Answer:
Sum of zeroes ,
sum of product of zeroes taken two at a time ,
product of zeroes ,
Substituting the values in the equation above , we get
hope helpful~
Step-by-step explanation:
C(500)=65×500+850
=32500+850
=33350
The required answer
Answer: Yes
Step-by-step explanation:
We will assume that each bulb is of 100 kW (kilowatt).
We will calculate how much energy in kilowatt-hours the light bulb will use per year in kilowatts, by the number of hours in a year.
We have, 100 kW = 0.1 kW so the energy consumed in one year is,
0.1 \times 8760=876.0\text {kWh} Since there are 8760 hours in one year. It is given that there are 9 bulbs so we need to have, 876 \times 9 = 7884 \text {kWh}
It is given that 1 ton of coal produces 2460 kWh, so 4 tons of coal will produce, 4\times 2460= 9840 \text {kWh}
We can observe that 4 tons of coal is producing 9840 kWh which is mroe than 7884 kWh. So, yes, 4 tons of coal can produce enough power to light 9 bulbs for a year
4^10 and x^4 are the right answers
Answer:
1,000 different combinations are possible
Step-by-step explanation:
Because if youre lock ckntained 3 single digit numbers it would have to be 0-9 so each digit would have 10 options so there would be 1,000 different combinations that are possible.
