Scientific Notation 4-9
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
Tuesday, April 12, 2011
Algebra (Period 1)
Introduction to Quadratic Equations 13-1
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
Math 6 Honors (Period 6 and 7)
Problem Solving: Using Proportion 7-8
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Mr. Kaiser bought 4 tires for his car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
tires ___________ = _____________tires
cost _____________=_____________cost
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Mr. Kaiser bought 4 tires for his car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
tires ___________ = _____________tires
cost _____________=_____________cost
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
Math 6 Honors (Period 6 and 7)
Ratio 7-6 Word Problems
Since all of the word problems from Ratios 7-6 are excellent examples of using Ratios, I thought I would create a blog posting with the solutions. We did all of the even numbered ones in class and went over the answers to the odd numbered ones—as they were assigned for homework. Although I will list the actually word problems, you might need to turn to the textbook and page 229 for any charts included.
Read each problem carefully and set up the solutions based on the question asked
Page 229
1. What is the cost of grapes in dollars per kilogram if 4.5 kg of grapes costs $ 7.56?
Since they want $/kg you need to have 7.56/ 4.5 = 1.68; so the grapes cost $ 1.68/kg.
2. This one looked confusing, but just read it a few times: The index of refraction of a transparent substance is the ratio of the speed of light in space to the speed of light in the substance. (Read it again—it is just a ratio). Using the table on Page 229, find the index of refraction of
A. Glass
B. Water
According to the table Speed of light in space is 300,000 km/sec
Speed of light in glass is 200,000 km/ sec.
So
300,000
200,000
or 3/2
Speed of Light in Water is 225,000
so that ratio is
300,000
225,000
This takes a little more time dividing but you can simplify it to 4/3 Since it is a ratio you want to LEAVE it as an improper fraction.
3. The mechanical advantage of a simple machine is the ratio of the weight lifted by the machine to the force necessary to lift it. Now—just read that again—you don’t need to truly understand physics to get this problem. It is a RATIO again!! What is the mechanical advantage of a jack that lifts a 3200-pound car with a force of 120 pounds?
3200 =
120
80/3; so the mechanical advantage is 80/3
4. The C-String of a cello vibrates 654 times in 5 seconds, How many vibrations per second is that? (We are finding unit rate)
It’s a rate of time/sec.
654/ 5 = 130 4/5 vibrations per second
5. A four cubic foot volume of water at sea level weighs 250 lb. What is the density of water in pounds per cubic foot? (Do you need to understand density to do this problem?) No—just unit rates!! We want lbs/cubic foot. Look at the information and place it with lbs/ft and then divide to get the unit rate.
250lb/ 4 ft3. That equals 62 1/2 lb/ft3 .
6. A share of stock that costs $88 earned $16 last year. What was the price-to-earnings ratio of this stock? (Wish that was happening now!!)
Price/earnings so look carefully 88/16 = 11/2 so the Ratio is 11/2
In exercises 7 and 8 you will need to look at the diagrams on page 230. You are finding the ratio in lowest terms.
Look at both the small and large triangles. For A you are finding the ratio between segment AB of the larger triangle and DE of the smaller triangle.
4.8
3.2
Simplify ( I like to clear decimals to make it easier but in this case you could easily divide
= 3/2
B.
Perimeter of Triangle ABC
Perimeter of Triangle DEF
First you need to find the perimeter of the larger triangle by adding up the three sides
6.3 + 4.8 + 3 = 14.1
and then finding the perimeter of the smaller triangle by adding up those three sides
4.2 + 3.2 + 2 = 9.4
14.1
9.4
I like to clear decimals by multiplying by 10
141
94
= 3/2
What did you notice?
8. Again look at Page 230 for the figure
We are finding the following Ratios:
PQ:TU 2.6:6.5 or 26:65 = 2:5
QR:UV 1.2:3 = 12:30 = 2:5
Perimeter of PQRS: Perimeter of TUVW
You need to add up all the sides of each of the figures
2.2 + 2.6 + 1.2 + 3 = 9
and 5.5 + 6.5 + 3 + 7.5 = 22.5
9:22.5 = 90:225 = 2:5
What did you notice here?
