hectictar

   (3x² + 9x - 1) + (-5x² + 6x - 10)

We can get rid of the parenthesees since there is only addition between them.

=  3x² + 9x - 1 + -5x² + 6x - 10

Now we can rearrange the terms so like terms are together.

=  3x² - 5x² + 9x + 6x - 1 - 10

Combine like terms. Just like  3 apples - 5 apples = -2 apples ....so does  3x² - 5x² = -2x²

=      -2x²     +  15x        - 11

cups of vinaigrette  =  cups of vinegar + cups of oil

Let's call the amount of additional cups of oil  " x " .

There are   3  cups of vinegar   , and there are   1 + x  cups of oil   in the batch so far.

cups of vinaigrette  =  (3) + (1 + x)

cups of vinaigrette  =  4 + x

We want there to be  25  cups of vinegar for every  100  cups of vinaigrette.   That is....

\( \frac{\text{number of cups of vinegar}}{\text{number of cups of vinaigrette}}=\frac{25}{100} \)       Plug in  3  for cups of vinegar and  4 + x  for cups of vinaigrette.

\(\frac{3}{4+x}\,=\,\frac{25}{100} \\~\\ (4+x)(25)\,=\,(3)(100) \\ 100+25x\,=\,300 \\ 25x\,=\,200 \\ x\,=\,8\)                 Now solve this for  x  . First, let's cross-multiply.

So....she needs to add  8  cups of oil .

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The final recipe will have  9  cups of oil. To find the price, we need to know how many ounces this is.

1 cup    =  8 ounces          Multiply both sides by  9  .

9 cups  =  72 ounces

The final recipe will have  3  cups of vinegar.

1 cup    =  8 ounces

3 cups  =  24 ounces

the total price, in dollars   =   0.50 * (ounces of olive oil)  +  0.35 * (ounces of vinegar)

                                         =   0.50 * 72   +   0.35 * 24

                                         =   36   +   8.4

                                         =   44.40

Try typing this into the bar:       asin(0.591)         and press enter.

You should get approximately 36.228º  .

f(x)  is defined only for  0 ≤ x ≤ 1  , and  f(x)  =  ax + b  for constants  a  and  b  where  a < 0  .

Since  a  is negative, the smallest possible value for  f(x)  will be when  x = 1  .

f(1)  =  a(1) + b  =  a + b

Then, the largest possible value for  f(x)  will be when  x = 0  .

f(0)  =  a(0) + b  =  b

The smallest possible value for  f(x)  is  a + b  , and the largest possible value for  f(x)  is  b  .

So......the range for  f(x)  is  [ a + b ,  b ]  .

I don't really know what it means to make the division complete, but if  a = -1  , then...

\(\frac{x^2+ax-2}{x-2}\,=\,\frac{x^2-x-2}{x-2}\,=\,\frac{(x+1)(x-2)}{x-2}\,=\,x+1\qquad\text{, when}\,x\neq2\)

The given intersection points lie on  y = h(x)  and  y = j(x)  .

For example...since  (8, 12)  is an intersection point....   12 = h(8)   and   12 = j(8)

When  x = 4  ,     h(2x)    =    h(2(4))    =    h(8)    =    12    =    y

When  x = 4  ,     2j(x)     =    2j(4)       =    2(6)    =    12    =    y

So... (4, 12)  is a point on  y = h(2x)  and on  y = 2j(x)  .

4 + 12  =  16

Sorry that this isn't a very good explanation.....If it doesn't make sense, it might help to try different numbers for  x  , such as 1, 2, or 6...to see why they don't work.  

-sin² x  ≤  -1

                       Multiply both sides by  -1  and swap the sign.

 sin² x  ≥  1

The biggest possible value of  sin x  is  1  , and the smallest possible value of  sin x  is  -1  .

So, the biggest possible value of  sin² x  is  1²  or  (-1)²  , which both equal  1  .

There is no way that  sin² x  can be bigger than  1  .

 sin² x   is not always   ≥  1

-sin² x   is not always   ≤  -1

\(\frac{-4\,\pm\,\sqrt{4-4(32)(5)}}{2(32)} \\~\\ =\,\frac{-4\,\pm\,\sqrt{4-640}}{64} \\~\\ =\,\frac{-4\,\pm\,\sqrt{-636}}{64} \\~\\ =\,\frac{-4\,\pm\,\sqrt{-1\cdot2\cdot2\cdot3\cdot53}}{64} \\~\\ =\, \frac{-4\,\pm\,\sqrt{-1}\,\cdot\,\sqrt{2\cdot2}\,\cdot\,\sqrt{3\cdot53}}{64} \\~\\ =\, \frac{-4\,\pm \,i\,\cdot\,2\,\cdot\,\sqrt{159}}{64} \\~\\ =\, \frac{-4\,\pm\,2i\sqrt{159}}{64}\)

If this is from a quadratic equation, is it    \(\frac{-4\pm\sqrt{4^2-4(32)(5)}}{2(32)} \)    ?

Also...

\(\frac{(-3)^{2n+1}}{27\,\cdot\,(-3)^{2n}}=\frac{(-3)^{2n+1}}{27}\,\cdot\,\frac{(-3)^{2n+1}}{(-3)^{2n}}\)

This is not true.

If you multiply the two fractions on the right side together, you will get   \(\frac{[ (-3)^{2n+1})]^2}{27\,\cdot\,(-3)^{2n}}\)     .

\(\frac{a}{bc}\,\neq\,\frac{a}{b}\,\cdot\,\frac{a}{c}\)

But..you can distribute division the same as you distribute multiplication, like this....

\(\frac{8 + 6 +10}{2}=\frac12(8+6+10)\,=\,(\frac12)(8)+(\frac12)(6)+(\frac12)(10)\,=\,4+3+5\,=\,12\)