Member: Rom

Quote:
If I have a problem where I have two congruent pentagons and I need to find an unknown side length but there aren't any variables, what should I do?

you need to give us more to work with than this.

do you know the length of a side in the first pentagon?

is there some way you can figure out the scale factor between the first and second pentagons?

RomFeb 10, 2014

+6251

Quote:
g(x) = 2x^2 −16x + 15 I need to know how to put that into standard form of a parabola which is g(x)= a(x − h)^2 + k if you can help(:

this is called completing the square.
learn it, love it

2x ² - 16x + 15 ; factor the x ² coefficient to the outside

2[size=150]([/size]x ² - 8x + 15/2[size=150])[/size] ; now note the coefficient of the x term is 8 and we take 1/2 of this to be h in our (x-h) ² term

2[size=150]([/size](x-4) ² - 4 ² + 15/2[size=150])[/size] ; study this a sec we added and subtracted 4 ², make sure you understand how

2[size=150]([/size](x-4) ² - 16 + 15/2[size=150])[/size] ; compute the square of 4

2[size=150]([/size](x-4) ² - 17/2[size=150])[/size] ; do the addition

2(x-4) ² - 17 ; distribute the multiplcation by 2

RomFeb 10, 2014

+6251

Quote:
what do you do when your have to sin 40degree30 minutes what do that do when you get the anwer

not real sure what you're asking here but I'm going to guess you want to know how to convert

40 degrees 30 minutes to just degrees so you can punch it into the calculator

think of 1 degree as 1 hour

there are 60 minutes to an hour, there are 60 minutes to a degree

there are 60 seconds to a minute in both cases.

so 30 minutes is 30/60 = 1/2 degree

so 40 degrees 30 minutes is 40 1/2 = 40.5 degrees

and you can just punch in

4 0 . 5 sin =

to get the answer

RomFeb 10, 2014

+6251

Ok I get it now.

If they want radians per second then the radius isn't needed.

If you are travelling 3pi/4 radians in 5 seconds you are travelling (3pi/4)/5 radians/s = 3pi/20 rad/s

If they wanted a linear speed then you would have to find the linear distance 3pi/4 radians is given a radius of 4km

D = (arc length in radians)*(radius) = 3pi/4 * 4km = 3pi km

v = D/t = 3pi/5 km/s

RomFeb 10, 2014

+6251

Quote:
What are the solutions of 3x2 + 6x + 6 = 0?

just plug and chug using the quadratic formula

ax ² + bx + c = 0 ==> x = (1/2a)(-b +/- sqrt(b ²-4ac))

In this case a=3, b=6, c=6

plugging and chugging we get

x = (1/2*3)(-6 +/- sqrt(6 ²-4*3*6))
x = (1/6)(-6 +/- sqrt(36-72))
x = (1/6)(-6 +/- sqrt(-36))
x = (1/6)(-6 +/- 6i)
x = -1 +/- i

RomFeb 10, 2014

+6251

Quote:
11pi/4 answer cannot be in decimal form

there's really no way to simplify this.

Maybe what you mean is what angle is it congruent to. If so
remember that an angle alpha is congruent to all angles (alpha + 2pi * k) where k is any integer

we note
2pi < 11pi/4 < 4pi

so
11pi/4 - 2pi =
11pi/4 - 8pi/4 =
3pi/4

and thus 11pi/4 is congruent to 3pi/4

RomFeb 10, 2014

+6251

Quote:
Rosa Poured 1 3/4 quarts of orange juice in container. She added 3 1/3 Quarts of Apple juice & then she poured 2 2/3 quarts of mixed juice in pitcher. How many quarts of of mix juice left in container.

Before pouring the juice into the pitcher there was

(1 3/4) + (3 1/3) quarts in the container

(1 3/4) + (3 1/3) = ; combine the whole and fractional parts separately
(1+3) + ( 3/4 + 1/3 ) = ; note that the least common multiple of 3 and 4 is 12 so you can convert the fractions to a common denominator
4 + ( 9/12 + 4/12 ) = ; do the addition
4 + 13/12 = ; convert 13/12 to whole and fraction pieces
4 + (1 1/12) = ; add the whole and fractional parts separately
(4+1) + 1/12 = 5 1/12

So there were 5 1/12 quarts in the container before she poured any into the pitcher

Then she poured 2 2/3 quarts into the pitcher. So

(5 1/12) - (2 2/3) quarts were left in the container

You do this exactly like we did above except that it's subtraction now and so I'm going to let you finish it.

RomFeb 10, 2014

+6251

sorry I don't understand what you are asking

what is s?

RomFeb 10, 2014

+6251

Quote:
How to solve 4sinx= 2sin2xcos2x

I don't see any clever ways of attacking this so I guess brute force is what's left

note:

sin(2x) = 2sin(x)cos(x)
cos(2x) = 2cos ²(x) - 1

4sin(x)= 2sin(2x)cos(2x) ; divide both sides by 2

2sin(x) = sin(2x)cos(2x) ; substitute in the the formulas for sin and cos of 2x from above

2sin(x) = 2sin(x)cos(x)[size=150]([/size]2cos ²(x) - 1[size=150])[/size] ; divide both sides by 2sin(x)

1 = cos(x)[size=150]([/size]2cos ²(x) - 1[size=150])[/size] ; multiply out and rearrange things to get a cubic polynomial equal to 0

2cos ³(x) - cos(x) - 1 = 0 ; let u=cos(x)

2u ³ - u - 1 = 0

You have a cubic equation in u now. In general not very easy to solve. Since the constant term is 1 you should try to see if it's divisible by (u-1) and/or (u+1)

Lo and Behold
2u ³ - u - 1 = (u-1)[size=150]([/size]2u ² + 2u + 1[size=150])[/size] and
2u ³ - u - 1 = (u+1)[size=150]([/size]2u ² - 2u + 1[size=150])[/size]

so either
(u-1) = 0, or [size=150]([/size]2u ² + 2u + 1[size=150])[/size] = 0 or
(u+1) = 0, or [size=150]([/size]2u ² - 2u + 1[size=150])[/size] = 0

If you try to solve either of the quadratic factors on the right you'll find neither has real roots so u=1 or u=-1 are our only real solutions. So

u = 1 ==> cos(x) = 1 ==> x = pi*2k where k is any integer and
u = -1 ==> cos(x) = -1 ==> x = pi*(2k+1) where k is any integer

so our final solution is x = pi*k where k is some integer

This is a pretty tricky problem.

RomFeb 10, 2014

+6251

Ok I laughed but if your idea of fun is trolling math boards you really need to get out more.

RomFeb 10, 2014

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