hectictar

(18/2)²   =   9²   =   81

81  needs to be added to both sides to complete the square on the left side.

3.

(a² + 3a - 7)(2a² - a + 4)

=  a²(2a²) + a²(-a) + a²(4) + 3a(2a²) + 3a(-a) + 3a(4) + (-7)(2a²) + (-7)(-a) + (-7)(4)

=  2a⁴  +  -a³  +  4a²  +  6a³  +  -3a²  +  12a  +  -14a²  +  7a  +  -28

=  2a⁴  +  5a³  -  13a²  +  19a  -  28

y  =  2x^2 - 4x + 4    and    y  =  -x^2 - 2x + 4

First let's find the  x  values of the intersection points by solving this equation:

2x^2 - 4x + 4   =   -x^2 - 2x + 4

                                                  Add  x^2  to both sides of the equation.

3x^2 - 4x + 4   =   -2x + 4

                                                  Add  2x  to both sides.

3x^2 - 2x + 4   =   4

                                                  Subtract  4  from both sides.

3x^2 - 2x  =  0

                                                  Factor  x  out of the two terms on the left side.

x(3x - 2)   =   0

                                                  Set each factor equal to zero.

x  =  0     or     3x - 2  =  0

                       3x  =  2

                         x  =  2/3

The  x  coordinates of the intersection points are  0  and  2/3 .

So   c = 2/3   and   a = 0

c - a   =   2/3 - 0   =   2/3

From the problem statement, we can make these two inequalities:

x + y  ≤  600            and            5x + 7y  ≥  3500

If  x = 330 , then...

330 + y  ≤  600

y  ≤  270

and

5(330) + 7y  ≥  3500

1650 + 7y  ≥  3500

7y  ≥  1850

y  ≥  264 + 2/7      To sell at least  264 and 2/7  tickets, we have to sell at least 265 tickets.

y  ≥  265

Is there a possible value of  y  such that   y  ≤  270   and   y  ≥  265   ?

yes, for example,  y = 265 , or  y  =  266 , or  y = 267

1.   The square root of a negative number is not real.  -9/16  has no real square roots.

2.   \(\sqrt[3]{-0.000125} \,=\,\sqrt[3]{(-0.05)^3}\,=\,-0.05\)

That's a good way MIRB, I didn't know of that!

You can also do it by typing:  abs(-2)

original area   =   L * W

new length   =   L + 20% * L   =   L + 0.2L   =   1.2L

new width   =   W + 10% * W   =   W + 0.1W   =   1.1W

new area   =   1.2L * 1.1W

new area   =   1.2 * 1.1 * L * W

new area   =   1.32 * L * W

new area   =   1.32 * original area

new area   =   (1 + 0.32) * original area

new area   =   original area + 0.32 * original area

new area   =   original area + 32% * original area

The new area is the original area increased by  32%  of the original area.

1.

The maximum  y  value is  1  and the minumum  y  value is  -7

The equation of the midline is   y  =  (1 + -7)/2   =   -6/2   =  -3

The equation of the midline is  y  =  -3

2.

The maximum  y  value is  6   and the minimum  y  value is  -4 .

The amplitude is   (6 - -4)/2   =   10/2   =   5        

f(x)  =  (x - 3)/(x - 4)

                                     First let's find  f^-1(x) . Instead of  f(x) , write  y .

y   =   (x - 3)/(x - 4)

                                     Now solve for  x . Multiply both sides of the equation by  (x - 4) .

y(x - 4)   =   x - 3

xy - 4y   =   x - 3

                                    Add  4y  to both sides and subtract  x  from both sides.

xy - x   =   4y - 3

                                    Factor  x  out of the terms on the left side.

x(y - 1)   =   4y - 3

                                    Divide both sides by  (y - 1) .

x   =   (4y - 3)/(y - 1)

                                    So...

f^-1(x)  =  (4x - 3)/(x - 1)

And...

(4x - 3)/(x - 1)  is undefined when  x - 1  =  0

f^-1(x)  is undefined when  x = 1