Question

The use of electricity in the US in 1902 was 6.03 x 10^9 kilowatt-hours.The use of electricity in the US in 1950 was 4.3 x 10^11 kilowatt-hours.What was the total kilowatt-hours of electricity used during these two years?(1 point)Responses(6.03×109)+(4.3×1011)=0.063×1011+(4.3×1011)=4.3603×1011open paren 6 point 0 3 times 10 to the 9th power close paren plus open paren 4 point 3 times 10 to the 11th power close paren is equal to 0 point 0 6 3 times 10 to the 11th power plus open paren 4 point 3 times 10 to the 11th power close paren is equal to 4 point 3 6 0 3 times 10 to the 11th power(6.03×109)×(4.3×1011)=25.929×102=2.6×101open paren 6 point 0 3 times 10 to the 9th power close paren times open paren 4 point 3 times 10 to the 11th power close paren is equal to 25 point 9 2 9 times 10 squared is equal to 2 point 6 times 10 to the 1th power(6.03×109)×(4.3×1011)=25.929×1020=2.6×1019open paren 6 point 0 3 times 10 to the 9th power close paren times open paren 4 point 3 times 10 to the 11th power close paren is equal to 25 point 9 2 9 times 10 to the 20th power is equal to 2 point 6 times 10 to the 19th power(6.03×109)+(4.3×1011)=10.33×1020=1.033×1019

Answer 1

To calculate the total kilowatt-hours of electricity used during the years 1902 and 1950, we need to add the two values given:

1. Use in 1902: \( 6.03 \times 10^9 \) kilowatt-hours
2. Use in 1950: \( 4.3 \times 10^{11} \) kilowatt-hours

First, we can adjust \( 6.03 \times 10^9 \) to match the exponent of \( 10^{11} \):

\[
6.03 \times 10^9 = 0.0603 \times 10^{11}
\]

Now, we can add:

\[
0.0603 \times 10^{11} + 4.3 \times 10^{11} = (0.0603 + 4.3) \times 10^{11} = 4.3603 \times 10^{11}
\]

Thus, the total kilowatt-hours of electricity used during these two years is:

\[
\boxed{4.3603 \times 10^{11}} \text{ kilowatt-hours}
\]