Please help this is a 30 60 90 triangle

Mathematics

2 answers:

natka813 [3]3 years ago

8 0

Answer:

See below

Step-by-step explanation:

$x = \frac{1}{2} \times \sqrt{6} \\ \\ x = \frac{ \sqrt{6} }{2} \\ \\ y = \frac{ \sqrt{3} }{2} \times \sqrt{6} \\ \\ y = \frac{ \sqrt{3 \times 6} }{2} \\ \\ y = \frac{ \sqrt{18} }{2} \\ \\ y = \frac{ 3\sqrt{2} }{2}$

Elden [556K]3 years ago

3 0

Answer:

x = $\frac{\sqrt{6}}{2}$ and y = $\frac{3\sqrt{2}}{2}$

Step-by-step explanation:

The ratio of the sides is a:a $\sqrt{3}$ :2a, and you are give the side that is equal to 2a. This means that x = $\frac{\sqrt{6}}{2}$ and y = $\frac{\sqrt{6} * \sqrt{3}}{2}$ = $\frac{3\sqrt{2}}{2}$

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Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$