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3 years ago

13

The temperature was 10°F. It dropped so that the temperature was 0°F. ___°F represents the change in temperature. Is it 10 or -

10

1 answer:

yan [13]3 years ago

4 0

Answer:

$Change = - 10^{\circ}F$

Step-by-step explanation:

Given

$Initial = 10^{\circ}F$

$New = 0^{\circ}F$

Required

Determine the change in temperature

This is calculated by:

$Change = New - Initial$

Substitute values for New and Initial Temperatures

$Change = 0^{\circ}F - 10^{\circ}F$

$Change = - 10^{\circ}F$

Hence, the change in temperature is: $- 10^{\circ}F$

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4 years ago

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;

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

$- A {e}^{ - 2\pi} ( 1 ) - 2 A {e}^{ - 2\pi} ( 0 ) + B {e}^{ - 2\pi}( 0) - 2B {e}^{ - 2\pi} ( 1 ) = 5$

$- A {e}^{ - 2\pi} - 2B {e}^{ - 2\pi} = 5$

But B=0

$- A {e}^{ - 2\pi} = 5$

$A = 5{e}^{2\pi}$

Therefore the particular solution is

$y(t) = - 5 {e}^{2\pi} {e}^{ - 2t} \cos(t) + 0 \times {e}^{ - 2t} \sin(t)$

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

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4 years ago