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#### EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS
Note:
• In order to solve for x, you must isolate x.
• In order to isolate x, you must remove it from under the radial.
• If there are two radicals in the equation,isolate one of the radicals.
• Then raise both sides of the equation to a power equal to the index of the isolated radical.
• Raise both sides of the equation to a power equal to the index of the isolated radical.
• You should now have a polynomial equation. Solve it.
• Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions. Therefore, you must check your answers.
Example 5:
First make a note of the fact that you cannot take the square root of a negative number. Therefore,the term is valid only if and the term is valid if . The restricted domain must satisfy both of these constraints. Therefore, the domain is the set of real numbers
Isolate the term by adding 3 to both sides of the equation.
Square both sides of the equation.
Isolate the term.
Square both sides of the equation.
Solve for x using the quadratic formula.
There are two approximate answers, 8163.162 and 18.8385.
Check the solution 8163.162 by substituting 8163.162 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left side:
• Right Side:
Since the left side of the original equation equals the right side of the original equation after we substituted our solution for x, we have verified that the solutions is x=8163.162.
Check the solution 18.8385 by substituting 18.8385 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left side:
• Right Side:
Since the left side of the original equation does not equal the right side of the original equation after we substituted our solution for x, the solution 18.8385 is not a valid solution.
You can also check the answer by graphing the equation:
.
The graph represents the right side of the original equation minus the left side of the original equation.. The x-intercept(s) of this graph is(are) the solution(s). Since the x-intercept is 8163.162, we have verified the solution.
If you would like to test yourself by working some problems similar to this example, click on Problem.
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### High School Mathematics - 29.9 The Point Slope Form of a Line
The equation of a line passing through a point P(x1 , y1) having a slope m is (y - y1) = m(x - x1). This form of a line is called the Point-Slope form. Example: Find the equation of a line passing through (3,5) having a slope 3. Solution: Given that slope is 3 i.e., m = 3 A point on the line is (3,5) Using the point slope form equation, i.e., (y - y1) = m(x - x1). y - 5 = 3(x - 3) y - 5 = 3x - 9 y = 3x - 9 + 5 y = 3x - 4 Therefore, y = 3x -4 is the required equation of the line. Directions: Solve the following questions. Also write at least 10 examples of your own.
Q 1: Find the equation of a line passing through (5,7) having a slope 4.y = 4x + 8y = 4x + 7y = 4x + 13y = 4x - 13 Q 2: Find the equation of a line passing through (5,8) having a slope 4.y = 4x + 12y = 4x + 6y = 4x - 12y = 4x - 6 Q 3: Find the equation of a line passing through (7,8) having a slope 3.y = 3x + 13y = 3x - 8y = 3x + 5y = 3x - 13 Q 4: Find the equation of a line passing through (3,8) having a slope 6.y = 6x + 8y = 6x - 10y = 6x + 10y = 6x - 7 Q 5: Find the equation of a line passing through (3,9) having a slope 3.y = 3xy = 3x + 9y = 3x - 9 Q 6: Find the equation of a line passing through (3,7) having a slope 2.y = 2x + 3y = 2x - 1y = 2x + 1 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! |
EMT 668 Assignment 3
Some Different Ways to Examine
James W. Wilson and Luke Rapley University of Georgia
It has now become a rather standard exercise, with the available technology, to construct graphs to consider the equation
and to overlay several graphs of
for different values of a, b, or c as the other two are held constant. From these graphs discussion of the patterns
for the roots of
can be followed. For example, if we set
for b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following picture is obtained.
We can now discuss the "movement" of a parabola as b is varied while keeping both a and c constant.
1. All these parabola's have one point in common, (0,1) (the point where each parabola crosses the y-axis).
2. For b<-2, the parabola intersects the x-axis in two points with positive x values (i.e. the original equation will have two real roots, both positive).
3. For b=-2, the parabola is tangent to the x-axis and so the original equation has one real(positive) root at the point of tangency.
4. For -2<b<2, the parabola does not intersect the x-axis -- the original equation has no real roots.
5. For b=2, the parabola is tangent to the x-axis and so the original equation has one real(negative) root at the point of tangency.
6. For b>2, the parabola intersects the x-axis in two points with negative x values (i.e. the original equation will have two real roots, both negative).
Let us now take a different approach to investigating the equation
Consider a version of the equation above
We will now graph this equation in the xb-plane. We get the following graph.
If we take any particular value of b, say b = 3, and overlay this equation on the graph above we add a line parallel to the
x-axis. If this line intersects the curve in the xb plane, the intersection points correspond to the root(s) of the original
equation for that value of b. The graph with b = 3 overlayed looks like the following.
For each value of b we select, we get a horizontal line. It is clear from this graph that we get two negative real roots of the original equation when b > 2, one negative real root when b = 2, no real roots for -2 < b < 2, one positive real root when b=-2, and two positive real roots when b < -2.
Viewing the quadratic in the xb-plane allows one to see more clearly that there are no real roots on the interval -2<b<2. The graph below illustrates this point by overlaying the equations b=1 and b=-1 onto the xb-plane which already shows the graph of the quadratic.
From this graph one can see that since the graphs of b=-1 and b=1 do not cross the graph of the quadratic that there are no real roots associated with the quadratic with b=-1 and b=1.
Now consider the case when c = - 1 rather than c = 1.
Overlaying graphs of this equation in the xy-plane for b={-3, -2, -1, 0, 1, 2, 3} gives you the following.
Now graphing the quadratic in the xb-plane yields
In contrast to the graph in the xy-plane, this graph makes it easy to see that for any value of b you get two real roots, one positive and one negative.
We will now move on to the case when c is varied and a and b are held constant. We will use the following equation in this example:
If the equation is graphed in the xc-plane,
it is easy to see that the curve will be a parabola. For each value of c considered, its graph will be a line crossing the parabola in 0, 1, or 2 points -- the intersections being at the roots of the orignal equation at that value of c. In the graph below, the graph of c = 1 is overlayed on the previous graph.
From the two previous graphs one can see, there is one value of c where the equation will have only 1 real root -- at c=6.25. Also, for c > 6.25 the equation will have no real roots and for c < 6.25 the equation will have two roots. Both roots will be negative for 0 < c < 6.25, one root will be negative and one 0 when c = 0 and one root will be negative and one positive when c < 0.
This equation,, will have two negative roots -- approximately -0.2 and -4.8.
The graphs in the xa-plane are nearly as easy to read as those in the xb and xc. We have not explored the xa-plane here; however, we leave it to the reader to do this exploration if his/her curiousity is running wild.
We have just demostrated some different ways of looking at the quadratic
for different values of a, b and c and determining what type(s) of roots the equation will yield. Viewing the quadratic in the xb or xc-plane, may make determining roots easier for some people. If you are comfortable with determining roots using the xy-plane that is great but you should still understand these other methods and vice-versa. It is worth understanding how many different methods work even if you prefer one over the other(s). |
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Finding Angle Measures
Grade 8 Math Worksheets
The straight lines OA and OB meet at the point ‘O’, which is called the vertex of the angle AOB. OA and OB are the arms of the angle AOB. The amount of turning from OA to OB is called the measure of ÐAOB. The measure of ÐAOB does not depend upon the length of arms OA, OB of the angle.
If the straight line segment OA rotates about ‘O’ through a complete rotation then it will come back to its own position. This complete rotation is divided into 360 equal parts. Each part is called a degree and it is denoted by ‘°’. The measure of half the rotation is 180°.
Right Angle
One fourth the rotation will have a measure of 90°, which is called a ‘right angle’.
Acute Angle
An angle with a measure less than 90° is called an ‘acute’ angle.
Obtuse Angle
An angle with its measure lying between 90° and 180° is called an ‘obtuse’ angle.
Reflex Angle
An angle with its measure lying between 180° and 360° is called a ‘reflex’ angle.
Adjacent Angles
Two angles which have a common vertex and have a common arm are called ‘adjacent’ angles.
∠ AOB and ∠ BOC are adjacent angles with OB as the common arm.
Example: If AOB = 90°, find the value of AOE, EOD and DOB.
∠AOE + EOD + DOB = 90° (Given AOB = 90°)
(2x – 3) + (2x + 4) + (x + 9) = 90°
5x + 10 = 90
5x = 80 => x = 16
So, AOE = 2x – 3 = (2 × 16) – 3 = 32 – 3 = 29°
∠EOD = 2x + 4 = (2 × 16) + 4 = 32 + 4 = 36°
∠DOB = x + 9 = 16 + 9 = 25°
Example: If (2x + 17)° and (x + 4)° are complementary, what is the value of x?
(2x + 17) + (x + 4) = 90° (Given as complementary angles)
3x + 21 = 90
3x = 69 implies x = 23
Check Point
1. Find the value of YOZ, if XOY is a straight line.
2. If (5a + 42)° and (22 – a)° are supplementary angles, find a.
3. Find the measure of DOC.
4. AB and CD intersect at O. If AOD = 146°, find BOD.
5. Find x in the given figure.
1. 118°
2. 29
3. 110°
4. 34°
5. 45°
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# How do you write an equation in point slope form that passes through Point: (1, –7); Slope: -2/3?
Jun 3, 2018
$y + 7 = - \frac{2}{3} \left(x - 1\right)$
#### Explanation:
Given a point:
$\left({x}_{1} , {y}_{1}\right) = \left(1 , - 7\right)$
and a slope:
$m = - \frac{2}{3}$
Point slope form is:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 7\right) = - \frac{2}{3} \left(x - 1\right)$
$\therefore y + 7 = - \frac{2}{3} \left(x - 1\right)$
graph{y+7=-2/3(x-1) [-14.96, 5.04, -8.44, 1.56]}
Jun 3, 2018
See a solution process below:
#### Explanation:
The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$
Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.
Substituting the information from the problem gives:
$\left(y - \textcolor{b l u e}{- 7}\right) = \textcolor{red}{- \frac{2}{3}} \left(x - \textcolor{b l u e}{1}\right)$
$\left(y + \textcolor{b l u e}{7}\right) = \textcolor{red}{- \frac{2}{3}} \left(x - \textcolor{b l u e}{1}\right)$ |
## COMBINATION OF SOLID SHAPES PROBLEMS
Problem 1 :
The radius of a solid sphere us 24 cm. It is melted and drawn into a long wire of uniform cross section. Find the length of the wire if its radius is 1.2 mm.
Solution :
Radius of the solid sphere = 24 cm
Radius of the wire = 1.2 mm
Now we have to convert the radius of wire from mm to cm.
r = 1.2/10 ==> 0.12 cm
Now we need to find the length of the wire that is height of the wire. after melted the sphere and drawn into a long wire of uniform cross section
Volume of sphere = Volume of cylindrical shaped wire
(4/3) Πr3 = Π r2 h
(4/3) r3 = r2 h
(4/3) (24)3 = (0.12)2 h
h = (4/3) (24)3/(0.12)2
h = (4/3) 24 ⋅ 24 ⋅ 24 ⋅ (1/0.12) ⋅ (1/0.12)
h = 4 ⋅ 8 ⋅ 200 ⋅ 200
h = 1280000 mm
Now we have to convert this from mm to km
h = 1280000/100000
h = 12.8 km
So, the required height height is 12.8 km.
Problem 2 :
A right circular conical vessel whose internal radius is 5 cm and height is 24 cm is full of water. The water is emptied into an empty cylindrical vessel with internal radius 10 cm. Find the height of the water level in the cylindrical vessel.
Solution :
Radius of the right circular cone = 5 cm
Height of right circular cone = 24 cm
Radius of cylindrical tank = 10 cm
Now we have to find the height of the cylindrical tank
Volume of water inside the conical container = Volume of water in the cylindrical tank
(1/3) Π r2 h = Π r2 h
(1/3) 524 = 10² h
(1/3) ⋅ ⋅ 5 ⋅ 24 = 10 ⋅ 10 h
h = (1/3) ⋅ ⋅ ⋅ 24/(10 ⋅ 10)
h = 8/4
h = 2 cm
So, height of water is 2 cm.
Problem 3 :
A solid sphere of diameter 6 cm is dropped into a right circular cylindrical vessel with diameter 12 cm, which is partly filled with water, If the sphere is completely submerged in water, how much does the water level in the cylindrical vessel.
Solution :
Diameter of sphere = 6 cm
Radius of sphere = 3 cm
diameter for cylindrical vessel = 12 cm
Radius of cylindrical vessel = 6 cm
Volume of water level increased in the cylindrical tank
= Volume of sphere dropped into the vessel
Πr2h = (4/3)Πr
r2h = (4/3) r3
6⋅ h = (4/3)(3)2
⋅ ⋅ h = (4/3) ⋅ ⋅ ⋅ 3
h = (4/3) ⋅ 3 ⋅ 3 ⋅ 3 / (6 ⋅ 6)
h = 1 cm
Therefore increased water level is 1 cm.
Problem 4 :
Though a cylindrical pipe of internal radius 7 cm, water flows out at the rate of 5 cm/sec. Calculate the volume of water (in liters) discharged through the pipe in half an hour.
Solution :
Now we need to find volume of water (in liters) discharged through the pipe in half an hour.
For that, let us convert speed into seconds.
half an hour = 30 minutes
30 ⋅ 60 = 1800 seconds.
Volume of water discharged in half an hour
= Area of cross section ⋅ time ⋅ speed
= Π r2 ⋅ ⋅ 1800
= (22/7) ⋅ 72 ⋅ ⋅ 1800
= (22/7) ⋅ ⋅ ⋅ ⋅ 1800
= 22 ⋅ ⋅ ⋅ 1800
= 1386000 cubic cm
But here we have to find volume of water discharged in liters.
For that we have to use the following conversion that is
1000 cm3 = 1 liter
= 1386000/1000
= 1386 liter
So, the quantity of water discharged is 1386 liters.
Problem 5 :
Water in a cylindrical tank of diameter 4 m and height 10 m is released through a cylindrical pipe of diameter 10 cm at the rate of 2.5 km/hr. How much time will it take to empty the half of the tank? Assume that the tank is full of water to begin with.
Solution :
Now we are going to convert every measurements into m
Diameter of the cylindrical tank = 4 m
radius of the cylindrical tank = 2 m
height of the tank = 10 m
Diameter of the cylindrical pipe = 10 cm
radius of the cylindrical pipe = 10/2 = 5 cm
= 5/100 m
speed of water = 2.5 km/hr
1000 m = 1 km
= 2.5 x 1000
Now we are going to convert every measurements into m
Diameter of the cylindrical tank = 4 m
radius of the cylindrical tank = 2 m
height of the tank = 10 m
Diameter of the cylindrical pipe = 10 cm
radius of the cylindrical pipe = 10/2 = 5 cm
= 5/100 m
speed of water = 2.5 km/hr
1000 m = 1 km
= 2.5 x 1000
= 2500 m/hr
Volume of water discharged from the cylindrical pipe
= (1/2) Volume of cylindrical tank
Area of cross section ⋅ time ⋅ speed = (1/2) Πr2 h
Π r2⋅ time ⋅ speed = (1/2) Π r2 h
(5/100)2⋅ time ⋅ 2500 = (1/2) ⋅ 22⋅ 10
(5/100) ⋅ (5/100) ⋅ time ⋅ 2500 = (1/2)⋅ 22⋅10
Time = 2 ⋅ 10 ⋅ (100/5) ⋅ (100/5) ⋅ 1(1/2500)
Time = 80/25
= 3.2 hour
1 hour = 60 minute
= 192 minute
= 180+12
3 hour and 12 minute
So, time taken to empty the half of the tank is 3 hour and 12 minute.
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# MAT 112 Integers and Modern Applications for the Uninitiated
## Section2.5The Loop repeat_until
Loops allow us to repeat sequences of instructions. In a repeat_until -loop a sequence of instructions is repeatedly followed until a specified statement is true.
A repeat_until -loop starts with a repeat instruction and ends with a until instruction. The instructions between repeat and until are followed (in order) until the statement after until is true. If you follow the algorithm and get to the instruction until and the statement after until is false, you jump back to repeat and execute the first instruction after repeat.
Every time we follow the instructions between repeat and _until we call an iteration of the ( repeat_until )-loop. We call the first time we follow these instructions the first iteration and so on.
Our first algorithm with a repeat_until -loop subtracts $$2$$ from a given natural number $$n$$ until we get number that is less than $$2\text{.}$$ When the number is even the output is $$0$$ , when the number the output is $$1\text{.}$$
We now use Algorithm 2.29 to determine whether $$5$$ is even or not.
### Example2.30.Algorithm 2.29 with input $$n=5$$.
We follow the instructions of Algorithm 2.29 for the input $$n=5\text{.}$$
Input: $$n=5$$
1. repeat: A repeat_until -loop starts here
1. let $$n:=n-2$$ : We have $$n=5\text{.}$$ We compute $$n-2=5-2=3$$ and assign this value to $$n\text{.}$$ So now $$n=3\text{.}$$
2. until $$n\lt 2\text{:}$$ We have $$n=3\text{.}$$ Since the statement $$n\lt 2$$ is false we repeat the instructions in the loop.
1. repeat: We continue in the repeat_until -loop.
1. let $$n:=n-2$$ : We have $$n=3\text{.}$$ We compute $$n-2=3-2=1$$ and assign this value to $$n\text{.}$$ So now $$n=1\text{.}$$
2. until $$n\lt 2\text{:}$$ We have $$n=1\text{.}$$ Since the statement $$n\lt 2$$ is true we leave the loop and continue with Item 3.
3. return $$n\text{:}$$ We return the value of $$n$$ which is 1.
Output: $$n=1$$
In our next example, given a natural number $$n$$ , computes the sum of the natural numbers up to $$n\text{.}$$ In the algorithm, $$s$$ is the sum so far and $$i$$ is incremented in each iteration of the repeat_until loop. The algorithm computes $$1+2+3+\dots+(n-2)+(n-1)+n\text{.}$$
In the Example 2.32 we compute the sum of the natural numbers up to $$4$$ with Algorithm 2.31.
### Example2.32.Algorithm 2.31 for the input $$n=4$$.
We follow the steps of Algorithm 2.31 for the input $$n=4\text{.}$$ For each step of the algorithm, we give the new values of the variables that change values in that step.
Input: $$n=4$$
1. let $$i:=0$$ : So the variable $$i$$ has the value $$0$$ now.
2. let $$s:=0$$ : So the variable $$s$$ has the value $$0$$ now.
3. repeat : A repeat_until -loop starts here.
3.a. let $$i:=i+1$$ : As the old value of $$i$$ was zero the new value of $$i$$ is $$0+1=1\text{.}$$
3.b. let $$s:=s+i$$ : As the old value of $$s$$ was zero and the value of $$i$$ is $$1$$ the new value of $$s$$ is $$0+1=1\text{.}$$
4. until $$i=n$$ : As $$i=1$$ and $$n=4$$ the statement $$i=n$$ is false. We repeat the loop and continue with step 3.a
3. repeat : We continue the A repeat_until -loop.
3.a. let $$i:=i+1$$ : As the old value of $$i$$ was $$1$$ the new value of $$i$$ is $$1+1=2\text{.}$$
3.b. let $$s:=s+i$$ : As the old value of $$s$$ was $$1$$ and the value of $$i$$ is $$2$$ the new value of $$s$$ is $$1+2=3\text{.}$$
4. until $$i=n$$ : As $$i=2$$ and $$n=4$$ the statement $$i=n$$ is false. We repeat the loop and continue with step 3.a.
3. repeat : We continue the A repeat_until -loop.
3.a. let $$i:=i+1$$ : As the old value of $$i$$ was $$2$$ the new value of $$i$$ is $$2+1=3\text{.}$$
3.b. let $$s:=s+i$$ : As the old value of $$s$$ was $$3$$ and the value of $$i$$ is $$3$$ the new value of $$s$$ is $$3+3=6\text{.}$$
4. until $$i=n$$ : As $$i=3$$ and $$n=4$$ the statement $$i=n$$ is false. We repeat the loop and continue with step 3.a.
3. repeat : We continue the A repeat_until -loop.
3.a. let $$i:=i+1$$ : As the old value of $$i$$ was $$3$$ the new value of $$i$$ is $$3+1=4\text{.}$$
3.b. let $$s:=s+i$$ : As the old value of $$s$$ was $$6$$ and the value of $$i$$ is $$4$$ the new value of $$s$$ is $$6+4=10\text{.}$$
4. until $$i=n$$ : As $$i=4$$ and $$n=4$$ the statement $$i=n$$ is true. We continue with step 5.
5. return $$s$$ : The algorithm returns $$s=10\text{.}$$
Output: 10
In Example 2.33 click your way through the steps of Algorithm 2.31 to see how the values of the variables change with each instruction. When following the loop you have to click on repeat as well as until.
### Example2.33.Algorithm sum interactive.
In the video in Figure 2.34 we introduce repeat_until loops and consider Algorithm 2.31 and Algorithm 2.29 once more.
A repeat_until -loop with a statement that is always false in the repeat instruction yields a never ending loop. A sequence of instructions (even if this is only the case fore certain input values) that contains such a loop is not an algorithm.
### Example2.35.Infinite loop.
We give an example of a sequence of instructions that gets caught in a never ending loop for certain input values.
Input: an integer $$a$$
1. let $$m:=1$$
2. repeat
1. let $$m:=m\cdot a$$
2. let $$a:=a-1$$
3. until $$a=0$$
4. return $$m$$
When the input $$a$$ is 0 or a negative integer, this sequence of commands does not end. It never reaches the instruction return $$m$$ since $$a$$ will never be 0. In this case we do not have a finite sequence of instructions, so it does not match the definition of an algorithm ( Definition 2.1 ).
### Problem2.36.
With the algorithm below answer the following questions:
1. Follow the steps of the algorithm for $$b=5$$ and $$n=3\text{.}$$
2. What does the algorithm return for the input $$b=-2$$ and $$n=6\text{.}$$
3. What does the algorithm compute?
Solution.
1. We follow the algorithm for the input $$b=5$$ and $$n=3\text{.}$$ For each step with an assignment we give the new value of the variable.
Input: $$b=5$$ and $$n=3$$
1. $$c=0$$
2. $$i=0$$
3. A repeat_until -loop starts
3.a. $$c=5$$
3.b. $$i=1$$
4. As $$i=1$$ and $$n=3$$ the statement $$i=n$$ is false. We repeat the loop and continue with step 3.a.
3. We continue in the repeat_until -loop.
3.a. $$c=10$$
2. 3.b. $$i=2$$
4. As $$i=2$$ and $$n=3$$ the statement $$i=n$$ is false. We repeat the loop and continue with step 3.a.
3. We continue in the repeat_until -loop.
3.a. $$c=15$$
3.b. $$i=3$$
4. As $$i=3$$ and $$n=3$$ the statement $$i=n$$ is false. We exit the loop and continue with step 5.
5. We return the value of $$c$$ which is $$15\text{.}$$
Output: 15
3. Proceeding as above we find that for the input $$b=-2$$ and $$n=6$$ the algorithm returns $$-12\text{.}$$
4. Until $$i\lt n$$ the algorithm adds $$b$$ to $$c$$ and adds $$1$$ to $$n\text{.}$$ Thus the number of times $$b$$ is added to $$c$$ is $$n\text{.}$$ As initially $$i$$ is set to 0 (see step 2 ) the number of times $$b$$ is added to $$0$$ is $$n\text{.}$$ So the output of the algorithm is $$n \cdot b\text{.}$$ Also compare Definition 1.31 where we had defined multiplication as repeated addition.
We refer to this algorithm in the following two problems.
### Problem2.39.Algorithm 2.38 with input $$n=3$$.
Follow the instruction in Algorithm 2.38 when the input is $$n=3\text{.}$$ We only list the steps in which the values of a variable changes.
Solution.
We follow the algorithm for the input $$n=3\text{.}$$
Input: $$n=3\text{.}$$
1. let $$f:=1$$
2.a. let $$f:=1\cdot 3=3$$
2.b. let $$n:=3-1=2$$
3. As $$n=2$$ the statement $$n=1$$ is false. To repeat the loop we continue with step 2.a.
2.a. let $$f:=3\cdot 2=6$$
2.b. let $$n:=2-1=1$$
3. As $$n=1$$ the statement $$n=1$$ is true. We exit the loop and continue with step 4.
4. We return the value of $$f$$ which is $$6\text{.}$$
Output: 6
Now that we have gone through the algorithm instruction by instruction for a specific input we have a better idea what the algorithm computes.
### Problem2.40.What does Algorithm 2.38 compute ?
What does Algorithm 2.38 compute ?
Solution.
For the input value $$n\text{,}$$ the output of the algorithm is
\begin{equation*} 1\cdot n\cdot (n-1)\cdot\ldots\cdot2. \end{equation*}
So, the algorithm computes the product of the natural numbers up to $$n\text{.}$$
The product of the natural numbers up to $$n$$ is important enough to have a name.
### Definition2.41.
Let $$n$$ be a natural number. The product
\begin{equation*} 1\cdot 2\cdot 3\cdot\dots \cdot(n-1)\cdot n \end{equation*}
is called $$n$$ factorial. We denote $$n$$ factorial by $$n!\text{.}$$
In Checkpoint 2.42 we unroll the loop in Algorithm 2.38. Follow the instructions to find the factorial of the input.
### Checkpoint2.42.Compute a factorial with Algorithm 2.38.
Consider the following algorithm.
Algorithm Factorial
Input: A natural number $$n$$
Output: $$n!$$
(1) let $$f:=1$$
(2) repeat
-- (a) let $$f:=f\cdot n$$
-- (b) let $$n:=n-1$$
(3) until $$n=0$$
(4) return $$f$$
Now use the algorithm to compute the factorial of $$n=5\text{.}$$
Input: A natural number $$n =$$ .
(1) let $$f:=1\text{.}$$
(2) repeat
-- (a) let $$f:=f\cdot n=$$
-- (b) let $$n:=n-1=$$
(3) Because the statement $$n=1$$ is false, the loop is repeated. We continue with step (2).
(2) repeat
-- (a) let $$f:=f\cdot n=$$
-- (b) let $$n:=n-1=$$
(3) Because the statement $$n=1$$ is false, the loop is repeated. We continue with step (2).
(2) repeat
-- (a) let $$f:=f\cdot n=$$
-- (b) let $$n:=n-1=$$
(3) Because the statement $$n=1$$ is false, the loop is repeated. We continue with step (2).
(2) repeat
-- (a) let $$f:=f\cdot n=$$
-- (b) let $$n:=n-1=$$
(3) Because the statement $$n=1$$ is true, the loop ende. We continue with step (4).
(4) return $$f$$
Output: $$f =$$ |
Homework 2 solutions
# Homework 2 solutions - AMS 361: Applied Calculus IV by Prof...
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Unformatted text preview: AMS 361: Applied Calculus IV by Prof Y. Deng Homework 2 (Problem 2-1, Prob 16, P. 41) Find the general solutions (implicit if necessary, explicit if convenient) of the differential equations in the following problem: 1 tan Solution: This is a separable equation 1 1 (implicit) or 1 explicit (Problem 2-2, Prob 28, P. 41) Find the explicit particular solution of the initial value problem 2 4 Solution: Please refer to the solution in the "Deng Notes" (Prob 2-2, HW2 Spring 2005, p 247) (Problem 2-3, Prob 16, P. 56) Solve the following differential equation. Primes denote derivatives with respect to x. 1 Solution: Please refer to the solution in the "Deng Notes" (Prob 2-4, HW2 Fall 2005, p 276) (Problem 2-4, Prob. 18, P.56) Find the general solution of the differential equation. Primes denote derivatives with respect to x. 2 cos Solution: Please refer to the solution in the "Deng Notes" (Prob 2-4, HW2 Spring 2005, p 247) (Problem 2-5, Prob. 27, P.56) Find the general solution of the differential equation (regarding x as depending variable and y as independent). Primes denote derivatives with respect to x: 1 cos 2 cos /4 AMS 361: Applied Calculus IV by Prof Y. Deng Homework 2 1 2 1 Solution: Please refer to the solution in the "Deng Notes" (Prob 2-5, HW2 Spring 2005, p 248) Method 1: Exact Equation Method: The original DE can be written as 1 1 2 0 Thus, 1 1 Then, we can have 1 Resulting in a new equation that's exact: 1 2 1 [ 1 Now, we can introduce the following two equations 1 1 1 1 Thus, we have 1 Plugging into the second equation above, we get 2 1 Thus, we get 1 or Finally, 1 Thus, the solution is , 1 1 Method 2: Change the equation to 1 2 1 as dependent variable and as independent: , 2 2 2 2 0 1 1 2 constant constant another constant and it becomes 1st order linear DE with - 2 ...
