Multiple Choice
Solve the problem.
-Suppose that a hair dryer operates on a voltage that can be represented by the relation and that it draws a current represented by the relation , where is time measured in seconds. The power consumed by the appliance is VI. Graph the power in by and use an identity to write the expression for the power in the form , where , and are constants.
A)
by ![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 ]](https://d2lvgg3v3hfg70.cloudfront.net/TB4654/11ecc6ec_2c3e_3aeb_a77c_578a4261f302_TB4654_11.jpg)
B)
by ![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 ]](https://d2lvgg3v3hfg70.cloudfront.net/TB4654/11ecc6ec_32c8_bacc_a77c_b3d93d5fa51a_TB4654_11.jpg)
C)
by ![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 ]](https://d2lvgg3v3hfg70.cloudfront.net/TB4654/11ecc6ec_1d43_1609_a77c_c7e75d12f00d_TB4654_11.jpg)
D)
by
Correct Answer:

Verified
Correct Answer:
Verified
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