I really like all your answers 

Middle and median have d's in the middle

Mode and most both have o's 

So mean must be  the other one ....

That Hey Diddle Diddle rhyme is great    

I thought this was so good I have added this thread to our Reference Material Sticky thread  

I aM going to think about this

5^2015

$$\\5^1=5\\
5^2=25\\
5^3=125\\
5^4=525\\$$

It seems that if the power is bigger than 1 that the last 2 digiits will always be 25

MAYBE SOMEONE WOULD LIKE TO PROVE THIS  ??

Above is the graph of y = sin x and the line y = (-1)/2 over the interval [-180^@, 180^@]. Where do the two graphs intersect? Give exact answers in degrees,

$$\small{\text{$ \begin{array}{rcl|rcl}\sin{(x_1)} &=& -\frac12 \qquad & \qquad \sin{(x_1)} &=& \sin{ (180\ensurement{^{\circ}} -x_2) } =-\frac12 \\&&&\\ x_1 &=& \arcsin{( -\frac12 )} \qquad & \qquad \sin{ (180\ensurement{^{\circ}} - x_2) } &=&-\frac12 \\&&&\\
x_1 &=& -30\ensurement{^{\circ}} \qquad & \qquad 180\ensurement{^{\circ}} - x_2 &=& \arcsin(-\frac12) \\&&&\\
&& &\qquad x_2 &=& 180\ensurement{^{\circ}}-\arcsin{( -\frac12 )} \\&&&\\&&\qquad & x_2 &=& 180\ensurement{^{\circ}} +30\ensurement{^{\circ}} - 360 \ensurement{^{\circ}} \\
&&&\\
&&\qquad & x_2 &=& -150\ensurement{^{\circ}}\\
\end{array}$}}$$

Thejamesmachine , you working was good but you forgot the - sign   

the graph of y = cos x and the line y = 1/2 over the interval [0 degree,360 degree]. Where do the two graphs intersect? Give exact answers in degrees

$$\small{\text{$ \begin{array}{rcll}\cos{(x)} &=& \frac12 \\\\\cos{(x)} &=& \frac12 & \qquad | \qquad \pm\arccos{} \\\\x_{1,2} &=& \pm \arccos{(\frac12 )}\\\\x_{1,2} &=& \pm 60 \ensurement{^{\circ}} \\\\x_{1} &=& 60 \ensurement{^{\circ}} \\\\\\
x_2 &=& -60\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\
x_2 &=& 300 \ensurement{^{\circ}}
\end{array}$}}$$

The plus sign (+) is also a mathematical operator,  so you've used two operators. No cookie for you!

wouldn't it be 65°??

The pyramid is square based. Therefore the diagonal between corners would be 12.73, because:

9² + 9² = 162

√162 = 12.73

So the adjacent side of the triangle would be 6.365 (half of 12.73) and the hypotenuse would be 15. 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{6.365}}}{{\mathtt{15}}}}\right)}$$ = 65 (to the nearest degree)

Is this right ^^ or am I doing something wrong?

is your answer 12.36%?

$${\frac{{\mathtt{445.41}}}{{\mathtt{21.21}}}} = {\mathtt{21}}$$

.255

100 percent is always 1. So anything less than 100 percent is 0. something. In this case, 0.255

Isn't the S.A. of a hemisphere:

pi*r^2 + (0.5)*4*pi*r^2 = S.A.

so . . .

3pi*r^2 = 618cm 

r^2 = 618/3pi

r = sqrt(618/3pi)

r approximately = 8.0976 cm 

2(x - 3) + 5x = 4(x + 2) -5 ..... Expand

2x - 6 + 5x = 4x + 8 -5 ...... Collect like terms

7x - 6 = 4x + 3 ....... Simplify

3x = 9

x = 3

To check:

2(3-3) + 5*3 = 4(3 + 2) - 5

0 + 15 = 4*5 -5

15 = 20 - 5

15 = 15

So the average of something is the sum of the various data points divided by the number of data points.

With two numbers it is also just the number in the middle of them eg. the average of 4 and 6 is 5.

$$\frac{61 + 81.5}{2}$$ = 71.25%

Hmm I have found a problem with this. Usually with grades different assessments have different weightings so without the total number of assessments and their respective % allocation of your total grade I don't believe this is possible. But the average of 61 and 81.5 is still 71.25.

If, for example, you had 4 assessments of equal weighting (25% each) you would collect them together as marks out of 25 and then add them to get a mark out of 100, or a percentage. 

lol ok this is why I don't like it when someone posts the same question twice. I just gave exactly the same answer this question's doppleganger 

$$\frac{34}{100} + \frac{5}{10}$$ = 

$$\frac{34}{100} + \frac{50}{100}$$ = 

$$\frac{84}{100}$$ =

$$\frac{21}{25}$$

Fingue out the following below:

Two ordered pairs: I'm assuming this means  (100, 14) and (200, 20) although I am not sure. 

