Notice All Classes:Test 5 on Thursday December 10 Test topics will include but not limited to the following: Multiplying and dividing monomials Mulitplying and dividing polynomials by monomials Scientific Notation conversion and operations Simplifying or evaluating algebraic expressions Solving equations with variables on single or both sides, evaluating expressions, + others ME12/92Homework: Handout, complete both sides.
Thursday, November 19, 2009 QUARTERLY TEST NOTICE NOTICE:Quarterly test will be administered next Tuesday, November 24. Topics to be included but not limited to are as follows; Motion problems Coin problems Consecutive integer problems Solving and graphing inequalities Solving e quations Ratio Problems Percent problems Percent change problems Properties of numbers Adding and subtracting polynomials
Opposite motion problems: Remember that Distance is = to the rate or speed times the time D = RT in all of the following problems D is replace by the values of RT when known or by one of the values and a variable either R or T.
Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph. In how many hours will the cars be 252 miles apart? Given is R for each car and the total distance travelled by both cars going in opposite directions. Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement: Let x = the number of hours traveled by each car. Both cars are travelling for the same time T. Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance 28mph x X + 35mph x X = 252 miles 28X + 35X = 252 miles. 63X = 252 X= 4 hrs
Lesson Review: Example problem: How far can a man drive out into the country at the average rate of 40 mph and return over the same road at the average rate 30 mph if he travels a total of 7 hours? Need to find T then substitute in RT to find distance in D=RT
Tout + T back = 7 hrs and Distance out is same as the Distance back
Tout + Tback = 7hrs. Let Tout = X X +Tback = 7hrs subtract X from both sides: Tback = 7hrs - X So: T out = X and T back = 7-X
Dout = RT = 40mph x X Dback = RT 30mph x (7-X) If Dout = Dback Then 40X = 30 (7-X) 40X = 210 -30X 40X + 30X =210 70X = 210 X=3 hrs Dout = RT = 40(3) = 120 miles as a check Dout should = Dback Dback = RT = 30 (7-x) = 30(7-3) = 30(4) = 120 miles
Opposite motion problems: Remember that Distance is = to the rate or speed times the time D = RT in all of the following problems D is replace by the values of RT when known or by one of the values and a variable either R or T.
Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph. In how many hours will the cars be 252 miles apart? Given is R for each car and the total distance travelled by both cars going in opposite directions. Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement: Let x = the number of hours traveled by each car. Both cars are travelling for the same time T. Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance 28mph x X + 35mph x X = 252 miles 28X + 35X = 252 miles. 63X = 252 X= 4 hrs
Opposite motion problems: Remember that Distance is = to the rate or speed times the time D = RT in all of the following problems D is replace by the values of RT when known or by one of the values and a variable either R or T.
Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph. In how many hours will the cars be 252 miles apart? Given is R for each car and the total distance travelled by both cars going in opposite directions. Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement: Let x = the number of hours traveled by each car. Both cars are travelling for the same time T. Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance 28mph x X + 35mph x X = 252 miles 28X + 35X = 252 miles. 63X = 252 X= 4 hrs
Homework:Complete handout side 2 above plus, textbook pg. 139/ # 35,and 37 and pg. 132/#'s 4,25,33 and
post test review handout above.
Lesson Review: Remember that the number of coins times the value of each coin in cents, is = total value of the coins in cents.
Example: Bill has 4 times as many quarters as dimes. In all he has $2.20. How many coins of each type does she have?
Solution: Let x= number of dimes (x is always = to the coin in the problems that follows the word "than" or"as")then,4x is = to the number of quarters
The value of a dime is 10 cents and the value of a quarter is 25 cents The total value of all the coins is $2.20 cents which is the same as 220 cents after we multiply $2.20 times 100 cents.
Now we write the equation: Dimes + Quarters = $2.20 x + 4x = $2.20 Now I can't add a dime and a quarter without converting then to cents, so:
1"Consecutive integer problems, all problems both sides
For consecutive only integers Aways start with a Let Statement and increment by 1.
Example Let: 1st # = x 2nd# = x+ 1 3rd #= x + 2 4th # = x + 3 set of numbers Then translate a verbal problem into an algebraic equation and solve
For consecutive odd or even integers, also always start with a Let Statement and increment by 2: Example Let: 1st # = x 2nd # = x+2 3rd # = x+4 4th # = x+6 Then translate a verbal problem into an algebraic equation and solve