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# Monthly Archives: November 2011
## Area of Shapes!
Finding the area of a square and rectangle are easy to find. The forulma for find the area of a square is 4S, and the forulma for the area of the rectangle is 2W x 2L. These are two basic forulmas that make finding area easy. However when you get into shapes like a triangle and dont know the parts that you need, you can make these shapes in to ones that we know.
When finding the area of a triangle can be difficult if you dont know all the parts or if it is part of another shape. For examlpe if the the triangle is inside a square and we can find the area of the square we can find the area of the triangle. The picture below is a square with a triangle inside it we want to know the area of the triangle.
We know the area of the square by counting the square inside of the big one. The area of this square is 12 square centimeters. After finding the area we can cut out the triangle which we want to find the area. The picture below is cutting out the triangle, notice that when you cut out the one triangle the two extra parts make the same triangle.
With two triangles that are the same, we can conclude that the area of one triangle is half the area of the square. The area of the trangle is 6 square centimeters. The formula for the area of a triangle is 1/2BxH. We can also use this to find other shapes that we do not know. If the shape can be made in to a square or a rectangle this is a good way to find the area of something!
Posted by on November 26, 2011 in Uncategorized
## Shapes in circles?
Have you even been told to draw a shape in side a circle and not know what they are talking about? Well it is very possible to do, you can do this by taking the central angle of the circle divided by the number of side the shape will have. For example a circle has 360 degrees and a regular hexagon has 6 sides; we would take 360/6= 60, each angle from the center of the circle would then be 60 degrees.
Next we can connect the verticies, creating the six side of the hexagon.
Now that we have our hexagon, another thing that we can find is an angle of the hexagon, we can find the vertex angle by taking 180-the central angle, in this case 180-60=120 degrees.
We can put any shape inside a circle. All you have to do is follow the steps, not only is it interesting but is also fun and can be made into different types of projects!
Posted by on November 18, 2011 in Uncategorized
## Feet to Centimeters….
In Math we can be given a problem in one unit and have to give the answer in another. This is called converting of units. We can go from feet to centimeters in a few easy step when knowing the converstions.
The Conversion units for Length are:
English Metric Bridges
12in=1ft 1km=1000m 1in=2.54cm
3ft=1yd 1m=1000mm 1mi=1.609
5280ft=1mi 1m=100cm
1cm=10mm
We need the bridges to be able to go from metric to english and viscera. There are also other conversions that can be helpful; Conversions for mass and time.
The Conversion units for Mass are:
English Metric Bridges
16oz=1lb 1000mg=1g 454g=1lb
2000lb=1tons 1000g=1kg 1kg=2.2Lb
The Conversion units for Time are:
60 sec=1 min 52 wk= 1 yr
60 min=1 hr 365 days=1 yr
24 hr= 1 day 10 yr= 1 decade
7 days= 1 wk 100 yr= 1 century
With these conversions units we can do conversions. We can take 2.6 meters and convert it to millimeters. We can do this by starting with 2.6 meters and finding a conversions that will take use to millimeters.
We can also convert 3.0 miles into inches.
Posted by on November 16, 2011 in Units of Measure
## Irrational Numbers
A rational number is a number that can be written as a ratio; which means it can be written as a fraction. 10 is a rational number because it can be written as 10/1. Any number that is not rational is considered an irrational number. An irrational number can be written as a decimal; however not a fraction. Irrational numbers also have a decimal that is not ending but it is not repeated. An example of an irrational number is pi. Pi is equal to 3.141592…….., another example is square root 2. Even though square root 2 is an irrational number, this doesn’t mean that all roots are irrational numbers.
We can also simplify roots with a higher index. An index is a number written in the “check mark” area of the radical, that indicates some other root besides a square root.
When we simplify we need to pay attention to the index. The Index is important when simplifing because that is the number of grouping that you need to take out.
We need to find groups of three when simplifying. We can simplify 162 by using a factor tree.
Then we can rewrite this under the ratical.
Pulling out the group of three and taking it to the front of the ratical leaving the 3*2 underneith the ratical.
Then we can simplify the 3*2 under the ratical.
Posted by on November 15, 2011 in Uncategorized
## Out of 100!
Finding a percent can be very intimidating, however if with start with an easy percent to find we can build off that. A percent that is easy to find is 10%, you just move the decimal one place to the left. For example 10% of 52 is 5.2, we also know that 10% of 6.3 is 0.6. Not only can we find 10% of a number but we can find 20 percent. To find 20 percent of a number we can find 10% and the double it. For example if we want to find 20%of 600, we can first find 10 percent which is 60 and then double that to 120, so 20% of 600 is 120.
We can also use this for finding the new price of an item, for example if we had an item that is \$200 and you get 20% off what is the new price? You can start this problem off by finding 10% of 200 which is 20 and then doubling 20 to get 40. The new problem is \$40 off \$200 so you will pay \$160 for the item.
We can also use these methods to find a smaller percent like 5, because 5 is half of 10 we can find 10% of a number and then split it in half. For example 5% of 4200. We can start this problem by finding 10 percent; 10% of 4200 is 420. Then we can take half of 420 and get 210. So 5% of 4200 is 210.
Another way to find a percent is to use a percent chart. Below are to images of percent charts.
These grids have a hundred squares to represent 100%. The problem for the first grid is 40% of 200. Which is 80.
The problem for the second grid is 20%of 40. Which is 8.
Posted by on November 5, 2011 in Percent
## Decimals to Fraction!
In math there are many things that go hand in hand, for example addition and subtraction, also multiplication and division. Another thing that goes hand in hand is fractions and decimals. We can take a fraction and make it decimal but we can also have a decimal and make it into a fraction. In order to do so we need to understand the term “rational numbers”. Rational numbers are simply decimals that are terminating or repeating.
We can take the word “rational” we can take out the word “Ratio” meaning of integers (fraction of integers).
Terminating decimals are numbers like 0.55, 0.205, and 6.1 (these are all decimals that have a clear ending). We can make these decimal into fractions by paying attention to the place values that the digits hold. For example 0.55 is in the hundredths place therefore 0.55 is 55/100.
Repeating Decimal are numbers like 0.333333….., 0.77777….,and 0.242424 (decimals that have a repetition in the number. it may start later in the number). We can make these decimals in to numbers by setting them equal to “X”. A good example is 0.333333…. because we know of the top of our heads that it is 1/3 but do we know why it is? Below are the steps to show why 0.3333…. is 1/3.
Step 1) X=0.333333….
Step 2) 10X=3.333333…. (Find a number of ten that will get rid of the repeating end of the decimal. Then subtract Step 1 and Step 2.)
Step 3) 9X=3
Step 4) 9X/9=3/9 or 1/3
We can use these steps on any repeating decimals. Lets try the decimal 0.242424….! We will start the problem just like we did for 0.333……!
Step 1) X = 0.242424
Step 2) 100x = 24.2424 (We want to set it equal to 100X because it will move the decimal two places to the right and it will get rid of the repetition)
Step 3) 99X = 24
Step 4) 99X/99 = 24/99 or 8/33
Posted by on November 3, 2011 in Decimals
## Decimals….
First we need to know the terminology for decimals before we jump in, the terminology is very important. We can start off by understanding the meaning of the Latin word decem, which mean ten. Some of the other terms that we need to understand are decimal points, and mixed decimals.
The Decimal point is the dot between the digits in a number. For example 17.63 the dot between the 17 and the 63 is the decimal point. 17.63 can also be considered a mixed decimal! Another important thing to pay attention to is the place values when dealing with decimals.
Above is a picture with Place Values for numbers with decimals. On the Left side of the decimal we can see that the ending is “s” and that on the right side the ending are “ths”. These are very important to pay attention too.
We can also write the number in an extended form.
5(10^3)+ 4(10^2)+ 7(10)+3(1)+ 2(1/10) +8(1/100) +6(1/1000)
When we read the number we read it in a different way than what we would if just reading a whole number. When reading a decimal we say the word “and” when we come across the decimal. Below is a example of this. |
# Invariant Points Worksheets, Questions and Revision
GCSE 6 - 7AQAEdexcelOCRWJECHigherWJEC 2022
## Invariant points
When we transform a shape – using translations, reflections, rotations, enlargements, or some combination of those $4$, there are sometimes points on the shape that end up in the same place that they started. These are known as invariant points.
You are expected to identify invariant points.
Make sure you are happy with the following topics before continuing:
Level 6-7 GCSE
## Example 1: Invariant Points – Reflections
Shape $A$ is shown below.
State the coordinates of any invariant points when shape $A$ is reflected in the line $x=1$.
[1 mark]
In order to see if there are any invariant points in a transformation, we need to do the transformation.
First, we draw the line $x=1$. It is a horizontal line that passes through $1$ on the $x$-axis.
Then, completing the reflection we get the picture shown below.
The reflected shape is shown in green.
As we can see, all points on the shape have moved to a different location except for one the one that was located on the reflection line.
Here we’ve marked this with a red X.
So, we can conclude that there is $1$ invariant point, and the coordinates of it are
$(1, 1)$
Level 6-7GCSE
## Example 2: Invariant Points – Translation
Perform the following two transformations to shape $A$, in the given order:
1. Rotate $180\degree$ about the point $(1, 0)$
2. Translate by the vector $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$.
Label the final shape $C$ and state the coordinates of any invariant points.
[2 marks]
Firstly, we need to perform the rotation. Here, we’ll mark a point at $(1, 0)$ to note the centre of rotation, and then we’ll call the resulting shape $B$.
Then, we have to translate by the vector
$\begin{pmatrix} 2 \\ 2 \end{pmatrix}$
This means moving shape $B$ $2$ to the right, and then $2$ up.
This completes the transformation, so we label the resulting shape $C$. The full picture is shown below.
We can now see one invariant point between $C$ and $A$, with coordinates
$(2, 1)$
Here, marked with a red X.
Level 6-7GCSE
## Example Questions
a) To do the enlargement, we draw lines for the origin through each of the corners of the shape, extending them twice as far to account for the scale factor 2. Then, joining the corners together, we get the enlarged shape as shown below.
b) Looking at the graph, we can see that none of the points on the triangle ended up being in their original spot. Therefore, the number of invariant points is zero.
A reflection in the line $x=0$ is just a reflection in the y-axis. The result of this is shown in grey below.
Then, taking this new shape and rotating it $180\degree$ about (0, 1) (using tracing paper or otherwise), we get the shape labelled B below.
b) Looking at the graph, we can see there is one corner of the shape that ends up in its original position. Reading off, we see that the coordinates of the invariant point are
$(-4, 1)$
a) The line $y=x$ is the straight line that passes through the origin, and points such as (1, 1), (2, 2), and so on. Reflecting the shape in this line and labelling it B, we get the picture below.
b) We want to perform a translate to B to make it have two point that are invariant (with respect to shape A). To do this, we want the two sharp corners at the bottom of the arrow shapes to meet. This can be done by moving shape B left by 1 and up by 1, as shown below.
Therefore, the required transformation to get two invariant points is a translation by the vector
$\begin{pmatrix} -1 \\ 1 \end{pmatrix}$
Level 6-7GCSE
Level 4-5GCSE
## Worksheet and Example Questions
### (NEW) Invariant points Exam Style Questions - MME
Level 6-7NewOfficial MME
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# Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q15
My Solution for "Establish that if $a$ is an odd integer, then for any $n \geq 1$$a^{2^{n}} \equiv 1 \pmod {2^{n + 2}}$$ [Hint: Proceed by induction on$n$.]" ## Table of Contents ## Background All theorems, corollaries, and definitions listed in the book's order: I will only use theorems or facts that are proved before this question. So, you will not see that I quote theorems or facts from the later chapters. ## Question Establish that if$a$is an odd integer, then for any$n \geq 1$$a^{2^{n}} \equiv 1 \pmod {2^{n + 2}}$$
[Hint: Proceed by induction on $n$.]
## Solution
We can prove the statement by using mathematical induction.
Base Case:
For $n = 1$, we need to prove $a^{2} \equiv 1 \pmod {8}$ when $a$ is an odd integer.
By Theorem 2.1 Division Algorithm, we can write $a$ in the form $4q + 1$ or $4q + 3$ for some integer $q$.
For $4q + 1$, $a^{2} \equiv 16q^{2} + 8q + 1 \equiv 1 \pmod {8}$. For $4q + 3$, $a^{2} \equiv 16q^{2} + 24q + 9 \equiv 1 \pmod {8}$.
Induction Hypothesis:
Assume that if $a$ is an odd integer, $a^{2^{k}} \equiv 1 \pmod {2^{k + 2}}$ for some integer $k \geq 1$.
Consider the case for $k + 1$:
From the inductive hypothesis, we have $a ^{2^{k}} - 1 = m2^{k + 2}$ for some integer $m$. Then for $a ^{2^{k + 1}} \pmod {2^{k + 3}}$:
$$$$\begin{split} a ^{2^{k + 1}} & \equiv a^{2^{k} \cdot 2} \\ & \equiv (a^{2^{k}})^{2} \\ & \equiv (1 + m2^{k + 2})^{2} \\ & \equiv m^{2} \cdot 2^{2k + 4} + 2 \cdot m \cdot 2^{k + 2} + 1 \\ & \equiv m^{2} \cdot 2^{k + 3} \cdot 2^{k + 1} + m \cdot 2^{k+3} + 1 \\ & \equiv 1 \pmod {2^{k + 3}} \end{split} \nonumber$$$$
Conclusion:
This proves the statement by induction.
Read More: All My Solutions for This Book
MathNumber TheorySolution |
## Pages
Showing posts with label factoring. Show all posts
Showing posts with label factoring. Show all posts
## Monday, November 5, 2012
### Chapter 6 Sample Test Questions
Click here for a worksheet containing 20 sample test questions with answers.
Example: Given the function:
a. Calculate f(0).
b. Find all values of x for which f(x) = 0.
Example: Find a quadratic equation with integer coefficients and the following solution set.
Example: The product of two consecutive odd positive integers is eleven less than 10 times the larger. Find the integers.
Example: The length of a rectangle is 6 centimeters longer than twice its width. If the area is 140 square centimeters, find the dimensions of the rectangle.
Example: The cost of a particular car in dollars can be approximated by the function:
Here x represents the age of the car in years.
a. Use the graph to determine the cost of the car when it was new?
b. How old will the car be when it reaches its minimum cost?
c. How much is this car worth when it reaches 5 years old?
## Sunday, November 4, 2012
### Solving Equations by Factoring
Previously we have learned how to solve linear equations, now we will outline a technique used to solve factorable quadratic equations that look like
In addition, we will revisit function notation and apply the techniques in this section to quadratic functions.
The above zero factor property is the key to solving quadratic equations by factoring. So far we have been solving linear equations, which usually had only one solution. We will see that quadratic equations have up to two solutions.
Solve:
Step 1: Obtain zero on one side and then factor.
Step 2: Set each factor equal to zero.
Step 3: Solve each of the resulting equations.
This technique requires the zero factor property to work so make sure the quadratic is set equal to zero before factoring in step 1.
Tip: You can always see if you solved correctly by checking your answers. On an exam it is useful to know if got the correct solutions or not.
Instructional Video: The Zero-Product Property
Solve.
When solving quadratic equations by factoring, the first step is to put the equation in standard form ax^2 + bx + c = 0, equal to zero.
Solve:
Obtain standard form and then factor.
Set each factor equal to zero and solve.
Check.
Important: We must have zero on one side of our equation for this technique to work.
Solve.
You can clear fractions from any equation by multiplying both sides by the LCD.
Solve:
Multiply both sides by the LCD 6 here to clear the fractions.
Solve.
Work the entire process in reverse to find equations given the solutions.
Find a quadratic equation with given solution set.
Remember that the notation y = f(x) reads, "y equals of x."
Evaluate the given function.
Given the graph of the quadratic function, find the x- and y- intercepts.
Video Examples on YouTube: |
# SAT Practice Test 4, Section 8: Questions 11 - 16
Given:
The sum of a list prices divided by the average of the prices = k
To find:
The value of k
Solution:
Topic(s): Statistics
So, k represents the number of prices.
Answer: (D) The number of prices
Given:
The figure
Area of square = 81
Perimeter of each of the 4 triangles = 30
To find:
The perimeter of the figure outlined by the solid line
Solution:
Topic(s): Area of square, perimeter of triangle
Given that the area of the square is 81, the length of any one its side would be = 9.
The side of the square is also of base of the triangle. So the length of the base of the triangle is 9.
Given that the perimeter of the triangle is 30, then the lengths of the other two sides of the triangle must be 30 – 9 = 21.
The perimeter of the figure is formed by the 4 of such triangles.
So perimeter = 4 × 21 = 84
Given:
The graph y = g(x)
g(2) = k
To find:
The value of g(k)
Solution:
We are given g(2) = k.
From the graph, we can obtain g(2) = 5. This means that k = 5.
From the graph, we find that g(5) = 2.5
Given:
0 ≤ x ≤ 8
–1 ≤ y ≤ 3
To find:
The smallest and largest possible values for xy
Solution:
The largest value for x is 8 and the largest value for y is 3. Since both the values are positive, the largest value for xy is 8 × 3 = 24
Finding the smallest value would be trickier. The smallest value for x is 0 and the smallest value for y is – 1. Since one of the values is negative, we need to logically consider what would be the smallest negative value for xy. The smallest value would be –1 multiplied with the biggest value of x, which is 8. –1 × 8 = –8
Answer: (E) –8 ≤ xy ≤ 24
Given:
The figure
To find:
The sum of the marked angles in terms of n
Solution:
Topic(s): Supplementary angles, sum of angles in a triangle
If you know the rule that the exterior angle of a triangle is equals to the sum of the opposite interior angles then you can work out that a + b = n and c + d = n. So, a + b + c + d = 2n.
If you did not know the rule, you can deduce that
n + p = 180 ⇒p =180 – n (supplementary angles)
a + b + p =180 (sum of angles in a triangle)
a + b + 180 – n = 180 ⇒ a + b = n
Similarly, we can deduce that c + d = n
So, a + b + c + d = 2n.
Given:
t is the first term of a sequence
t ≠ 0
After the first term, each term in the sequence is 3 greater than of the preceding term
To find:
The ratio of the 2 nd term to the 1 st term
Solution:
Topic(s): Number sequence problems
The first term is t.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Mathematics 2014 | Study Mode
Question 31
A
8x2 - 2x + 1
B
8x2 - 4x + 1
C
12x2 - 2x + 1
D
12x2 - 4x + 1
##### Explanation
If y = 4x3 - 2x2 + x, then;
$$\frac{\delta y}{\delta x}$$ = 3(4x2) - 2(2x) + 1
= 12x2 - 4x + 1
Question 32
A
$$\frac{1}{3} \sin 3x$$
B
$$-\frac{1}{3} \sin 3x$$
C
3 sin 3x
D
-3 sin 3x
##### Explanation
y = cos 3x
Let u = 3x so that y = cos u
Now, $$\frac{\delta y}{\delta x} = 3$$,
$$\frac{\delta y}{\delta x} = -sin u$$
By the chain rule,
$$\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}$$
$$\frac{\delta y}{\delta x} = (-\sin u) (3)$$
$$\frac{\delta y}{\delta x} = -3 \sin u$$
$$\frac{\delta y}{\delta x} = -3 \sin 3x$$
Question 33
A
4
B
1
C
-1
D
-4
##### Explanation
y = x2 - 2x - 3,
Then $$\frac{\delta y}{\delta x} = 2x - 2$$
But at minimum point,$$\frac{\delta y}{\delta x} = 0$$,
Which means 2x - 2 = 0
2x = 2
x = 1.
Hence the minimum value of y = x2 - 2x - 3 is;
ymin = (1)2 - 2(1) - 3
ymin = 1 - 2 - 3
ymin = -4
Question 34
A
cos 2x + k
B
$$\frac{1}{2}$$cos 2x + k
C
$$-\frac{1}{2}$$cos 2x + k
D
-cos 2x + k
##### Explanation
$$\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k$$
$$- \frac{1}{2} \cos 2x + k$$
Question 35
A
$$\frac{1}{12} (2x + 3)^6 + k$$
B
$$\frac{1}{3} (2x + 3)^{\frac{1}{2}} + k$$
C
$$\frac{1}{3} (2x + 3)^{\frac{3}{2}} + k$$
D
$$\frac{1}{12} (2x + 3)^{\frac{3}{4}} + k$$
##### Explanation
$$\int (2x + 3)^{\frac{1}{2}} \delta x$$
let u = 2x + 3, $$\frac{\delta y}{\delta x} = 2$$
$$\delta x = \frac{\delta u}{2}$$
Now $$\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}$$
$$= \frac{1}{2} \int u^{\frac{1}{2}} \delta u$$
$$= \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k$$
$$= \frac{1}{3} u^{\frac{3}{2}} + k$$
$$= \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k$$
Question 36
A
t
B
-t
C
2
D
-2
##### Explanation
Mean x = $$\frac{\sum x}{n}$$
= [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1) $$\div$$] 5
= [11 - 1 + 3t - 3t] $$\div$$ 5
= 10 $$\div$$ 5
= 2
Question 37
A
1
B
2
C
3
D
4
Question 38
A
6
B
5
C
4
D
3
##### Explanation
First arrange the numbers in order of magnitude;
1,2,3,3,4,5,5,5,5,6,7,8,9,9,10
Hence the median = 5
Question 39
A
$$\sqrt{2}$$
B
$$\sqrt{3}$$
C
$$\sqrt{6}$$
D
$$\sqrt{10}$$
##### Explanation
Mean x = $$\frac{\sum x}{n}$$
$$= \frac{5 + 4 + 3 + 2 + 1}{5}$$
$$= \frac{15}{5}$$
= 3
$$\begin{array}{c|c} x & d = x - 3 & d^2 \\ \hline 5 & 2 & 4 \\ 4 & 1 & 1 \\ 3 & 0 & 0 \\ 2 & -1 & 1 \\ 1 & -2 & 4 \\ \hline & & \sum d^2 + 10 \end{array}$$
Hence, standard deviation;
$$= \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}}$$
$$= \sqrt{2}$$
Question 40
A
$$\frac{7!}{3!}$$
B
$$\frac{7!}{4!}$$
C
$$\frac{7!}{3!4!}$$
D
$$\frac{7!}{2!5!}$$
##### Explanation
A team of 2 girls can be selected from 7 girls in $$^7C_3$$
$$= \frac{7!}{(7 - 3)! 3!}$$
$$= \frac{7!}{4! 3!} ways$$
Try this quiz in in E-test/CBT Mode
switch to |
# Speed, distance etc
• Jul 3rd 2007, 03:10 AM
Speed, distance etc
Fred cycled from home to his friends house and back again.
The distance is 20km.
On his way home, he cycled x km per hour.
On his way back, his speed decreased by 2km per hour.
It took him 4 hours altogether to cycle to his friends house and back.
1. Write down an equation for x
x = (20/4) - 2 ?
2. Show that the equation can be written as x^2 - 12x + 10 = 0
3. Solve the equation in question 2 to 1 decimal place
4. Only one of teh answers in question 3 can be Freds speed. Explain why.
• Jul 3rd 2007, 04:09 AM
earboth
Quote:
Fred cycled from home to his friends house and back again.
The distance is 20km.
On his way home, he cycled x km per hour.
On his way back, his speed decreased by 2km per hour.
It took him 4 hours altogether to cycle to his friends house and back.
1. Write down an equation for x
x = (20/4) - 2 ?
2. Show that the equation can be written as x^2 - 12x + 10 = 0
3. Solve the equation in question 2 to 1 decimal place
4. Only one of teh answers in question 3 can be Freds speed. Explain why.
Hello,
you know that $\displaystyle speed = \frac{distance}{time}$ that means $\displaystyle time = \frac{distance}{speed}$
So you can set up an equation to calculate the time:
$\displaystyle \frac{20}{x} + \frac{20}{x-2} = 4$ . Multiply both sides by x(x-2):
$\displaystyle 20(x-2) + 20x = 4x(x-2)$
$\displaystyle 20x - 40 + 20x = 4x^2 - 8x$
$\displaystyle 4x^2 - 48x + 40 = 0$
$\displaystyle x^2 - 12x + 10=0$
I assume that you know how to solve a quadratic equation. (I've got the solutions: x =11.1 or x = 0.901) It is obvious that the 2nd solution never satisfies the given conditions because Fred would have traveled back with a negative speed :mad:
• Jul 3rd 2007, 11:36 PM
I tried using the formula to solve it because I didn't think it could be factorised, bt got different answers:
x = ( -12 + or - sq.rt 12^2 - 4*1*-10 ) / 2
x = ( -12 + or - sq.rt 184 ) / 2
x= -5.22 or -18.78
Arghhhh!
• Jul 4th 2007, 01:58 AM
earboth
Quote:
I tried using the formula to solve it because I didn't think it could be factorised, bt got different answers:
x = ( -12 + or - sq.rt 12^2 - 4*1*-10 ) / 2
x = ( -12 + or - sq.rt 184 ) / 2
x= -5.22 or -18.78
Arghhhh!
Hello,
you are one sign away from the corect solution:
x = ( 12 + or - sq.rt (12^2 - 4*1*10 )) / 2
x = ( 12 + or - sq.rt 104 ) / 2
x= 11.1 or 0.901 |
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# For the Curve Y = 5x – 2x3, If X Increases at the Rate of 2 Units/Sec, Then Find the Rate of Change of the Slope of the Curve When X = 3 - CBSE (Science) Class 12 - Mathematics
ConceptProcedure to Form a Differential Equation that Will Represent a Given Family of Curves
#### Question
For the curve y = 5x – 2x3, if x increases at the rate of 2 units/sec, then find the rate of change of the slope of the curve when x = 3
#### Solution
The given curve is y = 5x – 2x3 and dx/dt = 2 units/sec
y = 5x – 2x3
Differentiating both sides w.r.t x, we get
Slope of the curve = dy/dx = 5 - 6x^2
Differentiating both sides w.r.t t, we get
=> d/dt (dy/dx) = 0 - 12x dx/dt
=> d/dt (dy/dx)_(x = 3) = 0 - 12 xx 3 xx2 = -72 "units/sec"
Thus, the slope of the curve is decreasing at the rate of 72 units/sec when x = 3
Is there an error in this question or solution?
#### APPEARS IN
Solution For the Curve Y = 5x – 2x3, If X Increases at the Rate of 2 Units/Sec, Then Find the Rate of Change of the Slope of the Curve When X = 3 Concept: Procedure to Form a Differential Equation that Will Represent a Given Family of Curves.
S |
GCF the 28 and 56 is the largest feasible number that divides 28 and also 56 precisely without any type of remainder. The components of 28 and also 56 are 1, 2, 4, 7, 14, 28 and also 1, 2, 4, 7, 8, 14, 28, 56 respectively. There space 3 typically used methods to find the GCF that 28 and also 56 - Euclidean algorithm, element factorization, and long division.
You are watching: What is the gcf of 28 and 56
1 GCF of 28 and also 56 2 List the Methods 3 Solved Examples 4 FAQs
Answer: GCF of 28 and also 56 is 28.
Explanation:
The GCF of 2 non-zero integers, x(28) and y(56), is the greatest positive integer m(28) the divides both x(28) and y(56) without any kind of remainder.
The methods to uncover the GCF the 28 and 56 are explained below.
Long department MethodListing usual FactorsUsing Euclid's Algorithm
### GCF that 28 and 56 by lengthy Division
GCF the 28 and also 56 is the divisor that we obtain when the remainder i do not care 0 after ~ doing long division repeatedly.
Step 2: due to the fact that the remainder = 0, the divisor (28) is the GCF of 28 and also 56.
The corresponding divisor (28) is the GCF the 28 and 56.
### GCF that 28 and 56 through Listing usual Factors
Factors of 28: 1, 2, 4, 7, 14, 28Factors that 56: 1, 2, 4, 7, 8, 14, 28, 56
There room 6 typical factors the 28 and also 56, that are 1, 2, 4, 7, 14, and 28. Therefore, the greatest common factor of 28 and also 56 is 28.
### GCF of 28 and also 56 by Euclidean Algorithm
As every the Euclidean Algorithm, GCF(X, Y) = GCF(Y, X mode Y)where X > Y and also mod is the modulo operator.
Here X = 56 and also Y = 28
GCF(56, 28) = GCF(28, 56 mode 28) = GCF(28, 0)GCF(28, 0) = 28 (∵ GCF(X, 0) = |X|, wherein X ≠ 0)
Therefore, the worth of GCF that 28 and also 56 is 28.
## GCF of 28 and also 56 Examples
Example 1: The product of two numbers is 1568. If their GCF is 28, what is their LCM?
Solution:
Given: GCF = 28 and product of number = 1568∵ LCM × GCF = product that numbers⇒ LCM = Product/GCF = 1568/28Therefore, the LCM is 56.
