proyaop

First, notice that it is a right triangle, so x and y could be 7. You want x to be as close to y as possible to get the shape as close to a square as possible (since it is the optimal area), thus x and y are 3, 4 so x + y = 7

\(1\over1+{1\over\sqrt{3}}+2\)

Multiply top and bottom by root3 and combine 1 + 2.

\(\sqrt{3}\over3\sqrt{3}+1\)

Multiply top and bottom by \(3\sqrt{3}-1\)

\(\sqrt{3}(3\sqrt{3}-1)\over(3\sqrt{3}+1)(3\sqrt{3}-1)\)

\(9 - \sqrt{3}\over26\)

Adding a negative number is parity: the minus sign is like a -1 x 58

The minus sign takes priority so +-58 = - 58.

Multiply everything by 5.

5/4 < x < 65/2

1.25

x = 2 to 32 which is 32 - 2  + 1 numbers = 31 integers

There are 5 Gs, so its 3 to the 3 to the 3 to the 3 to the 3 to the 1.

3 to the 3 to the 3 to the 3 to the 3

3 to the 3 to the 3 to the 27

3 to the 3 to the 7625597484987.... thats all i can go up to before calculators go like: invalid input.

\(f(x)=\sqrt{3 + \sqrt{5 + \sqrt{x}}}\)

\(x\ge0\)

Smallest value of the function would be \(\sqrt{3 + \sqrt{5}}\)

Maximum value is infinity.

\(a + b\sqrt{c} = \sqrt{3 + \sqrt{5}}\)

\(a + b\sqrt{5} = \sqrt{3 + \sqrt{5}}\)

(a + bsqrt(5))^2 = 3 + sqrt(5)

a^2 + 5b^2 + 2absqrt(5) = 3 + sqrt(5)

a^2 + 5b^2 = 3

sqrt(5) = 2absqrt(5)

2ab = 1

ab = 1/2

You can do it from here... Domain: [a + bsqrt(c), inf)

Get a and b, or just use the sqrt(3 + sqrt(5))

First we want to get y to cancel out.

Equation 1 can be rewritten as 12x - 18y = -6

Equation 2 can be rewritten as 2kx - 18y = 6

Subtracting equation 2 from equation 1:

(12 - 2k)x = -12

(6-k)x = -6

If k = 6, then 0 = -6 which is false.

Convert decimals to fractions and open parenthesis and do the minus distibution, do LCM of 3, 12, 15. Combine like terms.

40/60 - 25/60 - 28/60 - 12/60 = -25/60

-5/12