The power generated by an electrical circuit (in watts) as a function of its current c (in amperes) is modeled by: P(c)=-20(c-3)^(2)+180 Which currents will produce no power (i.e. 0 watts)? Enter the lower current first. Lower current: amperes Higher current: amperes

Given power function: We are given the power function P(c) = -20(c - 3)^2 + 180. To find the currents that produce no power, we need to solve the equation P(c) = 0.Set equal and solve: Set the power function equal to zero and solve for c: 0 = -20(c - 3)^2 + 180Move 180 and solve: Move 180 to the other side of the equation: -20(c - 3)^2 = -180Divide and isolate: Divide both sides by -20 to isolate the squared term: (c - 3)^2 = 9Take square root: Take the square root of both sides to solve for c: c - 3 = ±3Solve for c: Solve for c by adding 3 to both possible values of the square root: c = 3 ± 3Two solutions for c: This gives us two solutions for c: Lower current: c = 3 - 3 = 0 amperes Higher current: c = 3 + 3 = 6 amperes

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The power generated by an electrical circuit (in watts) as a function of its current
c (in amperes) is modeled by:

P(c)=-20(c-3)^(2)+180
Which currents will produce no power (i.e. 0 watts)?
Enter the lower current first.
Lower current: amperes
Higher current: amperes

The power generated by an electrical circuit (in watts) as a function of its current $c$ (in amperes) is modeled by: $\newline$ $P(c)=-20(c-3)^{2}+180$ $\newline$ Which currents will produce no power (i.e. $0$ watts)? $\newline$ Enter the lower current first. $\newline$ Lower current: $\text{amperes}$ $\newline$ Higher current: $\text{amperes}$

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Q. The power generated by an electrical circuit (in watts) as a function of its current

c

(in amperes) is modeled by:

\newline

P(c)=-20(c-3)^{2}+180

\newline

Which currents will produce no power (i.e.

0

watts)?

\newline

Enter the lower current first.

\newline

Lower current:

\text{amperes}

\newline

Higher current:

\text{amperes}

Given power function: We are given the power function $P(c) = -20(c - 3)^2 + 180$ . To find the currents that produce no power, we need to solve the equation $P(c) = 0$ .
Set equal and solve: Set the power function equal to zero and solve for $c$ : $0 = -20(c - 3)^2 + 180$
Move $180$ and solve: Move $180$ to the other side of the equation: $\newline$ $-20(c - 3)^2 = -180$
Divide and isolate: Divide both sides by $-20$ to isolate the squared term: $\newline$ $(c - 3)^2 = 9$
Take square root: Take the square root of both sides to solve for $c$ : $c - 3 = \pm 3$
Solve for c: Solve for c by adding $3$ to both possible values of the square root: $\newline$ $c = 3 \pm 3$
Two solutions for c: This gives us two solutions for c: $\newline$ Lower current: $c = 3 - 3 = 0$ amperes $\newline$ Higher current: $c = 3 + 3 = 6$ amperes