hectictar

If  f(x)  truly equals  20 / 5 + 6e ^ -0.1   for all possible values of  x , then

f(5)  =  20 / 5 + 6e ^ -0.1

f(5)  =  4 + 6e ^ -0.1

f(5)  =  4 + \(\dfrac{6}{e^{0.1}}\)

f(5)  ≈  9.4290

\(\ \phantom{=\quad}\dfrac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}\\~\\ =\quad\dfrac{10^{2000}(1+10^{2})}{10^{2000}(10+10)}\\~\\ =\quad\dfrac{(1+10^{2})}{(10+10)}\\~\\ =\quad\dfrac{(1+100)}{20}\\~\\ =\quad\dfrac{1}{20}+\dfrac{100}{20}\\~\\ =\quad\dfrac{1}{20}+5 \)

The whole number closest to  5  and  1/20th  is  5 .

15bx - 20  >  35

                                  We can factor  -5  out of  15bx - 20  because  -5( -3bx + 4 )  =  15bx - 20

-5( -3bx + 4 )  >  35

                                  Rearrange the terms inside the parenthesees...

-5( 4 - 3bx )  >  35

                                  Divide both sides of the inequality by  -5 , a negative number, so swap the sign.

4 - 3bx  <  -7

Does that help answer your question?

Let the x-coordinate  Q  =  a

Then the y-coordinate of  Q  =  -a + 6

slope of AQ  =  slope between  (10, -10)  and  (a, -a + 6)\(\ =\ \frac{(-a+6)-(-10)}{a-10}\ =\ \frac{16-a}{a-10}\)

slope of OQ  =  slope between  (0, 0)  and  (a, -a + 6)\(\ =\ \frac{(-a+6)-(0)}{a-0}\ =\ \frac{6-a}{a}\)

in order for  m∠OQA  to be  90°,  the slope of AQ  must be the negative reciprocal of the slope of  OQ . So...

\(\frac{16-a}{a-10}\ =\ -(\frac{a}{6-a})\\~\\ \frac{16-a}{a-10}\ =\ \frac{-a}{6-a}\\~\\ (16-a)(6-a)\ =\ (-a)(a-10)\\~\\ 96-22a+a^2\ =\ -a^2+10a\\~\\ 2a^2-32a+96\ =\ 0\\~\\ a^2-16a+48\ =\ 0\\~\\ (a-12)(a-4)\ =\ 0\\~\\ a=12\qquad\text{or}\qquad a=4\)

There are two points for  Q  that make  m∠OQA = 90° .   They are:   (12, -6)   and   (4, 2)

12 + -6  =  6     and     4 + 2  =  6

6 + 6  =  12

\(3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

                        Divide both sides of the equation by  15

\(\frac{3}{15}\ =\ \frac{\sqrt{24}}{\sqrt x}\)

                                        Square both sides of the equation.

\(\Big(\frac{3}{15}\Big)^2\ =\ \Big(\frac{\sqrt{24}}{\sqrt x}\Big)^2\)

                                        Simplify both sides using the rule  \(\big(\frac{a}{b}\big)^2\ =\ \big(\frac{a}{b}\big)\big(\frac{a}{b}\big)\ =\ \frac{a^2}{b^2}\)

.\(\frac{3^2}{15^2}\ =\ \frac{(\sqrt{24}\,)^2}{(\sqrt x\,)^2}\)

\(\frac{9}{225}\ =\ \frac{24}{x}\)

                              Now multiply both sides of the equation by  x

\(x\cdot\frac{9}{225}\ =\ 24\)

                                                  Multiply both sides of the equation by  \(\frac{225}{9}\)

\(x\cdot\frac{9}{225}\cdot\frac{225}{9}\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 600\)

=  10x² + 13x - 30

=  10x² + 25x - 12x  - 30

                                             Factor  5x  out of the first two terms.

=  5x(2x + 5) - 12x - 30

                                             Factor  -6  out of the last two terms.

=  5x(2x + 5) - 6(2x + 5)

                                             Factor  (2x + 5)  out of both remaining terms.

=  (2x + 5)(5x - 6)

Does that help answer your question?

\(\ h(x)\ =\ f(x)\cdot g(x)\)

                                          Divide both sides of the equation by  f(x)

\(\dfrac{h(x)}{f(x)}\ =\ g(x)\)

                                          Substitute  6x + 15  in for  h(x)  and substitute  2x + 5  in for  f(x)

\(\frac{6x+15}{2x+5}\ =\ g(x)\)

                                          That is the same as....

\(g(x)\ =\ \frac{6x+15}{2x+5}\)

                                          We can factor  3  out of  6x + 15  because  3(2x + 5)  =  6x + 15

\(g(x)\ =\ \frac{3(2x+5)}{2x+5}\)

                                          Cancel the common factor of  (2x + 5)  in the numerator and denominator.

\(g(x)\ =\ 3\)          and note that   2x + 5  ≠  0   so   x  ≠  -5/2

So we can say     g(x)  =  3     for all possible values of  x  except  x = -5/2

\(y\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

What is  x  when  y = 3 ?  Plug in  3  for  y  and solve for  x .

\( 3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

To solve for  x ,

divide both sides of the equation by  15

then square both sides of the equation

then multiply both sides of the equation by  x

then multiply both sides of the equation by  \(\frac{225}{9}\)

Can you do those things sinclairdragon?  If you get stuck let us know!

Thanks!!

EmeraldWonder, there are a couple of small errors in your work at the end because  \(\frac{8\,*\,5t^2}{5t\,*\,3}\ \neq\ \frac{8t^2}{3}\)

Here is another way to work this problem:

\(\ \phantom{=\quad}\dfrac{24t^3}{15t^4}\cdot \dfrac{5t^8}{3t^6}\)

\(=\quad\dfrac{24\,\cdot\,t^3\,\cdot\,5\,\cdot\,t^8}{15\,\cdot\,t^4\,\cdot\,3\,\cdot\,t^6}\)

\(=\quad\dfrac{120\,\cdot\,t^3\,\cdot\,t^8}{45\,\cdot\,t^4\,\cdot\,t^6}\)          because we can multiply numbers in any order and   24 · 5 = 120   and   15 · 3 = 45

\(=\quad\dfrac{120\,\cdot\,t^{11}}{45\,\cdot\,t^{10}}\)          because   t³ · t⁸  =  t t t · t t t t t t t t  =  t^{(3 + 8)}  =  t¹¹   and   t⁴ · t⁶  =  t^{(4 + 6)}  =  t¹⁰

\(=\quad\dfrac{120\,\cdot\,t^{10}\cdot\,t}{45\,\cdot\,t^{10}}\)          because   t¹¹  =  t t t t t t t t t t t  =  t¹⁰ · t

\(=\quad\dfrac{120\,\cdot\,t}{45}\)          when  t ≠ 0  because we can cancel the common factor of  t¹⁰  in the numerator and denominator.

\(=\quad\dfrac{8t}{3}\)          because   \(\frac{120}{45}\)   reduces to   \(\frac83\)_