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4 years ago

Use the trigonometric ratio sin A = 0.38 to determine which angle of the triangle is < A

Mathematics

1 answer:

-Dominant- [34]4 years ago

6 0

<span>According to trigonometric the formula for sin is sin = opposite/hypotenuse.Sin is defined as the ratio of sides of a right angled triangle.Hypotenuse is the is the longest side of a right-angled triangle,which is the side opposite of the right angle.So the the hypotenuse is clearly 1.3, so you divide 0.5/1.3 Hence the answer is around 0.38 The result is Angle2..</span>

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Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e

sergiy2304 [10]

Answer:

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$

Step-by-step explanation:

The given initial value problem is

$y''+4y'+5=0$

$y( \frac{ \pi}{2} ) = 0$

$y'( \frac{ \pi}{2} ) = 5$

The corresponding characteristic equation is

${m}^{2} + 4m + 5 = 0$

$m = - 2 \pm \: i$

The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$A {e}^{ - 2\pi} (0 ) + B {e}^{ - 2\pi} ( 1) = 0$

$B = 0$

Also;

$y'( \frac{\pi}{2} ) = - A {e}^{ - 2\pi} \sin( \frac{\pi}{2} ) - 2 A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) - 2B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$