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Are you posting every question of your homework on here? Your not going to learn anything by making us do it for you. Maybe you should have a go at your homework questions yourself first before asking us to do it for you. It's ok to ask for help if you truly get stuck, but you shouldn't be asking us to do your whole homework for you. Who knows; you might get all the answers to your questions yourself if you have a go at them. I'm sure whoever set you your homework won't have set you anything you couldn't already do. Have a go at them. Good luck 

16%=73.6

I'm not sure what you mean by "dabl" could you try and clarify.

Thanks!

Spongebob, 

It is really rude of you to give yourself 6 points and geno no points at all.  You gave him no thanks either.

Do you think this rude behaviour helps you in some way?  Do you think no one will notice?

thanks for your kind reply; but I wanted to know why the scientific calculator gives me error?

36E9:2=x:100

36E9 / 2 = x / 100, solve for x
cross multiply

36E11 = 2x
divide both sides by 2
x =36E11 / 2 =1,800,000,000,000

\(\frac{cos(-\theta)cot(-\theta)}{cosec(\theta)}\\ =cos(\theta)*\frac{cos(-\theta)}{sin(-\theta)}\div\frac{1}{sin(\theta)}\\ =cos(\theta)*\frac{cos(\theta)}{-sin(\theta)}*\frac{sin(\theta)}{1}\\ =-cos^2(\theta)\)

"Write the augmented matrix ..."

thhhhxxx

Use an Addition or Subtraction Formula to write the expression as a trigonometric function of one number.
\(\sin(12^\circ) \cos(18^\circ) + \cos(12^\circ) \sin(18^\circ) \)

\(\begin{array}{|rcll|} \hline && \sin(12^\circ) \cos(18^\circ) + \cos(12^\circ) \sin(18^\circ) \\ &=& \sin(12^\circ+18^\circ) \\ &=& \sin(30^\circ) \\ &=& \mathbf{0.5} \\ \hline \end{array}\)

Let \(A = (3, \theta_1)\) and \(B = (9, \theta_2)\) in polar coordinates. If  \(\theta_1 - \theta_2 = \dfrac{\pi}{2}\), then  find the distance \(AB\)

Formula:
We have two points \((r_1,\ \theta_1)\) and \(( r_2,\ \theta_2)\).
The distance between two polar coordinates \(\mathbf{d =\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1})}\)

\(\begin{array}{|rcll|} \hline d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1) } \\ d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos\Big(-(\theta_1 - \theta_2)\Big) } \\ d &=& \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1 - \theta_2) } \\ d &=& \sqrt{3^2+9^2+2*3*9\cos\left( \dfrac{\pi}{2} \right) } \\ d &=& \sqrt{9+81+54\cos\left( \dfrac{\pi}{2} \right) } \quad | \quad \cos\left( \dfrac{\pi}{2} \right) = 0 \\ d &=& \sqrt{9+81+0 } \\ d &=& \sqrt{90} \\ d &=& \sqrt{9*10} \\ d &=& \sqrt{3^2*10} \\ \mathbf{d} &=& \mathbf{3\sqrt{10}} \\ \hline \end{array} \)

As follows:

Let \(a,b,c,d\) be positive real numbers.

Find the minimum value of \(\dfrac{(a + b)(a + c)(b + c)}{abc}\)

\(\mathbf{\huge{AM \geq GM }}\)

\(\begin{array}{|rcll|} \hline \dfrac{a+b}{2} & \ge & \sqrt{ab} \\\\ \dfrac{a+c}{2} & \ge & \sqrt{ac} \\\\ \dfrac{b+c}{2} & \ge & \sqrt{bc} \\\\ \hline \left( \dfrac{a+b}{2} \right) \left( \dfrac{a+c}{2} \right) \left( \dfrac{b+c}{2} \right) & \ge & \sqrt{ab}\sqrt{ac} \sqrt{bc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{abacbc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{a^2b^2c^2} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & abc \\\\ \dfrac{(a + b)(a + c)(b + c)}{abc} & \ge & 8 \\\\ \mathbf{ 8 } & \le & \mathbf{\dfrac{(a + b)(a + c)(b + c)}{abc}} \\ \hline \end{array}\)

The minimum value is 8

ok, if you entered it into the LaTex box, all you probably needed to do was delete the dollar signs. 

Play around with it for a bit and you will work it out.

Find the number of ordered pairs \((a,b)\) of integers such that \(\dfrac{a + 2}{a + 5} = \dfrac{b}{3}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a + 2}{a + 5}} &=& \mathbf{\dfrac{b}{3}} \\\\ 3(a+2) &=& b(a+5) \\ 3a+6 &=& ab+5b \\ -3a+5b+ab &=& 6 \\ -3a+5b+ab+(-3*5) &=& (-3*5)+6 \\ -3a+5b+ab-15 &=& -15+6 \\ \mathbf{(a+5)(b-3)} &=& \mathbf{-9} \\ \mathbf{m*n} &=& \mathbf{-9} \\ && \text{find all integer products }\ m*n = -9, \\ && \text{where}\ m=a+5,\ \text{ and }\ n=b-3 \\ \hline \end{array} \)

all integer products

\(\begin{array}{|ccl|r|r|r|r|} \hline && m*n & m & n & a=m-5 & b=n+3 \\ \hline -9 &=& 1\times -9 & 1 & -9 & -4 & -6 \\ -9 &=& -1\times 9 & -1 & 9 & -6 & 12 \\ -9 &=& 3\times -3 & 3 & -3 & -2 & 0 \\ -9 &=& -3\times 3 & -3 & 3 & -8 & 6 \\ -9 &=& 9\times -1 & 9 & -1 & 4 & 2 \\ -9 &=& -9\times 1 & -9 & 1 & -14 & 4 \\ \hline \end{array} \)

There are 6 integer products.

The number of ordered pairs (a,b) are 6.

\((-4,-6),\ (-6,12),\ (-2,0),\ (-8,6),\ (4,2),\ (-14,4) \)

If

\(\dfrac{p}{q}=\dfrac{q}{r}\) ,

then prove the given problem.

\(\begin{array}{|rcll|} \hline \dfrac{p}{q} &=& \dfrac{q}{r} \\ pr &=& q^2 \\ \mathbf{2pr} &=& \mathbf{2q^2} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (p+q+r)(p-q+r) \\ &=& \Big((p+r)+q\Big) \Big((p+r)-q\Big) \\ &=& (p+r)^2-q^2 \\ &=& p^2+2pr+r^2-q^2 \quad &| \quad \mathbf{2pr=2q^2} \\ &=& p^2+2q^2+r^2-q^2 \\ \mathbf{(p+q+r)(p-q+r)} &=& \mathbf{p^2+q^2+r^2}\ \checkmark \\ \hline \end{array}\)

By polar coordinates, AB = sqrt(3^2 + 3*9 + 9^2) = 3*sqrt(13).

The sum of all possible values of z is 2 + 4 + 5 = 11.

1. The maximum possible value of |a - b| is 20.

2 The maximum possible value of |a - n| is also 20.

The area of triangle LPQ is 18.