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Looks like four values........although it's difficult to tell what they might be.....

In the xy-plane below, point C has coordinates (6,9). Which of the following is an equation of the line that contains point O and C?

An object of height 4 cm is placed upright in front of spherical concave mirror at a distance of 20 cm and forming an inverted image of 16 cm high. Find the radius of mirror

If the _____ of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

"diagonals"

=0.0000661501909415

what's the rest of the question?

1)  27x - 13y  =  0

2)  3x - y  =  4y/9

     Multiply each side by 9:             27x - 9y  =  4y

     Subtract 4y from both sides:     27x - 13y  =  0

     This is the same equation as equation1).

3)  9x - 4y  =  y/3

     Multiply each side by 3:           27x - 12y  =  y

     Subtract y from both sides:     27x - 13y  =  0

     This, also, is the same equation as equation 1).

4)  27x - 12y  =  y

     Subtract y from both sides:     27x - 13y  =  0

     This, also, is the same equation as equation 1).

There are an infinite number of solutions to  27x - 13y  =  0.

Two are:  (0, 0)  and  (13, 27).

(1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x) ??

\(\begin{array}{|rcll|} \hline \cot (x) = \frac{1+\cos (2x)}{\sin(2x)} \\ \hline \end{array}\)  see unit-circle: \(\begin{array}{|rcll|} \hline \cot (\frac{x}{2}) = \frac{1+\cos (x)}{\sin(x)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \frac { 1 + \cos(x) + \cos(2x)} { \sin(x) + \sin(2x) } &=& \cot(x) \\\\ \frac { 1 + \cos(2x) + \cos(x) } { \sin(x) + \sin(2x) } &=& \cot(x) \\\\ 1 + \cos(2x) + \cos(x) &=& \cot(x) \cdot [ \sin(x) + \sin(2x) ] \quad | \quad 1+\cos (2x) = \sin(2x)\cdot \cot(x)\\ \sin(2x)\cdot \cot(x) + \cos(x) &=& \cot(x) \cdot [ \sin(x) + \sin(2x) ] \\ \sin(2x)\cdot \cot(x) + \cos(x) &=& \cot(x) \cdot \sin(x) + \cot(x) \cdot \sin(2x) \\ \cos(x) &=& \cot(x) \cdot \sin(x) \\ \frac{\cos(x)}{\sin(x)} &=& \cot(x) \\ \hline \end{array}\)

E_p=E_k

v=sqrt(2gh)

E_P=mgh

mg=F_g

g=~10m/s²

E_k=mv²/2

What is the third term? It doesn't make sense.

\({6\over{\sqrt{32}}} = {6\over{\sqrt{16*2}}} = {6\over{4*\sqrt{2}}} = {3*\sqrt{2}\over{2*2}} = {{3\over{4}}*\sqrt{2}} \approx 0.75 * 1.41 \approx 1.06\)

I get the following:

(15/3)+ 2(4^2-5) = 27

y=x^2+4x-9         (1)

x+2y=3               (2)

Replace y in (2) using (1):

x + 2x^2 + 8x - 18 = 3

Rearrange:

2x^2 + 9x - 21 = 0

Solve for x using the quadratic formula to get:

x₊ = (-9 + sqrt(249))/4   and x_- =  (-9 - sqrt(249))/4

Use these in (2) and rearrange to get the corresponding values of y as:

y₊ = (21 - sqrt(249))/8   and y_- = (21 + sqrt(249))/8

.

C= x(Php 200.00)

5x Php 200.00 = Php 1000.00

Hi Ginger, 

I thought you would be a carpenter becasue you were going ot 'join' the harem  :)

It was good to have a laugh, thank you :)

I am pleased that your application to 'join' A.P.E. has met with the boards approval.  

They/we really do need a carpenter :))

CONGRATULATIONS !!

Maybe this is a coded message for someone.  

I hope so because it certainly does not translate properly.....

Simplify (1+3+5+...+199)/(2+4+6+...+200).

\(\begin{array}{|rcll|} \hline s_n &=& \frac{(a_1+a_n)}{2} \cdot n \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \frac{1+3+5+...+199}{2+4+6+...+200} \\\\ &=& \frac{\frac{(a_1+a_n)}{2} \cdot n }{\frac{(b_1+b_n)}{2} \cdot n} \\\\ &=& \frac{a_1+a_n}{b_1+b_n}\\\\ &=& \frac{1+199}{2+200}\\\\ &=& \frac{200}{202}\\\\ &=& \frac{100}{101}\\\\ &=& 0.99009900990\dots \\ \hline \end{array} \)

           10
find      ∑ (k+2)*3*2^k
         k=0

\(\begin{array}{|rcll|} \hline && \sum \limits_{k=0}^{10} {(k+2)\cdot 3 \cdot 2^k} \\ &=& 3\cdot \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} \\ &=& 3\cdot S_{10} \qquad |\qquad \text{We set } ~ S_{10} = \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline S_{10} = \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} &=& 2\cdot 2^0 + & 3\cdot 2^1+ 4\cdot 2^2+ \dots + 11\cdot 2^9 + 12\cdot 2^{10} \\ 2\cdot S_{10} &=& & 2\cdot 2^1 + 3\cdot 2^2 + \dots + 10\cdot 2^{9}+ 11\cdot 2^{10} + 12\cdot 2^{11} \\ \hline S_{10} -2\cdot S_{10} &=& 2\cdot 2^0 & +2^1+2^2+2^3+\dots + 2^9+2^{10}-12\cdot2^{11} \\\\ && & 2^0+2^1+2^2+2^3+\dots + 2^9+2^{10} = 2^{11} - 1 \\ && & 2^1+2^2+2^3+\dots + 2^9+2^{10} = 2^{11} - 2 \\\\ S_{10} -2\cdot S_{10} &=& 2\cdot 2^0 & +2^{11} - 2 -12\cdot2^{11} \\ (1-2)\cdot S_{10} &=& 2\cdot 2^0 & +2^{11} - 2 -12\cdot2^{11} \\ - S_{10} &=& & 2^{11} -12\cdot2^{11} \\ - S_{10} &=& & -2^{11}\cdot ( 12-1) \\ S_{10} &=& & 2^{11}\cdot 11 \\ S_{10} &=& & 11\cdot 2^{11} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \sum \limits_{k=0}^{10} {(k+2)\cdot 3 \cdot 2^k} \\ &=& 3\cdot S_{10} \\ &=& 3\cdot 11\cdot 2^{11} \\ &=& 33\cdot 2^{11} \\ &=& 67584 \\ \hline \end{array}\)

I want to add to Max's answer, alternatively you can use the power 1/2.

What I mean is       sqrt{number} = {number} \(^{1\over{2}}\)

I use this whenever I cannot find the square root function but make sure to use enough brackets.

15/(3+2)*(4^2-5)=27

I think that is the minimum number of brackets.

Although it would be more obvious when you write

(15/(3+2)) * ((4^2)-5) = 27

\({15\over{3+2}}*(4^{2}-5)=27\)

3 * 9 = 27