For exercised 9-12 you need to use the Table on Page 230 as well
9. The population of Centerville in 1980 to its population in 1970?
44/36 = 11/9
10. The growth in the population of Easton to its 1980 population.
You need to subtract the 1970 population from the 1980 population to get the growth so
28-16 = 12 12/18 = 3/7
11. The total population of both towns in 1970 to their total population in 1980
36 + 16 = 52 and 44 + 28 = 72 52/72 = 13/ 18
12. The total growth in the population of both towns to their total 1980 population
Like number 10 you need to subtract to find the growth
(44-36) + (28-16) = 8 + 12 = 20
44 + 28 = 72 (from before)
20/72 = 5/18
13. During a season a baseball player hit safely in 135times at bat; the player struck out or was fielded out in 340 times at bat. What is the player’s ratio of hits to times at bat?
You take the 135 + 340 = 475 (the total times at bat)
135/475 = 37/95
14. A fruit drink recipe requires fruit juice and milk in the ration 3:5. What fraction of the drink is milk? What fraction is juice? Remember we need to find the total first. 3 + 5 = 8
so the fraction that is milk is 5/8 and the fraction that is fruit juice is 3/8
Since all of the word problems from Ratios 7-6 are excellent examples of using Ratios, I thought I would create a blog posting with the solutions. We did all of the even numbered ones in class and went over the answers to the odd numbered ones—as they were assigned for homework. Although I will list the actually word problems, you might need to turn to the textbook and page 229 for any charts included.
Read each problem carefully and set up the solutions based on the question asked
Page 229
1. What is the cost of grapes in dollars per kilogram if 4.5 kg of grapes costs $ 7.56?
Since they want $/kg you need to have 7.56/ 4.5 = 1.68; so the grapes cost $ 1.68/kg.
2. This one looked confusing, but just read it a few times: The index of refraction of a transparent substance is the ratio of the speed of light in space to the speed of light in the substance. (Read it again—it is just a ratio). Using the table on Page 229, find the index of refraction of
A. Glass
B. Water
According to the table Speed of light in space is 300,000 km/sec
Speed of light in glass is 200,000 km/ sec.
So
300,000
200,000
or 3/2
Speed of Light in Water is 225,000
so that ratio is
300,000
225,000
This takes a little more time dividing but you can simplify it to 4/3 Since it is a ratio you want to LEAVE it as an improper fraction.
3. The mechanical advantage of a simple machine is the ratio of the weight lifted by the machine to the force necessary to lift it. Now—just read that again—you don’t need to truly understand physics to get this problem. It is a RATIO again!! What is the mechanical advantage of a jack that lifts a 3200-pound car with a force of 120 pounds?
3200 =
120
80/3; so the mechanical advantage is 80/3
4. The C-String of a cello vibrates 654 times in 5 seconds, How many vibrations per second is that? (We are finding unit rate)
It’s a rate of time/sec.
654/ 5 = 130 4/5 vibrations per second
5. A four cubic foot volume of water at sea level weighs 250 lb. What is the density of water in pounds per cubic foot? (Do you need to understand density to do this problem?) No—just unit rates!! We want lbs/cubic foot. Look at the information and place it with lbs/ft and then divide to get the unit rate.
250lb/ 4 ft3. That equals 62 1/2 lb/ft3 .
6. A share of stock that costs $88 earned $16 last year. What was the price-to-earnings ratio of this stock? (Wish that was happening now!!)
Price/earnings so look carefully 88/16 = 11/2 so the Ratio is 11/2
In exercises 7 and 8 you will need to look at the diagrams on page 230. You are finding the ratio in lowest terms.
Look at both the small and large triangles. For A you are finding the ratio between segment AB of the larger triangle and DE of the smaller triangle.
4.8
3.2
Simplify ( I like to clear decimals to make it easier but in this case you could easily divide
= 3/2
B.
Perimeter of Triangle ABC
Perimeter of Triangle DEF
First you need to find the perimeter of the larger triangle by adding up the three sides
6.3 + 4.8 + 3 = 14.1
and then finding the perimeter of the smaller triangle by adding up those three sides
4.2 + 3.2 + 2 = 9.4
14.1
9.4
I like to clear decimals by multiplying by 10
141
94
= 3/2
What did you notice?
