Math 1.1

Gauss's Problem

When Gauss was in early grade school, his teacher gave the class an assignment, intending it to take them a good deal of time. She asked the class to find the sum of numbers from 1 to 100. Gauss answered almost immediately. He took 1+2+3+,...+98+99+100 and flipped the numbers below it from 100+99+98+,...+3+2+1 the numbers added in the column is equal to 101 2s=101(100 because there are 100 terms) 2s=10,100 divide it by two S=5,050

Four-step Problem Solving Process

1. Understand the problem 2. Devise a plan 3. Carryout tge plan 4. Check your work

Making a Diagram

A store sells bicycles and tricycles. There are a total of 15 cycles in the store, and there are 35 wheels on those cycles. How many of each type of cycle are in the store? Draw 15 bicycles then count the number of wheels. =30 wheels so add wheels to the 15 bikes to get 35 wheels altogether. There should be 10 bicycles and 5 tricycles.

Examining a simpler case

If a soccer league has 10 teams and each team plays every other team once, how many games will the league manager need to schedule at the start of the season? Instead of starting with all 10 teams let's try using 3 teams. Team A Team B and Team C. Team A will play team B and Team C. Team B will play Team C. =3 games because AB is the same game as BA. So you'll take 10 teams multiply by 9 games and divide by 2. 10x9=90/2=45 games

Working backwards

One winter night the temperature fell by 15 degrees between midnight and 5am. By 9am the temperatures numerical value had doubled from what it was at 5am. By noon it had risen another 10 degrees to 32 degrees. What was the temperature at midnight. So we know here that the end temperature was 32 degrees we subtract 10 to equal the temp. at noon which is 22. We know that from 9am to 5am the temp. Doubled so we have to divide 22/2 = 11 so from midnight to 5am the temperature fell 15 degrees so we have to add 15+11=26degrees which is what the temperature is at midnight.

Guess and check

We have 48ft of fencing and we wish to make an enclosure along one side of an existing barn. What lengths would the sides need to be to have the greatest area enclosed for our chickens? First draw a barn and a rectangle that connects with the barn. We know that with a rectangle 2 pairs of congruent sides. Guess: 10,10,28 check 10x28=280 5,5,38 check 5x38=190 etc the largest number is the greatest area for the chicken coop. (Answer is 12,12,24 12x24=288ft squared)