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assuming degrees

arccos(cos230)

=arccos(-cos(230-180))

=arccos(-cos(50))

=arccos(cos(180-50))

=arccos(cos(130))

=130 degrees

Two guy wires are attached to the top of a telecommunications tower and anchored to the ground on opposite sides of the tower, as shown. The length of the guy wire is 20 m more than the height of the tower. The horizontal distance from the base of the tower to where the guy wire is anchored to the ground is one-half the height of the tower. How tall is the tower, to the nearest tenth of a metre?

\(\small{ \begin{array}{rcll} \mathbf{ \text{Pythagoras:} }\\ \left( \dfrac{h}{2} \right)^2 + h^2 &=& (h+20)^2 \\ \dfrac{h^2}{4} + h^2 &=& h^2+40h+400 \\ \dfrac{h^2}{4} &=& 40h+400 \\ \dfrac{h^2}{4} -40h-400&=& 0 \qquad & | \qquad \cdot 4 \\ h^2 -160h-1600&=& 0 \\ \hline \boxed{~ \begin{array}{rcll} ax^2+bx +c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~} \\ \hline x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \qquad a = 1 \quad b = -160 \quad c = -1600 \\ h_{1,2} &=& {160 \pm \sqrt{160^2-4\cdot(-1600)} \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160^2+40 \cdot 160 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot (160+40) } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{160\cdot 200 } \over 2} \\ h_{1,2} &=& {160 \pm \sqrt{1600\cdot 20 } \over 2} \\ h_{1,2} &=& {160 \pm 40 \sqrt{20 } \over 2} \\ h_{1,2} &=& {160 \pm 40\cdot 2\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& {160 \pm 80\cdot \sqrt{ 5 } \over 2} \\ h_{1,2} &=& 80 \pm 40\cdot \sqrt{ 5 } \\ h &=& 80 + 40\cdot \sqrt{ 5 } \\ h &=& 40\cdot( 2 + \sqrt{ 5 } ) \\ \mathbf{ h} & \mathbf{=} & \mathbf{169.442719100} \end{array} }\)

The tower is 169.4 m tall.

This is what I am imagining is the situation.

Now use Pythagorean theorum.

\((t+20)^2=t^2+(t/2)^2\\ t^2+40t+400=(4t^2+t^2)/4\\ 4t^2+160t+1600=5t^2\\ 0=t^2-160t-1600\\\)

\(x = {160 \pm \sqrt{160^2+4*1600} \over 2}\\ x = 169.4427\)

The height is 169.4 m

If you were to multiply them together ?

It is nice to meet you syakirah159.  

But you need to check this answer.

1.6f - 8.3f = ????

The circumference of a circle is 2πj (2 x 22/7 x radius).

x⁷/ x^-3

= x^7-(-3)

= x⁷⁺³

= x^{10  #}

If f(x) = 0.41*ln(x+1)+28 then

(i) f(0) = 0.41*ln(0+1)+28 → f(0) = 0.41*0 + 28 → 28      because ln(1) = 0

(ii) f(100) = 0.41*ln(100 + 1) + 28 → 0.41*ln(101) + 28    Use a calculator to evaluate this.

We have:

\(\frac{x}{y}=\frac{3}{7}\)

which means that \(y=\frac{7}{3}x\)

Use this in y - x = 56 to get \(\frac{7}{3}x-x=56 \rightarrow\frac{4}{3}x=56\rightarrow x=\frac{3\times56}{4}\rightarrow x=3\times14\rightarrow x=42\)

Use this in y - x = 56 to get  y - 42 = 56 or y = 98

Finallly I am happy with my answer.  

Density = Mass/Volume, so just divide one by the other,   Using your values the result will have units of kg/m³

^.

3+3 = 6

that's right, well done!!!

4.7 (2f - 0.5) = -6 (1.6f - 8.3f)

   9.4f - 2.35  = -6 (6.7f)

    9.4f - 2.35 = -40.2f

  40.2f + 9.4f = 2.35

            49.6f = 2.35

                   f = 2.35/49.6         

                     = 47/992

                     = 0.04737 #

Write an equation of the line that is perpendicular to the line through 9,10 and 3,-2 and passes through the x-intercept of that line.