Example 2: discover the greatest number the divides 28 and 56 exactly.
Solution:
The best number that divides 28 and 56 precisely is their greatest common factor, i.e. GCF of 28 and also 56.⇒ factors of 28 and 56:
Factors of 28 = 1, 2, 4, 7, 14, 28Factors that 56 = 1, 2, 4, 7, 8, 14, 28, 56
Therefore, the GCF that 28 and also 56 is 28.
Example 3: discover the GCF that 28 and 56, if your LCM is 56.
Solution:
∵ LCM × GCF = 28 × 56⇒ GCF(28, 56) = (28 × 56)/56 = 28Therefore, the greatest typical factor that 28 and also 56 is 28.
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## FAQs top top GCF of 28 and also 56
### What is the GCF that 28 and 56?
The GCF the 28 and also 56 is 28. To calculate the greatest usual factor (GCF) of 28 and also 56, we need to aspect each number (factors of 28 = 1, 2, 4, 7, 14, 28; determinants of 56 = 1, 2, 4, 7, 8, 14, 28, 56) and also choose the greatest variable that exactly divides both 28 and also 56, i.e., 28.
### If the GCF the 56 and 28 is 28, uncover its LCM.
GCF(56, 28) × LCM(56, 28) = 56 × 28Since the GCF that 56 and 28 = 28⇒ 28 × LCM(56, 28) = 1568Therefore, LCM = 56☛ GCF Calculator
### What is the Relation in between LCM and GCF the 28, 56?
The following equation deserve to be used to express the relation in between Least typical Multiple (LCM) and GCF that 28 and 56, i.e. GCF × LCM = 28 × 56.
### How to uncover the GCF the 28 and also 56 by element Factorization?
To find the GCF that 28 and 56, us will uncover the prime factorization of the given numbers, i.e. 28 = 2 × 2 × 7; 56 = 2 × 2 × 2 × 7.⇒ since 2, 2, 7 are typical terms in the element factorization of 28 and also 56. Hence, GCF(28, 56) = 2 × 2 × 7 = 28☛ What room Prime Numbers?
### How to discover the GCF of 28 and also 56 through Long department Method?
To uncover the GCF the 28, 56 making use of long division method, 56 is divided by 28. The matching divisor (28) when remainder amounts to 0 is taken together GCF.
See more: How Much Is 1 Oz Of Cheese Cubes In An Ounce Of Cheese? How Much Is 1 Oz Of Cheese In Cups
### What are the techniques to uncover GCF the 28 and 56?
There room three frequently used methods to uncover the GCF of 28 and also 56. |
#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (viii) maths
$-xcos\: x-2sin\: x$
Hint:
You must know about derivative of x cos x
Given:
x cos x
Solution:
$Let\: \: xcos\: x$
Use multiplicative rule
As UV=UV1+U1V
Where U=x & V=cos x
\begin{aligned} &\frac{d y}{d x}=x \frac{d}{d x} \cos x+\frac{d}{d x} x \cos x \quad\left(\frac{d}{d x} x=1, \frac{d}{d x} \cos x=-\sin x\right) \\ &\frac{d y}{d x}=-x \sin x+\cos x \end{aligned}
Use multiplicative rule
As UV=UV1+U1V
Where U=x & V=sin x
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=-\left(x \frac{d}{d x} \sin x+\frac{d}{d x} x \cdot \sin x\right)+\frac{d \cos x}{d x} \\ &\frac{d^{2} y}{d x^{2}}=-(x \cos x+\sin x)-\sin x \quad\left(\frac{d}{d x} \cos x=-\sin x\right) \\ &\frac{d^{2} y}{d x^{2}}=-x \cos x-\sin x-\sin x \\ &\frac{d^{2} y}{d x^{2}}=-x \cos x-2 \sin x \end{aligned} |
# Using Congruent Triangles (PowerPoint)
Document Sample
``` 4.5 Using Congruent
Triangles
Objectives/Assignment
Use congruent triangles to plan and write
proofs
Use congruent triangles to prove
constructions are valid
Assignment: 5-19 odd, 20, 21
Goal 1: Planning a Proof
Knowing that all pairs of corresponding
parts of congruent triangles are congruent
congruent figures.
Planning a Proof
For example, suppose you
want to prove that PQS ≅ Q
RQS in the diagram shown
at the right.
One way to do this is to
show that ∆PQS ≅ ∆RQS by
the SSS Congruence
Postulate. Then you can
use the fact that R
corresponding parts of P
congruent triangles are
congruent to conclude that S
PQS ≅ RQS.
Example 1: Planning & Writing a
Proof
Given: AB ║ CD, BC ║
Prove: AB≅CD
Plan for proof: Show
that ∆ABD ≅ ∆CDB.
Then use the fact that
corresponding parts of A D
congruent triangles are
congruent.
Example 1: Planning & Writing a
Proof
Solution: First copy the
diagram and mark it with the B C
given information.
information you can deduce.
Because AB and CD are
parallel segments intersected by
a transversal, and BC and DA
are parallel segments A D
intersected by a transversal, you
can deduce that two pairs of
alternate interior angles are
congruent.
Example 1: Paragraph Proof
from the Alternate Interior
Angles Theorem that ABD
B C
≅CDB. For the same
because BC║DA. By the
Reflexive property of
Congruence, BD ≅ BD. You
can use the ASA Congruence
Postulate to conclude that A
∆ABD ≅ ∆CDB. Finally D
because corresponding parts
of congruent triangles are
congruent, it follows that AB
≅ CD.
Example 2: Planning & Writing a
Proof
Given: A is the midpoint of
MT, A is the midpoint of SR.
M R
Prove: MS ║TR.
Plan for proof: Prove that
∆MAS ≅ ∆TAR. Then use
the fact that corresponding A
parts of congruent triangles
are congruent to show that
M ≅ T. Because these T
angles are formed by two S
segments intersected by a
transversal, you can
conclude that MS ║ TR.
M R
Given: A is the midpoint of MT, A is the
midpoint of SR.
Prove: MS ║TR. A
S T
Statements: Reasons:
1. A is the midpoint of MT, A 1. Given
is the midpoint of SR.
2. MA ≅ TA, SA ≅ RA 2. Definition of a midpoint
3. MAS ≅ TAR 3. Vertical Angles Theorem
4. ∆MAS ≅ ∆TAR 4. SAS Congruence Postulate
5. M ≅ T 5. Corres. parts of ≅ ∆’s are ≅
6. MS ║ TR 6. Alternate Interior Angles
Converse
Example 3: Using more than one
pair of triangles
Given: 1≅2, 3≅4
Prove ∆BCE≅∆DCE D
Plan for proof: The only
∆BCE and ∆DCE is that 1≅2 2 4
and that CE ≅CE. Notice, C 3 A
however, that sides BC and DC 1 E
are also sides of ∆ABC and
∆ADC. If you can prove that
∆ABC≅∆ADC, you can use the B
fact that corresponding parts of
congruent triangles are
congruent to get a third piece of
∆DCE.
Given: 1≅2, 3≅4. 2 4
Prove ∆BCE≅∆DCE C 1 E 3 A
B
Statements: Reasons:
1. 1≅2, 3≅4 1. Given
2. AC ≅ AC 2. Reflexive property of
Congruence
3. ∆ABC ≅ ∆ADC 3. ASA Congruence
Postulate
4. Corres. parts of ≅ ∆’s are ≅
4. BC ≅ DC
5. Reflexive Property of
5. CE ≅ CE Congruence
6. ∆BCE≅∆DCE 6. SAS Congruence Postulate
Goal 2: Proving Constructions are
Valid
In Lesson 3.5 you learned to copy an angle
using a compass and a straight edge. The
construction is summarized on pg. 159 and
on pg. 231.
Using the construction summarized above,
you can copy CAB to form FDE. Write
a proof to verify the construction is valid.
Plan for Proof
Show that ∆CAB ≅ ∆FDE. C
Then use the fact that
corresponding parts of
congruent triangles are
congruent to conclude that A
CAB ≅ FDE. By
construction, you can B
assume the following F
statements:
– AB ≅ DE Same compass
setting is used
– AC ≅ DF Same compass D
setting is used
– BC ≅ EF Same compass E
setting is used
C
Given: AB ≅ DE, AC ≅ DF, BC ≅ EF A
Prove CAB≅FDE B
F
D
E
Statements: Reasons:
1. AB ≅ DE 1. Given
2. AC ≅ DF 2. Given
3. BC ≅ EF 3. Given
4. ∆CAB ≅ ∆FDE 4. SSS Congruence Post
5. CAB ≅ FDE 5. Corres. parts of ≅ ∆’s
are ≅.
```
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Math worksheets in 11th grade on metric relations in any triangle.
Exercise 1 – Trigonometric equations
Solve in the following equations.
1.
2.
Exercise 2 – Determine the value of cosine
In this exercise, we give:
Calculate the exact value of and then of
Exercise 3 – Exercise on the tangent
In this exercise, we have the following data:
1. Let . Prove that
2. Deduce that :
Exercise 4 – Solving a trigonometric equation
Solve in the equation: sin(2x) = cos(x).
Exercise 5 – Solving two trigonometric equations
Solve in the following equations:
Exercise 6 – Solving a complex trigonometric equation
Solve in the equation :
Exercise 7 – Equilateral triangle and angle measurement
In the figure below, ABC is equilateral, BCI and ACJ are isosceles rectangles at B and J respectively.
1. Determine a measure of each of the following angles:
2. Show that the points A, I and J are aligned.
Exercise 8 – Trigonometric circle and points
Draw a trigonometric circle and place on this circle
the points A, M, N, P and Q marked by the following numbers :
Exercise 9 – Principal measure of an angle
Determine the principal measure of angles:
; ;
Exercise 10 – Metric relations in the triangle
ABC is a triangle with .
1. Prove that .
2. Calculate the exact values of AB and AC .
Exercise 11 – Graphical representation of trigonometric functions
Show that the graphical representation of the function defined on by :
is located between the lines of equation y = – 3 and y = 1 .
Exercise 12 – Solving a trigonometric equation
Show that, for any real :
Exercise 13 – Using addition formulas
Using the addition formulas, calculate the exact value of
Exercise 14 – The Al-Kashi formulas
Let ABC be any triangle.
We note:
– has the length of the segment [BC];
– b the length of the segment [AC];
– c the length of the segment [AB];
Show that :
Exercise 15 – Trigonometry formula
Show that:
Exercise 16 – Main measure and figure in the plane
Compute the principal measure of knowing that :
Exercise 17 – Metric relations in the triangle
ABC is a triangle with BC = 4, and .
1. Prove that .
2. Calculate the exact values of AB and AC.
3. Calculate the exact value of the area of ABC.
Exercise 18
A triangle ABC has area S = 5 cm².
Moreover, c=AB=13 cm and b=AC= 2 cm.
Calculate the possible length(s) of the third side a = BC.
Exercise 19
ABC is a triangle.
We know that AB = 7, AC= 4 and .
1. Calculate the exact value of BC.
2. Calculate the exact value of .
Exercise 20
Show that two supplementary angles have the same sine.
ABCD is a quadrilateral.We assume that the segments [AC] and [BD] are inside the quadrilateral.
Show that the area S of quadrilateral ABCD is given by :
( is the angle formed by the diagonals).
Exercise 21
A walker walks 5 km east, then 2 km northeast, surprised by the weather, he runs straight back to his starting point.
What distance did he run?
We will give the exact value and then the approximate value to the nearest 0.01 km.
Exercise 22
Prove the following property:
ABC is a right-angled triangle in A .
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 7 - Mathematics4.49 Special Products and Expansions - Formulas
Formulas:
(a+b)2 = (a+b)(a+b) =a2+2ab+b2
(a-b)2 = (a-b)(a-b) =a2-2ab+b2
(a+b)3 = (a+b)(a+b)(a+b) = (a+b)(a2+2ab+b2) =a3+3ab2+3a2b+b3
(a-b)3 = (a-b)(a-b)(a-b) = (a-b)(a2-2ab+b2) =a3+3ab2-3a2b+b3
Sum of squares: a2+b2 =(a+b)2- 2ab a2+b2 =(a-b)2+ 2ab Difference of squares: a2-b2 = (a+b)(a-b) Sum of cubes: a3 + b3 = (a2 - ab + b2)(a + b) Difference of cubes: a3 - b3 = (a2 + ab + b2)(a - b)
Directions:Solve the following problems. Also write two examples for each formula above.
Q 1: If x=2, then find the value of 8x2+6x+7.514947 Q 2: Multiply (7x2-6x+7) with 6x.42x3+36x2+42x42x3-36x2-42x42x3-36x2+42x Q 3: Multiply (x-7) with (x2+1).x3-7x2-x+7x3-7x2+x-7x3-7x2+x+7 Q 4: Multiply (6x+7) with (7x+3).42x2+65x+2142x2+67x+2142x2+67x+24 Q 5: Multiply (6x2+7x-1) with x2.6x4-7x3-x26x4+7x3+x26x4+7x3-x2 Q 6: Multiply (7x+7y) with (6x+3).42x2+21x+42xy+21y42x2+15x+42xy+21y42x2+21x+46xy+21y Q 7: Multiply (3x+2) with (3x-5).9x2-9x+109x2+9x-109x2-9x-10 Q 8: Multiply (6x+7) with (2y+3).12xy+18y+14x+2112xy+18x+14y+2112xy+18x+14y+20 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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## How does an area model show both the two factors and the product?
In mathematics, an area model is a rectangular diagram or model used for multiplication and division problems, in which the factors or the quotient and divisor define the length and width of the rectangle. Then we add to get the area of the whole, which is the product or quotient.
## What is the difference between an array and area model?
Draw 4 rows of 3 cars. Use your array to fill in the blanks. An area model is made of shaded squares organized into rows and columns.
## What does partial products look like?
The partial product method involves multiplying each digit of a number in turn with each digit of another where each digit maintains its place. (So, the 2 in 23 would actually be 20.) For instance, 23 x 42 would become (20 x 40) + (20 x 2) + (3 x 40) + (3 x 2).
## What are two partial products you would add to find 32 405 drag the partial products into the box?
• Given : 32 × 405.
• To Find : two partial products.
• Solution:
• = 12800 + 160.
• = 12150 + 810.
• 810 & 12,150 are the partial products.
## What two partial products would you add to find 513 x46?
We are going to multiply 513 with 46. So the two partial products that we are going to choose are 40 and 6.
## How do you multiply using partial products?
2. Solve each problem by multiplying each digit of one factor by each of the digits in the other factor, taking into account the place value of each digit. 3. Add the partial products to find the total product.
## What is an array model?
The array model for multiplication uses the number of rows and number of columns in an array to illustrate the product of two numbers.
Relate area to the operations of multiplication and addition. a Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths. Use area models to represent the distributive property in mathematical reasoning.
## Do you add or multiply for perimeter?
To find the perimeter of a rectangle, add the lengths of the rectangle’s four sides. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Multiply both the height and width by two and add the results.
## Do you add or multiply to find the perimeter?
When you add each side you get the perimeter. When you multiply the base times height you get area. For a 3-D figure you find the area of the base, then multiply it by the height, and you get the volume. Comment on Milo Shields’s post “When you add each side you get the perimeter.
## What is the formula for finding a perimeter?
The formula for the perimeter of a rectangle is often written as P = 2l + 2w, where l is the length of the rectangle and w is the width of the rectangle. The area of a two-dimensional figure describes the amount of surface the shape covers. You measure area in square units of a fixed size.
## Can you find the area of a rectangle if you know the perimeter?
Explanation: The area of a rectangle can be calculated by multiplying the lengths of two adjacent sides. The perimeter and the length of one side: Using the perimeter formula, you can find the length of an adjacent side, making this choice false.
## How do I find the length and width of a rectangle if I have the area?
A = L * W, where A is the area, L is the length, W is the width or breadth. NOTE: When multiplying the length by the width, always ensure you work in the same unit of length. If they are given in different units, change them to the same unit. Let’s work out a few example problems about the area of a rectangle. |
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# Is 68 A Prime Number?
• No the number 68 is not a prime number.
• Sixty-eight is a composite number. Because 68 has more divisors than 1 and itself.
## Prime Factorization Of 68
• Prime factors of number 68 are: 2, 17
• Equcation of 68 is: 2 * 2 * 17
• The smallest common factor of 68 is number 2
• Highest or greatest common factor GCF of 68 is number 17
## How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 6 8
• About Number 6. Six is the smallest composite number with two distinct prime factors, and the third triangular number. It is the smallest perfect number: 6 = 1 + 2 + 3 and the faculty of 3 is 6 = 3! = 1 * 2 * 3, which is remarkable, because there is no other three numbers whose product is equal to their sum. Similarly 6 = sqrt(1 ^ 3 + 2 + 3 ^ 3 ^ 3). The equation x ^ 3 + Y ^ 3 ^ 3 + z = 6xyz is the only solution (without permutations) x = 1, y = 2 and z = 3. Finally 1/1 = 1/2 + 1/3 + 1/6. The cube (from the Greek) or hexahedron (from Latin) cube is one of the five Platonic solids and has six equal areas. A tetrahedron has six edges and six vertices an octahedron. With regular hexagons can fill a plane without gaps. Number six is a two-dimensional kiss number.
• About Number 8. The octahedron is one of the five platonic bodies. A polygon with eight sides is an octagon. In computer technology we use a number system on the basis of eight, the octal system. Eight is the first real cubic number, if one disregards 1 cube. It is also the smallest composed of three prime number. Every odd number greater than one, raised to the square, resulting in a multiple of eight with a remainder of one. The Eight is the smallest Leyland number.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. Primes can thus be considered the basic building blocks of the natural numbers. There are infinitely many primes, as demonstrated by Euclid around 300 BC. The property of being prime (or not) is called primality.
In number theory, the prime number theorem describes the asymptotic distribution of the prime numbers among the positive integers. It formalizes the intuitive idea that primes become less common as they become larger.
Primes are used in several routines in information technology, such as public-key cryptography, which makes use of properties such as the difficulty of factoring large numbers into their prime factors.
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In this video, we’ll perform length conversions
using unit fractions. In the first example, we’re going to convert six feet to inches. We’ll begin by writing
six feet as a fraction with a denominator of one. And now looking at the conversions, notice one foot is equal to 12 inches, we can use this conversion
to form a unit fraction to convert feet to inches. Because we want feet to simplify out, and right now we have
feet in the numerator, when we form the unit fraction, in order for the units
of feet to simplify out, we need to have feet in the denominator and inches in the numerator. And since one foot equals 12 inches, the unit fraction is 12
inches over one foot. And again, we have feet divided by feet, and therefore the units
of feet simplify out leaving us with inches. And now we multiply. And notice how the denominator is one, and therefore we just have
six times 12 which is 72, giving us a product of 72 inches. We now know six feet equals 72 inches. Next, we’re going to
convert 126 inches to yards. We begin by writing 126
inches as a fraction with a denominator of one. And our goal is to convert to yards. Looking at the conversions,
there’s not a direct conversion from inches to yards, we
can use the conversion of one foot equals 12 inches
to convert inches to feet, and then use the conversion
one yard equals three feet to convert feet to yards. Because we have to use two conversions, we have to multiply by two unit fractions. Let’s first convert inches to feet. Because we have inches
in the numerator here, and we want inches to simplify out, we must have inches in the denominator and feet in the numerator. The conversion is one
foot equals 12 inches, which means the unit fraction
is one foot over 12 inches. Because we have inches divided by inches, the units of inches simplify out, and now we have feet, but our goal is to have yards not feet, so now we convert feet to yards using a second unit fraction. Because we want feet to simplify out, we have to have feet in the denominator. Because we have feet
in the numerator here, and we’re converting feet to yards, and therefore we have
yards in the numerator. The conversion is one
yard equals three feet, and therefore the second unit fraction is one yard over three feet. And notice the units of feet simplify out leaving us with yards. So now we multiply. In the numerator we have 126, in the denominator we have
12 times three which is 36, and now the units are yards. To finish, we need to find this quotient, let’s use the calculator. 126 divided by 36 is equal to 3.5. 126 inches is equal to 3.5 yards. Next, we want to convert
2,200 yards to miles. We begin by writing 2,200
yards as a fraction, with a denominator of one. There is not a direct conversion given from yards to miles. What we can do is convert yards to feet, that’s one yard equals three feet, and then convert feet to miles, since one mile equals 5,280 feet. So the first unit fraction, we
want to convert yards to feet. We want yards to simplify out, and therefore we have to
have yards in the denominator and feet in the numerator. Since one yard equals three feet, the first unit fraction is
three feet over one yard. Because we have yards divided yards, units of yards simplify out, leaving us with feet. And now we convert feet to miles. So we multiply by another unit fraction. We want feet to simplify out, and therefore we must have
feet in the denominator, and miles in the numerator. The conversion is one
mile equals 5,280 feet, and therefore the second unit fraction is one mile over 5,280 feet. Notice we have feet divided by feet, the units of feet simplify
out, leaving us with miles. Now we multiply. Notice how the units are miles. Multiplying in the numerator, we have 2,200 times three which is 6,600. In the denominator we have 5,280. And now we need to find this quotient to determine how many
miles in 2,200 yards. So going back to the calculator, 6,600 divided by 5,280 is equal to 1.25, giving us 1.25 miles. 2,200 yards equals 1.25 miles. And now for the last conversion, we want to convert 1.5 miles to inches. We begin by writing 1.5
miles as a fraction, with a denominator of one. Once again, there’s no direct conversion from miles to inches. We can use the conversion,
one mile equals 5,280 feet, convert miles to feet, and then use the conversion
one foot equals 12 inches to convert the feet into inches. Because we’re using two conversions, we need two unit fractions. Let’s first convert miles to feet, we want miles to simplify out, and therefore miles must
be in the denominator and feet in the numerator. One mile equals 5,280 feet, and therefore the first unit fraction is 5,280 over one mile. The units of miles simplify
out, leaving us with feet. And now we convert feet to inches. We want feet to simplify out, and therefore feet must
be in the denominator and inches in the numerator. The conversion of one
foot equals 12 inches, and therefore the unit fraction
is 12 inches over one foot. And notice the units of feet simplify out, leaving us with inches. So now we multiply, and again we now know the units are inches. Notice how the denominator is one, and therefore the product is
just 1.5 times 5,280 times 12, which gives us 95,040, and therefore there are
95,040 inches in 1.5 miles. I hope you found this helpful. |
### Section HSE Homogeneous Systems of Equations
From A First Course in Linear Algebra
Version 2.02
http://linear.ups.edu/
In this section we specialize to systems of linear equations where every equation has a zero as its constant term. Along the way, we will begin to express more and more ideas in the language of matrices and begin a move away from writing out whole systems of equations. The ideas initiated in this section will carry through the remainder of the course.
#### Subsection SHS: Solutions of Homogeneous Systems
As usual, we begin with a definition.
Definition HS
Homogeneous System
A system of linear equations, ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) is homogeneous if the vector of constants is the zero vector, in other words, b = 0.
Example AHSAC
Archetype C as a homogeneous system
For each archetype that is a system of equations, we have formulated a similar, yet different, homogeneous system of equations by replacing each equation’s constant term with a zero. To wit, for Archetype C, we can convert the original system of equations into the homogeneous system,
\eqalignno{ 2{x}_{1} − 3{x}_{2} + {x}_{3} − 6{x}_{4} & = 0 & & \cr 4{x}_{1} + {x}_{2} + 2{x}_{3} + 9{x}_{4} & = 0 & & \cr 3{x}_{1} + {x}_{2} + {x}_{3} + 8{x}_{4} & = 0 & & }
Can you quickly find a solution to this system without row-reducing the augmented matrix?
As you might have discovered by studying Example AHSAC, setting each variable to zero will always be a solution of a homogeneous system. This is the substance of the following theorem.
Theorem HSC
Homogeneous Systems are Consistent
Suppose that a system of linear equations is homogeneous. Then the system is consistent.
Proof Set each variable of the system to zero. When substituting these values into each equation, the left-hand side evaluates to zero, no matter what the coefficients are. Since a homogeneous system has zero on the right-hand side of each equation as the constant term, each equation is true. With one demonstrated solution, we can call the system consistent.
Since this solution is so obvious, we now define it as the trivial solution.
Definition TSHSE
Trivial Solution to Homogeneous Systems of Equations
Suppose a homogeneous system of linear equations has n variables. The solution {x}_{1} = 0, {x}_{2} = 0,…, {x}_{n} = 0 (i.e. x = 0) is called the trivial solution.
Here are three typical examples, which we will reference throughout this section. Work through the row operations as we bring each to reduced row-echelon form. Also notice what is similar in each example, and what differs.
Example HUSAB
Homogeneous, unique solution, Archetype B
Archetype B can be converted to the homogeneous system,
\eqalignno{ − 11{x}_{1} + 2{x}_{2} − 14{x}_{3} & = 0 & & \cr 23{x}_{1} − 6{x}_{2} + 33{x}_{3} & = 0 & & \cr 14{x}_{1} − 2{x}_{2} + 17{x}_{3} & = 0 & & }
whose augmented matrix row-reduces to
\left [\array{ \text{1}&0&0&0 \cr 0&\text{1}&0&0 \cr 0&0&\text{1}&0 } \right ]
By Theorem HSC, the system is consistent, and so the computation n − r = 3 − 3 = 0 means the solution set contains just a single solution. Then, this lone solution must be the trivial solution.
Example HISAA
Homogeneous, infinite solutions, Archetype A
Archetype A can be converted to the homogeneous system,
\eqalignno{ {x}_{1} − {x}_{2} + 2{x}_{3} & = 0 & & \cr 2{x}_{1} + {x}_{2} + {x}_{3} & = 0 & & \cr {x}_{1} + {x}_{2}\quad \quad & = 0 & & }
whose augmented matrix row-reduces to
\left [\array{ \text{1}&0& 1 &0 \cr 0&\text{1}&−1&0 \cr 0&0& 0 &0 } \right ]
By Theorem HSC, the system is consistent, and so the computation n − r = 3 − 2 = 1 means the solution set contains one free variable by Theorem FVCS, and hence has infinitely many solutions. We can describe this solution set using the free variable {x}_{3},
S = \left \{\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} } \right ]\mathrel{∣}{x}_{1} = −{x}_{3},\kern 1.95872pt {x}_{2} = {x}_{3}\right \} = \left \{\left [\array{ −{x}_{3} \cr {x}_{3} \cr {x}_{3} } \right ]\mathrel{∣}{x}_{3} ∈ {ℂ}^{}\right \}
Geometrically, these are points in three dimensions that lie on a line through the origin.
Homogeneous, infinite solutions, Archetype D
Archetype D (and identically, Archetype E) can be converted to the homogeneous system,
\eqalignno{ 2{x}_{1} + {x}_{2} + 7{x}_{3} − 7{x}_{4} & = 0 & & \cr − 3{x}_{1} + 4{x}_{2} − 5{x}_{3} − 6{x}_{4} & = 0 & & \cr {x}_{1} + {x}_{2} + 4{x}_{3} − 5{x}_{4} & = 0 & & }
whose augmented matrix row-reduces to
\left [\array{ \text{1}&0&3&−2&0 \cr 0&\text{1}&1&−3&0 \cr 0&0&0& 0 &0 } \right ]
By Theorem HSC, the system is consistent, and so the computation n − r = 4 − 2 = 2 means the solution set contains two free variables by Theorem FVCS, and hence has infinitely many solutions. We can describe this solution set using the free variables {x}_{3} and {x}_{4},
\eqalignno{ S & = \left \{\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} } \right ]\mathrel{∣}{x}_{1} = −3{x}_{3} + 2{x}_{4},\kern 1.95872pt {x}_{2} = −{x}_{3} + 3{x}_{4}\right \} & & \cr & = \left \{\left [\array{ −3{x}_{3} + 2{x}_{4} \cr −{x}_{3} + 3{x}_{4} \cr {x}_{3} \cr {x}_{4} } \right ]\mathrel{∣}{x}_{3},\kern 1.95872pt {x}_{4} ∈ {ℂ}^{}\right \} & & \cr & & }
After working through these examples, you might perform the same computations for the slightly larger example, Archetype J.
Notice that when we do row operations on the augmented matrix of a homogeneous system of linear equations the last column of the matrix is all zeros. Any one of the three allowable row operations will convert zeros to zeros and thus, the final column of the matrix in reduced row-echelon form will also be all zeros. So in this case, we may be as likely to reference only the coefficient matrix and presume that we remember that the final column begins with zeros, and after any number of row operations is still zero.
Example HISAD suggests the following theorem.
Theorem HMVEI
Homogeneous, More Variables than Equations, Infinite solutions
Suppose that a homogeneous system of linear equations has m equations and n variables with n > m. Then the system has infinitely many solutions.
