<h3>Given</h3>
P = 3r + 2s
<h3>Find</h3>
the corresponding equation for s
<h3>Solution</h3>
First of all, look at how this is evaluated in terms of what happens to a value for s.
- s is multiplied by 2
- 3r is added to that product
To solve for s, you undo these operations in reverse order. The "undo" for addition is adding the opposite. The "undo" for multiplication is division (or multiplication by the reciprocal).
... P = 3r + 2s . . . . . . starting equation
... P - 3r = 2s . . . . . . -3r is added to both sides to undo addition of 3r
... (P -3r)/2 = s . . . . . both sides are divided by 2 to undo the multiplication
Note that the division is of everything on both sides of the equation. That is why we need to add parentheses around the expression that was on the left—so the whole thing gets divided by 2.
Your solution is ...
... s = (P - 3r)/2
Answer:
By using Carl Gauss's clever formula, (n / 2)(first number + last number) = sum, where n is the number of integers, we learned how to add consecutive numbers quickly. We now know that the sum of the pairs in consecutive numbers starting with the first and last numbers is equal.
<u>Yes the first one is correct because n squared is n with the 2 above it and adding it to three gets your first answer and then the three on the outside means to multiply so your first response is correct.</u>
A. How many kilowatt hours of electricity did the Smiths use during February?
Kilowatt hours of electricity the Smiths used during February:
Meter read on March 1 - meter read on February 1 =
20,288 kilowatt hours - 19,423 kilowatt hours =
865 kilowatt hours
Answer: The Smiths used 865 kilowatt hours of electricity during February
b. How many kilowatt hours did they use during March?
Kilowatt hours of electricity the Smiths used during March:
Meter read on April 1 - meter read on March 1 =
21,163 kilowatt hours - 20,288 kilowatt hours =
875 kilowatt hours
Answer: The Smiths used 875 kilowatt hours of electricity during March
