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acsc(30/360) = Error: invalid argument

asin(360/30) = asin(12)

The arguement for asin must be between -1 and 1

=   log (  (7*15)/5)    = log (105/5) = log 21 = 1.3222

Sorry typo     x, y = 16,74     NOT 10,74 as I summarized earlier

oOOPS......here is how

x+y = 4 (sqrt3)

x*y = 3     so   y = 3/x  (sub this into the first equation)

x + 3/x = 4 sqrt3     Multiply through by 'x'

x^2 + 3 = 4 sqrt 3  x     re-arrange

x^2 - 4 sqrt3  x   + 3 = 0          Now use the quadratic formula to find   2 (sqrt 3)-3   and   2 (sqrt 3) + 3

Quadratic Formula:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

challenge accepted.

I am only looking at the semicircle on the right.

The coloured in brown bit plus the two little sectors must be half the size of the segment on the far right.

$cos\theta = \frac{h}{r}\\ h=rcos\theta\\ Area\;of\;brown\; triangle=0.5r^2sin(2\theta)\\ Area \;of\;one\;little\;sector=\frac{(0.5\pi-\theta)}{2\pi}\pi r^2=\frac{(0.5\pi-\theta)r^2}{2}\\ \text{add these three areas together and get}\\ \text{half of middle slice is }\\ 0.5r^2sin(2\theta)+2*\frac{(0.5\pi-\theta)r^2}{2}\\ =0.5r^2sin(2\theta)+(0.5\pi-\theta)r^2\\ =r^2[0.5sin(2\theta)+0.5\pi-\theta]\\~\\ \text{area of segment }=0.5\pi r^2-r^2[0.5sin(2\theta)+0.5\pi-\theta]\\ \text{area of segment }=r^2[0.5\pi-0.5sin(2\theta)-0.5\pi+\theta]\\ \text{area of segment }=r^2[\theta-0.5sin(2\theta)]\\~\\ r^2[\theta-0.5sin(2\theta)]=2*r^2[0.5sin(2\theta)+0.5\pi-\theta]\\ \theta-0.5sin(2\theta)=sin(2\theta)+\pi-2\theta\\ 3\theta-1.5sin(2\theta)=\pi\\ $

2(sqrt 3) - 3     and     2(sqrt 3) +3

Hope you can get through my scrawl...and it is in a manner to which you are familiar !

A tesseract has three diagonals and you didn’t specify which diagonal. This is not a problem. I’m trained to read minds and find missing questions and missing parts of questions asked by the lower level primates, like you, that visit this forum.  I can do this even if I’ve had too many banana daiquiris. Here’s a sample of my work. 

Scary, huh?     Probably scares Melody more than me!   

Here is the parabola

=8x^3 * 2^(-15) =1

8x^3 = 1/(2^-15)

8x^3 = 2^(15)

2^3 x^3 = 2^15

x^3 = 2^12            Take cube root of both sides

x = 16

Alright, I re-read your question....  for this to 'BE DEFINED'  ,   the value of

x^2 - 4x+1     HAS to be POSITIVE (greater than 0-- because there is no logs of NEGATIVE numbers)

so

x^2 - 4x +1 > 0

Quadratic formula results in  ZEROES at   x = 2 +-sqrt3

If you graph this fxn you will see a parabola which is POSITIVE outside of these values and NEGATIVE between them

so       2+sqrt3 <   x < 2-sqrt3

EP: Great minds think alike!. I got this from Melody!.

The 1/3 full one which is twice as large equals  2/3 of a smaller one

so you have    2/3 + 1/2 = 7/6 of  smaller one of 3 yr old wine

you basically are filling THREE casks (the smaller one PLUS the 2x bigger one)

the part that will be 3 yr old will be 7/6  out of 18/6   (three casks)

(7/6) / (18/6)    = 7/18ths of the new mixture will be 3 yr old wine.

You have pretty good sense of humour! However, it is very interesting question and I will take a crack at it!!. Here we go:

A cask half-full of 3-year old wine is 1/4 of a cask twice its size. Therefore: 1/4 + 1/3=7/12 of the large cask contains 3-year old wine. After filling both casks with 1-year old wine, the small cask is half full of 1-year old wine and the large cask is 2/3 full of 1-year old wine. Therefore: 1/4 + 2/3 =11/12 of the large cask contains 1-year old wine. Since 7/12 of the large cask contains 3-year old wine and 11/12 of it contains 1-year old wine, it follows that the total amount of wine, both new and old, 18/12. Thus (7/12) / (18/12) =7/18. Therefore, 7/18 of the mixture is 3-year old wine! That is my take on it!.

Solve for x:
(log(x^2 - 4 x + 1))/(log(10)) = 0

Multiply both sides by log(10):
log(x^2 - 4 x + 1) = 0

Cancel logarithms by taking exp of both sides:
x^2 - 4 x + 1 = 1

Subtract 1 from both sides:
x^2 - 4 x = 0

Factor x from the left hand side:
x (x - 4) = 0

Split into two equations:
x - 4 = 0 or x = 0

Add 4 to both sides:
Answer: |x = 4               or                  x = 0

This is not an EQUATION...this is a RELATION.

to solve for x, you need to have an = 'something'    to make it ann equation.     Sorry

Where they intercept, they will obviously be equal....so set them equal to each other:

5x-6 = 4x + 10

x = 16                  then sub this value of 'x' to find 'y'

y = 5(16) -6  = 74       x,y = 10,74

Solve for x over the real numbers:
2^(x^2 + 2 x - 15) = 1

Take the logarithm base 2 of both sides:
x^2 + 2 x - 15 = 0

The left hand side factors into a product with two terms:
(x - 3) (x + 5) = 0

Split into two equations:
x - 3 = 0 or x + 5 = 0

Add 3 to both sides:
x = 3 or x + 5 = 0

Subtract 5 from both sides:
Answer: |x = 3           or          x = -5

x=10-5

x=5

1inch ≈ 2.54 cm

1 ft ≈ 30.48 cm

So......  120 / 30.48  ≈ 3.937 ft

And .... 160 / 30.48 ≈  5.249 ft

So  120 cm * 160 cm  =  [3.937 ft  * 5.249 ft ]  ≈  20.665 sq ft

Thanks for the brownie, EJ.......just in time for lunch !!!!

That's right:

(and yeah, I still give those brownie pictures for right answers).

Let the number of members in "A"  = N.......then the number of members of "B"  = 30 - N

And we know that

A + 1   =  2 [ B - 1 ]    or

N + 1  = 2 [ (30 - N) - 1]    simplify

N + 1  = 2 [ 29 - N ]

N + 1   = 58 - 2N     add 2N to both sides, subtract 1 from each side

3N  = 57      divide both sides by 3

N  = 19 = number in "A"    and 30 - N    = 11  =  number in "B"

Proof : If one more member leaves "B" and joins "A"...."A" will have 20 members and "B" will  have 10  and "A" will have twice as many members as "B"