Proof We are assuming the system is homogeneous, so Theorem HSC says it is consistent. Then the hypothesis that n > m, together with Theorem CMVEI, gives infinitely many solutions.
Example HUSAB and Example HISAA are concerned with homogeneous systems where n = m and expose a fundamental distinction between the two examples. One has a unique solution, while the other has infinitely many. These are exactly the only two possibilities for a homogeneous system and illustrate that each is possible (unlike the case when n > m where Theorem HMVEI tells us that there is only one possibility for a homogeneous system).
#### Subsection NSM: Null Space of a Matrix
The set of solutions to a homogeneous system (which by Theorem HSC is never empty) is of enough interest to warrant its own name. However, we define it as a property of the coefficient matrix, not as a property of some system of equations.
Definition NSM
Null Space of a Matrix
The null space of a matrix A, denoted N\kern -1.95872pt \left (A\right ), is the set of all the vectors that are solutions to the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ).
(This definition contains Notation NSM.)
In the Archetypes (Appendix A) each example that is a system of equations also has a corresponding homogeneous system of equations listed, and several sample solutions are given. These solutions will be elements of the null space of the coefficient matrix. We’ll look at one example.
Example NSEAI
Null space elements of Archetype I
The write-up for Archetype I lists several solutions of the corresponding homogeneous system. Here are two, written as solution vectors. We can say that they are in the null space of the coefficient matrix for the system of equations in Archetype I.
\eqalignno{ x = \left [\array{ 3 \cr 0 \cr −5 \cr −6 \cr 0 \cr 0 \cr 1 } \right ] & &y = \left [\array{ −4 \cr 1 \cr −3 \cr −2 \cr 1 \cr 1 \cr 1 } \right ] & & & & }
However, the vector
z = \left [\array{ 1 \cr 0 \cr 0 \cr 0 \cr 0 \cr 0 \cr 2 } \right ]
is not in the null space, since it is not a solution to the homogeneous system. For example, it fails to even make the first equation true.
Here are two (prototypical) examples of the computation of the null space of a matrix.
Example CNS1
Computing a null space, #1
Let’s compute the null space of
A = \left [\array{ 2&−1& 7 &−3&−8 \cr 1& 0 & 2 & 4 & 9 \cr 2& 2 &−2&−1& 8 } \right ]
which we write as N\kern -1.95872pt \left (A\right ). Translating Definition NSM, we simply desire to solve the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ). So we row-reduce the augmented matrix to obtain
\left [\array{ \text{1}&0& 2 &0&1&0 \cr 0&\text{1}&−3&0&4&0 \cr 0&0& 0 &\text{1}&2&0 } \right ]
The variables (of the homogeneous system) {x}_{3} and {x}_{5} are free (since columns 1, 2 and 4 are pivot columns), so we arrange the equations represented by the matrix in reduced row-echelon form to
\eqalignno{ {x}_{1} & = −2{x}_{3} − {x}_{5} & & \cr {x}_{2} & = 3{x}_{3} − 4{x}_{5} & & \cr {x}_{4} & = −2{x}_{5} & & \cr & & }
So we can write the infinite solution set as sets using column vectors,
N\kern -1.95872pt \left (A\right ) = \left \{\left [\array{ −2{x}_{3} − {x}_{5} \cr 3{x}_{3} − 4{x}_{5} \cr {x}_{3} \cr −2{x}_{5} \cr {x}_{5} } \right ]\mathrel{∣}{x}_{3},\kern 1.95872pt {x}_{5} ∈ {ℂ}^{}\right \}
Example CNS2
Computing a null space, #2
Let’s compute the null space of
C = \left [\array{ −4&6&1 \cr −1&4&1 \cr 5 &6&7 \cr 4 &7&1 } \right ]
which we write as N\kern -1.95872pt \left (C\right ). Translating Definition NSM, we simply desire to solve the homogeneous system ℒS\kern -1.95872pt \left (C,\kern 1.95872pt 0\right ). So we row-reduce the augmented matrix to obtain
\left [\array{ \text{1}&0&0&0 \cr 0&\text{1}&0&0 \cr 0&0&\text{1}&0 \cr 0&0&0&0 } \right ]
There are no free variables in the homogeneous system represented by the row-reduced matrix, so there is only the trivial solution, the zero vector, 0. So we can write the (trivial) solution set as
N\kern -1.95872pt \left (C\right ) = \left \{0\right \} = \left \{\left [\array{ 0 \cr 0 \cr 0 } \right ]\right \}
1. What is always true of the solution set for a homogeneous system of equations?
2. Suppose a homogeneous system of equations has 13 variables and 8 equations. How many solutions will it have? Why?
3. Describe in words (not symbols) the null space of a matrix.
#### Subsection EXC: Exercises
C10 Each Archetype (Appendix A) that is a system of equations has a corresponding homogeneous system with the same coefficient matrix. Compute the set of solutions for each. Notice that these solution sets are the null spaces of the coefficient matrices.
Archetype A
Archetype B
Archetype C
Archetype D/Archetype E
Archetype F
Archetype G/ Archetype H
Archetype I
and Archetype J
Contributed by Robert Beezer
C20 Archetype K and Archetype L are simply 5 × 5 matrices (i.e. they are not systems of equations). Compute the null space of each matrix.
Contributed by Robert Beezer
C30 Compute the null space of the matrix A, N\kern -1.95872pt \left (A\right ).
A = \left [\array{ 2 & 4 & 1 & 3 &8 \cr −1&−2&−1&−1&1 \cr 2 & 4 & 0 &−3&4 \cr 2 & 4 &−1&−7&4 } \right ]
Contributed by Robert Beezer Solution [190]
C31 Find the null space of the matrix B, N\kern -1.95872pt \left (B\right ).
\eqalignno{ B & = \left [\array{ −6& 4 &−36& 6 \cr 2 &−1& 10 &−1 \cr −3& 2 &−18& 3 } \right ] & & }
Contributed by Robert Beezer Solution [192]
M45 Without doing any computations, and without examining any solutions, say as much as possible about the form of the solution set for corresponding homogeneous system of equations of each archetype that is a system of equations.
Archetype A
Archetype B
Archetype C
Archetype D/Archetype E
Archetype F
Archetype G/Archetype H
Archetype I
Archetype J
Contributed by Robert Beezer
For Exercises M50–M52 say as much as possible about each system’s solution set. Be sure to make it clear which theorems you are using to reach your conclusions.
M50 A homogeneous system of 8 equations in 8 variables.
Contributed by Robert Beezer Solution [193]
M51 A homogeneous system of 8 equations in 9 variables.
Contributed by Robert Beezer Solution [193]
M52 A homogeneous system of 8 equations in 7 variables.
Contributed by Robert Beezer Solution [193]
T10 Prove or disprove: A system of linear equations is homogeneous if and only if the system has the zero vector as a solution.
Contributed by Martin Jackson Solution [193]
T20 Consider the homogeneous system of linear equations ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ), and suppose that u = \left [\array{ {u}_{1} \cr {u}_{2} \cr {u}_{3} \cr \mathop{\mathop{⋮}} \cr {u}_{n} } \right ] is one solution to the system of equations. Prove that v = \left [\array{ 4{u}_{1} \cr 4{u}_{2} \cr 4{u}_{3} \cr \mathop{\mathop{⋮}} \cr 4{u}_{n} } \right ]is also a solution to ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ).
Contributed by Robert Beezer Solution [194]
#### Subsection SOL: Solutions
C30 Contributed by Robert Beezer Statement [186]
Definition NSM tells us that the null space of A is the solution set to the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ). The augmented matrix of this system is
\left [\array{ 2 & 4 & 1 & 3 &8&0 \cr −1&−2&−1&−1&1&0 \cr 2 & 4 & 0 &−3&4&0 \cr 2 & 4 &−1&−7&4&0 } \right ]
To solve the system, we row-reduce the augmented matrix and obtain,
\left [\array{ \text{1}&2&0&0& 5 &0 \cr 0&0&\text{1}&0&−8&0 \cr 0&0&0&\text{1}& 2 &0 \cr 0&0&0&0& 0 &0 } \right ]
This matrix represents a system with equations having three dependent variables ({x}_{1}, {x}_{3}, and {x}_{4}) and two independent variables ({x}_{2} and {x}_{5}). These equations rearrange to
\eqalignno{ {x}_{1} & = −2{x}_{2} − 5{x}_{5} &{x}_{3} & = 8{x}_{5} &{x}_{4} & = −2{x}_{5} & & & & & & }
So we can write the solution set (which is the requested null space) as
N\kern -1.95872pt \left (A\right ) = \left \{\left [\array{ −2{x}_{2} − 5{x}_{5} \cr {x}_{2} \cr 8{x}_{5} \cr −2{x}_{5} \cr {x}_{5} } \right ]\mathrel{∣}{x}_{2},{x}_{5} ∈ {ℂ}^{}\right \}
C31 Contributed by Robert Beezer Statement [187]
We form the augmented matrix of the homogeneous system ℒS\kern -1.95872pt \left (B,\kern 1.95872pt 0\right ) and row-reduce the matrix,
\eqalignno{ \left [\array{ −6& 4 &−36& 6 &0 \cr 2 &−1& 10 &−1&0 \cr −3& 2 &−18& 3 &0 } \right ] &\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ \text{1}&0& 2 &1&0 \cr 0&\text{1}&−6&3&0 \cr 0&0& 0 &0&0 } \right ] & & }
We knew ahead of time that this system would be consistent (Theorem HSC), but we can now see there are n − r = 4 − 2 = 2 free variables, namely {x}_{3} and {x}_{4} (Theorem FVCS). Based on this analysis, we can rearrange the equations associated with each nonzero row of the reduced row-echelon form into an expression for the lone dependent variable as a function of the free variables. We arrive at the solution set to the homogeneous system, which is the null space of the matrix by Definition NSM,
\eqalignno{ N\kern -1.95872pt \left (B\right ) = \left \{\left [\array{ −2{x}_{3} − {x}_{4} \cr 6{x}_{3} − 3{x}_{4} \cr {x}_{3} \cr {x}_{4} } \right ]\mathrel{∣}{x}_{3},\kern 1.95872pt {x}_{4} ∈ ℂ\right \} & & }
M50 Contributed by Robert Beezer Statement [188]
Since the system is homogeneous, we know it has the trivial solution (Theorem HSC). We cannot say anymore based on the information provided, except to say that there is either a unique solution or infinitely many solutions (Theorem PSSLS). See Archetype A and Archetype B to understand the possibilities.
M51 Contributed by Robert Beezer Statement [188]
Since there are more variables than equations, Theorem HMVEI applies and tells us that the solution set is infinite. From the proof of Theorem HSC we know that the zero vector is one solution.
M52 Contributed by Robert Beezer Statement [188]
By Theorem HSC, we know the system is consistent because the zero vector is always a solution of a homogeneous system. There is no more that we can say, since both a unique solution and infinitely many solutions are possibilities.
T10 Contributed by Robert Beezer Statement [188]
This is a true statement. A proof is:
() Suppose we have a homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ). Then by substituting the scalar zero for each variable, we arrive at true statements for each equation. So the zero vector is a solution. This is the content of Theorem HSC.
() Suppose now that we have a generic (i.e. not necessarily homogeneous) system of equations, ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) that has the zero vector as a solution. Upon substituting this solution into the system, we discover that each component of b must also be zero. So b = 0.
T20 Contributed by Robert Beezer Statement [188]
Suppose that a single equation from this system (the i-th one) has the form,
{a}_{i1}{x}_{1} + {a}_{i2}{x}_{2} + {a}_{i3}{x}_{3} + \mathrel{⋯} + {a}_{in}{x}_{n} = 0
Evaluate the left-hand side of this equation with the components of the proposed solution vector v,
\eqalignno{ {a}_{i1}\left (4{u}_{1}\right ) & + {a}_{i2}\left (4{u}_{2}\right ) + {a}_{i3}\left (4{u}_{3}\right ) + \mathrel{⋯} + {a}_{in}\left (4{u}_{n}\right ) & & & & \cr & = 4{a}_{i1}{u}_{1} + 4{a}_{i2}{u}_{2} + 4{a}_{i3}{u}_{3} + \mathrel{⋯} + 4{a}_{in}{u}_{n} & &\text{Commutativity} & & & & \cr & = 4\left ({a}_{i1}{u}_{1} + {a}_{i2}{u}_{2} + {a}_{i3}{u}_{3} + \mathrel{⋯} + {a}_{in}{u}_{n}\right ) & &\text{Distributivity} & & & & \cr & = 4(0) & &\text{$u$ solution to $ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right )$} & & & & \cr & = 0 & & & & }
So v makes each equation true, and so is a solution to the system.
Notice that this result is not true if we change ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) from a homogeneous system to a non-homogeneous system. Can you create an example of a (non-homogeneous) system with a solution u such that v is not a solution? |
# How to Calculate The Capacity of Water Tank And Construct Any Shape of Tank For A Specific Capacity
Updated on August 26, 2011
## simple calculations
Generally water is stored either in a cylindrical tank or in a square tank. We can easily find out the capacity of water storage tank of any size in liters. For eg. if you want to find out the capacity of a square water tank measuring 1m length, 1m breadth, 1m height, first convert meter to cm. As the formula to find out the volume of cube is a^3 we can get answer as 1000,000 cc (1 m = 100 cm)
1000 cc is 1 liter
Therefore above said cubic tank can hold 1000 liters.
By this same method we can easily calculate the capacity of tanks in various sizes.
## How to Construct Any Type of a Water Tank for a Specific Capacity
We can easily construct any type of water tank like cylindrical, cubical and rectangular for a specific capacity. For instance, if you want to get measurements of tank in any shape to store 5000 liters, you will feel easy with following examples. But, you need to count only inner measurements of tanks to get accurate answers.
Let us try to find out the side of a cubical tank to store 5000 liters.
Formula for the volume of a cube = a^3, a = the side of cubical tank
Specific Capacity = 5000 liters (i.e.) 5000,000 cc
Therefore a^3 = 5000,000 cc
The side 'a' = (5000,000)^1/3 = 170.99 cm
Similarly we can find out the measurements of a rectangular tank holding 5000 liters
Formula for the volume of a rectangular tank is l*b*h
Say the length is 1.5 times breadth and the height is equal to breadth
Now the volume of the tank can be written as 1.5*b*b*b = 1.5*b^3
Specific Capacity = 5000 liters (i.e.) 5000,000 cc
Therefore 1.5*b^3 = 5000,000 cc
Now 'b' will come as [5000,000/1.5]^[1/3] = 149.38 cm
Now we can easily get length, breadth and height of the tank as follows, l = 224.07 cm, b = 149.38 cm, h = 149.38 cm which can hold 5000 liters of water
The same way we can easily design a cylindrical tank to store 5000 liters
Formula for the volume of a cylinder is 3.14*r^2*h
Say height is 2 times radius of the tank
Now the volume of the tank can be written as 3.14*r^2*2*r = 6.28*r^3
Specific Capacity = 5000 liters (i.e.) 5000,000 cc
Therefore 6.28*r^3 = 5000,000 cc
Now 'r' will come as [5000,000/6.28]^[1/3] = 92.68 cm
Now we can easily get height and radius of the cylindrical tank holding 5000 liters as follows, h = 185.36 cm and r = 92.68 cm
You can also try to find out the measurements of tanks for different capacities.
## Popular
28
57
• ### Math Project Ideas: Examples of Project-Based Learning
10
Submit a Comment
• Deepak8426
3 years ago
Dear Sir, i have to create 3000 ltr of water tank with 5 ft length, 3 ft breadth, i cant understand how much it's depth to store 3000 ltr water. please suggest.
• PRAVEEN
4 years ago from BANGALORE
Hi,
I need find out the my UGR tank capacity Width: 24f.4'' ,Lang ht 32f.6'' and Hight 8f.1'' pls replay me
• pragneshprajapati
4 years ago
hello sir,
how r you,
i want to calculate thickness of of h.d.p.e plastics tank as per capacity. my tank capacity is 60 kl = 60,000 litre, so how much wall thickness get for this tank, tank size: approx 4 mtr dia x approx 4.8 mtr heigh.
but its thickness = ? , i dont know
• AUTHOR
ratnaveera
7 years ago from Cumbum
Hello sdhiman111, Thanks for visiting this page. Thickness is depending upon the material used for the construction. 5 mm wall thickness will be enough for the ms tank.
• sdhiman111
7 years ago
I want calculate the thickness of body plate to store the 5000 ltr. water in tank
• ARIF JAAN
7 years ago
Dear i want to calculate the cost of water tank e.g.
its capacity is 50000 gallons i want to know that if i sell a 20000 gallons the remaining 30000 gallons know i need a per cost of that anyone please help me thanks
• shamskpm
8 years ago
Hi, Ratna,
Im Shams, our house is under construction. we prefer to use ceramic disc shower mixers in the bath rooms. I wish to know the ideal height for an Overhead Tank from the roof height of +680Lvl. or please advise us for an economical solution to get the right pressure in all mixers especially in first floor.
shahana.shams@gmail.com
• subhash7
8 years ago from KSA
Hi, Ratna ..
i am working in fiber glass company as draughtsman. And i want to know TANK calculations each and everything in simple way our metric in meters and letters if can mail send a mail to me sureshgiri.gs@gmail.com.
So i hope u can help me out of this.
8 years ago
8 years ago
i am a welder based in tank construction i just need a simple formular for calculating the volume of a tank.kindly send it to my email ayadiariyo@yahoo.com
• shaik
8 years ago
I need to known the tank capacity for my fire pump which is 120gmp. can any help me regarding this
• AUTHOR
ratnaveera
8 years ago from Cumbum
Dear sujay, It's very easy to calculate the size of a rectangular tank that can hold 27000 liter. The formula to the volume of a rectangular tank is l*b*h. Say the length is 2 times breadth and breadth is equal to the height. Now 2b*b*b = 27000 liter
b^3 = 27000/2
b^3 = 13500*1000 cc
Therefore
b = 238.11 cm
l = 476.22 cm
h = 238.11 cm
That's all.
• sujay
8 years ago
dear sir i have 27000ltr capcity how to find size of rectangular tank
• Maldonagus
8 years ago
How do i know the pump on and pump off level of an existing hydro tank and level on level off of a water storage tank
• satish
8 years ago
if i have 0.02cc cartridge then how much fluid will pass through that cartridge. pressure variation can be effect on volume of fluid
• faheem
8 years ago
hello ma'am...i need to design an overhead steel square water tank of size 2m x 2m x 2m at a height of 3mts...
• bhanu
8 years ago
Mam, I want to calculate the capacity of concrete bucket having upper diameter 700mm, lower diameter 400mm & height 550mm. please tell me as soon as possible.
• AUTHOR
ratnaveera
8 years ago from Cumbum
Dear trix, Thank you for coming here and your nice question. Here is your answer
1 GALLON = 4.54 LITERS
Your required tank capacity is 20000*4.54 liters = 90800000 cc
your cylinderical tank height is 15 ft = 450 cm
Formula for the volume of a cylinder is 3.14*r*r*h
Now substitute the values
90800000 cc = 3.14*r*r*450 cm
r*r = 90800000/[3.14*450 cm]
r*r = 64260.438
r = 253.49 cm
You can check out it now! The r value is correct to my knowledge. Thanks again! You are always welcome friend!
• trix
8 years ago
i used your method to calculate the height of my cylindrical tank if i want to calculate the head how would i go about this i got 15 feet for the height and the capacity of the tank is approximately 20000gallons
• Tom
9 years ago
Here is a easy way to figure it out! http://volumeformula.org/cylinder.html
• AUTHOR
ratnaveera
9 years ago from Cumbum
Dear Sanjay, Thank you for your question. Could you tell me what shape of tank to be constructed?
• sanjay
9 years ago
Dear Ratnaveera, I want to construct 100000 Liters water tank for our nursery can you suggest me dimensions for the tank and design?
• bhanu
9 years ago
thanxxxxx a ton mam.... !!
• AUTHOR
ratnaveera
9 years ago from Cumbum
Dear bhanu, Say height is 2 times r. Then formula to find out the volume of the cylinder is 3.14*r*r*2r which equals
20000000 cc. Therefore 20000000/6.28 = r^3. Now radius of the tank can be calculated as 147 cm and height as 294 cm
• bhanu
9 years ago
hello, i want to make 20tons capacity cylindrical tank. so what is the formula and dimensions, radius and height.
• Santhosh Kumar
9 years ago
Dear sir Wat is CFM
• chandu2373
9 years ago
Dear Ratna can u help me up with this how can i calculate
volume of an elliptical cylinder in litters where i am having all the three sides i.e length of tank is L-289.5cm, width of tank is W-193cm, height of tank is H-134.6cm
• Karthikeyan
9 years ago
Good and useful..
• AUTHOR
ratnaveera
9 years ago from Cumbum
Dear Shadab, Thanks for your nice question. CC means cm cube. That is cm x cm x cm.
9 years ago
Ratna What Is CC?
• 143
9 years ago
8x6x15 sized tank=? liters
• dheeraj
9 years ago
I have an underground tank and want to calculate that how much liters of water it contains. So, please tell me how can I get to know its capacity. For your help I have taken measurement as breadth=4 inch length=5 inch and height=6 inch. Please reply on id as dheeraj142@yahoo.co.in
• noman
9 years ago
thnx for the help.. and can u tell me water flow mainaining formula?? eg i have to get 2 ltr per second water from over head tank to ground tank..and same flow for ground tank to over head tank..
• AUTHOR
ratnaveera
9 years ago from Cumbum
Dear noman, Happy to know that my answers are helpful to you. I am a chemical Engineer. I am not sure about helping you on mechanical field doubts. Any way you ask your questions and let us try!
• noman
9 years ago
Thnx Buddy..
which field u r related to?
if i need to ask any formulae related to mechanical field..
can u help me?
Thnx again
• AUTHOR
ratnaveera
9 years ago from Cumbum
Dear noman, Thanks for your questions. The formula to find out the volume of a cylindrical tank is 3.14*r*r*h, r=radius of the cylindrical tank and whereas h=height of it.
First you need to convert measurement of tank into cm. Then only you get easily the quantity by volume. For eg. if r=100 cm and h=150 cm. The quantity of that cylindrical tank is coming as 4710000 cc that is 4710 litres.
The formula for the square tank is a^3, therefore a^3=5000000 cc, a=170.9 cm Once again thanks for your questions. Hope this would be useful for you!
• noman
9 years ago
Hi..
if i need to take out the capicity of cylindrical tank then which formula i shud use??
and 2nd thing.. for eg if i have to design a square tank for 5000 ltrs water then which formula i have to use??
• Amrith
9 years ago
Nice stuff !! This helped me .. Thank you very much !
• AUTHOR
ratnaveera
10 years ago from Cumbum
Dear Ragu, Thanks for your question. First you need to convert 200000 litres into cc and it'll come as 200000000 cc. If it be a square the side of the tank would come as 584.8 cm. If it be a rectangular tank with length 1.5 and height 1 times breadth, then the formula to find out the measurement is 200000000 cc= 1.5b*1b*1b
= 1.5b^3
Therefore b= 510.8 cm l= 766.2 cm h= 510.8 cm
• Ragu
10 years ago
Hi,
I just like to know how to calculate the capacity for 200000 ltrs in rectangular tank. kindly reply please
• AUTHOR
ratnaveera
10 years ago from Cumbum
Dear khurshed, First you need to convert the measurement into cm. For eg. if you have 2 m breadth and 1 m height and 3 m long tank. You need to convert them into cm as 200 cm, 100 cm and 300 cm. Now you will get the volume multiplying them l*b*h, 200*100*300=6000000 cc (i.e) 6000 litres
• khurshed
10 years ago
hi i have 6000x3000x3000 water tank i want to know the volume of total please reply
• sks
11 years ago
nice info thanks
working |
NCERT Class 6 Solutions: Whole Numbers (Chapter 2) Exercise 2.3–Part 2
Q-4 Find using distributive property:
Solution:
Q-5 Study the pattern:
Equation Equation Equation Equation Equation
Write the next two steps. Can you say how the pattern works?
Solution:
The pattern works in such a way that in each step the digit added to the least significant digit is one less than previous digit. Therefore we can write the numbers as follows:
Let’s understand why such a pattern emerges with an example:
• , this is true for all numbers so we can generalize it
• Now subtract 1234 from our example we get, or number . Applying the distributive property we get . In general
• Now subtract 1234 again, we will get . Now we can rewrite, 11110 – 1234 = (10000 + 1000 + 100 + 10) – (1000 + 200 + 30 + 4).
Again using the distributive property we can write,
,
which we know would be
= = 9000 + 800 + 70 + 6=9876. Therefore, generalizing this argument we can write, .
Explore Solutions for Mathematics |
# How do you solve 6/ (x + 3) = 4/(x - 3)?
Oct 8, 2015
$x = 15$
#### Explanation:
Given
[1]$\textcolor{w h i t e}{\text{XXX}} \frac{6}{x + 3} = \frac{4}{x - 3}$
Multiplying both sides by $\left(x + 3\right) \left(x - 3\right)$
[2]$\textcolor{w h i t e}{\text{XXX}} \frac{6 \cancel{x + 3} \left(x - 3\right)}{\cancel{x + 3}} = \frac{4 \left(x + 3\right) \left(\cancel{x - 3}\right)}{\cancel{x - 3}}$
[3]$\textcolor{w h i t e}{\text{XXX}} 6 \left(x - 3\right) = 4 \left(x + 3\right)$
Simplifying by distribution
[4]$\textcolor{w h i t e}{\text{XXX}} 6 x - 18 = 4 x + 12$
Add $\left(18 - 4 x\right)$ to both sides
[5]$\textcolor{w h i t e}{\text{XXX}} 2 x = 30$
Divide both sides by $2$
[6]#color(white)("XXX")x=15 |
# math4b-16 - Math 4B Lecture 16 Doug Moore Electrical...
• Notes
• 23
This preview shows page 1 - 7 out of 23 pages.
Math 4B Lecture 16 May 22, 2013 Doug Moore
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Electrical circuits give rise to complicated systems of differen- tial equations. Circuits which include linear resistors, inductors, and capacitors give rise to linear systems of equations. Other circuit elements, such as transistors or vacuum tubes, are often nonlinear, and lead to nonlinear systems. We might be interested in designing a radio from various circuit elements. To design such a circuit, we would divide the circuit into various stages, oscillators, tuners, amplifiers and so forth.
Here we have pictured a relatively simple circuit. To find the equations for the currents flowing through an electric circuit such as this, we need to know the voltage drops associated with the various linear circuit elements.
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The voltage drops are: A current x ( t ) flowing through a resistor of resistance R causes a voltage drop E R = Rx ( t ). Here the resistance R is measured in ohms, the current x in amperes, and the voltage drop E R in volts. A current y ( t ) flowing through an inductor of inductance L causes a voltage drop E L = L ( dy/dt )( t ), where the inductance L is measuure in henrys. A charge q ( t ) accumulated on a capacitor of capacitance C causes a voltage drop E C = (1 /C ) q ( t ), where the capacitance C is measured in farads. Thus if z ( t ) is the current flowing through the capacitor, z = dq dt q ( t ) = Z t 0 z ( τ ) + constant E C = (1 /C ) Z t 0 z ( τ ) + constant .
To determine the equations for the currents, we then apply Kirchhoff’s laws: Kirchhoff’s voltage law: The sum of voltage drops around any loop in the circuit must equal the voltage which is applied to that loop. Kirchhoff’s current law: The sum of currents flowing into a node must equal the the sum of currents flowing out of the node.
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According to Kirchhoff’s current law, if x ( t ) is the current flowing up through the left hand wire, and y ( t ) is the current flowing down through the middle wire, then the current flowing down through the right hand wire must be ( x - y )( t ). Hence, applying Kirchhoff’s voltage law to the left hand loop, we
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## Poisson Distribution
### Learning Outcomes
• Recognize the Poisson probability distribution and apply it appropriately
There are two main characteristics of a Poisson experiment.
1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
2. The Poisson distribution may be used to approximate the binomial if the probability of success is “small” (such as 0.01) and the number of trials is “large” (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a “success.”
The random variable $X=$ the number of occurrences in the interval of interest.
### Example
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?
## Notation for the Poisson: $P=$ Poisson Probability Distribution Function
$X{\sim}P(\mu)$
Read this as “X is a random variable with a Poisson distribution.” The parameter is $\mu$ (or $\lambda$); $\mu$ (or $\lambda$)$=$ the mean for the interval of interest.
### Example
Leah’s answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
### Example
According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let $X=$ the number of emails an email user receives per day. The discrete random variable X takes on the values $x=$0, 1, 2 …. The random variable has a Poisson distribution: $X{\sim}P(147)$. The mean is 147 emails.
