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#1
+118725
0
2/3 times 3
Draw some cakes. divide them all into thirds.
Colour in 2/3 of a cake 3 times. Don't start on a new cake until all of the current cake is coloured. See how much cake/s you end up colouring.
After you have done this think about how you could have got the answer just with the numbers.
You can of course use your calculator to check the answer.
Melody
Nov 2, 2013
#1
+118725
0
"find the angular size of a circular object with a 1 inch diameter viewed from a distance of 2 yards"
I am assuming the simplest scenario.
Your body and the circle are in parallel planes. Your eye to the centre of the circle are at right angles to those planes.
start by considering an isosceles triangle, the circle diameter is one side (1 inch ), the other 2 equal sides mean at the centre of your eye.
Cut the isosceles triangle in half to make 2 right angled triangles.
Now the centre side is 2 yards = 72 inches long and each little side is 0.5 inch long.
Now use trigonometry to get half the required angle (that is, the little angle in one of the right angled triangles)
then duble it to get the whole angle.
Draw a picture to try and work out what I am saying. if you still need help post again and I will try to explain more.
Melody
Nov 2, 2013
#1
+2
0
It is important to choose right one since the first time. Nice post to hang on. I really loved it the way of the stuff provided in this article.
tarbcl
Nov 2, 2013
Nov 1, 2013
#1
0
yes its the real Beyonce and i like to help kids out and it is 6
Guest
Nov 1, 2013
#1
+118725
0
I think you mean
trying to figure out what the tax on a chair is the chair is 178.90 and the tax rate is 5.75%
Different countries do this differently and I don't know your countries system so I will assume
you pay $178.90 plus tax
tax = 5.75/100*178.90
Melody
Nov 1, 2013
#3
+118725
0
5X
2
- X - 2 = 4X
2
then solve this question by using log both sides, right??
NO not right.
now you take it all to one side and solve it as quadratic equation.
Melody
Nov 1, 2013
#2
0
then solve this question by using log both sides, right??
Guest
Nov 1, 2013
#1
+118725
0
to find the time it takes for a dropped object to hit the ground, you can use the equation, h=16t^2, where h, is the height in feet and t, is the time , in seconds. if an object is dropped from 48 feet, what is the approximate falling time, in seconds, of this object.
just substitute h=48 and solve for t
Melody
Nov 1, 2013
#1
+118725
0
x-(0.44x) = 0.1369
x - 0.44x = 0.1369
x ( 1- 0.44) = 0.1369
can you do it from here?
Melody
Nov 1, 2013
#2
+118725
0
this question has received an enormous amount of attention in the second forum is anyone is interested.
Melody
Nov 1, 2013
#2
+3146
0
[input]lim( (x^2 + x^3 - x) / (x^4 - x^2 + 1), x=inf )[/input]
admin
Nov 1, 2013
#2
+118725
0
A cottage factory with 178 workers computed its mean wage bill and found it to be K£ 49.95 and its modal wage was K£ 44.50. If the variance was 78.5625, calculate;
i. The total wage bill 178*49.95
ii. The median wage I don't think that there is enough information given
iii. The coefficient of variation sqr(78.5625) / 49.95
"The coefficient of variation represents the ratio of the standard deviation to the mean, and it is a useful statistic for comparing the degree of variation from one data series to another, even if the means are drastically different from each other."
Melody
Nov 1, 2013
#1
+118725
0
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
Use L'Hopital's Rule
This is not exactly the question you wrote, but I think it is what you intended.
If I got the intention wrong then I am sorry.
lim [x
2
+ x
3
- x] / [x
4
- x
2
+1]
x--> infinity
= lim [2x + 3x
2
- 1] / [4x
3
- 2x]
x--> infinity
= lim [2 + 6x ] / [12x
2
- 2 ]
x--> infinity
= lim 6/(24x)
x--> infinity
=0
Melody
Nov 1, 2013
#1
+118725
0
log
2
(5X
2
- X - 2 ) = 2 + 2 log
2
X
First I am going to assume that this is the question that you intended.
Now I am going to look at the RHS
RHS= 2 + 2 log
2
X
=2(1+log
2
X)
=2( log
2
2 + log
2
X)
=2 log
2
(2X)
= log
2
(2X)
2
=log
2
(4X
2
)
So 5X
2
- X - 2 = 4X
2
Now you should be able to solve it yourself.
