Logarhythm

\(2(w + \frac{w}{3}) = 24\)

\(\frac{4w}{3} = 12 \)

\(4w = 36 \)

\(w = 9\)

Area = \(9 \cdot \frac{1}{3} (9) = 9 \cdot 3 = 27\)

Area = 27 cm²

\((x - 1 + 2)^2 = (2x - 4)^2 + (x - 1)^2\)

\(x^2 + 2x + 1 = 4x^2 - 16x + 16 + x^2 - 2x + 1\)

\(2x = 4x^2 - 18x + 16\)

\(x = 2x^2 - 9x + 8\)

\(2x^2 - 10x + 8 = 0\)

\(2(x^2 - 5x + 4) = 0\)

2(x - 1)(x - 4) = 0

(x - 1)(x - 4) = 0

x = 4 

x = 1

However, side WU is (x - 1) so x can't be 1 since the side will be 0, which is impossible. 

Which means your answer is 4

\(\sin\theta = \frac{3}{5}\)

\(\cos \theta = \frac{4}{5}\)
 

Hypotenuse of the second triangle is: \(\sqrt{24^2+7^2} = 25\)

\(\sin \alpha = \frac{7}{25}\)

\(\tan \alpha = \frac{7}{24}\)

This means: 

\(\theta = \arcsin \frac{3}{5} ≈ 36.87\)

\(\alpha = \arcsin \frac{7}{25} ≈ 16.26\)

Now:
1. \(\sin 2\theta ≈ \sin 2(36.87) ≈ \sin73.74 = 0.96\)

2. \(\cos 2\theta ≈ \cos 2(36.87) ≈ \cos 73.74 = 0.28\)

3. \(\sin 2\alpha ≈ \sin 2(16.26) ≈ \sin 32.52 = 0.5376\)

4. \(\tan 2\alpha ≈ \tan 2(16.26) ≈ \tan 32.52 ≈ 0.6376\)

Total marbles in the bag: 6 + 5 + 8 = 19

The probability that all three marbles drawn will be green is = \(\frac{5}{19} \cdot \frac{4}{18} \cdot \frac{3}{17} = \frac{10}{969}\)

≈ 1.03%

For simplicity sake, let a = adult ticket and s = student ticket.

a + s = 1132 tickets
 

(5)a + (1)s = 2552

5a + s = 2552

s = 2552 - 5a

a + (2552 - 5a) = 1132

a - 5a = 1132 - 2552

-4a = -1420

a = 355 adult tickets

a + s = 1132

355 + s = 1132

s = 777 student tickets

5) The hypotenuse of the "12 by 16" triangle, using pythagoras' theorem, is \(\sqrt{12^2 + 16^2} = \sqrt{400} = 20\)

Then, the hypotenuse of the "12 by x" triangle, using the big triangle, is \(\sqrt{(x+16)^2 - 20^2}\)

But we also know that that side is \(\sqrt{12^2 + x^2}\)

So we can make an equation since the two are equivalent:
\(\sqrt{(x + 16)^2 - 20^2} = \sqrt{12^2 + x^2}\)

\((x + 16)^2 - 20^2 = 12^2 + x^2\)

\((x + 16)(x + 16) - x^2 = 144 + 400\)

\(x^2 - x^2 + 32x + 256 = 544\)

\(32x = 288\)

\(x = 9\)

6) This question is similar to the previous one so I won't explain in depth:
\(\sqrt{60^2 - 36^2} = 48\)

\(\sqrt{(x - 36)^2 + 48^2} = \sqrt{x^2 - 60^2}\)

\((x - 36)(x - 36) - x^2 = - 60^2 - 48^2\)

\(x^2 - x^2 - 72x + 1296 = -5904\)

\(-72x = -4608\)

\(x = 64\)

side a = 18√2 ÷ √2 = 18

This means side b = 18 ÷ √3 = 6√3 

Finally, side c = 2(6√3) = 12√3

since 0.54545454... is just \(\frac{6}{11}\)

we then know that 0.00545454... is \(\frac{6}{1100}\) and 0.43 is just \(\frac{43}{100}\),

0.43545454... = \(\frac{(43)(\frac{1100}{10})}{1100}+\frac{6}{1100} = \frac{473 + 6}{1100} = \frac{479}{1100}\)

Start by simplifying the terms:

\(\frac{2a}{a-2b} - \frac{6b}{a+2b} = 3\)

\(\frac{(2a)(a+2b)-(6b)(a-2b)}{(a-2b)(a+2b)} = 3\)

\(\frac{2a^2 -2ab +12b^2}{(a-2b)(a+2b)} = 3\)

\(\frac{2a^2 -2ab +12b^2}{a^2-4b^2} = 3\)

\(\frac{a}{a-2b}+\frac{6b}{a+2b}=-1\)

\(\frac{(a)(a+2b)+(6b)(a-2b)}{(a-2b)(a+2b)}=-1\)

\(\frac{a^2+8ab-12b^2}{(a-2b)(a+2b)}=-1\)

\(\frac{a^2+8ab-12b^2}{a^2-4b^2}=-1\)

Can you take it from there?