CPhill

Let 2N be the first even number

2N + 2 the next

2N + 4 the next

2N + 6  the last   .......so we have

Average  =   [sum of the numbers ] / [ how many numbers we have]

17  = [ 2N + ( 2N + 2) + ( 2N + 4) + (2N + 6)] /  4      simplify

17  =  [ 8N + 12]  / 4    multiply both sides by 4

68  = 8N + 12     subtract 12 from both sides

56  = 8N    divide both sides by 8

7 = N   

So the first even number =  2(N)  = 2(7)  = 14

Which means that the next three are  16, 18 and 20

Mmmm....this is an interesting one !!!

The volume of the snow needed for the snowman is given by :

(4/3)pi [ 5^3 + 3^3 + 2^3 ]   = about  670 ft^3   of snow

The volume of snow in the yard is given by :

[  50 ft * 25 ft * (2/3) ft ] =  about 833.33 ft^3   of snow

Looks like there's more than enough......!!!!

Sorry, Max.....I believe you made an error....look at your second step....you have.....

l 2x + 6  l  + 16 = 0       (1)

If we claim that the answer is -11, then we have

l 2(-11) + 6 l + 16 = 0

l -22 + 6 l  + 16  = 0

l -16 l + 16   = 0

16 + 16   = 0

32  = 0     .......which is clearly false

We have no solution because the left side of (1)   will always be >  0

6(3-x)=3x    distribute the 6 across the terms in the parenteses on the left

18 - 6x   = 3x         add 6x to both sides

18  = 9x               divide both sides by 9

2  = x

y = x^2  - 1       add 1 to both sides

y + 1  = x^2       take the positive/negative square root of both sides

± √ [y - 1 ]  =    x

Assumng that we are talking about rational functions, it is not possible for a function to have a horizontal asymptote and an oblique asymptote at the same time. The reason for this is that a horizontal asymptote will occur when we have the same degree polynomial in the numerator as in the denominator. A slant asymptote will occur when the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator.......thus we can have the following combo possibilities :

vertical asymptote / horizontal asyptote  ....example      [ x+ 3] / [x - 1]

vertical asymptote / slant asymptote  ...... example    [ x^2] / [ x - 1]

5. (x^2y^4)^4/(y^-2)^4 =

x^8 * y^16 / y^-8

But remember....a negative exponent "throws" the result back  across the fraction bar  and makes the exponent positive........so we have

x^8 * y^16 * y^8   =

x^8 * y^24

6. (16x^2y^-4z^-1)^0/y^2 =   1 / y^2      [assuming x, y and z are not = 0 ]

7. (3ab^-1)^2  =

3^2 * a^2 * b^-2  =

9a^2 / b^2

l 2x + 6 l   - 4  = -20   add 4 to both sides

l 2x + 6 l   = -16

Note that this has no solution because absolute value is always positive (or 0).......thus, the left side [ a positive, or zero] cannot equal the right side [ a negative]

Comparing the length of the model with the actual length, the scale must be :   2 ft   = 800 meters

So.....the actual  width can be foung thusly :

800/2  = width/ 1.4     multiply both sides by 1.4

[800] * [1.4] / [2]  = width = 560 meters

And the area is length * width.......so the area  = 

800m * 560m =  448,000 square meters

And the perimeter  =

2 [width + length ]  = 

2 [ 560 + 800]   =  2720 meters

Let x be the height  .....  then the base = 2x + 2

So we have

Area = (1/2) (base) (height)

20  = (1/2)(2x + 2)(x)    simplify

20 = x (x + 1)

20 = x^2 + x            subtract 20 from both sides

x^2 + x - 20   = 0        factor

(x + 5) ( x - 4)  = 0

Only the second factor makes sense

x - 4 =  0      add 4 to both sides

x = 4 ft      and this is the height

And the base =   2(4) + 2   = 10 ft