All Answers 355948 Answers

5.
3x - x (x-4) = - 3x^2

\(\begin{array}{|lrcll|} \hline & 3x - x (x-4) &=& - 3x^2 \\ & 3x - x^2+4x &=& - 3x^2 \\ & 2x^2+7x &=& 0 \\ & x(7+2x) &=& 0\\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 7+2x &=& 0 \\ & 2x &=& -7 \\ & \mathbf{x}&\mathbf{=}& -\mathbf{\dfrac{7}{2}} \\ \hline \end{array} \)

4.
(x+2)^2 + 5 = 8

\(\begin{array}{|rcll|} \hline (x+2)^2 + 5 &=& 8 \\ (x+2)^2 &=& 8-5 \\ (x+2)^2 &=& 3 \\ x+2 &=& \pm\sqrt{3} \\ x &=& -2 \pm\sqrt{3} \\\\ \mathbf{x_1}&\mathbf{=}& \mathbf{-2 +\sqrt{3}} \\ \mathbf{x_2}&\mathbf{=}& \mathbf{-2 -\sqrt{3}} \\ \hline \end{array}\)

3.
9x^2 + 6x + 1 = 0

\(\begin{array}{|rcll|} \hline 9x^2 + 6x + 1 &=& 0 \\ (3x+1)^2 &=& 0 \\ 3x+1 &=& 0 \\ 3x &=& -1 \\ \mathbf{x}&\mathbf{=}& \mathbf{-\dfrac{1}{3}} \\ \hline \end{array}\)

2.
(4x+12)(x-6) = 0

\(\begin{array}{|lrcll|} \hline &(4x+12)(x-6) &=& 0 \\ 1.& 4x+12 &=& 0 \\ & 4x &=& -12 \\ & \mathbf{x}&\mathbf{=}& \mathbf{-3} \\\\ 2. & x-6 &=& 0 \\ & \mathbf{x}&\mathbf{=}& \mathbf{6} \\ \hline \end{array}\)

1.
-6x^2 + 24x = 0

\(\begin{array}{|lrcll|} \hline & -6x^2 + 24x &=& 0 \\ & x(24-6x) &=& 0 \\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 24-6x &=& 0 \\ & 6x &=& 24 \\ & \mathbf{x}&\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

Solve the inequality

\(\large{\dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} < \dfrac{1}{30}} \)

\(\small{ \begin{array}{|rcll|} \hline \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} &<& \dfrac{1}{30} \quad | \quad x\ne 1,\ x\ne 2,\ x\ne 3,\ x\ne 4 \\\\ \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} -\dfrac{1}{30} &<& 0 \\\\ -\dfrac{x^4-10x^3+5x^2+100x+84}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \\\\ \boxed{\small{x^4-10x^3+5x^2+100x+84 = (x - 7)(x - 6)(x + 1)(x + 2)}} \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot 30 \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot(-1)! \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \quad | \quad \cdot \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2) \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)^2(x - 2)^2(x - 3)^2(x - 4)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \mathbf{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)(x - 2)(x - 3)(x - 4)} & \mathbf{>}& \mathbf{0} \\ \hline \end{array} } \)

\(\begin{array}{|r|c|c|c|c|c|c|c|c|} \hline &(-\infty,-2) & (-2,-1) & (-1,1) & (1,2) & (2,3) & (3,4) & (4,6) & (6,7) & (7,\infty) \\ \hline x+2 & -& +& +& +& +& +& +& +& +\\ \hline x+1 & -& -& +& +& +& +& +& +& +\\ \hline x-1 & -& -& -& +& +& +& +& +& +\\ \hline x-2 & -& -& -& -& +& +& +& +& +\\ \hline x-3 & -& -& -& -& -& +& +& +& +\\ \hline x-4 & -& -& -& -& -& -& +& +& +\\ \hline x-6 & -& -& -& -& -& -& -& +& +\\ \hline x-7 & -& -& -& -& -& -& -& -& +\\ \hline \text{sign} & \\ \text{of} & \\ \text{all} & \mathbf{\large{+}} & -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}} \\ & \mathbf{>0} & & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0} \\ \hline \end{array}\)