8. Again look at Page 230 for the figure
We are finding the following Ratios:
PQ:TU 2.6:6.5 or 26:65 = 2:5
QR:UV 1.2:3 = 12:30 = 2:5
Perimeter of PQRS: Perimeter of TUVW
You need to add up all the sides of each of the figures
2.2 + 2.6 + 1.2 + 3 = 9
and 5.5 + 6.5 + 3 + 7.5 = 22.5
9:22.5 = 90:225 = 2:5
What did you notice here?
For exercised 9-12 you need to use the Table on Page 230 as well
9. The population of Centerville in 1980 to its population in 1970?
44/36 = 11/9
10. The growth in the population of Easton to its 1980 population.
You need to subtract the 1970 population from the 1980 population to get the growth so
28-16 = 12 12/18 = 3/7
11. The total population of both towns in 1970 to their total population in 1980
36 + 16 = 52 and 44 + 28 = 72 52/72 = 13/ 18
12. The total growth in the population of both towns to their total 1980 population
Like number 10 you need to subtract to find the growth
(44-36) + (28-16) = 8 + 12 = 20
44 + 28 = 72 (from before)
20/72 = 5/18
13. During a season a baseball player hit safely in 135times at bat; the player struck out or was fielded out in 340 times at bat. What is the player’s ratio of hits to times at bat?
You take the 135 + 340 = 475 (the total times at bat)
135/475 = 37/95
14. A fruit drink recipe requires fruit juice and milk in the ration 3:5. What fraction of the drink is milk? What fraction is juice? Remember we need to find the total first. 3 + 5 = 8
so the fraction that is milk is 5/8 and the fraction that is fruit juice is 3/8
Monday, April 11, 2011
Math 6 Honors (Period 6 and 7)
Proportions 7-7
WE used fictional numbers for the middle schools in our district
AEWright 6th grade 160 students and 10 teachers
ACStelle 6th grade 144 students and 9 teachers
We looked at the teacher to student ratios of both schools and found
10/160 = 1/16
and
9/144 = 1/16
so 10/160 = 9/144
An equation that states two ratios are equal is called a proportion/10, 160, 9, and 144 are called the terms of the proportion.
Let's say LCMS has 192 students. How many teachers will be needed if the teacher- strudent ratio needs to be the same.
Let t = the number of teachers
10/160 = t/192
Using cross multiplying we discovered that
160t = 192 (10)
160t = 192 (10)
160 160
t = 192/16
or t = 12
12 teachers
we found that when
a/b = c/d and b is not equal to 0 and d is not equal to zero
then
ad = bc
3/8 = 12/n
3n = 8 (12)
3n = 96
3n/3 = 96/3
n = 32
what about 4/3 = n/7
3n = 4(7)
3n = 28
3n/3 = 28/3
n = 9 1/3
3/75 = 2/n
n = 50
3/m = m/27
m2 = 3 (27)
m2 = 81
so m = 9
49/n = n/4
n2 = 4(49)
n2 = 196
n = 14
WE used fictional numbers for the middle schools in our district
AEWright 6th grade 160 students and 10 teachers
ACStelle 6th grade 144 students and 9 teachers
We looked at the teacher to student ratios of both schools and found
10/160 = 1/16
and
9/144 = 1/16
so 10/160 = 9/144
An equation that states two ratios are equal is called a proportion/10, 160, 9, and 144 are called the terms of the proportion.
Let's say LCMS has 192 students. How many teachers will be needed if the teacher- strudent ratio needs to be the same.
Let t = the number of teachers
10/160 = t/192
Using cross multiplying we discovered that
160t = 192 (10)
160t = 192 (10)
160 160
t = 192/16
or t = 12
12 teachers
we found that when
a/b = c/d and b is not equal to 0 and d is not equal to zero
then
ad = bc
3/8 = 12/n
3n = 8 (12)
3n = 96
3n/3 = 96/3
n = 32
what about 4/3 = n/7
3n = 4(7)
3n = 28
3n/3 = 28/3
n = 9 1/3
3/75 = 2/n
n = 50
3/m = m/27
m2 = 3 (27)
m2 = 81
so m = 9
49/n = n/4
n2 = 4(49)
n2 = 196
n = 14
Pre Algebra (Period 2 & 4)
Scientific Notation 4-9
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
Subscribe to:
Posts (Atom)
