(dy)/(dt)=2t+3" and "y(1)=6 What is t when y=0 ? Choose all answers that apply: A t=-3 B t=-1 c t=1 D t=0 ㅌ t=-2 F t=-4

Integrate and Find Solution: Given the differential equation (dy)/(dt) = 2t + 3 and the initial condition y(1) = 6, we need to integrate the differential equation to find the general solution for y(t).Use Initial Condition: Integrate (dy)/(dt) = 2t + 3 with respect to t to find y(t). ∫(dy) = ∫(2t + 3) dt y(t) = t^2 + 3t + C, where C is the constant of integration.Substitute Constant and Solve: Use the initial condition y(1) = 6 to find the value of the constant C. 6 = (1)^2 + 3(1) + C 6 = 1 + 3 + C C = 6 - 4 C = 2Find t when y = 0: Substitute the value of C back into the general solution to get the particular solution for y(t). y(t) = t^2 + 3t + 2Factor and Solve Quadratic Equation: Now we need to find the value of t when y = 0. 0 = t^2 + 3t + 2 We have a quadratic equation to solve: t^2 + 3t + 2 = 0.Identify Values of t: Factor the quadratic equation. (t + 1)(t + 2) = 0Set each factor equal to zero and solve for t. t + 1 = 0 => t = -1 t + 2 = 0 => t = -2We have found two values of t when y = 0: t = -1 and t = -2. These correspond to options B and E.

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(dy)/(dt)=2t+3" and "y(1)=6
What is
t when
y=0 ?
Choose all answers that apply:
A
t=-3
B
t=-1
c
t=1
D
t=0
ㅌ
t=-2
F
t=-4

$\frac{d y}{d t}=2 t+3 \text { and } y(1)=6.$ $\newline$ What is $t$ when $y=0$ ? $\newline$ Choose all answers that apply: $\newline$ (A) $t=-3$ $\newline$ (B) $t=-1$ $\newline$ (C) $t=1$ $\newline$ (D) $t=0$ $\newline$ (E) $t=-2$ $\newline$ (F) $t=-4$

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Algebra 2

Solve complex trigonomentric equations

Full solution

\frac{d y}{d t}=2 t+3 \text { and } y(1)=6.

\newline

What is

t

when

y=0

\newline

Choose all answers that apply:

\newline

(A)

t=-3

\newline

(B)

t=-1

\newline

(C)

t=1

\newline

(D)

t=0

\newline

(E)

t=-2

\newline

(F)

t=-4

Integrate and Find Solution: Given the differential equation $\frac{dy}{dt} = 2t + 3$ and the initial condition $y(1) = 6$ , we need to integrate the differential equation to find the general solution for $y(t)$ .
Use Initial Condition: Integrate $\frac{dy}{dt} = 2t + 3$ with respect to $t$ to find $y(t)$ . $\int(dy) = \int(2t + 3) dt$ $y(t) = t^2 + 3t + C$ , where $C$ is the constant of integration.
Substitute Constant and Solve: Use the initial condition $y(1) = 6$ to find the value of the constant $C$ .
$6 = (1)^2 + 3(1) + C$
$6 = 1 + 3 + C$
$C = 6 - 4$
$C = 2$
Find $t$ when $y = 0$ : Substitute the value of $C$ back into the general solution to get the particular solution for $y(t)$ . $\newline$ $y(t) = t^2 + 3t + 2$
Factor and Solve Quadratic Equation: Now we need to find the value of $t$ when $y = 0$ . $\newline$ $0 = t^2 + 3t + 2$ $\newline$ We have a quadratic equation to solve: $t^2 + 3t + 2 = 0$ .
Identify Values of $t$ : Factor the quadratic equation. $(t + 1)(t + 2) = 0$
Identify Values of $t$ : Factor the quadratic equation. $(t + 1)(t + 2) = 0$ Set each factor equal to zero and solve for $t$ . $t + 1 = 0 \Rightarrow t = -1$ $t + 2 = 0 \Rightarrow t = -2$
Identify Values of t: Factor the quadratic equation. $\newline$ $(t + 1)(t + 2) = 0$ Set each factor equal to zero and solve for t. $\newline$ $t + 1 = 0 \Rightarrow t = -1$ $\newline$ $t + 2 = 0 \Rightarrow t = -2$ We have found two values of t when $y = 0$ : $t = -1$ and $t = -2$ . These correspond to options B and E.