1. What is the probability that an email user receives exactly 160 emails per day?
2. What is the probability that an email user receives at most 160 emails per day?
3. What is the standard deviation?
## Review
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is “small” (less than or equal to 0.05) and the number of trials is “large” (greater than or equal to 20).
## Formula Review
$X{\sim}P(\mu)$ means that X has a Poisson probability distribution where $X=$ the number of occurrences in the interval of interest.
X takes on the values $x=$0, 1, 2, 3, …
The mean μ is typically given.
The variance is $σ^{2}=\mu$, and the standard deviation is $\sigma=\sqrt{\mu}$.
When P(μ) is used to approximate a binomial distribution, $\mu=np$ where n represents the number of independent trials and represents the probability of success in a single trial. |
# Solving Two-Step Equations With Context [Part 2]
Visual math talk prompts with context used to help emerge the writing and solving of two-step algebraic equations involving partitive division.
## In This Set of Math Visual Prompts…
Students will explore modelling and solving two-step algebraic equations involving multiplication and division.
## Intentionality…
This set of visual math talk prompts is taken from the Math Talk section of Day 4 in the Make Math Moments Problem Based Unit called Planting Flowers Revisited. The purpose of the Math Talk is to reinforce key concepts and big ideas from this problem based math unit including:
• There are two types of division;
• Partitive division is when the total quota is known (the dividend), and the number of parts or groups (the divisor) is known;
• Partitive division reveals a rate;
• In partitive division, the dividend and the divisor often have different units;
• The dividend from any division sentence can be decomposed into smaller parts to allow for friendlier division by the divisor. This strategy is known as partial quotients. (i.e.: 85 ÷ 5 = 45 ÷ 5 + 40 ÷ 5 = 9 + 8 = 17);
• Division is the inverse operation of multiplication;
• Variables are used to represent changing or unknown quantities;
• When solving an equation where the coefficient of the unknown variable is not equal to 1, division is required to determine the value of the unknown variable and the context of the problem determines which type of division is required.
• A two-step equation is an algebraic equation that takes you two steps to solve.
Unsure about some of the terminology listed above?
## Preparing to Facilitate
Present the following equations one at a time. Encourage students to describe a context that could be represented by each. All of today’s equations and contexts should be partitive. Meaning, the number of groups (parts) is known, the rate is unknown.
For example:
Luisa has 4 large containers with an equal number of monarch caterpillars in each. She also has two swallowtail caterpillars in a smaller container. If she has 30 caterpillars altogether, how many caterpillars are in each of the large containers?
Write an equation and solve using a model of your choice.
Students should use a model of their choice to solve and/or represent their thinking.
Four (4) additional visual math prompts are shared in the Teacher Guide from Day 4 of the Planting Flowers Revisited problem based math unit
Although you could lead this math talk by orally sharing the context and representing student thinking on a chalkboard/whiteboard, consider leveraging the following visual math talk prompt videos to ensure accessibility for all students. We would recommend that you still model student thinking to ensure student voice is honoured prior to sharing the visual silent solution shared in the visual prompt video.
Visual Math Talk Prompt #1
Show the visual math talk prompt video and be ready to pause once prompted.
In the video, students are presented with visuals and a text prompt to set the context involving an equal quantity of caterpillars stored in large containers plus an additional two caterpillars in a smaller container.
You’ll notice that this first prompt attempts to make this context very visual and scaffolds students toward an algebraic equation that could be used to represent this situation, 4m + 2 = 30.
It is worth noting that this context is going to require the use of partitive division as the unknown variable represents the rate of caterpillars per large container while the number of large containers is known.
Some students might immediately begin using trial and error to determine the value of m. For example, some students might try m = 5 and will skip count to 20 (or multiply 4 large containers by 5 caterpillars) plus 2 caterpillars is 22. Therefore, they might adjust up until they land on the value of m that will result in a total of 30 caterpillars.
Other students might leverage a linear model like a stacked bar model beginning with 30 caterpillars and working backwards to reveal how many caterpillars must be in the 4 large containers. By subtracting the two additional caterpillars, students will quickly realize that there must be 28 caterpillars divided equally across the 4 large containers.
Now that the 2 additional caterpillars have been removed from the equation, we now have a simple one step equation involving partitive division to solve. Some students might choose to scale down by halving twice, or other students might immediately partition the quantity into 4 groups representing the 4 large containers.
Ensure that students are able to connect how multiplying by 1 fourth (or by 1 half and again by 1 half) is equivalent to dividing by 4. This idea can be helpful for students to solidify their understanding of how dividing by a quantity is equivalent to multiplying by the reciprocal.
Facilitator Note:
Be sure to ask students about what they notice when they analyze the equation 4(m) + 2 = 30 and the operations they used to solve for the unknown number of caterpillars per large container, m. Some students may be able to recognize that we are using opposite operations to undo or work backwards to find the value of m. Ensure this is explicitly highlighted.
Visual Math Talk Prompt #2
In this next visual math talk prompt, we are encouraging students to start with an equation and attempt to craft a scenario using the same context.
Some students may quickly realize that this equation is equivalent to the equation from the first visual math talk prompt and we can now highlight the use of opposite operations leveraged in the previous prompt (subtracting 2 caterpillars and then dividing by 4 large containers).
Encourage students to articulate what each of the quantities in this equation could represent in this scenario.
In the visual math talk prompt video, you’ll see the 30 caterpillars being partitioned by subtracting 2 caterpillars to be stored in the single small container and the remaining 28 caterpillars fair shared (divided partitively) across 4 large containers.
We can also leverage our stacked bar model from the previous prompt to show how this equivalent equation can be solved for m by performing the same operations.
## Want to Explore These Concepts & Skills Further?
Four (4) additional math talk prompts are available in Day 4 of the Planting Flowers Revisited problem based math unit that you can dive into now.
Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page.
Did you use this in your classroom or at home? How’d it go? Post in the comments!
Math IS Visual. Let’s teach it that way. |
# 3.2 Vector addition and subtraction: graphical methods (Page 4/15)
Page 4 / 15
## Multiplication of vectors and scalars
If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk , or 82.5 m, in a direction $\text{66}\text{.}0\text{º}$ north of east. This is an example of multiplying a vector by a positive scalar . Notice that the magnitude changes, but the direction stays the same.
If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector $\mathbf{A}$ is multiplied by a scalar $c$ ,
• the magnitude of the vector becomes the absolute value of $c$ $A$ ,
• if $c$ is positive, the direction of the vector does not change,
• if $c$ is negative, the direction is reversed.
In our case, $c=3$ and $A=27.5 m$ . Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.
## Resolving a vector into components
In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x - and y -components, or the north-south and east-west components.
For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction $\text{29}\text{.0º}$ north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion , and much more when we cover forces in Dynamics: Newton’s Laws of Motion . Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.
## Phet explorations: maze game
Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.
## Summary
• The graphical method of adding vectors $\mathbf{A}$ and $\mathbf{B}$ involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector $\mathbf{R}$ is defined such that $\mathbf{\text{A}}+\mathbf{\text{B}}=\mathbf{\text{R}}$ . The magnitude and direction of $\mathbf{R}$ are then determined with a ruler and protractor, respectively.
• The graphical method of subtracting vector $\mathbf{B}$ from $\mathbf{A}$ involves adding the opposite of vector $\mathbf{B}$ , which is defined as $-\mathbf{B}$ . In this case, $\text{A}–\mathbf{\text{B}}=\mathbf{\text{A}}+\left(\text{–B}\right)=\text{R}$ . Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector $\mathbf{R}$ .
• Addition of vectors is commutative such that .
• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
• If a vector $\mathbf{A}$ is multiplied by a scalar quantity $c$ , the magnitude of the product is given by $\text{cA}$ . If $c$ is positive, the direction of the product points in the same direction as $\mathbf{A}$ ; if $c$ is negative, the direction of the product points in the opposite direction as $\mathbf{A}$ .
what is physics
a15kg powerexerted by the foresafter 3second
what is displacement
movement in a direction
Jason
hello
Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
while
Hosea
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
hi
Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
hi guys
thomas |
Pythagoras and his theorem
Monica Kochar
Most math students know the following:
Pythagoras theorem
If there is a right triangle, the sum of the squares of the lengths of the two sides will equal the square of the length of the hypotenuse, which is the longest side of the triangle.
Hence, a2 + b2 = c2
In fact, this applies to any three numbers a, b, and c that satisfy this condition. Hence, a2 + b2 = c2, will form a right triangle with ‘c’ as the hypotenuse.
For example:
[3, 4, 5], for 52 = 32 + 42
[5, 12, 13], for 132 = 52 + 122
[9, 40, 41], for 412 = 402 + 92
Try yourself!
See how the Pythagoras theorem helps you solve this problem!
A removal truck comes to pick up a pole of length 6.5m. The dimensions of the truck are 3m, 3.4m, and 4.1m. Will the pole fit in the truck?
But why does the theorem work?
The theorem was known to the Babylonians, Chinese, and Indians long before Pythagoras. Pythagoras did not discover it, he only proved it!
Many proofs exist for this theorem. The simplest proof is shown below.
For the [3, 4, 5] right triangle, 32, 42 and 52 can be seen as areas of the squares on the sides of lengths 3, 4, and 5.
Now 25 = 9 + 16, and hence 52 = 32 + 42
In general, for a triangle with sides [a, b, c], a2, b2, c2 represent areas of the squares on the sides.
According to Pythagoras, this is the relation between the areas.
Or
a2 + b2 = c2
Several other proofs can be checked from the website http://www.cut-the-knot.com/pythagoras/
Pythagoras triples
Such three numbers a, b, and c are called Pythagorean triples. The smallest triple is [3, 4, 5]. It is very easy to get more triples by scaling up these numbers. If [a, b, c] is a Pythagorean triple, so is any triple obtained by multiplying all numbers of the triple.
So, from [3,4,5], we get [6,8,10] or [9,12,15] or [12,16,20] and so on… Hence there are infinite Pythagorean triples!
In fact, [3n, 4n, 5n] are always Pythagorean triples for an integer ‘n’.
Try yourself!
• Check if a Pythagorean triple – scaled up also gives another Pythagorean triple by taking your own examples!
Check the Pythagorean triples given below, scaled using a basic triple
[3,4,5] [6,8,10] [9,12,15] [12,16,20] [15,20,25] [8,15,17] [16,30,34] [24,45,51] [32,60,68] [40,75,85]
Look at them carefully. Do you notice that in a triple either:
a. All three numbers are even? or
b. Two are odd and one is even?
These are the only two combinations possible. A Pythagorean triple can never be made of three odd numbers or two even numbers and one odd!
• Why not three odd numbers? The square of an odd number is odd and the sum of two odd numbers is even! Hence it is not possible.
• Why not two even and one odd? The square of even numbers is even and the sum of even numbers is even. Further, the square of an odd number is odd. The sum of an odd and an even number is even. Hence, it is not possible.
Try yourself!
• Check if the rules given above work by taking your own examples!
It will be very difficult however to keep scaling in order to generate the triples. Euclid, the Greek mathematician, discovered a formula for creating the Pythagorean triples. It says:
If m and n are two positive integers with m>n, then [a, b, c] where a = k (m2-n2), b = k (2mn) and c = k (m2+n2) will always be a Pythagorean triple for a positive integer ‘k’.
Try yourself!
• Check for yourself whether a = k (m2-n2), b = k (2mn) and c = k (m2+n2) will always be a Pythagorean triple for a +ve integer ‘k’.
Who was Pythagoras?
Pythagoras was a Greek philosopher who lived between 580 BC and 500 BC. He made significant contributions to mathematics, astronomy, and the theory of music. He spent much of his life studying mathematics and built a special school, the Pythagoras Society, whose members followed strict rules. He believed that everything in the world could be explained by numbers. He identified numbers as male or female, ugly or beautiful, or had a special meaning attached to every number.
Some ideas that Pythagoras worked on are interesting for they are things you still learn about in school. For example:
Odd numbers – 1, 3, 5, 7, 9, 11
Even numbers – 2, 4, 6, 8, 10, 12
Triangular numbers – 1, 3, 6, 10, 15
Square numbers – 1, 4, 9, 16, 25
He also studied shapes and was interested in triangles. One theorem they worked on is this famous one:
The sum of the angles of a triangle is equal to two right angles or 1802.
This means that if you take any triangle, tear off the corners and fit them together on a line, you will make a straight line (that is the same as two right-angles or 180°).
As we can see, Pythagoras contributed a lot more to mathematics than just the Pythagoras theorem.
However, Pythagoras worked with the members of the Society he formed and therefore it is difficult to decide whether all the ideas attributed to Pythagoras are really his. The Society was always secretive about its work and individuals were never given credit. All of the Society’s findings were credited always to Pythagoras.
Pythagoras also related music to mathematics. He had long played the seven string lyre, and discovered musical harmony on his own.
There are many different stories around Pythagoras’ death.
• Killed by an angry mob
• Caught up in a war between Agrigentum and Syracusans
• Killed by the Syracusans
• Thrown out of his school, he went to Metapontum where he starved to death;
• Refused to trample a crop of bean plants in order to escape and was caught.
Whatever the reason for his death, Pythagoras lives on through the Pythagorean theorem. For this is a cornerstone of mathematics. It continues to fascinate mathematicians. There are more than 400 different proofs of the theorem so far!
The author creates various kinds of multi-sensory learning environments for mathematics classes whereby learning the subject becomes a fun-filled experience. Her USP is her capacity to see the child’s learning perspective in any situation. Now, she uses this extremely important element while working with adults in the zones of teacher training, content development, testing, and writing. She can be reached through her website www.humanemaths.com. |
# Multiplication Table Chart Vector
Learning multiplication after counting, addition, and subtraction is perfect. Youngsters learn arithmetic via a normal progression. This growth of understanding arithmetic is truly the pursuing: counting, addition, subtraction, multiplication, and lastly section. This statement leads to the concern why learn arithmetic in this series? Most importantly, why find out multiplication soon after counting, addition, and subtraction just before department?
## These facts respond to these queries:
1. Children find out counting initially by associating graphic objects with their hands. A concrete case in point: The number of apples are there inside the basket? Much more abstract example is how outdated are you presently?
2. From counting phone numbers, the next plausible move is addition combined with subtraction. Addition and subtraction tables can be quite helpful training tools for children as they are visible tools producing the transition from counting less difficult.
3. That ought to be discovered after that, multiplication or division? Multiplication is shorthand for addition. At this moment, children possess a firm understanding of addition. For that reason, multiplication will be the after that rational type of arithmetic to understand.
## Assess basic principles of multiplication. Also, evaluate the fundamentals the way you use a multiplication table.
We will evaluation a multiplication instance. Utilizing a Multiplication Table, increase several times about three and have a solution twelve: 4 by 3 = 12. The intersection of row 3 and column four of a Multiplication Table is a dozen; 12 is definitely the response. For children starting to discover multiplication, this really is straightforward. They may use addition to eliminate the trouble as a result affirming that multiplication is shorthand for addition. Case in point: 4 by 3 = 4 4 4 = 12. It is an superb guide to the Multiplication Table. The additional benefit, the Multiplication Table is visual and displays to understanding addition.
## Exactly where can we begin studying multiplication while using Multiplication Table?
1. Initially, get informed about the table.
2. Get started with multiplying by a single. Start at row number one. Proceed to column number 1. The intersection of row one particular and column the initial one is the perfect solution: 1.
3. Recurring these steps for multiplying by 1. Flourish row 1 by posts 1 through a dozen. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Repeat these steps for multiplying by two. Grow row two by columns one through 5. The answers are 2, 4, 6, 8, and 10 respectively.
5. Allow us to bounce forward. Repeat these actions for multiplying by 5 various. Flourish row 5 various by posts 1 through 12. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us improve the amount of issues. Repeat these techniques for multiplying by three. Multiply row 3 by columns one by means of 12. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be comfortable with multiplication so far, try out a analyze. Resolve these multiplication difficulties in your thoughts after which assess your responses to the Multiplication Table: multiply 6 and two, grow nine and a few, grow 1 and eleven, increase a number of and 4, and increase 7 and 2. The situation replies are 12, 27, 11, 16, and 14 correspondingly.
Should you received 4 from several troubles proper, create your personal multiplication exams. Compute the replies in your mind, and appearance them while using Multiplication Table. |
# How to solve sss triangle
College algebra students learn How to solve sss triangle, and manipulate different types of functions. We can solve math problems for you.
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We can do your math homework for you, and we'll make sure that you understand How to solve sss triangle. A linear solver is defined as a method that can be used to solve for a linear equation or linear system. A linear solver is a mathematical algorithm that takes a set of input values and generates an output value. It is often used to calculate the best line from two points, such as a straight line between two cities. A linear solver is most often used when the problem involves only one variable, or when there are no constraints on the solution. There are two main types of linear solvers: iterative and recursive. An iterative solver starts with some starting value and works towards a solution using smaller and smaller steps until the final solution is reached. The drawback to an iterative solver is that it can take longer to find the solution because it must start at some initial value and then repeat this process several times before finding the correct answer. A recursive solver works by repeating the same process over and over again until it reaches a solution. This type of solver is much faster than an iterative solver because it does not have to start at any arbitrary point in order to begin calculating the next step in solving the problem. Regardless of which type of linear solvers you decide to use, make sure they are implemented correctly so they will work properly on your specific problem. In addition, make sure you understand how each type of linear solvers works before you rely
One of the best ways to improve at math is by learning how to solve problems. Knowing how to set up equations, work with fractions and percentages, and use arithmetic are essential skills that underlie all math. Solving problems is also a great way to challenge yourself and practice your problem-solving skills. Solving problems can be challenging at times, but it's never impossible. With practice and patience, you'll get better at solving problems every time you sit down at the table.
First determine the y intercept. The y intercept is the value where the line crosses the Y axis. It is sometimes referred to as the "zero" point, or reference point, along the line. The y intercept of an equation can be determined by drawing a vertical line down through the origin of each graph and placing a dot at the intersection of the two lines (Figure 1). When graphing a parabola, the y intercept is placed at the origin. When graphing a line with a slope 1, then both y-intercepts are placed at 0. When graphing a line with a slope >1, then both y-intercepts are moved to positive infinity. In order to solve for x intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find x-intercept. In order to solve for y intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find y-intercept. Example: Solve for x-intercept of y = 4x + 10 Solution: Substitute 4x + 5 = 0 into original problem: y = 4x + 10 => y = 4(x + 5) => y =
It's also a good tool for students who want to learn how to solve equations on their own, without having to rely on someone else. The steps can be simplified or complex depending on your needs. You can also save the steps you've solved so you can refer back to them later.
Do you have a math phobia? Do you hate math? Do you dread having to do math homework? If so, then the math picture app is the perfect solution for you. The math picture app is a unique app that allows kids to draw equations and shapes on their iPhone or iPad screens. The app then converts these drawings into equations and shapes, thereby providing a fun and engaging way for kids to learn about math. Best of all, it doesn’t matter how good or bad your drawing skills are—the app will convert your drawings into equations and shapes automatically! How to use: 1) Open the math picture app. 2) Draw an equation or shape using the drawing tools on your phone or iPad screen. 3) Tap “Share” to convert your drawing into an equation or shape.
100/10 It's really helpful in understanding various math problems and how they are solved. There are also books with already provided solutions to problems available only if you are a Plus subscriber, so I highly recommend subscribing! Hopefully, they include the word problems in their next updates
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Math problem solver picture Direct variation solver Camera math solver free Solve the equation symbolically Solve the system of linear equations by graphing solver Quadratic formula solver |
# How do you find the domain of k(a)=a^2/(2a^2+3a-5)?
Aug 25, 2017
Domain of $f \left(a\right) = \left(- \infty , - \frac{5}{2}\right) \cup \left(- \frac{5}{2} , 1\right) \cup \left(1 , + \infty\right)$
#### Explanation:
$k \left(a\right) = {a}^{2} / \left(2 {a}^{2} + 3 a - 5\right)$
$= {a}^{2} / \left(\left(2 a + 5\right) \left(a - 1\right)\right)$
Note that $k \left(a\right)$ is defined for all $a \in \mathbb{R}$ except where $\left(2 a + 5\right) \left(a - 1\right) = 0$
I.e. where $a = - \frac{5}{2} \mathmr{and} 1$
Hence $k \left(a\right)$ is defined $\forall a \in \mathbb{R} : a \ne \left\{- \frac{5}{2} , 1\right\}$
$\therefore$ the domain of $k \left(a\right)$ is all $a \in \mathbb{R} : a \ne \left\{- \frac{5}{2} , 1\right\}$
Or in interval notation: $\left(- \infty , - \frac{5}{2}\right) \cup \left(- \frac{5}{2} , 1\right) \cup \left(1 , + \infty\right)$
graph{x^2/((2x+5)(x-1)) [-10, 10, -5, 5]}
This is demonstrated by the graph of $k \left(a\right)$ below - Where the axes are $a$ and $k \left(a\right)$ replacing the conventional $x$ and $y$. |
# NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.1
Chapter 3, Trigonometric Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. The chapter contains questions and concepts revolving around Trigonometric functions. Exercise 3.1 of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions is based on the following topics:
1. Introduction
2. Angles
1. Degree measure
3. Relation between radian and real numbers
4. Relation between degree and radian
These solutions are helpful for the students to get an idea of how to answer the questions from the board exam perspective. View online or download the NCERT Solutions for Class 11 to get a hold of all the concepts covered in the chapter.
## NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Exercise 3.1
### Access other exercise solutions of Class 11 Maths Chapter 3 – Trigonometric Functions
To access the other exercise answers from NCERT Class 11 Maths Solutions Chapter 3, click on the links below.
Exercise 3.2 Solutions 10 Questions
Exercise 3.3 Solutions 25 Questions
Exercise 3.4 Solutions 9 Questions
Miscellaneous Exercise on Chapter 3 Solutions 10 Questions
### Access Solutions for Class 11 Maths Chapter 3.1 exercise
1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°
Solution:
(iv) 520°
2. Find the degree measures corresponding to the following radian measures (Use π = 22/7).
(i) 11/16
(ii) -4
(iii) 5Ï€/3
(iv) 7Ï€/6
Solution:
(i) 11/16
(ii) -4
(iii) 5Ï€/3
We get
= 300o
(iv) 7Ï€/6
We get
= 210o
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution:
It is given that
No. of revolutions made by the wheel in
1 minute = 360
1 second = 360/60 = 6
We know that
The wheel turns an angle of 2Ï€ radian in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian
Therefore, in one second, the wheel turns at an angle of 12Ï€ radian.
4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).
Solution:
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Solution:
The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB as the chord of the circle, i.e., length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = π/3 radian
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre,
We get θ = 1/r
Therefore, the length of the minor arc of the chord is 20Ï€/3 cm.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Solution:
7. Find the angle in the radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
Solution:
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r
We know that r = 75 cm
(i) l = 10 cm
So we get
By further simplification,
(ii) l = 15 cm
So, we get
By further simplification,
(iii) l = 21 cm
So, we get |
# Quick Answer: What Times Table Does 42 Go Into?
## What is the table of 36?
36 Times Tables36 Addition36 Subtraction36 Multiplication1 + 36 = 3737 – 36 = 11 x 36 = 362 + 36 = 3838 – 36 = 22 x 36 = 723 + 36 = 3939 – 36 = 33 x 36 = 1084 + 36 = 4040 – 36 = 44 x 36 = 14416 more rows.
## What is the table of 30?
Table of 30 charts30X3030X6030X9030X12030X15015 more rows
## What is the table of 26?
Tables are the basics of Mathematics based on which many multiplication calculations could be simplified quickly. Table of 26 gives the repeated addition of number 26, repeated for a certain number of times. For example, 4 baskets of 26 apples each, gives the sum of 26 + 26 + 26 + 26 = 104.
## What is the table of 35?
Multiplication Table for 3535x3535x14035x17535x21035x24545 more rows
## What multiplication makes 36?
36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, or 6 x 6. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Prime factorization: 36 = 2 x 2 x 3 x 3, which can also be written 36 = 2² x 3².
## What table does 63 come in?
63 Times Tables63 Addition63 Subtraction63 Multiplication1 + 63 = 6464 – 63 = 11 x 63 = 632 + 63 = 6565 – 63 = 22 x 63 = 1263 + 63 = 6666 – 63 = 33 x 63 = 1894 + 63 = 6767 – 63 = 44 x 63 = 25216 more rows
## What times tables have 13 in them?
Multiplication Times Tables Chart 13 to 2413 Times Table14 Times Table15 Times Table1 x 13 = 131 x 14 = 141 x 15 = 152 x 13 = 262 x 14 = 282 x 15 = 303 x 13 = 393 x 14 = 423 x 15 = 454 x 13 = 524 x 14 = 564 x 15 = 606 more rows
## What is the 31 times table?
31 Times Tables31 Addition31 Subtraction31 Multiplication2 + 31 = 3333 – 31 = 22 x 31 = 623 + 31 = 3434 – 31 = 33 x 31 = 934 + 31 = 3535 – 31 = 44 x 31 = 1245 + 31 = 3636 – 31 = 55 x 31 = 15516 more rows
## What are the multiples of 42?
Table of Factors and MultiplesFactorsMultiples1, 2, 3, 6, 7, 14, 21, 42422101, 43432151, 2, 4, 11, 22, 44442201, 3, 5, 9, 15, 454522541 more rows
## What is a table of 13?
13 × 4 = 52. 13 × 5 = 65. 13 × 6 = 78. 13 × 7 = 91. 13 × 8 = 104.
## What are all the factors for 48?
48 is a composite number. 48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12, or 6 x 8. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Prime factorization: 48 = 2 x 2 x 2 x 2 x 3, which can also be written 48 = 2⁴ x 3.
## What is the 42 times table?
42 Times Tables42 Addition42 Subtraction42 Multiplication2 + 42 = 4444 – 42 = 22 x 42 = 843 + 42 = 4545 – 42 = 33 x 42 = 1264 + 42 = 4646 – 42 = 44 x 42 = 1685 + 42 = 4747 – 42 = 55 x 42 = 21016 more rows
## What times tables does 48 go into?
48 Times Tables Sheets x 48 = 48. 48 ÷ 48 = 1. 2 + 48 = 50. … x 48 = 96. 96 ÷ 48 = 2. 3 + 48 = 51. … x 48 = 144. 144 ÷ 48 = 3. 4 + 48 = 52. … x 48 = 192. 192 ÷ 48 = 4. 5 + 48 = 53. … x 48 = 240. 240 ÷ 48 = 5. 6 + 48 = 54. … x 48 = 288. 288 ÷ 48 = 6. 7 + 48 = 55. … x 48 = 336. 336 ÷ 48 = 7. 8 + 48 = 56. … x 48 = 384. 384 ÷ 48 = 8.More items…
## At what age should a child know their times tables?
In maths, it says pupils should be introduced to the two, five and 10 times tables by year two – at the age of six and seven. Between the age of seven and eight, children should start to learn the three, four and eight times tables, the document says.
## What can equal 48?
The factor pairs of the number 48 are: 1 x 48, 2 x 24, 3 x 16, 4 x 12, and 6 x 8. When you multiply each of these factor pairs together, you get 48….
## What order are times tables taught in?
The main messages: Take each multiplication table one at a time. There is a logical order which usually works; 2s, 5s and 10s first (usually around Year 2), 3s, 4s and 8s next (usually around Year 3), then 11s, 6s, 9s, 12s and then 7s come later (usually around Year 4).
## What is the table of 19?
Multiplication Table of 1919x1919x13319x15219x17119x19015 more rows
## What times tables should YEAR 1 know?
In year 1, children do not need to learn any of their times tables, however, they are expected to understand some very basic multiplication facts. They should know the doubles and corresponding halves up to the number 10.
## IS 42 in the 3 times table?
Forward counting by 3’s: 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, ……
## What times table should YEAR 2 know?
Learning the 2, 5 and 10 times tables, plus division facts. It is essential that your child knows these times tables by the end of Year 2. They also need to know the division facts for these times tables, for example if 4 x 5 = 20, then 20 / 4 = 5 and 20 / 5 + 4. |
# How Do You Find The Square Root Of 16?
## Is 16 a perfect square?
Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers..
## Why is 4 square root 16?
For example the number 16 has a square root of 4 because 4 x 4 = 16. So when you multiply the divisor or the quotient together (or by itself – the two factors being the same) the product is the dividend (i.e. the number you are seeking to find the root of).
## Is Root 16 a SURD?
To simplify a surd, write the number under the root sign as the product of two factors, one of which is the largest perfect square. Note that the factor 16 is the largest perfect square. Recall that the numbers 1, 4, 9, 16, 25, 36, 49, … are perfect squares.
## What is the square of the 4?
4x4The square of 4 is 4×4. To show that a number is squared, a small 2 is placed to the top right of the number. Like this: These signs are the same as saying “3 squared, 4 squared, and x squared.”
