CPhill

WollframAlpha actually is capable of performing these operations.....I suppose there are some algorithms that one could learn, but....I'm not sure it would be worth the trouble to remember them !!!

Just for fun...here is the answer to the third one :

B TW...the last one can't be represented by a Roman Numeral....it produces a decimal approximation

Nice use of that identity, Melody.....!!!

I think you made a slight error here, HSC......let's go from here......

-8(8 - x) = 4/5(x + 10)

-64 + 8x  = (4/5)x + 8    multiply through by 5 to clear the fraction

-320 + 40x = 4x + 40    add 320 to each side, subtract 4x from each side

36x  = 360    divide both sides by 36

x = 10

Note: the two rules we are using here are :  [ a^m]^n  = a^(m * n)

and  a^m * a^n   = a^(m +n)

(3x^2)^3  =  3^3 * [x^2]^3  =   27 x^6  [note that the exponent was just "distributed" over the terms in the parentheses

7x^2*3x^5  = 7 * 3 * x^2 * x^5   =  21 x^7    [note that we could arrange the terms in any order we want to since we are just multiplying them together ]

(5x)^2*x^7   = 5^2 * x^2 * x^7   =  25 x^9

I assume we are using the Exterior Angle Theorem here....if so, we have

13x - 3   = [ 3x + 2] + 55     simplify

13x - 3  = 3x + 57    add 3 to both sides, subtract 3x from both sides

10x  = 60      divide both sides by 10

x = 6 [degrees]

We would have to know how many terms we have to give a precise answer.

The general form for a geometric sum, S, is given by;

S =   a₁ [ 1 - r^n] / [ 1 - r]     where a₁ is the first term, r is the common ratio, and n is the number of terms

Notice, however  that

39360  = a₁ [ 1 - 3^n] / [ 1 - 3]   is true for n= 2 when  a₁ = 9840

But  it's also true for n= 4 when a₁  = 984

And it's also true for n = 8  when a₁ = 12

In fact, we could generate an endless list of possible first terms depending upon our choices for n !!!

6*(4.56)   = 27.36 oz   = 27.36/16  = 1.71 lbs.

Here's an example....suppose we want to change 39 base 10 to a binary number....we divide repetitively as follows until we have a fraction < 1......

39 / 2   =    19  +  remainder 1

19/2    =       9  + remainder 1

9/2     =        4  + reminader 1

4/2  =           2  +  remainder 0

2/2  =           1   +  remainder 0

1/2  =                    remainder 1

Now......write  the remainders  from bottom to top in order  =  100111

This is the base 2 representation of 39.....

Proof

1*2^5  + 0 *2^4 + 0*2^3 + 1*2^2 + 1*2^1  + 1*2^0  =

32 + 0 + 0 + 4 + 2 + 1  =

32 + 4 + 2 + 1  =

39

y = 7x

Just substitute the x value of each point into the above equation

A (-2, -14)      B ( 4, 28)

I don't see any diagram, but just taking a guess, we want to find the area of two congruent isosceles triangles in a circle whose equal sides are 8 and the included angle between these sides is 125°. Also......I'm assuming that these vertical angles are central angles in the circle.....if so, the area  =

(8^2) * sin(125)  = about 52.426 in^2