\(\mathbf{(-\infty,-2) \cup (-1,1) \cup (2,3) \cup (4,6) \cup (7,\infty)}\)

Compute 

\(\mathbf{\huge{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } } } \\\)

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} \quad | \quad \sqrt{n+\sqrt{n^2-1}}\sqrt{n-\sqrt{n^2-1}}=1 \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{n-\sqrt{n^2-1}} \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{\dfrac{1}{2}\left( 2n-2\sqrt{n^2-1} \right) } \\\\ &=& \dfrac{1}{\sqrt{2}}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad n^2-1 = (n+1)(n-1) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n+1}\sqrt{n-1} } \quad | \quad 2n-2\sqrt{n+1}\sqrt{n-1} = \left( \sqrt{n+1}-\sqrt{n-1} \right)^2 \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ \Big( \sqrt{n+1}-\sqrt{n-1} \Big)^2 } \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1}-\sqrt{n-1} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big)-\sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) = 0+1+ \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)\\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) -\left(0+1+\sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) \right) ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) = \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \underbrace{\sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)- \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)}_{=0} +\sqrt{9800} +\sqrt{9801} -1 ~\right] \\\\ &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ 98.9949493661 + 99 -1 ~\right] \\ &=& \dfrac{\sqrt{2}}{2}\cdot 196.994949366 \\ &\mathbf{=}& \mathbf{139.296464556} \\ \hline \end{array} \)

Sorry....my mistake you are correct!    the opposite side WOULD be another leg    D'Oh!

tan 44 = opp/adj = opp/45     opp = 43.46 cm

Same thing with no 2    break the figure up in to a triangle and a semicircle...calc the areas and add them together.....

#3 is too fuzzy for me to read.....

for No 1.   break the polygon into triangles ...the area of a triangle = 1/2 b h   add the triangles' areas to get total area:

Why would the opposite side be the hypotenuse? If the opposite side was the hypotenuse that would mean that the given angle would have to be 90 degrees (this is a right triangle), which it isn't. That would mean that you would use tangent: opposite / adjacent.

tan(44) = x/45

x = 43.46cm

Deleted ....error.  time to quit for the night!

Thanks :)

A caretaker uses a 10 foot ladder to reach 8 feet up the wall. What is the angle made with the floor? A Sketch will be apart of my solution

Consider the ladder leaning against the wall as the hypotenuse of a right triangle formed with the wall and the ground. 

The sine of the angle of interest is the opposite over the hypotenuse.  So

sine of the angle  =  8/10  = 0.8 

a look at the table shows that the sine of 53^o is 0.7986 so the angle is approximately 53^o

using a calculator shows that the sine^–1 of 0.8 is 53.1301^o if you want to be that exact 

.

You're welcome ! 

Thanks so much!! Really appreciate the help!

Here is a desmos graph:

wdieb

that's hysterical :P

k forms an infinite sequence with:

First term =1 / 5

Common ratio =1/5

Sum =F / (1 - R)

Sum =1/5 / (1 - 1/5)

Sum =1/5 / (4/5)

Sum =1/5 x 5/4

Sum = 5 / 20

Sum = 1 / 4

8 divided by 9 =

    0.88888888.......

2.    f(10) =  -5(10) + 7 = -50 +7 = - 43

  f(x) =32

        32 = -5x+7

        25 = -5x

         x = -5

oh thats what you mean

1.  f(1)= -2(1) +5=  -2+5 =3

     f(2)= -2(2) +5= -4+5 = 1

     f(3)= -2(3) +5= -6+5 = -1

     f(4)= -2(4) +5= -8+5 = -3

     f(5)= -2(5) +5= -10+5 = -5

⇒ range is { 3,1,-1,-3,-5 }

2.

     a) f(10)=  -5(10) +7= -50 +7= 43

    b) x=32/f