost gotten you home......the y intercept is at -4......this is the point (0, -4)....plot this point on the graph.....

Now...the slope is (1/3)......this means that y changes by 1 and x changes by 3. So, from (0, -4).....go up 1 unit on y and over 3 units on x.  You should be at the point (3, -3). Draw a line through these two points......there's your graph...!!!!

HAPPY 7 - you need to rethink this answer!

He only reads 5 pages in 1 hour - so it is  200/5 hours