Member: Rom

Quote:
A girl has some beads. She use 1/3 of the beads to make a bracelet. She uses 3/5 of the beads to make a necklace. She used 36 more beads for the necklace. How many beads did she use for the bracelet?

Please give me a method, not the answer!

And no algebra!

sorry you need algebra to do this sort of problem. It's not that bad

Let N be the number of beads for the necklace
Let B be the number of beads for the bracelet

Let T be the total number of beads she has

B = T/3 ; from the first statement
N = (3/5)T ; from the 2nd statement
N = B + 36 ; from the 3rd statement

plugging (1) and (2) into (3) we get

(3/5)T = T/3 + 36
(3/5)T - T/3 = 36
(9/15)T - (5/15)T = 36
(4/15)T = 36
4T = 540
T = 135

B = T/3 = 135/3 = 45

she used 45 beads for the bracelet

RomFeb 13, 2014

+6251

Quote:
What is fraction 2 2/5 x 80

(2 2/5) x 80 =
2x80 + (2/5)x80 =
160 + (2x80)/5 =
160 + 160/5 =
160 + 32 = 192

RomFeb 13, 2014

+6251

I don't see how this is solvable.

If I happen to point to the jack the answer you give me contains zero information. There's no question I could ask that would change that since the answer is completely at random.

Further If I point to an ace, based on your answer there is still a 1/6 probability that I could be pointing to a jack and the random answer happened to be correct. Again this is independent of any question I might come up with.

RomFeb 13, 2014

+6251

African or European woodchuck?

RomFeb 12, 2014

+6251

Quote:
A projectile is fired with initial velocity 5umeters/second at an angle of projection tan inverse of (4/3).
The projectile is halfway below its maximum height at times t1 and t2.
Show that:
a) t1 + t2 = (8u/g)
b) t1* t2 = (8u^2/g^2)

ok first off we note that the angle of projection is the midsized angle of a 3/4/5 triangle and thus it's cosine is 3/5 and it's sine is 4/5

so the projectiles initial vertical velocity is 5u*4/5 = 4u m/s

we can find it's maximum height by cheating a bit. The projectile has kinetic energy due to it's vertical component of (1/2)mv _vert ² = (1/2)16u ²m = 8u ²m
we know at the projectiles maximum height the potential energy due to gravity is equal to this kinetic energy. The potential at height h is mgh where g is the gravitational acceleration.
so the max height is at 8u ²m = mgh, h = 8u ²/g

You want to find the two times t1 and t2 where the height is half of this i.e. h(t1) = h(t2) = 4u ²/g

h(t) = (4u)t - g t ²/2

so just solve (4u)t - g t ²/2 = 4u ²/g

and this will get you your t1 and t2. Then just add them and multiply them and show that (a) and (b) are true.

RomFeb 12, 2014

+6251

Quote:
the 2 cylinders below are similar. cylinder A has a height that is half of cylinder B. The volume of cylinder A is 12 cubic inches. which is the closest to the volume of cylinder B ?

ok, cylinders can be described by two numbers, their radius and their height.

two cylinders are similar if for some number K>0 both the radius and height of one cylinder is K times those of the other.

the volume of a cylinder is V=pi R ² H

we are told that
H _A = (1/2) H _B
2*H _A = H _B

the cylinders are similar so
R _A = (1/2) R _B
2*R _A = R _B

Volume _B = pi R _B ² H _B =
pi (2 R _A) ² (2 H _A) = 8 pi R _A ² H _A =
8 Volume _A = 8 * 12 in ³ = 96 in ³

RomFeb 12, 2014

+6251

Quote:
What are the solutions of the equation x^2-3x-4=0

this one can be factored pretty easily by inspection

x ²-3x-4 = (x-4)(x+1)

so your solutions are x = 4 and x = -1

if you're no good with factoring by eye you can always use the trusty quadratic equation which says if

ax ² + bx + c = 0

then the solutions are given by

x = [size=200]([/size]-b +/- sqrt[size=150]([/size]b ² - 4ac[size=150])[/size][size=200])[/size][size=200]/[/size][size=150]([/size]2a[size=150])[/size]

in this case we've got a=1, b=(-3), c=(-4)

plugging and chugging we get

x = (3 +/- sqrt(9+16))/2 = (3 +/- 5)/2 = 8/2 and -2/2 = 4 and -1 which agree with what we've got above

RomFeb 12, 2014

+6251

Quote:
These are the questions-
13. C=49pi
14. A= 81pi
15. C= 10pi
16. A= 25pi
17. C= 17pi
18. A= 9pi

I'm going to go out on a limb and guess that C is circumference and A is area.

let r be the radius of your circle
C = pi 2r
A = pi r ²

you should be able to figure these out now

RomFeb 12, 2014

+6251

Quote:
ONE wrote::?: what could you say about the vectors 3c-d and 3/2d-1/2c

If c and d are linearly independent, then those vectors are also independent.

what vectors? are c and d vectors?

RomFeb 12, 2014

+6251

Einstein

The man revolutionized physics 100 or so years ago. You can at least spell his name correctly.

RomFeb 12, 2014

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