Solve the problem.
-Suppose that a hair dryer operates on a voltage that can be represented by the relation

\mathrm { V } ( \mathrm { t } ) = 160 \cos ( 120 \pi \mathrm { t } )

and that it draws a current represented by the relation

\mathrm { I } ( \mathrm { t } ) = 7 \cos ( 120 \pi \mathrm { t } )

, where

t

is time measured in seconds. The power consumed by the appliance is

P =

VI. Graph the power in

[ 0,0.04,0.01 ]

[ - 200,2000,200 ]

and use an identity to write the expression for the power in the form

\mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } ) + \mathrm { d }

, where

\mathrm { a } , \mathrm { k }

, and

\mathrm { d }

are constants.
A)

P = 560 \cos ( 240 \pi t ) + 560

[ 0,0.04,0.01 ]

[ - 200,2000,200 ]

$Solve the problem. -Suppose that a hair dryer operates on a voltage that can be represented by the relation \mathrm { V } ( \mathrm { t } ) = 160 \cos ( 120 \pi \mathrm { t } ) and that it draws a current represented by the relation \mathrm { I } ( \mathrm { t } ) = 7 \cos ( 120 \pi \mathrm { t } ) , where t is time measured in seconds. The power consumed by the appliance is P = VI. Graph the power in [ 0,0.04,0.01 ] by [ - 200,2000,200 ] and use an identity to write the expression for the power in the form \mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } ) + \mathrm { d } , where \mathrm { a } , \mathrm { k } , and \mathrm { d } are constants. A) P = 560 \cos ( 240 \pi t ) + 560 [ 0,0.04,0.01 ] by [ - 200,2000,200 ] B) P = 1120 \cos ( 240 \pi t ) + 1120 [ 0,0.04,0.01 ] by [ - 200,2000,200 ] C) \mathrm { P } = - 560 \cos ( 240 \pi \mathrm { t } ) + 560 [ 0,0.04,0.01 ] by [ - 200,2000,200 ] D) P = 560 \cos ( 120 \pi t ) + 560 [ 0,0.04,0.01 ] by [ - 200,2000,200 ]$

B)