## What is the square of negative 16?
√-16 = 4 i Note: since negative times negative equals positive, one could therefore conclude that -4 i is also a correct answer to the square root of negative 16.
## How do you find the square root of a number?
How to find the square root of a number and calculate it by handSTEP 1: Separate The Digits Into Pairs. To begin, let’s organize the workspace. … STEP 2: Find The Largest Integer. … STEP 3: Now Subtract That Integer. … STEP 4: Let’s Move To The Next Pair. … STEP 5: Find The Right Match. … STEP 6: Subtract Again.Feb 6, 2020
## What are two square roots of 16?
Thus, 16 has two square roots, +4 and -4.
## What is a square of 16?
256The square of 16 is 256.
## What is the positive and negative square root of 16?
So any positive real number, like 16, actually has 2 square roots, one positive, the other negative. The positive square root is termed the “principal square root,” the negative square root is unfortunately unnamed.
## Does a square root have 2 answers?
It has multiple answers so why do we pick the positive one? if x2=16⟹x=√16 or x=−√16 for respectively the positive and negative solution. This implies that the square root function has a single answer and we must negate its answer to obtain the second solution.
## What is the square of 12?
“Is it possible to recreate this up to 12×12?”0 Squared=09 Squared=8110 Squared=10011 Squared=12112 Squared=1448 more rows
## What is the number whose principal root is 16?
4This is because the ‘principal square root’ means the positive square root of a certain number, and so √16 = 4 × 4, making 4 the answer.
## What is the square roots of 16?
Table of Squares and Square RootsNUMBERSQUARESQUARE ROOT141963.742152253.873162564.000172894.12396 more rows
## What are the roots of 4?
But the roots could be positive or negative or we can say there are always two roots for any given number. Hence, root 4 is equal to ±2 or +2 and -2 (positive 2 and negative 2). You can also find square root on a calculator….Square Root From 1 to 50.NumberSquare Root Value31.7324252.23662.44946 more rows
## Is square root of 16 a whole number?
The square root of 16 is a rational number. The square root of 16 is 4, an integer.
## Is 4 a perfect square?
In mathematics, a square is a product of a whole number with itself. … In this case, 4 is termed as a perfect square. A square of a number is denoted as n × n. Similarly, the exponential notation of the square of a number is n 2, usually pronounced as “n” squared.
## Why is 16 a perfect square?
16 is a perfect square because it can be expressed as 4 * 4 (the product of two equal integers). |
# Introduction to Fractions
Fraction is derived from the Latin word “fractus” which means “a part of” or “broken”. Fractions are integers and are used to express parts of a whole. We shall discuss this concept here in detail with some solved examples.
## Introduction to Fractions
Fractions are types of integers and are used to represent a part of the whole. For example, if I have to say I have one half of a cake, it means that I divided 1 cake into 2 equal and smaller parts and I possess one of the 2 parts. To put it simply, I have 1/2 part of the cake.
They are represented in the format of “p/q” where the “p” part is called the Numerator and the “q” part is called the Denominator. P/q means ‘p’ parts out of ‘q’ number of total parts. The value of a fraction in decimal form can be obtained by dividing the numerator by denominator.
For example ‘3/5’ means 3 parts out of 5, 1/3 means 1 part out of 3. Fractions are also used to denote division or ratios. For example, 3/5 can be read as ‘3 divided by 5’ or as a ratio of ‘3 is to 5’.
(Source: Wikimedia)
## Equivalence
2 sets of fractions can be the same although their numerators and denominators do not match. Consider the fractions 3/6, 1/2 and 9/18. In all the 3 fractions we see that the numerator is one half or 1/2 of the denominator. Thus all the fractions are easily reducible to the common form 1/2. This method is called Simplifying or reducing the fraction.
If we observe the numerator and denominator to have a common factor we eliminate that factor to reduce the fraction to a simpler form. Some of the important ones are given below.
Addition of 2 fractions can take place only when they have the same denominator. For example, if we add 3/6 and 5/6 then this is possible as both the fractions have the same denominator i.e. 6. in such additions we add only the numerators and the denominator stays the same. Consider this as adding 3 parts out of 6 to another 5 parts out of 6. So,
3/6 + 5/6 = (3 + 5)/6 = 8/6
Consider the case when denominators of the fractions are not the same. In such a case we make the denominators of both the fractions same by multiplying the denominators with a suitable factor to form an l.c.m of the denominators and then add them. Consider the fractions 3/6 and 1/2
3/6 + 1/2
In this case, 2 and 6 have common LCM as 6. So, we make both the denominators as 6. To do this we multiply the numerator and denominator of 1/2 with 3 so as to not change the value of the fraction.
so 3/6 + 1/2 = 3/ 6 + 3/6 =6/6 =1
### Subtraction
Subtraction of fractions follows the same rule as the addition of fractions where it is necessary to have the same denominator for carrying out the operation. Consider the example 5/6 – 3/6= (5 – 3) / 6 = 2/6 = 1/3. Similarly, 5/6 -1/4 = (5 x 4) / (6 x 4) – (1 x 6) / (4 x 6) = 20/24 – 6/24 = (20 – 6) / 24 = 14 / 24 = 7/12.
### Multiplication
Multiplication of 2 fractions is performed by multiplying the numerators of the 2 numbers and multiplying the denominators of the 2 numbers separately and obtaining the final result. For example, 3/9 x 4/5 = (3 x 4) / (9 x 5) = 12/45
### Division
Division of 2 fractions again, is carried out by dividing the 2 numerators and dividing the 2 denominators separately and obtaining the final result. For example, 6/18 divided by 2/9 = (6 divided by 2) / (18 divided by 9) = 3/2
## Solved Example for You
Question 1: Obtain the result if (2/3) is added to (4/9) and is multiplied by 3
= 3x ( 6/9 + 4/9)
= 3x ( 10/9 )
= 30/9
= 10/3
Question 2: Explain what is fraction with example?
Answer: A fraction informs us how many parts of a whole we have. One can recognize a fraction by the slash that is between the two numbers. We have the numerator, a bottom number, a top number, and the denominator. For example, 1/2 happens to be a fraction. Also, 1/2 of a pie will be half a pie.
Question 3: Name the three types of fractions?
Answer: The three types of fractions are proper fraction, improper fraction, and mixed fraction.
Question 4: Explain how one can multiply fractions?
Answer: Below are the steps for carrying out multiplication of fractions:
• Carry out simplification of the fractions if not in lowest terms.
• Carry out multiplication of the numerators of the fractions in order to get the new numerator.
• Finally, carry out multiplication of the denominators of the fractions in order to get the new denominator.
Question 5: Can we say that 1 is a proper fraction?
Answer: No, 1 is not a proper fraction. This is because, in a proper fraction, the numerator happens to be smaller compared to the denominator. So, 1 and other integers are not proper fractions.
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# Constructing Parallel Lines - Problem 3
Brian McCall
###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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The alternate exterior angles theorem states that the alternate exterior angles of two parallel lines with a transversal are congruent. So, in order to create a parallel line, first draw a transversal through the given point p and the given line to create an angle. Placing the compass end at the vertex of the angle, swing an arc through one of the exterior angles. Place the compass at point p and swing an arc on the opposite side as the first arc, outside of the parallel lines (since the aim is to draw alternate exterior angles).
Find the distance between the arc and the intersections on the first angle with the compass. Then, keeping the compass the same size, place it on the arc drawn around point p and mark the intersection. Connecting this point of intersection with point p creates a line parallel to line l.
In this problem we’re being asked to construct a line parallel to a given line L through a point P using alternate exterior angles. The other two methods are alternate interior angles and corresponding angles, but why does this work?
Well if you come over here where I’ve sketched this problem, the idea behind this is to create a transversal that passes through point P and then say well if I duplicate alternate exterior angles then we would have created two parallel lines. So I’m going to draw my transversal that’s going to be my first step, then I’m going to duplicate this angle out here this exterior angle up at point P. So let’s grab our compass and get started.
The first step is to draw a transversal, I was going to swing an arc, but I won’t have an angle to start with, so we’re going to draw a line through point p intersecting our given line. So this angle right here you could also choose this obtuse angle it doesn’t matter they’re both going to be exterior angles, I think it’s easier to duplicate an acute angle. So I’m going to swing an arc from that vertex and I’m going to come up to point P and I know that if this angle is on the left side of the transversal, it’s alternate exterior angle will be on the right side, so I’m going to swing another arc from point p a little mistake there, but I think it will work.
Next I need to measure that point of intersection using my compass and I’m going to make a mark right here which is what your teacher is looking for. Come back up here and I’m going to swing that identical arc so now we have two points of intersection. We have P and that point. Connect them with your straightedge and we’ll have our parallel line.
The reason why these two lines have to be parallel is because the alternate exterior angles are congruent, which is the converse of the parallel lines there, so I’m going to mark these two lines as parallel. |
# Lesson 22
Subtract from Teen Numbers
## Warm-up: Number Talk: Subtract from a Teen Number (10 minutes)
### Narrative
The purpose of this Number Talk is to elicit strategies and understandings students have for finding the difference of two numbers. These understandings develop fluency and will be helpful later in this lesson when students subtract from teen numbers.
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$14 - 4$$
• $$14 - 5$$
• $$17 - 7$$
• $$17 - 9$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “How can you use $$14 - 4 = 10$$ to find the difference in $$14 - 5$$?” (I know $$14 - 4$$ is 10, so if I subtract 1 more, the answer is 9.)
• “How can you use $$17-7= 10$$ to find the difference in $$17-9$$?” ($$17 - 7$$ is 10, so I am subtracting 2 more in $$17 - 9$$. So I can take 2 from 10 to get the answer.)
## Activity 1: Subtraction Methods (20 minutes)
### Narrative
The purpose of this activity is for students to solve a Take From, Result Unknown problem, which requires decomposing a ten, in a way that makes sense to them. The problem is presented with an image that encourages students to think about 16 as a ten and 6 ones. They create a poster to share how they solved the problem, and participate in a gallery walk to see how their classmates solved. As students are working, the teacher monitors for methods to analyze during the activity synthesis. As they do so, students connect addition and subtraction and see how either can be used to solve the problem (MP7).
This activity uses MLR7 Compare and Connect. Advances: representing, conversing
### Launch
• Groups of 2
• Give students access to double 10-frames and connecting cubes or two-color counters.
### Activity
• 5 minutes: partner work time
• Monitor for students who
• take away 7 from 16
• count on from 7 to 16
• take away 6 to get 10 and then subtract 1 more
• count on from 7 to 10 and then add 6 more
### Student Facing
Elena has 16 crayons.
She gives 7 crayons to Diego.
How many crayons does she have left?
Show your thinking using drawings, numbers, or words.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
MLR7 Compare and Connect
• Give each group tools for creating a visual display.
• “Create a poster that shows your thinking about the problem. Make sure to show your thinking in a way others will understand.”
• 5 minutes: partner work time
• “As you walk around and look at the posters, think about how the work is the same and different.”
• 5 minutes: gallery walk
• “What is the same and what is different about the representations?” (They all showed 16 and 7. They used math tools to represent he problem. Some people used addition facts they knew, some counted up, some took away.)
## Activity 2: Number Card Subtraction (20 minutes)
### Narrative
The purpose of this activity is for students to subtract a one-digit number from a teen number. Students choose a teen number card, then choose a number card to subtract from the teen number. They write an equation to represent each round. The equations students write may involve subtraction, or addition with a missing addend. During the synthesis, students share their methods. When explaining, students have opportunities to revise their language to make their explanations more precise and clear (MP6).
Students play this game again in the next lesson with a double 10-frame, so keep the materials organized for future use.
Action and Expression: Internalize Executive Functions. Check for understanding by inviting students to rephrase directions in their own words.
Supports accessibility for: Memory, Organization
### Required Preparation
• Each group of 2 needs a set of Number Cards 0–10 and a set of Number Cards 11–20 used in a previous lesson.
### Launch
• Groups of 2
• Give each group a set of number cards 0–10, a set of number cards 11–20, and access to double 10-frames and connecting cubes or two-color counters.
• “We’re going to play a game to practice subtracting. To play this game you pick a teen number card. Then you pick a number card 0–9 and subtract that number from your teen number. Write an equation to show the difference.”
• “At the end of the game, you are going to pick one turn and show how you found the difference.”
### Activity
• 10 minutes: partner work time
• Monitor for students who:
• take away
• count on
• take away to get to 10, then take away some more
• “On your own, pick your favorite round. Show how you found the value of the difference using drawings, numbers, or words.”
• 3 minutes: independent work time
### Student Facing
1. Choose a teen number card.
2. Choose a number card to subtract.
3. Find the difference.
4. Write an equation.
My equations:
Show how you found the value of the difference using drawings, numbers, or words.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Invite previously identified students to share.
• “What method did you see today that you would like to try?”
## Lesson Synthesis
### Lesson Synthesis
Show 14 on the double 10-frame and the number card 6.
“Today we subtracted from teen numbers. How can you find the difference between 14 and 6?”
## Cool-down: Unit 3, Section D Checkpoint (0 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. |
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# Chapter 4.8: Determine if the Relation is a Function
Chapter 4.8: Determine if the Relation is a Function. Lesson 4.8: Function & Relations. Definition. Relation: Any set of ordered pairs. Function: A rule that establishes a relationship between 2 quantities, the input and the output. For every x there is exactly one y. Domain: x values.
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## Chapter 4.8: Determine if the Relation is a Function
E N D
### Presentation Transcript
1. Chapter 4.8: Determine if the Relation is a Function
2. Lesson 4.8: Function & Relations Definition Relation: Any set of ordered pairs Function: A rule that establishes a relationship between 2 quantities, the input and the output. For every x there is exactly one y Domain: x values Input: x values Range: y values Output: y values Set Notation {1,2,3} Only write the number once, even if it repeats
3. Lesson 4.8: Determining if the relation is a function To determine if a given relation is a function, we use the vertical line test. If the vertical line intersects the relation at one point as it moves across the graph, then that graph is a function. If the vertical line intersects the relation at more than one point as it moves across the graph, then that graph is NOT a function.
4. y x Lesson 4.8: Determining if the relation is a function- Vertical Line Test Run a vertical line across the page from left to right Since the line only touched once, there is only one x value for every y, it is a function.
5. y x Lesson 4.8: Determining if the relation is a function- Vertical Line Test Run a vertical line across the page from left to right Since the line touched more than once, there is more than one x value for every y, it is a function.
6. Lesson 4.8: Determining if the relation is a function given a table of values Step 1: Graph each point Input Output xy {1,2,3,4} Domain: 1 3 4 1 2 3 4 1 1 2 3 {1,3,4} Range: 3 3 4 4 Step 2: Conduct vertical line test.
7. y x Lesson 4.8: Determining if the relation is a function given a table of values Step 1: Graph each point Input Output Step 2: Vertical Line Test 1 2 3 4 1 3 4 Since the line only touched once, there is only one x value for every y, it is a function.
8. Lesson 4.8: Determining if the relation is a function given a table of values y x Step 1: Graph each point Input Output Step 2: Vertical Line Test 1 3 4 4 Since the line touched more than once, there is more than one x value for every y, it is a function.
9. Lesson 4.8: Determining if the relation is a function given a table of values Input Output Input Output 1 3 4 1 2 3 4 1 3 4 4
More Related |
# Multiplication
(Redirected from Dot operator)
Four bags with three marbles per bag gives twelve marbles (4 × 3 = 12).
Multiplication can also be thought of as scaling. Here we see 2 being multiplied by 3 using scaling, giving 6 as a result
Animation for the multiplication 2 × 3 = 6.
4 × 5 = 20, the rectangle is composed of 20 squares, each having dimensions of 1 by 1.
Area of a cloth 4.5m × 2.5m = 11.25m2; 4½ × 2½ = 11¼
Multiplication (often denoted by the cross symbol "×", by a point "·", by juxtaposition, or, on computers, by an asterisk "") is one of the four elementary, mathematical operations of arithmetic; with the others being addition, subtraction and division.
The multiplication of whole numbers may be thought as a repeated addition; that is, the multiplication of two numbers is equivalent to adding as many copies of one of them, the multiplicand, as the value of the other one, the multiplier. Normally, the multiplier is written first and multiplicand second,[1] though this can vary, especially in languages with different grammatical structures, such as Japanese, Japanese elementary schools teach writing the multiplicand first, and answers that reverse that order are marked as incorrect.[2] The distinction is not very meaningful:
${\displaystyle a\times b=\underbrace {b+\cdots +b} _{a}=\underbrace {a+\cdots +a} _{b}=b\times a}$
For example, 4 multiplied by 3 (often written as ${\displaystyle 3\times 4}$ and said as "3 times 4") can be calculated by adding 3 copies of 4 together:
${\displaystyle 3\times 4=4+4+4=12}$
Here 3 and 4 are the "factors" and 12 is the "product".
One of the main properties of multiplication is the commutative property, adding 3 copies of 4 gives the same result as adding 4 copies of 3:
${\displaystyle 4\times 3=3+3+3+3=12}$
The multiplication of integers (including negative numbers), rational numbers (fractions) and real numbers is defined by a systematic generalization of this basic definition.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have given lengths. The area of a rectangle does not depend on which side is measured first, which illustrates the commutative property. The product of two measurements is a new type of measurement, for instance multiplying the lengths of the two sides of a rectangle gives its area, this is the subject of dimensional analysis.
The inverse operation of multiplication is division. For example, since 4 multiplied by 3 equals 12, then 12 divided by 3 equals 4. Multiplication by 3, followed by division by 3, yields the original number (since the division of a number other than 0 by itself equals 1).
Multiplication is also defined for other types of numbers, such as complex numbers, and more abstract constructs, like matrices. For these more abstract constructs, the order that the operands are multiplied sometimes does matter. A listing of the many different kinds of products that are used in mathematics is given in the product (mathematics) page.
## Notation and terminology
The multiplication sign ×
In arithmetic, multiplication is often written using the sign "×" between the terms; that is, in infix notation.[3] For example,
${\displaystyle 2\times 3=6}$ (verbally, "two times three equals six")
${\displaystyle 3\times 4=12}$
${\displaystyle 2\times 3\times 5=6\times 5=30}$
${\displaystyle 2\times 2\times 2\times 2\times 2=32}$
The sign is encoded in Unicode at U+00D7 × MULTIPLICATION SIGN (HTML × · ×).
There are other mathematical notations for multiplication:
• Multiplication is also denoted by dot signs,[4] usually a middle-position dot (rarely period):
${\displaystyle 5\cdot 2\quad {\text{or}}\quad 5\,.\,2}$
The middle dot notation, encoded in Unicode as U+22C5 dot operator, is standard in the United States, the United Kingdom, and other countries where the period is used as a decimal point. When the dot operator character is not accessible, the interpunct (·) is used. In other countries that use a comma as a decimal mark, either the period or a middle dot is used for multiplication.[citation needed]
• In algebra, multiplication involving variables is often written as a juxtaposition (e.g., xy for x times y or 5x for five times x). The notation can also be used for quantities that are surrounded by parentheses (e.g., 5(2) or (5)(2) for five times two). This implicit usage of multiplication can cause ambiguity when the concatenated variables happen to match the name of another variable, when a variable name in front of a parenthesis can be confused with a function name, or in the correct determination of the order of operations.
• In matrix multiplication, there is a distinction between the cross and the dot symbols. The cross symbol generally denotes the taking a cross product of two vectors, yielding a vector as the result, while the dot denotes taking the dot product of two vectors, resulting in a scalar.
In computer programming, the asterisk (as in 5*2) is still the most common notation. This is due to the fact that most computers historically were limited to small character sets (such as ASCII and EBCDIC) that lacked a multiplication sign (such as • or ×), while the asterisk appeared on every keyboard. This usage originated in the FORTRAN programming language.
The numbers to be multiplied are generally called the "factors". The number to be multiplied is called the "multiplicand", while the number of times the multiplicand is to be multiplied comes from the "multiplier". Usually the multiplier is placed first and the multiplicand is placed second,[1] however sometimes the first factor is the multiplicand and the second the multiplier.[5] Additionally, there are some sources in which the term "multiplicand" is regarded as a synonym for "factor".[6] In algebra, a number that is the multiplier of a variable or expression (e.g., the 3 in 3xy2) is called a coefficient.
The result of a multiplication is called a product. A product of integers is a multiple of each factor. For example, 15 is the product of 3 and 5, and is both a multiple of 3 and a multiple of 5.
## Computation
The common methods for multiplying numbers using pencil and paper require a multiplication table of memorized or consulted products of small numbers (typically any two numbers from 0 to 9), however one method, the peasant multiplication algorithm, does not.
Multiplying numbers to more than a couple of decimal places by hand is tedious and error prone. Common logarithms were invented to simplify such calculations. The slide rule allowed numbers to be quickly multiplied to about three places of accuracy. Beginning in the early twentieth century, mechanical calculators, such as the Marchant, automated multiplication of up to 10 digit numbers. Modern electronic computers and calculators have greatly reduced the need for multiplication by hand.
### Historical algorithms
Methods of multiplication were documented in the Egyptian, Greek, Indian and Chinese civilizations.
The Ishango bone, dated to about 18,000 to 20,000 BC, hints at a knowledge of multiplication in the Upper Paleolithic era in Central Africa.
#### Egyptians
The Egyptian method of multiplication of integers and fractions, documented in the Ahmes Papyrus, was by successive additions and doubling. For instance, to find the product of 13 and 21 one had to double 21 three times, obtaining 2 × 21 = 42, 4 × 21 = 2 × 42 = 84, 8 × 21 = 2 × 84 = 168. The full product could then be found by adding the appropriate terms found in the doubling sequence:
13 × 21 = (1 + 4 + 8) × 21 = (1 × 21) + (4 × 21) + (8 × 21) = 21 + 84 + 168 = 273.
#### Babylonians
The Babylonians used a sexagesimal positional number system, analogous to the modern day decimal system. Thus, Babylonian multiplication was very similar to modern decimal multiplication. Because of the relative difficulty of remembering 60 × 60 different products, Babylonian mathematicians employed multiplication tables. These tables consisted of a list of the first twenty multiples of a certain principal number n: n, 2n, ..., 20n; followed by the multiples of 10n: 30n 40n, and 50n. Then to compute any sexagesimal product, say 53n, one only needed to add 50n and 3n computed from the table.
#### Chinese
38 × 76 = 2888
In the mathematical text Zhoubi Suanjing, dated prior to 300 BC, and the Nine Chapters on the Mathematical Art, multiplication calculations were written out in words, although the early Chinese mathematicians employed Rod calculus involving place value addition, subtraction, multiplication and division. These place value decimal arithmetic algorithms were introduced by Al Khwarizmi to Arab countries in the early 9th century.
### Modern methods
Product of 45 and 256. Note the order of the numerals in 45 is reversed down the left column. The carry step of the multiplication can be performed at the final stage of the calculation (in bold), returning the final product of 45 × 256 = 11520. This is a variant of Lattice multiplication.
The modern method of multiplication based on the Hindu–Arabic numeral system was first described by Brahmagupta. Brahmagupta gave rules for addition, subtraction, multiplication and division. Henry Burchard Fine, then professor of Mathematics at Princeton University, wrote the following:
The Indians are the inventors not only of the positional decimal system itself, but of most of the processes involved in elementary reckoning with the system. Addition and subtraction they performed quite as they are performed nowadays; multiplication they effected in many ways, ours among them, but division they did cumbrously.[7]
#### Grid Method
Grid method multiplication or the box method, is used in primary schools in England and Wales to help teach and understanding of how multiple digit multiplication works. An example of multiplying 34 by 13 would be to lay the numbers out in a grid like:
30 4
10 300 40
3 90 12
### Computer algorithms
The classical method of multiplying two n-digit numbers requires n2 simple multiplications. Multiplication algorithms have been designed that reduce the computation time considerably when multiplying large numbers. In particular for very large numbers methods based on the Discrete Fourier Transform can reduce the number of simple multiplications to the order of n log2(n) log2 log2(n).
## Products of measurements
One can only meaningfully add or subtract quantities of the same type but can multiply or divide quantities of different types. Four bags with three marbles each can be though of as:[1]
[4 bags] × [3 marbles per bag] = 12 marbles.
When two measurements are multiplied together the product is of a type depending on the types of the measurements. The general theory is given by dimensional analysis. This analysis is routinely applied in physics but has also found applications in finance.
A common example is multiplying speed by time gives distance, so
50 kilometers per hour × 3 hours = 150 kilometers.
Other examples:
${\displaystyle 2.5{\text{ meters}}\times 4.5{\text{ meters}}=11.25{\text{ square meters}}}$
${\displaystyle 11{\text{ meters/second}}\times 9{\text{ seconds}}=99{\text{ meters}}}$
## Products of sequences
### Capital Pi notation
The product of a sequence of terms can be written with the product symbol, which derives from the capital letter Π (Pi) in the Greek alphabet. Unicode position U+220F (∏) contains a glyph for denoting such a product, distinct from U+03A0 (Π), the letter. The meaning of this notation is given by:
${\displaystyle \prod _{i=1}^{4}i=1\cdot 2\cdot 3\cdot 4,}$
that is
${\displaystyle \prod _{i=1}^{4}i=24.}$
The subscript gives the symbol for a dummy variable (i in this case), called the "index of multiplication" together with its lower bound (1), whereas the superscript (here 4) gives its upper bound. The lower and upper bound are expressions denoting integers. The factors of the product are obtained by taking the expression following the product operator, with successive integer values substituted for the index of multiplication, starting from the lower bound and incremented by 1 up to and including the upper bound. So, for example:
${\displaystyle \prod _{i=1}^{6}i=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=720}$
More generally, the notation is defined as
${\displaystyle \prod _{i=m}^{n}x_{i}=x_{m}\cdot x_{m+1}\cdot x_{m+2}\cdot \,\,\cdots \,\,\cdot x_{n-1}\cdot x_{n},}$
where m and n are integers or expressions that evaluate to integers. In case m = n, the value of the product is the same as that of the single factor xm. If m > n, the product is the empty product, with the value 1.
### Infinite products
One may also consider products of infinitely many terms; these are called infinite products. Notationally, we would replace n above by the lemniscate ∞. The product of such a series is defined as the limit of the product of the first n terms, as n grows without bound. That is, by definition,
${\displaystyle \prod _{i=m}^{\infty }x_{i}=\lim _{n\to \infty }\prod _{i=m}^{n}x_{i}.}$
One can similarly replace m with negative infinity, and define:
${\displaystyle \prod _{i=-\infty }^{\infty }x_{i}=\left(\lim _{m\to -\infty }\prod _{i=m}^{0}x_{i}\right)\cdot \left(\lim _{n\to \infty }\prod _{i=1}^{n}x_{i}\right),}$
provided both limits exist.
## Properties
Multiplication of numbers 0–10. Line labels = multiplicand. X axis = multiplier. Y axis = product.
Extension of this pattern into other quadrants gives the reason why a negative number times a negative number yields a positive number.
Note also how multiplication by zero causes a reduction in dimensionality, as does multiplication by a singular matrix where the determinant is 0. In this process, information is lost and cannot be regained.
For the real and complex numbers, which includes for example natural numbers, integers and fractions, multiplication has certain properties:
Commutative property
The order in which two numbers are multiplied does not matter:
${\displaystyle x\cdot y=y\cdot x.}$
Associative property
Expressions solely involving multiplication or addition are invariant with respect to order of operations:
${\displaystyle (x\cdot y)\cdot z=x\cdot (y\cdot z)}$
Distributive property
Holds with respect to multiplication over addition. This identity is of prime importance in simplifying algebraic expressions:
${\displaystyle x\cdot (y+z)=x\cdot y+x\cdot z}$
Identity element
The multiplicative identity is 1; anything multiplied by one is itself. This is known as the identity property:
${\displaystyle x\cdot 1=x}$
Property of Zero
Any number multiplied by zero is zero. This is known as the zero property of multiplication:
${\displaystyle x\cdot 0=0}$
Negation
Negative one times any number is equal to the additive inverse of that number.
${\displaystyle (-1)\cdot x=(-x)}$
Negative one times negative one is positive one.
${\displaystyle (-1)\cdot (-1)=1}$
The natural numbers do not include negative numbers.
Inverse element
Every number x, except zero, has a multiplicative inverse, ${\displaystyle {\frac {1}{x}}}$, such that ${\displaystyle x\cdot \left({\frac {1}{x}}\right)=1}$.
Order preservation
Multiplication by a positive number preserves order:
if a > 0, then if b > c then ab > ac.
Multiplication by a negative number reverses order:
if a < 0 and b > c then ab < ac.
The complex numbers do not have an order predicate.
