hectictar

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12((3x-4)/(3)-(3-4x)/(4)-(6-2x)/(6))

\(=\,12(\frac{3x-4}{3}-\frac{3-4x}{4}-\frac{6-2x}{6})\)

\(=\,\frac{(12)(3x-4)}{3}-\frac{(12)(3-4x)}{4}-\frac{(12)(6-2x)}{6} \\~\\ =\,\frac{(4)(3x-4)}{1}-\frac{(3)(3-4x)}{1}-\frac{(2)(6-2x)}{1} \\~\\ =\,(4)(3x-4)-(3)(3-4x)-(2)(6-2x) \\~\\ =\,12x-16 - 9+12x -12+4x \\~\\ =\,28x-37\)            Reduce the fractions.

The cost of a telephone call is $0.75 + $0.25 times the number of minutes.

The cost of a telephone call   =   $0.75  +  $0.25  *  the number of minutes.

The number of minutes  =  t , so...

The cost of a telephone call  =  $0.75 + $0.25t

Or we can say

The cost of a telephone call, in dollars  =  0.75 + 0.25t        

-2  is less than  1

-2  <  1

    \(1\frac25-\frac48\)

                                  \(\frac48\)  is the same as  \(\frac12\)

=  \(1\frac25-\frac12\)

                                  \(1\frac25\)  is the same as   \(1+\frac25\)

=  \(1+\frac25-\frac12\)

                                  Now we need a common denominator. Multiply and divide  1  by  10 .

=  \(\frac{10}{10}+\frac25-\frac12\)

                                  Multiply and divide  \(\frac25\)  by  2 .

=  \(\frac{10}{10}+\frac{2*2}{5*2}-\frac12\)

=  \(\frac{10}{10}+\frac{4}{10}-\frac12\)

                                  Multiply and divide  \(\frac12\)  by  5 .

=  \(\frac{10}{10}+\frac{4}{10}-\frac{1*5}{2*5}\)

=  \(\frac{10}{10}+\frac{4}{10}-\frac{5}{10}\)

10 apples + 4 apples - 5 apples  =  9 apples   So

10 tenths + 4 tenths - 5 tenths  =  9 tenths

= \(\frac9{10}\)

Let the base of triangle ABC  be  " x ".

Now, let's draw a height from side  x  to angle  C .

This makes a right triangle with the hypotenuse of  y .

sin 70°  =  opposite / hypotenuse

sin 70°  =  height / y

y sin 70°  =   height

area of triangle ABC  =  (1/2)(base)(height)

area of triangle ABC  =  (1/2)(x)(y sin 70°)       This equals 30 because the problem says so.

Now let's look at triangle  PQR. Let the base be  x  .

This makes its height  =  y sin 110°

And the area of PQR  =  (1/2)(x)(y sin 110°)

Let's say

30  =  (1/2)(x)(y sin 70°)

a  =  (1/2)(x)(y sin 110°)        We want to know what  a  equals.

30 / a  =  [ (1/2)(x)(y sin 70°) ] / [ (1/2)(x)(y sin 110°) ]

30 / a  =  sin 70° / sin 110°

30 / a  =  1

30  =  a          

a)     average rate of change  =

 \(\frac{\text{change in f(x)}}{\text{change in x}}\,=\,\frac{f(1)-f(2)}{1-2}\,=\,\frac{(\,2(1)^3-3(1)\,)\,-\,(\,2(2)^3-3(2)\,)}{1-2}\,=\,\frac{(-1)-(10)}{-1}\,=\,\frac{-11}{-1}\,=\,11\)

like helperid said.

b)     We want an equation of a line that has a slope of  11  and passes through  the point (1, f(1) )

And  (1, f(1)  =  (1, -1)

So..   the equation in point-slope form is     y + 1  =  11(x - 1)

And in slope intercept form, it is   y =  11x - 12

c)     Find the average rate of change between the points  x = a  and  x = a + h .

Thie is the same as part  a  , just with letters in place of numbers.   

average rate of  change  =

\(=\,\frac{\text{change in f(x)}}{\text{change in x}}\\~\\ =\,\frac{f(a+h)-f(a)}{(a+h)-(a)} \\~\\ =\,\frac{(\,2(a+h)^3-3(a+h)\,)\,-\,(\,2(a)^3-3(a)\,)}{h} \\~\\ =\,\frac{2(a+h)^3-3(a+h)-2a^3+3a}{h} \)

I don't know if you need it more simplified than that.

x + 5  <  8

If for some reason we cant solve this inequality the normal way, then we can do it like this

x + 5  =  8

x  =  3

And then we know that

the values that make the original inequality,  x + 5 < 8 , true will be either  x < 3  or  x > 3 .

I thought about it because when you are trying to find the values that make a quadratic function greater or less than zero, we have to do it like this and test points in an interval.

....Is that a good explanation??

The domain of  f(x)  is  -1 < x < 1

So the domain of  f( \(\frac{x+1}{x-1}\) )  is  -1 < \(\frac{x+1}{x-1}\) < 1

*edit* (I was unhappy with my previous answer.)

No!  0/0  is undefined.