CPhill

We don't really need a calculator, here...

Note, sin2x  = 2sinxcosx.......so we have....

2sinxcox - sinx  = 0       factor out sinx

sinx (2cosx - 1)  =  0

So, setting each factor to 0, we have that  

sinx = 0      and this happens at 0 radians and pi radians on the interval  ....or....

2cosx - 1  = 0       add 1 to both sides

2cosx  = 1        divide both sides by 2

cosx  =  1/2       and this happens at pi/3  radians and 5pi/3  radians  on the interval

[ These solutions are exact values....]

Sorry, guest......

x cannot be -100  because it would mean that we would be taking the log of a negative number on both sides in the original equation....!!!

log(x/10) = log(10/x)+2           [   2 = log (100) ]

log(x/10)  = log (10/x) + log (100)

log(x / 10)  = log ( 10/x * 100)

log(x/10)  = log ( 1000/x )     and......equating logs, we have

x/10  = 1000/x       cross-multiply

x^2  = 10000

x = 100

Thanks, Melody.......!!!  ......here's another way to see this...

This must be a linear equation if the derivative is just some integer at all x values...so....

At  0, f(x)  = -5

So....at f(2).......x  has changed by 2  units

Thus, using the formula for slope :

 [ f(2) - (-5) ] / [ 2  - 0]  <=  3     simplify

[ f(2)  + 5]  / 2  <= 3     multiply both sides by 2

f(2) + 5 <= 6      subtract 5 from both sides

f(2)  <=  1

Thus f(2) cannot  = 2   because this is greater than 1

Let  one number  = -15    and the other - 2

So

  -15 *  - 2   = 30

And

-15 +  -2   = -17

Problem 2

Put 3 jewels on one side and 3 on the other :

1st scenario - they balance

Take them all off

Pick any two of the remaining three jewels

Put them on the scale.....if they balance....the remaining one is the real one

If they don't balance......the real one has to be the heavier

Result......real one is found after just two weighings

2nd scenario - they don't balance and one side is heavier

Take the three jewels from the heavier side

Pick any two of these

Put them on the scale.....if they balance....the remaining one is the real one

If they don't balance......the real one has to be the heavier

Result......real one is found after just two weighings

Trapezoid: b:x B:x+2 h:3

Area of the Trapezoid =   (3/2)[x + x + 2]  = (3/2)[2x + 2]   = (3/2)*2[x + 1]  = 3[x + 1]

Parallelogram: b:x+4 h:2

Area of the Parallelogram  =  [x + 4]*2

So....setting these equal, we have......

3*[x + 1]  =  [x + 4] *2

3x + 3  = 2x + 8       subtract 2x, 3 from both sides

x  =  5

The area is 18 sq units for each figure.....

5 + 13/15 =   88/15

2 + 1/5  =  11/5

Let n be the number

"of" means "multiply"   and " what" means "equal"     .......so....

88/15 * n   =   11/5       multiply both sides by 15/88

[15/88] * [88/15] n  =  [15/88] * [11/5]

n  = [15/88] * [11/5 ]  =  [15/5] * [ 11/88]  =  3/1 * 1/8  =   3/8

Very nice, heureka.....particularly, the proof part.....!!!!

100 - ((2 * 25-50 * 20/100))  =

100 - ((50 - 50*20/100))  =

100 - ((50 - 1000/100)) =

100 - ((50 - 10))  =

100 - (40)  =

60