The power utilised by frank in the month of march is 470 kilowatts - per hour.
The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.
Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.
Given that Franks's electric bill for the month of March was $85.78. The electric company charged a flat monthly fee of $20.00 for service plus $0.14 per kilowatt-hour of electricity used.
The equation will be written as,
B = 20 + 0.14K
85.78 = 20 + 0.14k
k = ( 80.78 - 20 ) / 0.14
K = 65.78 / 0.14
K = 470 Kilowatt-hour
Therefore, the power utilised by frank in the month of march is 470 kilowatts - per hour.
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Answer:
I got 469.8 kilowatt-hours. I got this by taking the total of Frank's bill, which was $85.78, and subtracting the flat monthly fee of $20.00. I did this because I need to find out the number of kilowatt-hours Frank used. Then, I divided $65.78 by $0.14 since that is the price per kilowatt-hour and got about 469.8 kilowatt-hours used by Frank.
Answer:
Step-by-step explanation:
A. <
B. =
C. >
2. Compare the fraction.
4/7 (?) 1/6
A. >
B. =
C. <
D. cannot be compared
Answer: 120
Step-by-step explanation: 3/10=0.3
0.3*400=120
Answer:
41) 10
42) 65
43) 26
44) 33
45)65
41.)10
42.) 65
43.)26
44.)33
45.)65
hoped that helped:)
A. 10a^4b^3
B. 3a^4b^4
C. 10a^4b^4
D. 3a^4b^3