Other mathematical systems that include a multiplication operation may not have all these properties. For example, multiplication is not, in general, commutative for matrices and quaternions.
## Axioms
In the book Arithmetices principia, nova methodo exposita, Giuseppe Peano proposed axioms for arithmetic based on his axioms for natural numbers.[8] Peano arithmetic has two axioms for multiplication:
${\displaystyle x\times 0=0}$
${\displaystyle x\times S(y)=(x\times y)+x}$
Here S(y) represents the successor of y, or the natural number that follows y. The various properties like associativity can be proved from these and the other axioms of Peano arithmetic including induction. For instance S(0). denoted by 1, is a multiplicative identity because
${\displaystyle x\times 1=x\times S(0)=(x\times 0)+x=0+x=x}$
The axioms for integers typically define them as equivalence classes of ordered pairs of natural numbers. The model is based on treating (x,y) as equivalent to xy when x and y are treated as integers. Thus both (0,1) and (1,2) are equivalent to −1. The multiplication axiom for integers defined this way is
${\displaystyle (x_{p},\,x_{m})\times (y_{p},\,y_{m})=(x_{p}\times y_{p}+x_{m}\times y_{m},\;x_{p}\times y_{m}+x_{m}\times y_{p})}$
The rule that −1 × −1 = 1 can then be deduced from
${\displaystyle (0,1)\times (0,1)=(0\times 0+1\times 1,\,0\times 1+1\times 0)=(1,0)}$
Multiplication is extended in a similar way to rational numbers and then to real numbers.
## Multiplication with set theory
The product of non-negative integers can be defined with set theory using cardinal numbers or the Peano axioms. See below how to extend this to multiplying arbitrary integers, and then arbitrary rational numbers. The product of real numbers is defined in terms of products of rational numbers, see construction of the real numbers.
## Multiplication in group theory
There are many sets that, under the operation of multiplication, satisfy the axioms that define group structure. These axioms are closure, associativity, and the inclusion of an identity element and inverses.
A simple example is the set of non-zero rational numbers. Here we have identity 1, as opposed to groups under addition where the identity is typically 0. Note that with the rationals, we must exclude zero because, under multiplication, it does not have an inverse: there is no rational number that can be multiplied by zero to result in 1. In this example we have an abelian group, but that is not always the case.
To see this, look at the set of invertible square matrices of a given dimension, over a given field. Now it is straightforward to verify closure, associativity, and inclusion of identity (the identity matrix) and inverses. However, matrix multiplication is not commutative, therefore this group is nonabelian.
Another fact of note is that the integers under multiplication is not a group, even if we exclude zero. This is easily seen by the nonexistence of an inverse for all elements other than 1 and −1.
Multiplication in group theory is typically notated either by a dot, or by juxtaposition (the omission of an operation symbol between elements). So multiplying element a by element b could be notated a ${\displaystyle \cdot }$ b or ab. When referring to a group via the indication of the set and operation, the dot is used, e.g., our first example could be indicated by ${\displaystyle \left(\mathbb {Q} \smallsetminus \{0\},\cdot \right)}$
## Multiplication of different kinds of numbers
Numbers can count (3 apples), order (the 3rd apple), or measure (3.5 feet high); as the history of mathematics has progressed from counting on our fingers to modelling quantum mechanics, multiplication has been generalized to more complicated and abstract types of numbers, and to things that are not numbers (such as matrices) or do not look much like numbers (such as quaternions).
Integers
${\displaystyle N\times M}$ is the sum of M copies of N when N and M are positive whole numbers. This gives the number of things in an array N wide and M high. Generalization to negative numbers can be done by
${\displaystyle N\times (-M)=(-N)\times M=-(N\times M)}$ and
${\displaystyle (-N)\times (-M)=N\times M}$
The same sign rules apply to rational and real numbers.
Rational numbers
Generalization to fractions ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}}$ is by multiplying the numerators and denominators respectively: ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}={\frac {(A\times C)}{(B\times D)}}}$. This gives the area of a rectangle ${\displaystyle {\frac {A}{B}}}$ high and ${\displaystyle {\frac {C}{D}}}$ wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.
Real numbers
Real numbers and their products can be defined in terms of sequences of rational numbers.
Complex numbers
Considering complex numbers ${\displaystyle z_{1}}$ and ${\displaystyle z_{2}}$ as ordered pairs of real numbers ${\displaystyle (a_{1},b_{1})}$ and ${\displaystyle (a_{2},b_{2})}$, the product ${\displaystyle z_{1}\times z_{2}}$ is ${\displaystyle (a_{1}\times a_{2}-b_{1}\times b_{2},a_{1}\times b_{2}+a_{2}\times b_{1})}$. This is the same as for reals, ${\displaystyle a_{1}\times a_{2}}$, when the imaginary parts ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ are zero.
Equivalently, denoting ${\displaystyle {\sqrt {-1}}}$ as ${\displaystyle i}$, we have ${\displaystyle z_{1}\times z_{2}=(a_{1}+b_{1}i)(a_{2}+b_{2}i)=(a_{1}\times a_{2})+(a_{1}\times b_{2}i)+(b_{1}\times a_{2}i)+(b_{1}\times b_{2}i^{2})=(a_{1}a_{2}-b_{1}b_{2})+(a_{1}b_{2}+b_{1}a_{2})i.}$
Further generalizations
See Multiplication in group theory, above, and Multiplicative Group, which for example includes matrix multiplication. A very general, and abstract, concept of multiplication is as the "multiplicatively denoted" (second) binary operation in a ring. An example of a ring that is not any of the above number systems is a polynomial ring (you can add and multiply polynomials, but polynomials are not numbers in any usual sense.)
Division
Often division, ${\displaystyle {\frac {x}{y}}}$, is the same as multiplication by an inverse, ${\displaystyle x\left({\frac {1}{y}}\right)}$. Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain x may have no inverse "${\displaystyle {\frac {1}{x}}}$" but ${\displaystyle {\frac {x}{y}}}$ may be defined. In a division ring there are inverses, but ${\displaystyle {\frac {x}{y}}}$ may be ambiguous in non-commutative rings since ${\displaystyle x\left({\frac {1}{y}}\right)}$ need not the same as ${\displaystyle \left({\frac {1}{y}}\right)x}$.
## Exponentiation
When multiplication is repeated, the resulting operation is known as exponentiation. For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 23, a two with a superscript three. In this example, the number two is the base, and three is the exponent. In general, the exponent (or superscript) indicates how many times the base appears in the expression, so that the expression
${\displaystyle a^{n}=\underbrace {a\times a\times \cdots \times a} _{n}}$
indicates that n copies of the base a are to be multiplied together. This notation can be used whenever multiplication is known to be power associative. |
Bullet Projectile Motion Experiment
This is a classic physics experiment which counter to our intuition. We have a situation where 1 ball is dropped from a point, and another ball is thrown horizontally from that same point. The question is which ball will hit the ground first?
(diagram from School for Champions site)
Looking at the diagram above you might argue that the ball that is dropped falls to the floor quicker as it has a shorter path. Or, you might think that the ball thrown sideways would travel faster to the ground because of its initial horizontal velocity. Both of these views are wrong however – as both balls will land at exactly the same time. To understand why, let’s look at the 2 situations in turn.
The ball launched sideways
To show that both balls would hit the ground at the same time we need to split the motion into its x and y components. We have
$x = v t \cos \theta$
$y = vt \sin \theta - \frac{1}{2} g t^2$
Where the angle theta is the angle of launch, v is the initial velocity, g is the gravitational constant 9.8 m/s. If we have a launch from the horizontal direction, then this angle is 0, which gives the simplified equations:
x = vt
y = 0.5gt2
if we relabel y as the vertical distance (d), then we have:
$\ t =\ \sqrt {\frac{2d}{g}}$
which is the time taken (ignoring air resistance etc) for an object launched horizontally to fall a distance d, where g is the gravitational constant 9.8 m/s.
So if we have a ball launched at a speed of 1 m/s from a height of 1m, it would hit the ground when:
t = (2/9.8)0.5 = 0.45 seconds
So we can use this value of t to see how far in the x direction it has travelled:
x = vt
x = 1(0.45)
x = 0.45m.
The ball dropped vertically
$x = v t \cos \theta$
$y = vt \sin \theta - \frac{1}{2} g t^2$
But this time we have no initial velocity as so we simply get:
x = 0
y = 0.5gt2
or as before, if we relabel y as the vertical distance (d), then we have:
$\ t =\ \sqrt {\frac{2d}{g}}$
So with a ball dropped from a height of 1m, it would also hit the ground when:
t = (2/9.8)0.5 = 0.45 seconds
But this time the distance in the x direction will of course be 0.
Showing this graphically
We can also show this graphically using the tracker software. This allows you to track the motion of objects in videos. So using the video above we can set the axis, and the height of the table
and then the motion capture software actually plots the parabola of the ball’s motion.
This first graph shows the change in the y direction with respect to time for the ball launched horizontally. We have large steps because the video was in super slow motion, so there were frames of very little movement. Nevertheless we can clearly see the general parabola, with equation:
y = -0.43x2 -1.2x + 107
The second graph shows the change in y direction with respect to time for the ball dropped vertically down. As before we have a clear parabola, with equation:
y = -0.43x2 -1.2x + 106
Which is a remarkably close fit. So, there we go, we have shown that the vertical motion of our 2 objects are independent of their horizontal motion. |
GreeneMath.com - Parallel & Perpendicular Lines Test
Parallel & Perpendicular Lines Test
When we have two parallel lines, the slopes will be the same, but the y-intercepts will be different. When we have perpendicular lines, the product of the slopes will be -1. To determine if we have parallel or perpendicular lines, place each line in slope-intercept form and inspect the slopes.
Test Objectives:
•Demonstrate an understanding of parallel and perpendicular lines
•Demonstrate the ability to determine if a pair of lines are parallel
•Demonstrate the ability to determine if a pair of lines are perpendicular
Parallel & Perpendicular Lines Test:
#1:
Instructions: Determine if each pair of lines is parallel, perpendicular, or neither.
a) 7x + 2y = 10 : 4x - 14y = 42
Watch the Step by Step Video Solution
|
View the Written Solution
#2:
Instructions: Determine if each pair of lines is parallel, perpendicular, or neither.
a) 2x - 5y = 0 : 6x - 15y = -30
Watch the Step by Step Video Solution
|
View the Written Solution
#3:
Instructions: Write the standard form of the equation of the line described.
a) through (-3,1) : parallel to:
y = -1x - 2 3
Watch the Step by Step Video Solution
|
View the Written Solution
#4:
Instructions: Write the standard form of the equation of the line described.
a) through (1,5) : parallel to:
y = -1x - 2 6
Watch the Step by Step Video Solution
|
View the Written Solution
#5:
Instructions: Write the standard form of the equation of the line described.
a) through (4,-5) : perpendicular to:
y = 8x - 1 5
Watch the Step by Step Video Solution
|
View the Written Solution
Written Solutions:
#1:
Solution:
a) perpendicular
Watch the Step by Step Video Solution
#2:
Solution:
a) parallel
Watch the Step by Step Video Solution
#3:
Solution:
a) x + 3y = 0
Watch the Step by Step Video Solution
#4:
Solution:
a) x + 6y = 31
Watch the Step by Step Video Solution
#5:
Solution:
a) 5x + 8y = -20
Watch the Step by Step Video Solution |
# How to Find Quartiles in Even and Odd Length Datasets
Quartiles are values that split up a dataset into four equal parts.
To find the first and third quartile for a dataset with an even number of values, use the following steps:
• Identify the median value (the average of the two middle values)
• Split dataset in half at the median
• Q1 is the median value in the lower half of the dataset (not including median)
• Q3 is the median value in the upper half of the dataset (not including median)
To find the first and third quartile for a dataset with an odd number of values, use the following steps:
• Identify the median value (the middle value)
• Split dataset in half at the median
• Q1 is the median value in the lower half of the dataset (not including median)
• Q3 is the median value in the upper half of the dataset (not including median)
The following examples show how to calculate quartiles for both types of datasets.
Note: When calculating quartiles, some formulas do include the median value. As noted by Wikipedia, there is actually no universal agreement on how to calculate quartiles for discrete distributions. The formulas shared here are used by TI-84 calculators, which is why we have chosen to use them.
## Example 1: Calculate Quartiles for Even Length Dataset
Suppose we have the following dataset with ten values:
Data: 3, 3, 6, 8, 10, 14, 16, 16, 19, 24
The median value is the average of the middle two values, which is (10 + 14) / 2 = 12.
We will not include this median value when calculating the quartiles.
The first quartile is the median of the lower half of values, which turns out to be 6:
Q1 = 3, 3, 6, 8, 10
The third quartile is the median of the upper half of values, which turns out to be 16:
Q3 = 14, 16, 16, 19, 24
Thus, the first and third quartiles for this dataset are 6 and 16, respectively.
## Example 2: Calculate Quartiles for Odd Length Dataset
Suppose we have the following dataset with nine values:
Data: 3, 3, 6, 8, 10, 14, 16, 16, 19
The median value is the value located directly in the middle: 10.
We will not include this median value when calculating the quartiles.
The first quartile is the median of the lower half of values. Since there are two values in the middle, we will take the average which turns out to be (3 + 6) / 2 = 4.5:
Q1 = 3, 3, 6, 8
The third quartile is the median of the upper half of values. Since there are two values in the middle, we will take the average which turns out to be (16 + 16) / 2 = 16:
Q3 = 14, 16, 16, 19
Thus, the first and third quartiles for this dataset are 4.5 and 16, respectively.
The following tutorials explain how to find the quartiles of a dataset using different statistical software:
## 3 Replies to “How to Find Quartiles in Even and Odd Length Datasets”
1. Sydney says:
I learned that Q1 is the lower and Q3 is the upper part of the data set. Also if you want to find Q1 and Q3 in a number line you have two different ways to to solve it depending if it is an odd number or even number.
2. Emma says:
The results of Q1 and Q3 for the odd data set are different compared to the excel formula. It follows this formula: Divide the data set into two halves, a bottom half and a top half. If n is odd, include the median value in both halves. Then the lower quartile is the median of the bottom half and the upper quartile is the median of the top half.
which one is correct?
1. James Carmichael says:
Hi Emma…To find quartiles in datasets, the process differs slightly depending on whether the dataset has an even or odd length. Here’s a step-by-step guide for both scenarios:
### Definitions
– **Quartiles**: These are values that divide your dataset into four equal parts.
– **Q1 (First Quartile)**: The median of the first half of the data.
– **Q2 (Second Quartile/Median)**: The median of the entire dataset.
– **Q3 (Third Quartile)**: The median of the second half of the data.
### Steps to Find Quartiles
#### For Even Length Datasets
1. **Sort the Data**: Arrange the data in ascending order.
2. **Find Q2 (Median)**:
– If the dataset has $$n$$ elements, the median is the average of the $$\frac{n}{2}$$th and $$\frac{n}{2} + 1$$th elements.
3. **Divide the Dataset into Two Halves**:
– First half: From the first element to the $$\frac{n}{2}$$th element.
– Second half: From the $$\frac{n}{2} + 1$$th element to the last element.
4. **Find Q1**:
– Q1 is the median of the first half.
5. **Find Q3**:
– Q3 is the median of the second half.
#### For Odd Length Datasets
1. **Sort the Data**: Arrange the data in ascending order.
2. **Find Q2 (Median)**:
– If the dataset has $$n$$ elements, the median is the $$\frac{n + 1}{2}$$th element.
3. **Divide the Dataset into Two Halves**:
– First half: From the first element to the $$\frac{n – 1}{2}$$th element.
– Second half: From the $$\frac{n + 1}{2} + 1$$th element to the last element.
4. **Find Q1**:
– Q1 is the median of the first half.
5. **Find Q3**:
– Q3 is the median of the second half.
### Examples
#### Even Length Dataset
Suppose the dataset is: [6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49, 50]
1. **Sort the Data**: It is already sorted.
2. **Find Q2 (Median)**:
– $$n = 12$$
– Median (Q2) = Average of the 6th and 7th elements = $$\frac{40 + 41}{2} = 40.5$$
3. **Divide the Dataset**:
– First half: [6, 7, 15, 36, 39, 40]
– Second half: [41, 42, 43, 47, 49, 50]
4. **Find Q1**:
– Q1 = Median of [6, 7, 15, 36, 39, 40] = $$\frac{15 + 36}{2} = 25.5$$
5. **Find Q3**:
– Q3 = Median of [41, 42, 43, 47, 49, 50] = $$\frac{43 + 47}{2} = 45$$
#### Odd Length Dataset
Suppose the dataset is: [6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49]
1. **Sort the Data**: It is already sorted.
2. **Find Q2 (Median)**:
– $$n = 11$$
– Median (Q2) = 6th element = 40
3. **Divide the Dataset**:
– First half: [6, 7, 15, 36, 39]
– Second half: [41, 42, 43, 47, 49]
4. **Find Q1**:
– Q1 = Median of [6, 7, 15, 36, 39] = 15
5. **Find Q3**:
– Q3 = Median of [41, 42, 43, 47, 49] = 43
### Summary
– **Even Length**: Calculate Q2 as the average of the middle two values, then find Q1 and Q3 as the medians of the lower and upper halves, respectively.
– **Odd Length**: Calculate Q2 as the middle value, then find Q1 and Q3 as the medians of the lower and upper halves, excluding the median.
This systematic approach ensures accurate calculation of quartiles for both even and odd length datasets. |
## Find Nth Factorial
Difficulty: Easy
#### Understanding The Problem
Problem Description
Write a program to find the factorial of a given number ``` n ``` .
Problem Note
• ``` n ``` is a non-negative integer.
• Factorial of a non-negative integer ``` n ``` is the multiplication of all integers smaller than or equal to ``` n ```
Example
``````Input: n = 6
Output: 720
Explanation: Factorial of 6 is 6*5*4*3*2*1 which is 720.``````
#### Solution
1. Recursive Solution : For f(x), ``` return 1 ``` when x is ``` 0 ``` , otherwise ``` return x*f(x-1) ```
2. Iterative Solution : Iterate from 1 to x and maintain a ``` factorial ``` variable.
You can try the problem here.
#### 1. Recursive Solution
We have to find the factorial of a number. Mathematically, the factorial of a number is the product from 1 to that number.
i.e. ``` factorial(Z) = 1 x 2 x 3 x 4 . . . x (Z-2) x (Z-1) x Z ```
Looking over the above equation, we can conclude that the ``` factorial(Z) = factorial(Z-1) x Z ```
Now, the equation seems like a recursive structure, where ``` factorial ``` is a function and we have to find the factorial of ``` Z ``` . Now, We know that the factorial of ``` 0 ``` is ``` 1 ``` and factorial doesn’t exist for -ve numbers. We can use this as a base case of the recursive function.
Base Case
Now think about the base case, where the function will directly give you results without breaking it again into sub-problem.
``````fact(5)
= 5 * fact(4)
= 5 * 4 * fact(3)
= 5 * 4 * 3 * fact(2)
= 5 * 4 * 3 * 2 * fact(1)
= 5 * 4 * 3 * 2 * 1 * fact(0)``````
The factorial of a negative number is not defined, so fact(0) is the smallest version of the factorial problem where our solution will terminate. Here n = 0 is the base case which will return 1. Since 0! = 1.
To understand more about recursion, refer here .
Solution Steps
1. Create a recursive function ``` factorial(Z) ```
2. Return 1 when ``` Z ``` is 0.
3. Otherwise, return ``` Z * factorial(Z-1) ```
Pseudo Code
``````int findFactorial(int n) {
if(n is 0)
return 1
return n * (findFactorial(n-1))
}``````
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
Critical Ideas To Think
• How did we define the recursive structure?
• Why we returned 1 when n is 1?
• Why is the space complexity is of O(n)? Answer : The recursive function uses stack space and in this case, we are recursively calling the function until n reaches to 0 and that will create n stack frames in the memory.
• Can you draw the recursion tree for this case?
#### 2. Iterative Solution
Instead of going recursive, we can use the logic of factorial in the way we do it in our notebooks and that is multiplying numbers from 1 to n.
So, the straight forward way is to use a for loop and iterate till n while maintaining a variable that will store the factorial.
Solution Steps
• Create a ``` factorial ``` variable and initialize it with ``` 1 ```
• Create a for loop and iterate from 1 to ``` n ```
• In each iteration, multiply ``` factorial ``` with ``` i ```
• Return ``` factorial ```
Pseudo Code
``````int findFactorial(int n) {
int factorial = 1
for(int i = 1 to i <= n){
factorial = factorial * i
}
return factorial
}``````
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
Critical Ideas To Think
• Why we started the iteration from 1 while the base case in recursive function was ``` n is 0 ``` ?
• According to you, what should be the maximum number for which you can find the factorial if you are using a single processor system and using some data structure like Big integer?
#### Suggested Problems To Solve
If you have any more approaches or you find an error/bug in the above solutions, please comment down below.
Happy Coding!
Enjoy Algorithms! |
Class 11 – Mathematics – Chapter 5 – Complex Numbers Quadratic Equations
Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a)
Sol: a = cos θ + i sin θ
Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a circle.
Sol: Let z = x + iy
Q11. Solve the equation |z| = z + 1 + 2i.
Sol: We have |z| = z + 1 + 2i
Putting z = x + iy, we get
|x + iy| = x + iy + 1+2i
Q12. If |z + 1| = z + 2( 1 + i), then find the value of z.
Sol: We have |z + 1
1 = z + 2(1+ i)
Putting z = x + iy, we get
Then, |x +
iy + 11 = x + iy + 2(1 + i)
⟹|x + iy + l|=x + iy + 2(1 +i)
Q13. If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
Sol: We have arg (z – 1) = arg (z + 3i), where z = x + iy
=> arg (x + iy – 1) = arg (x + iy + 3i)
=> arg (x – 1 + iy) = arg [x + i(y + 3)]
Q14. Show that | z-2/z-3| = 2 represents a circle . Find its center and radius .
Sol:
We have | z-2/z-3| = 2
Puttingz=x + iy, we get
Q15. If z-1/z+1 is a purely imaginary number (z ≠1), then find the value of |z|.
Sol: Let z = x + iy
Q17. If |z1 | = 1 (z1≠-1) and z2 = z1 – 1/ z1 + 1 , then show that real part of z2 is zero .
Q18. If Z1, Z2 and Z3, Z4 are two pairs of conjugate complex numbers, then find arg (Z1/ Z4) + arg (Z2/ Z3)
Sol. It is given that z1 and z2 are conjugate complex numbers.
Q20. If for complex number z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1 – z2| = | z1|- |z2 |
Q21. Solve the system of equations Re (z2) = 0, |z| = 2.
Sol: Given that, Re(z2) = 0, |z| = 2
Q22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.
Fill in the blanks
True/False Type Questions
Q26. State true or false for the following.
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z, the minimum value of |z| + |z – 11 is 1.
(iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1).
(v) If z is a complex number such that z ≠0 and Re(z) = 0, then Im (z2) = 0.
(vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z1 and Z2 be two complex numbers such that |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.
Sol:(i) False
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.
(ii) False
Let z = x + iy, where x, y > 0
i.e., z or point A(x, y) lies in first quadrant. Now, —iz = -i(x + iy)
= -ix – i2y = y – ix
Now, point B(y, – x) lies in fourth quadrant. Also, ∠AOB = 90 °
Thus, B is obtained by rotating A in clockwise direction about origin.
Matching Column Type Questions
Q24. Match the statements of Column A and Column B.
Column A Column B (a) The polar form of i + √3 is (i) Perpendicular bisector of segment joining (-2, 0) and (2,0) (b) The amplitude of- 1 + √-3 is (ii) On or outside the circle having centre at (0, -4) and radius 3. (c) It |z + 2| = |z – 2|, then locus of z is (iii) 2/3 (d) It |z + 2i| = |z – 2i|, then locus of z is (iv) Perpendicular bisector of segment joining (0, -2) and (0,2) (e) Region represented by |z + 4i| ≥ 3 is (v) 2(cos /6 +I sin /6) (0 Region represented by |z + 4| ≤ 3 is (Vi) On or inside the circle having centre (-4,0) and radius 3 units. (g) Conjugate of 1+2i/1-I lies in (vii) First quadrant (h) Reciprocal of 1 – i lies in (viii) Third quadrant
Q28. What is the conjugate of 2-i / (1 – 2i)2
Q29. If |Z1| = |Z2|, is it necessary that Z1 = Z2?
Sol: If |Z1| = |Z2| then z1 and z2 are at the same distance from origin.
But if arg(Z1) ≠arg(z2), then z1 and z2 are different.
So, if (z1| = |z2|, then it is not necessary that z1 = z2.
Consider Z1 = 3 + 4i and Z2 = 4 + 3i
Q30.If (a2+1)2 / 2a –i = x + iy, then what is the value of x2 + y2?
Sol:
(a2+1)2 / 2a –i = x + iy
Q31. Find the value of z, if |z| = 4 and arg (z) = 5Ï€/6
Q34. Where does z lies, if | z – 5i / z + 5i | = 1?
Sol:
We have | z – 5i / z + 5i |
Instruction for Exercises 35-40: Choose the correct answer from the given four options indicated against each of the Exercises.
Q35. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for
Q41. Which of the following is correct for any two complex numbers z1 and z2? |
Page 1 of 1
### Partial Fractions
Posted: June 2nd, 2011, 4:46 pm
$\frac{x+7}{x^2-x-6} = \frac{2}{x-3}-\frac{1}{x+2}$
Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition on the left side.
Steps
1. Factor the Denominator.
2. Express factored form as a sum of denominators using A and B as numerators and set this equal to the original problem.
3. Eliminate the denominators by multiplying each side by the LCD.
4. Find the excluded values of x. (What would make the denominator equal 0?)
5. Substitute the excluded values into the equation and solve for A and B.
6. Substitute the values into step 2.
A. $\frac{x+7}{x^2-x-6} = \frac{A}{x-3}-\frac{B}{x+2}$
$\left [\frac{x+7}{x^2-x-6} = \frac{A}{x-3}-\frac{B}{x+2} \right ](x-3)(x+2)$
$x+7 = A(x-2)+B(x-3)$
x = −2
$-2+7=A(-2+2)+B(-2-3)$
5 = −5B
B = −1
x = 3
$3+7=A(3+2)+B(3-3)$
10 = 5A
A = 2
$\frac{2}{x-3}-\frac{1}{x+2}$
B. $\frac{x+8}{x^2+6x+8} = \frac{A}{x+4}-\frac{B}{x+2}$
$\left [\frac{x+8}{x^2+6x+8} = \frac{A}{x+4}-\frac{B}{x+2} \right ](x+4)(x+2)$
$x+8 = A(x+2)+B(x+4)$
x = −4
$-4+8=A(-4+2)+B(-4+4)$
4 = −2A
A = −2
x = −2
$-2+8=A(-2+2)+B(-2+4)$
6 = 2B
B = 3
$\frac{-2}{x+4}-\frac{3}{x+2}$ |
# Unit 7: Sequences and Functions
## 7.1 Arithmetic Sequences
• Screen 2 defines arithmetic sequence and common difference, and then screen 3 has some questions: (1) What is the common difference, (2) which is an arithmetic sequence?
• Screen 4 introduces terms. First term, second term, $n$th term, etc.
• Screen 5 introduces $a_n$ notation, starting with the recursive formula $a_n = a_{n-1} + d$, with (screen 6) $a_1$ as the first term.
• Screen 7 has a student write the recursive formula for a given arithmetic sequence.
• Screen 10 introduces the explicit formula, $a_n = a_1 + \left(n-1\right)d$.
• Screen 11 has a student find the $n$th term of a sequence, given the explicit formula.
• Screen 13 introduces some word problems, and 14.
• Screen 15 discusses converting between the two forms of formulas here (recursive and explicit).
Arithmetic Sequences
Consider the sequence of numbers $3, 7, 11, 15, 19, \cdots$.
The first term of that sequence is $a_1 = 3$.
The second term of that sequence is $a_2 = 7$. We also know that $a_3=11$, $a_4=15$, and $a_5=19$.
From each term to the next we are adding $4$. This means that the common difference of this sequence is $d=4$. We can find it by subtracting any two terms: $19-15=4$, $15-11=4$, $11-7=4$, and $7-3=4$.
Consider the sequence of numbers $2, 6, 18, 54, \cdots$.
This sequence does not have a common difference, because $6-2=4$ but $18-6=12$. This is not an arithmetic sequence.
Recursive Form of an Arithmetic Sequence
The recursive form of an arithmetic sequence says where the sequence starts, and what happens from one term to the next.
Consider the sequence of numbers $3, 7, 11, 15, 19, \cdots$.
Where does the sequence start? $a_1=3$.
