All Answers 358087 Answers

Thanks Alan,

I can see that a denominator of 5 works for your example but I still don't comprehend where 

$$\int1 dx$$   comes from.  Why is it 1?  What are you finding the sum of?

I obvioulsy have a major gap in my understanding.

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Actually I am just thinking about the units as rates The average of anything can be thought of as whatever per unit or item.  That makes sense. So thinking of it like this I absolutely get the difference on the bottom.

For example (for the benefit of others) The average of 6m,3m and 2m = 9m/3items = 3m/item

So  $$\int1 dx$$  is the sum of units on the x axis which is 5.   

So for the given question the average is     sum of the gradients/unit of x

Does that make sense, sort of anyway?   

Maybe I have got it!  

Melody, does this help

 average x distance = area  so average = area/distance.

I guess you're meant to equate clockwise and anticlockwise moments:

So d ≈ 1.663 m

note that cos(30°) = (√3)/2

You should double check that the moments given above are correct - I did this rather quickly!

solution:

The name of the app is listed in the post. Scientific calculator by scalea software for android. 

I do not know what this notation means.

Do you mean ?

$${{log}}_{{\mathtt{100}}}{\left({\mathtt{0.16}}\right)} = -{\mathtt{0.397\: \!940\: \!008\: \!672\: \!037\: \!6}}$$      

$${{log}}_{{\mathtt{0.16}}}{\left({\mathtt{100}}\right)} = -{\mathtt{2.512\: \!941\: \!594\: \!732\: \!06}}$$

$${log}_{10}\left({\frac{{\mathtt{0.16}}}{{\mathtt{100}}}}\right) = -{\mathtt{2.795\: \!880\: \!017\: \!344\: \!075}}$$

I am sorry, i do not know what you want.

$$c^2=a^2+b^2\\\\
\Rightarrow a^2 = c^2 -b^2 = (c-b)(c+b)=(85-53)(85+53)=32*138\\
=16*2*2*69\\\\
a=\sqrt{16*4*69}=\sqrt{16}\sqrt{4}\sqrt{69}=4*2*\sqrt{69}=8\sqrt{69}\approx66.5$$

I have asked you politely to stop dragging these old posts up.  Do you get the idea now!

This is very old garbage that you have dragged back up and i do not a appreciate you doing so!

PLEASE ADD A TITLE TO YOU QUESTION!

You have just found the area of the middle bit

( 10-2x )^2 

You would have to add 4 lots of  x(10-2x)  to make your answer correct.

Look at my diagram and see if you can see this.  

i think by this way 

we have a square piece of paper with side-length 10 cm 

if we cut a small square from each corner the side lenght will be 10-2X ..... then the remaining area will be ( 10-2x )^2 

what is the mistake in my way ?

You are short a bracket.

What is between the 6 and the 3?

When you work out the answers to these you can just put it onto the calculator on the home page. 

Because you had a square peice of paper 10cm * 10cm.  That was 100cm^2

then you cut a square out of each of the 4 corners that was [xcm * xcm]*4 = 4x^2 cm^2

so you were left with  [100-4x²] cm²  of paper   

You need to give this question a title please.

This is interesting Alan,

You use calculus in a way that I have not seen it used before (or I don't remember anyway)

I have just enough of knowlege to attempt to understand

2x^2 -2x +5

tangent function : 4x-2

Average = sum of all tangents(numerator makes sense) over number of tangents (okay I am having trouble with the denominator)

Maybe I just need to think it through.  Can you help me get my mind around it?   

why it 100 - 4x^2 

$$\\It\; is\; 100-4x^2\\\\
\mbox {But this is not one of the answers so we need to look some more}\\
100=10^2\\
4x^2=(2x)^2\\
\mbox{So we have a difference of two squares}\\
100-4x^2=(10-2x)(10+2x)\\
\mbox{So the answer is c}\\$$

I have heard this term before 'decimal fraction' but it does not really make sense.

Maybe you would like to explain what they are?

The meaning of your question is not clear.

What percentage is 15 of 200:  7.5%

What is 15% of 200:  15/100*200=30

Thanks 15x3

Flattery and thumbs up will get you a very long way with me.  

DS, 

15x3 expressely asked for this to be solved WITHOUT a calculator.  So, did you really answer the question?

This is a rhetorical question - don't dare post an answer!

Okay. If you do the question first, that can help. 

 $$\frac{1}{\sqrt2+\sqrt3}\\\\=0.5176380902050415$$

Now, let's do C.

$$\sqrt2-\sqrt3=0.5176380902050415$$

Compare the answers.

$$0.5176380902050415=0.5176380902050415$$

So, the equation is true, and the answer is C. And yes, you can use the calculator too.

You can do it by not using the calculator.

(That helped me get the answer.)

But, there is another way, too. Do A, B, C, D, and E. That helps too, you know. That can really help.

hey melody ! 

you are amaiznnnnnnnnnnnnnng !!!! 

thanx my friend 

Yes you have to rationalise the denominator.

remember  $$(a-b)(a+b)=a^2-b^2$$      This is a difference of 2 squares.

oh, a+b and a-b are called a congugates of one another.  That is because they are the same except for the +- sign in the middle.

When rationalizing denominators like this you use this fact.   because it a or b is a surd then a² and b² are not!

$$\frac{1}{\sqrt2+\sqrt3}\\\\
=\frac{1}{\sqrt2+\sqrt3} \times \frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\\\\
=\frac{(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)} \\\\
=\frac{(\sqrt2-\sqrt3)}{(\sqrt2)^2-(\sqrt3)^2} \\\\
=\frac{(\sqrt2-\sqrt3)}{2-3} \\\\
=\frac{(\sqrt2-\sqrt3)}{-1} \\\\
=\frac{(\sqrt3-\sqrt2)}{+1} \\\\
=\sqrt3-\sqrt2\\$$

I talk about the question !