Melody
Nov 1, 2013
#1
+118725
0
the diagram shows a cross section of a circular washer of radius, 2.8 cm with a rectangular hole of length,3 cm and breath,2.5 cm.Find the area of the cross section of the washer
the circle minus the rectangle.= pi*r*r - (3*2.5) cm squared
Melody
Nov 1, 2013
#3
+118725
0
16+25 - (2*20*CosC)=36
42- (40cosC) = 36
6 = 40cosC
6/40 = cosC
cosC = 0.15
inverse cos 0.15 = 81.37 degrees
It is really good that you recognised that you needed an angle bigger than 60degrees.
Can you work out what you have been doing wrong?
Melody
Nov 1, 2013
#2
0
what confuzes me is that when i solve for C , I end up with 47.55 which makes no sense considering angle C is the largest angle in the triangle
Guest
Nov 1, 2013
#1
+118725
0
if 16+25+-2(20)CosC=36 , what is the angle measure of C ?
Before you even start you need to make sure you calculator is in degrees or radians - whichever one you need
Why are you confused ?
It would be more helpful if you put the brackets like this, you will be less likely to make mistakes.
16+25 - (2*20*CosC)=36
Once you solve for CosC then put the answer in you calc and press inv cos to get the value of C
Is that enough help?
Melody
Nov 1, 2013
#4
+118725
0
I just couldn't leave it alone.
What is the largest integer N such that 7^N = (1000!)/(500!)^2?
I experimented with 20!/(10!10!) with 3^N
and 30!/(15!15!) with 3^N
If you experiement with thes smaller ones hopefully you will be able to see where I got my logic from.
and this is what I came up with
numerator
1000/(7^1) = 142 (i don't care about the left overs)
1000/(7^2) = 20
1000/(7^3) = 2
1000/(7^4) = 0
Add them all up and you get 164 So 7^164 is a factor of the top
denominator
500/(7^1)= 71
500/(7^2)= 10
500/(7^3)= 1
500/(7^4)= 0
add them up and multiply by 2 = 2*82 = 164 so 7^164 is a factor of the bottom
When you cancel all the 7s cancel out and you get 1
So the biggest power of 7 is 0. (It is not divisable by 7)
I am pretty sure that is correct.
Melody
Nov 1, 2013
#3
+118725
0
This question is making my mind boggle.
I suggest that you try doing something very similar only with much smaller numbers and see if you can make sense of it that way.
Maybe you have already done this.
I haven't given up on it but I have other things to do now.
Melody
Nov 1, 2013
#2
0
Looks like this relates to Miaaa's other question.
12.15g of magnesium is half a mole.
the other number 1.5E23 is close to a quarter of a mole.
Guest
Nov 1, 2013
#3
0
Chemistry is not a strong point for me, but here's my take on this rather cryptic question.
3.01E23 is half a mole worth of atoms. If you go to the periodic table and select an element, say magnesium, you will find its atomic mass as 24.31. That means that one mole of magnesium atoms weighs 24.31 grams. So half a mole would weigh half that, like 12.155 grams.
Need that atomic mass to make the question work, me thinks Miaaaa
Cheers
Guest
Nov 1, 2013
#2
0
A lovely morning here in Adelaide. Thought I'd try my hand at this to see if it can be done without adding water, just mixing the 10 and 17's together. Bit hungover. Hmmmm.
14 gallons of 12%. Lets multiply them together and forget about the percentages for the moment....
14x12+168. We want 168 by mixing 10a plus 17b where a and b are the gallons we are trying to work out.
We can solve a and b by remembering that a plus b = 14 gallons, so
a = 14-b
So, Lets put it all together....
168=10(14-b) +17b
168=140 -10b+17b
Simplify
28 = 7b
b=4 gallons, thats the strong stuff, and
a=10 gallons of the weak stuff.
Lets check if correct.
10x10 + 17x4=168
Yipee!!!!!!!
Cheers
Guest
Nov 1, 2013
#2
0
Sorry! I meant "What is the largest integer N such that 7^N divides (1000!)/(500!)^2.
Guest
Nov 1, 2013
Oct 31, 2013
#1
+118725
0
What is the largest integer N such that 7^N = (1000!)/(500!)^2?
I think there is something missing.
You only have one unknown and an = sign therefore there is just one answer.
maybe it should be > or < ?
Also, (1000!)/(500!)^2 is a huge number. I've tried to do it on a couple of calculators and they just give error messages or do nothing at all. I've never seen that before. I am assuming that they just won't handle factorials of this size. The question can still be answered but not simplified.
This might not matter if the question is a bit different.
Melody
Oct 31, 2013
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