Slope: Also called the gradient, or m, the slope is how much y increases in relation to x, in the form rise/run. In these two points; x increases by 100, and y increases by 6. So the slope is 6/100, or 3/50.

m = 3/50

Find b: b is the y intercept, which can be found when x = 0. If y goes down by 6 every time x decreases by 100 then the y-intercept has to be 8, according to the point (100, 14).

b = 8

Equation: y = mx + b is the slope-intercept form.

By substituting our values:

y = 3/50 x + 8

Lets check this for point (300, 26)

26 = 3/50 * 300 +8

26 = 900/50 + 8

18 = 900/50

18 = 18

yay :)

Two airplanes are spotted on a radar map located at the coordinates (2√2 , - √5 ) and (3√2 , 5√5 ). Determine the distance the airplanes are apart using a calculator to state your answer in decimal form to the nearest hundreds. Assume the coordinates are given so that the distance is in units of miles

$$\binom{x_1}{y_1}=\binom{2\sqrt{2}}{-\sqrt{5}} \qquad
\binom{x_2}{y_2}=\binom{3\sqrt{2}}{5\sqrt{5}}$$

$$\boxed{\rm{distance}~&=& \sqrt { (x_2-x_1)^2 + (y_2-y_1)^2 }}\\\\
\small{\text{$
\begin{array}{rcl}
\rm{distance}~&=& \sqrt { ( 3\sqrt{2} - 2\sqrt{2})^2
+ [ 5\sqrt{5} - (-\sqrt{5}) ]^2 } \\\\
\rm{distance}~&=& \sqrt { ( 3\sqrt{2} - 2\sqrt{2})^2
+ [ 5\sqrt{5} + \sqrt{5} ]^2 } \\\\
\rm{distance}~&=& \sqrt { (\sqrt{2})^2 + ( 6\sqrt{5} )^2 } \\\\
\rm{distance}~&=& \sqrt { 2 + 36\cdot 5 } \\\\
\rm{distance}~&=& \sqrt { 182} \\\\
\rm{distance}~&=& 13.4907375632 \\\\
\rm{distance}~&\approx& 13.49 ~\rm{miles}
\end{array}
$}}$$

For 2t = 4/7 you just divide both sides by 2 to get rid of the 2 on the t

t = 2/7

Ok MG.

It is really good to work in advance but it is even more important to make sure you understand and can do all the work you have already covered.  This basic work is the building blocks that you will need for always.

The more thoroughly you understand them the easier future work will be and the further you will be able to go with mathematics in your future :)

Use forum questions - like you have been - to not only discover new and exciting mathematics but also to practice all the basic maths that you already have.

I still have not homed in properly on your level.

Here are a few questions.   I want to work out what you do and do not understand.

You can copy my questions and then put the answers with the questions.  

Don't get help, I want to see what you can do by yourself :)

$$\\1) 3x-2x^2 - 8xy+3yx-7\\
2) 5x*6xy\\
3) 3(x+4)\\
4) -2(x+4)\\
5) 17-4x-x+5\\
6) 3-4(5x-6)\\
7) \frac{3x}{2}+\frac{4x}{5} \qquad $(Make an entire fraction)$\\
8) 0.06*0.3 \qquad $ (No calculator)$\\
9) 0.06\div 0.2 \qquad $ (No calculator)$\\
10) 3x^2+2x+5\\
11) \frac{5x+2}{6y}\times 9xy^2$$

\\1) 3x-2x^2 - 8xy+3yx-7\\ 2) 5x*6xy\\ 3) 3(x+4)\\ 4) -2(x+4)\\ 5) 17-4x-x+5\\ 6) 3-4(5x-6)\\ 7) \frac{3x}{2}+\frac{4x}{5} \qquad $(Make an entire fraction)$\\ 8) 0.06*0.3 \qquad $ (No calculator)$\\ 9) 0.06\div 0.2 \qquad $ (No calculator)$\\ 10) 3x^2+2x+5\\11) \frac{5x+2}{6y}\times 9xy^2

I think you may mean 25.5%, in which case:

25.5% = 25.5/100 = 51/200

Ooops didn't answer the question :)

It is 0.255. As a fraction it is: 25.5/100, so divide by 100. This is easily done by moving the decimal place over 2 to the left.

---delete---

$$\frac{1}{3} - \frac{1}{12} = \frac{1}{x}$$

$$\frac{4}{12} - \frac{1}{12} = \frac{1}{x}$$

$$\frac{3}{12} = \frac{1}{x}$$ 

Now by cross multiplying:

$$3x = 12$$

$$x = 4$$

Make 1 parcel = X

4X = 124 kg

X = 124/4 kg

X = 31 kg