hectictar

volume of box   =   (width)(length)(height)

width   =   15 - x - x

width   =   15 - 2x

length   =   22 - x - x

length   =   22 - 2x

height   =   x

volume of box   =   (width)(length)(height)

volume of box   =   (15 - 2x)(22 - 2x)x

The expression  (15 - 2x)(22 - 2x)x  can be used to determine the volume of the box.

(  And by looking at a graph we can see that the greatest possible volume is about  432 in³ .

Let's call the length of the rectangle, in yards,  L  .

Let's call the width of the rectangle, in yards,  W .

The perimeter of the rectangle is 596 yards, so....

L + L + W + W   =   596

                                         Let's solve this equation for  L . Comine like terms.

2L + 2W   =   596

                                         Subtract  2W  from both sides of the equation.

2L   =   596 - 2W

                                         Divide through by  2 .

L   =   298 - W

Its length is 2 yards less than 4 times its width, so...

L   =   4W - 2

                                   Since   L  =  298 - W   we can replace  L  with  298 - W .

298 - W   =   4W - 2

                                   Add  2  to both sides, and add  W  to both sides.

298 + 2   =   4W + W

                                   Combine like terms.

300   =   5W

                                   Divide both sides by  5 .

60   =   W

Now we can use this value for the width to find the length.

L   =   298 - W

L   =   298 - 60

L   =   238

So the width is 60 yards and the length is 238 yards.

The sum of the digits for each of the first ten multiples of 9  is  9 .

So the first 10 positive multiples of 9 must be lucky integers.

etc.

But....

9 * 11  =  99     and     99/( 9 + 9 )  =  5.5

So

99  is the least positive multiple of  9  that is not a lucky integer.

Okay....

\(\ \quad\sqrt{\frac38}\\ =\\ \quad\sqrt{\frac38\cdot\frac88}\\ =\\ \quad\sqrt{\frac{24}{8^2}}\\ =\\ \quad\frac{\sqrt{24}}{\sqrt{8^2}}\\ =\\ \quad\frac{\sqrt{24}}{8}\\ =\\ \quad\frac{\sqrt{4\cdot6}}{8}\\ =\\ \quad\frac{\sqrt4\cdot\sqrt6}{8}\\ =\\ \quad\frac{2\cdot\sqrt6}{8}\\ =\\ \quad\frac{\sqrt6}{4}\)

BTW...

Whenever you use a radical, that is, this symbol:  √ 

it is a good idea to always include parenthesees after it, like this:  √( )

and then put all of the numbers that go under the radical in the parenthesees.

So for this expression it would be  √( 3/8 )  .

That way, there is no confusion.

Notice that the expression  √3/8  can also be interpreted as  \(\frac{\sqrt3}{8}\)  

(which would actually be the correct interpretation in this case).

Just to make sure....  Is this the expression in your question:

\(\sqrt\frac38\)

??  

Orange crackers

x - 2y  =  1

                         Add  2y  to both sides of the equation.

x  =  1 + 2y

x² - 3y²  =  13

                                               Since  x  =  1 + 2y  we can replace  x  with  1 + 2y .

(1 + 2y)² - 3y²  =  13

                                               Multiply out  (1 + 2y)²  .

(1 + 2y)(1 + 2y) - 3y²  =  13

(1)(1) + (1)(2y) + (2y)(1) + (2y)(2y) - 3y²  =  13

1 + 2y + 2y + 4y² - 3y²  =  13

1 + 4y + y²  =  13

y² + 4y + 1  =  13

                                    Subtract  13  from both sides.

y² + 4y - 12  =  0

                                    Factor the left side.

(y + 6)(y - 2)  =  0

                                    Set each factor equal to zero and solve for  y .

y + 6  =  0       or       y - 2  =  0

  y  =  -6          or         y  =  2

Now let's use the equation   x  =  1 + 2y   to find  x .

If   y  =  -6   then   x  =  1 + 2(-6)  =  1 + -12  =  -11

So one solution is   x  =  -11   and   y =  -6  .

If   y  =  2   then   x  =  1 + 2(2)  =  1 + 4  =  5

So another solution is   x  =  5   and   y  =  2  .

The solution set is   { (-11, -6), (5, 2) }

There is probably a better way to do this but here's one way....

x² + bx + c  >  0   for  x  in the interval  (-∞, -2) U (3, ∞)

This implies that  x² + bx + c  =  0  when  x  =  -2  and when  x  =  3

The equation of a parabola with zeros at  -2  and  3  is

y  =  a(x + 2)(x - 3)

y  =  a(x² - x - 6)

y  =  ax² - ax - 6a

Since the coefficient of  x²  in  x² + bx + c  is  1  then  a = 1

x² + bx + c   =   x² - x - 6

b  =  -1     and     c  =  -6

b + c   =   -1 + -6   =   -7

Line  \(l\)  appears to have a slope of  \(-\frac53\)  and a y-intercept of  5 .

So the equation of line  \(l\)  is           \(y\,=\,-\frac53x+5\)

On line  \(l\) ,  when  y  =  15 .....

\(15\,=\,-\frac53x+5\\~\\ 10\,=\,-\frac53x\\~\\ -\frac35\cdot10\,=\,x\\~\\ -6\,=\,x\)

So line  \(l\)  passes through the point  (-6, 15) .

Line  \(m\)  appears to have a slope of  \(-\frac27\)  and a y-intercept of  2 .

So the equation of line  \(m\)  is        \(y\,=\,-\frac27x+2\)

On line  \(m\) ,  when  y = 15 ....

\(15\,=\,-\frac27x+2\\~\\ 13\,=\,-\frac27x\\~\\ -\frac72\cdot13\,=\,x\\~\\ -\frac{91}{2}\,=\,x\)

So line  \(m\)  passes through the point  (\(-\frac{91}{2}\), 15)  .