\(\small{ \begin{array}{rcll} \hline \text{line } 1\quad m(\text{slope})=\ ?\\ m &=& \dfrac{y_2-y_1}{x_2-x_1} \qquad ( x_1 = 3,\ y_1=-2 ) \quad ( x_2=9,\ y_2 = 10 )\\ m &=& \dfrac{10-(-2)}{9-3} \\ m &=& \dfrac{12}{6} \\ \mathbf{m} &\mathbf{=}& \mathbf{2} \\\\ m_{\text{perpendicular}} &=& -\dfrac{1}{m} \qquad m=2\\ \mathbf{m_{\text{perpendicular}}} &\mathbf{=}& \mathbf{-\dfrac{1}{2}} \\ \hline \text{line } 1\quad x_{(intercept)}=x_i=\ ?\\ m &=& \dfrac{y-y_1}{x-x_1} \\ m &=& \dfrac{y_i-y_1}{x_i-x_1} \qquad ( x_1 = 3,\ y_1=-2 ) \quad m=2 \quad y_i = 0\\ 2 &=& \dfrac{0-(-2)}{x_i-3} \\ 2 &=& \dfrac{0-(-2)}{x_i-3} \\ 2 &=& \dfrac{2}{x_i-3} \qquad & | \qquad :2 \\ 1 &=& \dfrac{1}{x_i-3} \\ x_i-3 &=& 1 \\ x_i &=& 1+3 \\ x_i &=& 4 \\ \mathbf{\text{Point x-intercept}:\ ( x_i = 4,\ y_i=0 )}\\ \hline \text{line perpendicular:} \\ m_{\text{perpendicular}} &=& \dfrac{y-y_i}{x-x_i} \qquad ( x_i = 4,\ y_i=0 ) \quad m_{\text{perpendicular}} =-\dfrac12 \\ -\dfrac12 &=& \dfrac{y-0}{x-4} \\ -\dfrac12 &=& \dfrac{y}{x-4} \\ y &=& -\dfrac12 \cdot (x-4) \\ y &=& -\dfrac12 \cdot x - ( -\dfrac12 )(4) \\ \mathbf{y} &\mathbf{=}& \mathbf{-\dfrac12 \cdot x +2} \\ \end{array} }\)

x:y = 3:7

y-x = 56

The answer is;

y-x = 7-3

      = 4

y-x = 56/4

      = 14

x = 14x3

      =42 #

2+1=3

    x⁷          =    x¹⁰   because when dividing indices, we subtract them. Therefore x^7-(-3) = x⁷⁺³ = x¹⁰

    x^-3

@@ What is Happening?  [Wrap4]   Thurs 19/11/15   Sydney, Australia Time 5:40pm  ♪ ♫

Hello everyone,

We have had many great answers supplied by Heureka, Alan, CPhill, TMga2yb, nicolas0122, bubbleman, rarinstraw1195, LordEnedor, Rom, Saseflower, Omi67, EinsteinJr and Syakirah159. Thanks all.  

Forum Issues:

Mr Massow has replied to my message. He agrees with me that the access problems some people are having is the most important issue at the moment.  He is having trouble reproducing it but that is where his efforts are being directed at the present time.  He said he may be able to address some of the lesser issues over the weekend.

Thank you Mr Massow.  

Interest Posts:

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

1)  Quarters and nickels

a manufacturer is cutting fabric into rectangles that are 18.55in. long by 36.75in. wide. Each rectangle's length and width must be within 0.05 in. of the desired size. Write and solve inequalities to find the acceptable for the length l and the width w

\(18.50" \le length\le 18.60" \\ 36.7"5-0.05" \le\ width \le 36.75"+0.05"\)

You can tidy up the second one :))

What is the inverse of [2x+1/3]

\(y=2x+1/3\\ y-1/3=2x\\ x=\frac{1}{2}(y-\frac{1}{3})\\ x=\frac{1}{2}(\frac{3y-1}{3})\\ x=\frac{3y-1}{6}\\ \mbox{The inverse is }\\ y=\frac{3x-1}{6}\\\)

how do we solve: sqrt(x^2-10x+25)=4 ?

\(x^2-10x+25 = (x-5)^2\)

\(\begin{array}{rcll} \sqrt{x^2-10x+25} &=& 4 \\ \sqrt{(x-5)^2} &=& 4 \\ x-5 &=& 4\\ x &=& 4+5\\ x &=& 9 \end{array}\)

-3b+4 if b=4

=(-3x4)+4

=-12+4

=-8

14 and 2

because; 14x2 = 28 

because; 14+2 = 16

75+15x > 140

      15x > 140-75

      15x > 140-75

      15x > 65

          x > 65/15

          x > 65/15

          x > 13/3

          x > 4.3333 #