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3 years ago

13

Please help!! need to pass this class!!

1 answer:

Bezzdna [24]3 years ago

3 0

Answer:

ADB and DBC

Step-by-step explanation:

Angles ADB and DBC are both marked with three curved lines which shows that their angles are congruent to each other.

Hope this helps! :)

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First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

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How do you solve 5y+34=-2 (1-7y)

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Factor the following polynomial completely.<br><br>y squared plus 5 zy plus 4 z squared<br><br>

sweet [91]

Answer:

(y + z)(y + 4z)

Step-by-step explanation:

${y}^{2} + 5zy + 4 {z}^{2} \\ = {y}^{2} + zy + 4zy + 4 {z}^{2} \\ = y(y + z) + 4z(y + z) \\ = (y + z)(y + 4z)$

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Schach [20]

Option c would be correct

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