CPhill

y=x+1/x

Taking the derivative, we have....

y '  =  1 -  1/x^2     set to 0

1 - 1/x^2  = 0

1/x^2 = 1      and we have two critical points...... x = -1  or x = 1

Since x > 0 , let's take the second derivative

y "  = 2/x^3      and evaluating this at  x =1, we have 2/1^3  = 2 .......so......we have an "minimum" at (1 , 2). Note that this is just a minimum on the interval of interest, i.e.,  (0, ∞). This function actually has no absolute minimum  (or maximum) on its domain.

1 / 8  = .125

 a-(4+5)8=3.4    simplify

a -(9)8 = 3.4

a - 72  = 3.4      add 72 to both sides

a = 75.4

This represents the absolute value of a number.....i.e., the distance from 0 to the number on the number line. This is always either a positive quantity or 0.

If the scale factor is 1/12, the true depth must be 12 times as great as the depth of the model.

So 12 x 32cm = ????

I get something a little different here than geno.....see the following pic........

All  points inside triangle ADF will have x values less than their associated y values. And the area of this triangle = 1/2 sq units

And the area of the whole rectangle = (4)(1)  = 4 sq units

So....the probability that a random point has an x value less than its y value = (1/2) / 4   = 1/8

Nice, geno.....!!!!

Nice proof,  Melody......!!!!

I have a little different answer than Melody......see the following pic.....{I'm assuming the small square is 2 inches on each side....I''m also assuming that the distances between the squares are measured from edge to edge}

The area between the  medium square and the small square {the shaded area}  = 8^2 - 2^2  = 64 - 4 = 60 square units 

And the area of the whole dartboard = 16^2  = 256

So....the probability that the dart lands in the shaded area between the small and medium square = 60/256  = 15/64

To find the x coordinate of the midpoint, add the x values and divide by 2.

Do the same for the y values to find the y coordinate of the midpoint.