All Answers 358095 Answers

@fiora:/  Melody's answered  both questions of yours.

1st answer:     The length of  PM could be either  1  or  3 ;

2nd answer:    IF a point N is on a line segment PQ,  the angle NOP is 30^o , and the length of PM is 1, THEN the angle POM = 18.435^o , and the area of a triangle NOM = 0.464u² !

Bravo, Melody! 

*Cheers drink*

To Americas Independence!!!

...Especially considering I'm from England...!...

Question 2

What is this shape?

Triangular prism

If you cut it anywhere, parallel to the front triangle side.  What will be the shape of the cross section?

A triangle

What is the area of the front triangle? (Show working)

6.4*8.2 = 52.48 . As we are finding the area of a triangle we do the formula L*W/2.   52.48/2 = 26.24cm^2

If this was made of mm^3 blocks (some would need to be cut up) How many cubes would be needed to cover the front face? 

???

How many rows of block triangles would you need to make the whole shape?

15.7

What is the volume of this shape? (Show working)

Formula = (L*H)/2*W

(8.2*6.4)/2=26.24cm^2

26.24*15.7= 411.968cm^2

I don't get the 'mm/cm^3' bit...

I don't mind points!

Question 1

Look at this shape.  This is made up of cubic blocks  Each side is 1cm. 

What is the area of the front face? (include units) 

36cm^2

How many cubes are in the front row?

30

How many rows of blocks are there counting back from the front row?

3 rows

What is the volume of the prism?

108cm

Write a sentence as to why.

The formula is (L*H)*W

Very nice use of PEMDAS, TR !!!

3 points from me.......

This is pemdas, everyone!

parenthesees? No

exponents? No

multiplication? Yes

division? Yes

addition? Yes

subtraction? Yes

$${\frac{{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{786}}$$

So, first we do multiplication

24*4=96

now the equation is:

$${\frac{{\mathtt{96}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{786}}$$

96/2=48

now its

$${\mathtt{48}}{\mathtt{\,\small\textbf+\,}}{\mathtt{786}}$$

48+786=834

overall:

$${\frac{{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{4}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{786}} = {\mathtt{834}}$$

Thank you thank you

*audience applause

👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻

Hey,Melody!

I had mention that remenber "If" at the end of my question.That mean the given in question one is not vaild, when you doing number two.

I guess a few picture can help you understand my question.

Angle NOM is moving around in triangle QOP,what is the possible mininum area of triangle NOM?and find the measure of angle MOP.Sorry Melody.

Youre welcome

No :)

If you factorize it you get

4x(x^2+4x-2)

Your heading and your question are different equation. I am answering the actual question.

would the answer be 4x(x^5+4x-2)

$$4x^3 + 16x^2 - 8x$$

This cannot be simplified.  

It can be factorised - but not simplified 

Thanks Titanium, I will fix it up :))

Melody, i notice you spelled someones name wrong......you spelled cphills name incorrectly

@@ End of Day Wrap  Sat 4/7/15   Sydney, Australia Time   11:57pm  ♪ ♫

Hi all,

Great answers today from TitaniumRome, Radix, CPhill, Fiora and Civoamzuk.  Thank you 

12*5-15-16

$$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\
\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\
\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\
\angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{10}}}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$

$$\\\angle NOM_1=30-18.434948822922=\;11.565051177078$$

$${\mathtt{30}}{\mathtt{\,-\,}}{\mathtt{18.434\: \!948\: \!822\: \!922}} = {\mathtt{11.565\: \!051\: \!177\: \!078}}$$

Now Find N

N is the intersection of 

$$y=(tan30)x = \frac{x}{\sqrt3} \qquad and \qquad y=-x+2\sqrt2$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\
{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\
{\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\
{\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\
\end{array} \right\}$$

$$N(3\sqrt2-\sqrt6,\;\sqrt6-\sqrt2)$$

Distance ON

$${\sqrt{{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{6}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$ 

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$ 

NOW If I haven't made any stupid errors (which I probably have) that should be correct :)

Thanks Radix,  here is another way to look at it :/

$$\\9.5\%*X=733.55\\\\
so\\\\
X=733.55\div 9.5\%$$

$${\frac{{\mathtt{733.55}}}{\left({\frac{{\mathtt{9.5}}}{{\mathtt{100}}}}\right)}} = {\mathtt{7\,721.578\: \!947\: \!368\: \!421\: \!052\: \!6}}$$

Doesn't make sense

1×√(1-(719.31÷89602783250817600)

$${\mathtt{1}}{\mathtt{\,\times\,}}{\sqrt{\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{719.31}}}{{\mathtt{89\,602\,783\,250\,817\,600}}}}\right)\right)}} = {\mathtt{0.999\: \!999\: \!999\: \!999\: \!996}}$$

It worked fine    

For my college grades my second year was worth 30% and third year was worth 70%...

If I got 53.92 in second year and 58.92 in third year what's my overall percentage? 

I will assume that those marks are out of 100.

$${\frac{{\mathtt{53.92}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{30}} = {\frac{{\mathtt{2\,022}}}{{\mathtt{125}}}} = {\mathtt{16.176}}$$

$${\frac{{\mathtt{58.92}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{70}} = {\mathtt{41.244}}$$

$${\mathtt{16.176}}{\mathtt{\,\small\textbf+\,}}{\mathtt{41.244}} = {\frac{{\mathtt{2\,871}}}{{\mathtt{50}}}} = {\mathtt{57.42}}$$      percent