What happens as we move from one term to the next? We add $d=4$. $a_n=a_{n-1}+4$.
\begin{aligned}a_1&=3\\a_n&=a_{n-1}+4\end{aligned}
Explicit Form of an Arithmetic Sequence
...
Converting between Recursive and Explicit Forms of an Arithmetic Sequence
...
Converting from Recursive to Explicit Forms of an Arithmetic Sequence
...
Converting from Explicit to Recursive Forms of an Arithmetic Sequence
... |
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3.
Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 1 Chapter Name Number Systems Exercise Ex 1.3 Number of Questions Solved 9 Category NCERT Solutions
## NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3
Question 1.
Write the following in decimal form and say what kind of decimal expansion each has
Solution:
Question 2.
You know that $$\frac { 1 }{ 7 }$$ = $$\bar { 0.142857 }$$. Can you predict what the decimal expansions of $$\frac { 2 }{ 7 }$$ , $$\frac { 13 }{ 7 }$$ , $$\frac { 4 }{ 7 }$$ , $$\frac { 5 }{ 7 }$$ , $$\frac { 6 }{ 7 }$$ are , without actually doing the long division? If so, how?
Solution:
Question 3.
Express the following in the form $$\frac { p }{ q }$$where p and q are integers and q ≠ 0.
(i) 0.$$\bar { 6 }$$
(ii) 0.4$$\bar { 7 }$$
(iii) 0.$$\overline { 001 }$$
Solution:
(i)Let x= 0.$$\bar { 6 }$$ = 0.666… ….(i)
Multiplying Eq. (i) by 10, we get
10x = 6.666.. ….(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(10x- x)=(6.666…) – (0.666…)
9x = 6
x= 6/9
⇒ x=2/3
(ii)
Let x = 0.4$$\bar { 7 }$$ = 0.4777… …(iii)
Multiplying Eq. (iii) by 10. we get
10x = 4.777… . …(iv)
Multiptying Eq. (iv) by 10, we get
100x = 47.777 ….. (v)
On subtracting Eq. (v) from Eq. (iv), we get
(100 x – 10x)=(47.777….)-(4.777…)
90x =43
⇒ x = $$\frac { 43 }{ 90 }$$
(iii)
Let x = 0.$$\overline { 001 }$$= 0.001001001… …(vI)
Multiplying Eq. (vi) by (1000), we get
1000x = 1.001001001… .. .(vii)
On subtracting Eq. (vii) by Eq. (vi), we get
(1000x—x)=(1.001001001….) – (0.001001001……)
999x = 1
⇒ x = $$\frac { 1 }{ 999 }$$
Question 4.
Express 0.99999… in the form $$\frac { p }{ q }$$Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.99999… ………..(i)
Multiplying Eq. (i) by 10, we get
10x = 9.99999… …(ii)
On subtracting Eq. (ii) by Eq. (i), we get
(10 x – x) = (9.99999..) – (0.99999…)
9x = 9
⇒ x = $$\frac { 9 }{ 9 }$$
x = 1
Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $$\frac { 1 }{ 17 }$$? Perform the division to check your answer.
Solution:
The maximum number of digits in the repeating block of digits in the decimal expansion of $$\frac { 1 }{ 17 }$$ is 17-1 = 16 we have,
Thus,$$\frac { 1 }{ 17 }$$ = 0.$$\overline { 0588235294117647….., }$$a block of 16-digits is repeated.
Question 6.
Look at several examples of rational numbers in the form $$\frac { p }{ q }$$ (q≠ 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
Consider many rational numbers in the form $$\frac { p }{ q }$$ (q≠ 0). where p and q are integers with no common factors other that 1 and having terminating decimal representations.
Let the various such rational numbers be $$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 4 }$$, $$\frac { 5 }{ 8 }$$, $$\frac { 36 }{ 25 }$$, $$\frac { 7 }{ 125 }$$, $$\frac { 19 }{ 20 }$$, $$\frac { 29 }{ 16 }$$ etc.
In all cases, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10.
From the above, we find that the decimal expansion of above numbers are terminating. Along with we see that the denominator of above numbers are in the form 2m x 5n, where m and n are natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.
Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
0.74074007400074000074…
0.6650665006650006650000…
0.70700700070000…
Question 8.
Find three different irrational numbers between the rational numbers $$\frac { 5 }{ 7 }$$ and $$\frac { 9 }{ 11 }$$ .
Solution:
To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
so,
Question 9.
Classify the following numbers as rational or irrational
Solution:
(i) $$\sqrt{23}$$ (irrational ∵ it is not a perfect square.)
(ii) $$\sqrt{225}$$ = 15 (rational) (whole number.)
(iii) 0.3796 = rational (terminating.)
(iv) 7.478478… =7.$$\bar { 478 }$$ = rational (non-terminating repeating.)
(v) 1.101001000100001… = irrational (non-terminating non-repeating.)
We hope the NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3, drop a comment below and we will get back to you at the earliest. |
Help with Opening PDF Files
## Chapter 12
### Part 1: For the problem in the Teacher's Edition, page 300
Provide your students with a copy of the Problem Worksheet (PDF file).
If students need help on this activity, use the following questions to guide them.
• What is the expression for two times the number of trails that Antonio visited? (2a)
• What should we do with that expression so that it represents the number of trails that Brian visited? (Subtract five from it.)
• What is the expression for the number of trails that Brian visited? (2a − 5)
2a - 5
Antonio: 6, Brian: 7
Antonio: 7, Brian: 9
Antonio: 8, Brian: 11
### Part 2: Be an Investigator
A good time to do this investigation is after Lesson 1 on writing expressions.
#### Introducing the Investigation
Introduce the investigation by reading aloud the assignment at the top of the first page of the Description of Investigation and Student Report (PDF file), by having one of your students read aloud the assignment, or by having the students read the assignment individually.
Put students in groups of two to four students to work on the investigation.
#### Doing the Investigation
When students get to the last question, ask the following questions to help guide them.
• When the number of hours is less than four, do you think it is better to choose Plan A or Plan B? (Plan B)
• When the number of hours is greater than six, do you think it is better to choose Plan A or Plan B? (Plan A)
• What number of hours would you like to try to see if we get the same cost for both plans? (Try whatever the students suggest. Since five is greater than four and less than six, five would be a good number of hours to try.)
• What happens when we use five hours in both the Plan A and the Plan B expressions? (We get \$60 either way.)
Here are the expressions for each plan.
x is the number of hours of a bike trip.
Plan A: 50 + 2x
Plan B: 12x
Cost for Each Plan and Recommendation
Trip L: 2 hours
Plan A: 50 + (2 × 2) = 54 \$54
Plan B: 12 × 2 = 24 \$24
Plan B is the best choice.
Trip M: 4 hours
Plan A: 50 + (2 × 4) = 58 \$58
Plan B: 12 × 4 = 48 \$48
Plan B is the best choice.
Trip N: 6 hours
Plan A: 50 + (2 × 6) = 62 \$62
Plan B: 12 × 6 = 24 \$72
Plan A is the best choice.
Trip O: 8 hours
Plan A: 50 + (2 × 8) = 66 \$66
Plan B: 12 × 8 = 24 \$96
Plan A is the best choice.
If you take a bike trip of five hours it does not matter which plan you choose.
#### Student Report
The letter back to Susan Chung gives students an opportunity to show the expressions they have written and make recommendations. |
Question
# In the diagram of \triangle GKJ below, LH\parallel KJ, GL=4, LK=24, and HJ=30. What is the lenght of GJ?
Similarity
In the diagram of $$\displaystyle\triangle{G}{K}{J}$$ below, $$\displaystyle{L}{H}\parallel{K}{J}$$, GL=4, LK=24, and HJ=30. What is the lenght of GJ?
2021-08-09
Step 1
Here, the objective is to find the length of the side using the similarity of the triangle.
Step 2
In the figure, check the similarity of the triangles $$\displaystyle\triangle{G}{L}{H}\ {\quad\text{and}\quad}\ \triangle{G}{K}{J}$$.
Here, $$\displaystyle{L}{H}{p}{a}{r}{a}{l}\le{l}{K}{J}$$, therefore,
$$\displaystyle\angle{G}{L}{H}=\angle{G}{K}{J}$$
$$\displaystyle\angle{G}{H}{L}=\angle{G}{J}{K}$$
$$\displaystyle\angle{L}{G}{H}=\angle{K}{G}{J}$$
Therefore, both the triangles $$\displaystyle\triangle{G}{L}{H}\approx\triangle{G}{K}{J}$$
So, the ratio of sides of the triangles will be equal.
Step 3
Now, use the ratio of the sides as,
$$\displaystyle{\frac{{{G}{L}}}{{{L}{K}}}}={\frac{{{G}{H}}}{{{H}{J}}}}$$
$$\displaystyle{\frac{{{4}}}{{{24}}}}={\frac{{{G}{H}}}{{{30}}}}$$
GH=5
The length of the side GJ is,
GJ=GH+HJ
=5+30
=35
Therefore, the length of side GJ=35 |
Pronunciation: /əˈdɪ.ʃən/ Explain
Addition is joining two or more quantities together to make a sum. Figure 1 gives an example of integer addition. In the addition statement 4 + 3 = 7, 4 and 3 are addends and 7 is the sum. Addends are what is being added together, and the sum is the result of the addition. The symbol '+' is called the addition sign.
Figure 1: Representation of 4 + 3 = 7
Real numbers can also be added. Figure 2 gives a representation of 2 + 1 = 3.
Figure 2: Representation of 2 + 1 = 3
Manipulative 3 represents the addition of real numbers. Click on the blue points and drag them to change the figure.
Click on the points A and B near the top to change the figure. The blue and red arrows on the number line show A added to B. The sum A+B is below them in purple. Manipulative 1 - Addition Created with GeoGebra.
Addition is also defined for many types of math entities such as vectors and matrices.
### Subtraction
Mathematicians usually define subtraction as adding the additive inverse of a value. Subtraction is defined this way so that subtraction of vectors and matrices and other math entities makes more sense. Stated mathematically, a - ba + -b. A difference is the result of subtracting one number from another. For example, in the equation 7 - 4 = 3, 3 is the difference.
A example of this is 5 - 4 = 5 + -4 = 1
### Properties of Addition of Real and Complex Numbers
Property NameExampleDescription
a + 0 = 0 + a = a Any number plus zero equals the original number. 0 is the additive identity for real and complex numbers.
Additive inverse a + (-a) = 0 The additive inverse of any real or complex number is the negative of that numbers.
Associative property of addition a + ( b + c ) = ( a + b ) + a The order of in which multiple additions of real and complex numbers are performed does not change the result.
Commutative property of addition a + b = b + a It doesn't matter which of two numbers come first in addition of real and complex numbers.
Distributive Property of Multiplication over Addition and Subtraction a ( b + c ) = ab + ac, and a ( b - c ) = ab - ac Multiplication is distributive over addition and subtraction.
Additive property of equality If a = b then a + c = b + c. The additive property of equality states that any number can be added to both sides of an equation without changing the truth value of the equation.
Subtractive property of equality If a = b then a - c = b - c. The subtractive property of equality states that any number can be subtracted from both sides of an equation without changing the truth value of the equation.
Addition facts are two operands and the result of adding those two operands. The following table gives the addition facts for 0 through 10.
+012345678910
0012345678910
11234567891011
223456789101112
3345678910111213
44567891011121314
556789101112131415
6678910111213141516
77891011121314151617
889101112131415161718
9910111213141516171819
101011121314151617181920
### How to Add Complex Numbers
To add two complex numbers, add the corresponding parts. Given two complex numbers a + bi and c + di, (a + bi) + (c + di) = (a + c) + (b + d)i. Example: (3 - 2i) + (-1 + 3i) = (3 + (-1)) + (-2 + 3)i = 2 + 1i = 2 + i.
### References
1. McAdams, David E.. All Math Words Dictionary, addition. 2nd Classroom edition 20150108-4799968. pg 10. Life is a Story Problem LLC. January 8, 2015. Buy the book
### Revision History
4/5/2019: Updated formating of equations. (McAdams, David E.)
12/21/2018: Reviewed and corrected IPA pronunication. (McAdams, David E.) |
# Equation Formula
Quadratic equation is one of the fundamental concept of algebra. By using equations, this quadratic form can be solved and solutions will be resolved. In this case there are two solutions. The formula for equation is given as,
Let ax2+ bx + c = 0 is the quadratic equation, the solution will be,
## Equation Problems
Let us discuss the solutions of quadratic equations.
### Solved Examples
Question 1: Find out the roots of the given equation,x2 + 2x – 3 = 0 ?
Solution:
The given quadratic equation is,
x2 + 2x – 3 = 0
Here a = 1, b = 2 and c = -3
The formula for the solution,
x = $\frac{-bpm \sqrt{b^{2}-4ac}}{2a}$
x = $\frac{-2pm \sqrt{2^{2}-4\times1\times-3}}{2\times1}$
x = $\frac{-2pm \sqrt{4+12}}{2}$
x = $\frac{-2+4}{2}$ = 1 or x = $\frac{-2-4}{2}$ = -3
Question 2: Find out the roots of 2x2 – 3x – 5 = 0?
Solution:
The given quadratic equation is,
2x2 – 3x – 5 = 0
Here a = 2, b = -3 and c = -5
The formula for the solution,
x = $\frac{-bpm \sqrt{b^{2}-4ac}}{2a}$
x = $\frac{3\pm \sqrt{3^{2}-4\times2\times-5}}{2\times2}$
x = $\frac{3\pm \sqrt{41}}{4}$
x = $\frac{3+7}{4}$ = $\frac{5}{2}$ or x = $\frac{3-7}{4}$ = -1
More topics in Equation Formula Linear Equations Formula
Related Formulas Double Angle Formulas Formula for Gross Profit Equilateral Triangle Formula Function Notation Formula Hyperbolic Function Formula Graphs of Trigonometric Functions Formula Interest Formula Height of a Parallelogram Formula |
Instruction
1
Rectangle.Task: compute the area of a rectangleif you know its perimeter is 40 and the length b is 1.5 times the width a.
2
Solution.Use the well-known formula of the perimeter, it is equal to the sum of all sides of the figure. In this case, P = 2•a + 2•b. From the initial data of the problem, you know that b = 1,5•a, therefore P = 2•a + 2•1,5•a = 5•a where a = 8. Find the length of b = 1.5 x 8 = 12.
3
Write down the formula for the area of the rectangle:S = a•b,Substitute the known values:S = 8•*12 = 96.
4
Square.Task: find the area of a square if the perimeter is 36.
5
Solution.A square is a special case of rectangle where all sides are equal, so the perimeter is 4•a where a = 8. The area of a square define by the formula S = a2 = 64.
6
Triangle.Problem: suppose that we are given an arbitrary triangle ABC with a perimeter of 29. Find out the magnitude of its area, if it is known that the height BH, descended on the AC side, and divides it into segments of lengths 3 and 4, see
7
Solution.First recall the formula for area of a triangle:S = 1/2•c•h, where c is the base and h is the height of the figure. In our case, the basis is the AC side, which is known according to the problem: AC = 3+4 = 7, it remains to find the height BH.
8
The height is a perpendicular drawn to the side from the opposite vertex, therefore, it is to divide the triangle ABC into two right-angled triangles. Knowing this property, consider the triangle ABH. Remember the formula of Pythagoras that:AB2 = BH2 + AH2 = BH2 + 9 → AB = √(h2 + 9).In the triangle BHC on the same principle record:BC2 = BH2 + HC2 = BH2 + 16 → BC = √(h2 + 16).
9
Apply the formula for perimeter:P = AB + BC + Aspositive values expressed through the height:P = 29 = √(h2 + 9) + √(h2 + 16) + 7.
10
Solve the equation:√(h2 + 9) + √(h2 + 16) = 22 → [replace t2 = h2 + 9]:√(t2 + 7) = 22 - t, lift both sides of the equality to the square:t2 + 7 = 484 – 44•t + t2 → t≈10,84h2 + 9 = 117,5 → h ≈ 10,42
11
Find the area of triangle ABC:S = 1/2•7•10,42 = 36,47. |
abondantQ
2021-05-11
Evaluate the line integral, where C is the given curve
C xy ds
C: $x={t}^{2},y=2t,0\le t\le 5$
Dora
Integral solution:
Jeffrey Jordon
To evaluate the line integral ${\int }_{C}xyds$, where $C$ is the curve $C:x={t}^{2},y=2t,0\le t\le 5$, we first need to parameterize the curve in terms of a single parameter.
We can parameterize the curve by letting $x={t}^{2}$ and $y=2t$, which gives us the parametric equations $x={t}^{2}$ and $y=2t$, with $0\le t\le 5$.
Next, we need to express $ds$ in terms of $dt$. We have
$ds=\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt.$
Substituting $x={t}^{2}$ and $y=2t$ into this equation, we get
$ds=\sqrt{\left(2t{\right)}^{2}+\left(2{\right)}^{2}}dt=\sqrt{4{t}^{2}+4}dt=2\sqrt{{t}^{2}+1}dt.$
Now we can evaluate the line integral:
${\int }_{C}xyds={\int }_{0}^{5}\left({t}^{2}\right)\left(2t\right)\left(2\sqrt{{t}^{2}+1}\right)dt=4{\int }_{0}^{5}{t}^{3}\sqrt{{t}^{2}+1}dt.$
To evaluate this integral, we can use the substitution $u={t}^{2}+1$, which gives us $du=2tdt$. Substituting this into the integral, we get
${\int }_{0}^{5}{t}^{3}\sqrt{{t}^{2}+1}dt=\frac{1}{2}{\int }_{1}^{26}\left(u-1{\right)}^{3/2}du.$
Using the power rule for integration, we can evaluate this integral to get
$\frac{1}{2}{\int }_{1}^{26}\left(u-1{\right)}^{3/2}du=\frac{1}{2}·\frac{2}{5}\left(u-1{\right)}^{5/2}{|}_{1}^{26}=\frac{1}{5}\left(26\sqrt{677}-8\sqrt{2}-5\right).$
Therefore, the value of the line integral is $\overline{)\frac{1}{5}\left(26\sqrt{677}-8\sqrt{2}-5\right)}$.
RizerMix
The given line integral can be expressed as:
where $C$ is the curve defined by $x={t}^{2}$, $y=2t$, and $0\le t\le 5$.
To evaluate the line integral, we need to parameterize the curve in terms of a single variable. We can choose $t$ as the parameter, so that $x={t}^{2}$ and $y=2t$.
Then, the differential element of arc length $ds$ is given by:
Substituting for $x$, $y$, and $ds$ in the integral, we get:
Simplifying the integrand, we get:
To evaluate this integral, we can make the substitution $u={t}^{2}+1$, so that and ${t}^{2}=u-1$. Then, the integral becomes:
Evaluating the definite integral, we get:
Therefore, the value of the line integral is:
Do you have a similar question? |
### Scribe Post for April 23 2009
Friday, April 24, 2009
Hi everyone! During yesterdays math class we were given a word problem about parks in New York City. The parks were Carol Gardens and Flatbush. People think that the children of New York City should get a place to play, so they decided to make some lots have a playground and some blacktop so that children were able to play blacktop games, like basketball for instance.
Both places have the same area, 100 M by 50 M. 3/4 of the lot will be playground and 2/5 of the playground will be hardtop. Our goal is to find out which park will have more hardtop. Note that both lots are rectangular.
Flatbush
The image above is a representation of the Flatbush lot. 2/5 of the whole lot is the playground. Above is a representation of how much of the lot will be playground. Now 3/4 of that 2/5 will blacktop.
We're trying to find how many times 3/4 will fit into 2/5 evenly, since 2/5 of the lot is playground and 3/4 of that (2/5) is hardtop. To do so we divide 2/5 by 3/4.
One of the most easiest ways to compare fractions is to divide the numerator by the denominator, or finding the decimal. 8 divided by 15 is 0.53 repeated.
Carol Gardens
Above is a representation of Carol Gardens. As you can see it is identical to Flatbush. 3/4 of the whole lot is the playground.
2/5 of the playground (3/4 of the whole lot) is hardtop.
The darker shaded area is the hardtop. We're trying to find how many times 2/5 can fit into 3/4 evenly, because 3/4 of the lot is the playground and 2/5 of that is the hardtop. To do that you divide 3/4 by 2/5.
1.875 is 15/8 in decimal form.
Which brings us back to our main goal... Which park gets more blacktop than the other?
As mentioned previously, one of the easiest ways to compare fractions is to convert the fraction into a decimal. Flatbush gets 8/15 of hardtop, or 0.53 repeated. Carol Gardens gets 15/8 of hardtop, or 1.875. So, Carol Gardens gets more hardtop than Flatbush. I hope this helps and that you're all well! (:
Please feel welcome to comment if I've made a mistake, or have done this word problem incorrectly!
The next scribe will be Nikki.
1. Giselle 8-17 said...
i like how you explained the problem with pictures and the way your explanation was easy to understand. i didn't find any mistakes except that Carol Gardens is spelled with two "R"s (according to my book), but that doesn't really matter. great job linda! :)
April 26, 2009 at 4:50 PM
2. Karen8-17 said...
WOW! Good Job Linda! I completely understand now. Your explanation made it seem so simple and easy. Gizzle pointed out your mistake so I have nothing. Good job and again and yes, you explanation is easy to understand!
-karebear ;P
April 26, 2009 at 7:52 PM |
Sales Toll Free No: 1-855-666-7446
# Linear Function
Top
Sub Topics Calculus is based upon the linear Functions. Linear Functions are the functions whose graphs consist of sections of one straight line throughout the function's Domain. Linear function Statistics can be defined as the functions that have ‘x’ as the input variable and ‘x’ has an exponent of only 1. Linear functions can refer to the following concepts · A first degree polynomial function having one variable. · A map between two vector spaces that refers vector addition and Scalar multiplication. Linear function are said to be linear because the graphs of these functions in the Cartesian co- ordinate plane is Straight Line. For example there are some functions below whose graph is a straight line · g (x) = 2x + 4 · g (x) = x/2 - 3 Linear functions can be written in the following format f (x) = mx + b, (y – y1) = m (x - x1), 0 = Ax + By + C, In vector Algebra a linear function means a linear map i.e. a map between two vectors spaces that refers vector addition and Scalar Multiplication. The linear functions are those functions ‘f’ that can be expressed as, f (x) = Kx, where ‘K’ is a matrix. A function g (x) = mx + b is called a linear map if and only if b = 0. The form y = mx + b is named as the 'slope intercept form' of linear function. For a common graph (x,y) usually expressed as y = mx + b and in a formal function definition a linear function written as f (x) = mx + b. We can derive this fact for the linear functions. Suppose a linear function which takes value ‘g’ (c1) at c1 and g (c2) at c2 by the following formula- g (x) = g (c1) x – c2 / c1 – c2 + g (b) x – c1 / c2 – c1, The first term will be zero when ‘x’ is ‘c2’ and is ‘g (c1)’ when ‘x’ is ‘c1’, while the second term is zero when ‘x’ is ‘c1’ and is g(c2) when ‘x’ is ‘c2’. More convenient form for this function is- g (x) = x g (c2) – g (c1) / c2 – c1] + [c2 f(c1) – c1 f (c2) / c2 – c1]. g (x) = mx + c, Here the ‘m’ shows the Slope of this line. If we plot the graph on the ‘y’ axis then ‘c’ is called here the ‘y’ intercept of the line. The description of this equation can be provided as a Set of or locus of (x, y) points and these points lie along a straight line. The variable m refers to the Slope of this line and the variable ‘b’ refers to the ‘y’ co-ordinate where the line crosses the y- axis named as 'y-intercept'. Point slope form for the linear functions shows the equation of a line i.e. (y – y1) = m (x - x1).
## Statistical Modelling
In the form of mathematical equations the formalization of relationships between variables is known as Statistical model. This shows that how one or more random variables are related to the other random variables. A statistical model can be taken as a pair (Z, P) where ‘Z’ is the Set of possible observations and ‘P’ is the set of possible Probability distributions on ‘Z’. ‘P’ is here is a set of distinct elements which generates the observed data.
Statistical tests can be described in the form of statistical model. There is a similarity between tests and model.
In a formal way Statistical model can be defined as
"Statistical model ‘P’ is a collection of Probability Distribution functions or probability density Functions that is a collection of distributions, each of which is tabulated by a unique finite dimensional parameter ‘P’. The statistical model is that variability is represented using probability distributions which make the building blocks from which the model is constructed.
Statistical modeling has two main functions that are prediction and explanation.
Prediction is the calculation of the output when there is a given set of input values or the measurement of the output if when there is a change in a particular input.
Explanation is the measurement of the relationship of the variables. It is the measurement of the variation of the output on the dependent variable.
There are several different statistical model techniques.
Regression model: Regression model involves estimating the mathematical relationship between one variable and one or more explanatory variables or independent variables. The single variable is known as the response variable or dependent variable. The object of regression models is to forecast time series data. Regression analysis is used for the purpose of prediction and forecasting. It is also used to understand among the independent variables that which are related to the dependent variable. The performance of regression analysis methods in practice depends on the form of the process of data generation.
Some assumptions are there to understand regression those are-
The sample is representative of the population for the prediction of inference.
The error that is made is a Random Variable with a Mean of zero conditional on the explanatory variables.
Generally the independent variables are measured with less error or no error.
The predictors are linearly independent means it, is not possible to express any predictor as a linear Combination of the others.
Non parametric model is also the part of the statistical models but they are different from the parametric models because they are driven rather than structured with the parameters to be estimated.
Semi parametric models-
An example of semi parametric models is the regression model with a smoother component for greater flexibility.
Bayesian modeling is based on Probabilities rather than frequencies like the probability of ‘P’, given ‘Q’, ‘R’ and ‘S’, these are based on Baye's theorem i.e.
P (P | Q) = P (Q | A) P(P) / P(Q).
## Linear Correlation Coefficient
A Linear Correlation coefficient in Statistics is basically a measure of the strength of association in between two or more variables. To measure the strength of the linear correlation and association between two variables the most commonly used correlation coefficient is Pearson Product Moment Correlation Coefficient. Here we will denote the linear correlation coefficient for sample by “r” and denote linear correlation coefficient for a population by R.
The sign of the linear correlation coefficient with an absolute value always defines the actual direction and the magnitude of the relation between two variables. Here are some important facts about the linear correlation coefficient:
1. The range for the linear correlation coefficients is between -1 to 1.
2. The linear relationship between two variables will be stronger if the absolute value of the coefficient is high.
3. Strongest linear relationship is determined by the linear correlation coefficient which is either -1 or 1.
4. Weakest relationship id determines by the linear correlation coefficient which is 0.
5. If the linear correlation coefficient is positive than it means that if one variable is high than the other variable tends to get higher.
6. Similarly in case of negative linear correlation coefficient if one variable is bigger than the other variable tends to get smaller.
The Pearson Product Moment Correlation Coefficient is used to calculate the linearity of relationship so if the coefficient is zero that’s simple Mean that the relationship is not linear.
Here is the formula for calculating the Linear Correlation Coefficient (r):
Linear correlation coefficient formula, r = ∑ (a*b) / sqrt (∑ a^2) (∑b^2)
Here the sign ∑ is a summation symbol where a is (ai - a) and b is (bi - b).
ai and bi are the ith value of the observation a and b respectively and a and b are the mean of all the values.
In case of population correlation coefficient the formula is:
R = [1/n] * ∑ [(ai - µa) / σa] * [(bi - µb) / σb]
n is the number of observations, ∑ is a symbol for sum up, ai and bi are the values for the ith observations, µa and µb are the population mean for the variables a and b respectively. σa and σb are the population Standard Deviation of a and b.
Similarly in case of Sample correlation coefficient the formula is:
r = [1/(n – 1)] * ∑ [(ai - a) / Sa] * [(bi - b) / Sb]
where n is the number of observations, ∑ is a symbol for sum up, ai and bi are the values for the ith observations, Sa is the sample standard deviation of a and Sb is sample standard deviation of b.
The Sample correlation coefficient always depends on how we collect the sample data. It is basically an unbiased estimation of the population correlation coefficient which is slightly different from this.
The formulas for the Coefficient Of Linear Correlation of Sample and Populations are derived by the first one formula. These days many correlation software packages and graphic calculators are available in the market.
## Linear Regression Coefficient
For finding the relationship between two variables, we calculate linear regression coefficient. This coefficient of Linear Regression is useful for finding the strength and direction between two linear variables and we use following formula for evaluating the coefficient of linear regression-
Correlation coefficient r = N ∑x y - (∑x)(∑y) / √([N ∑x2 - (∑x)2]) . ([N ∑y2 - (∑y)2]),
Where N = data points in a Set,
∑x y = summation of all values of x y,
∑x = summation of all values of ‘x’,
∑y = summation of all values of ‘y’,
∑x2 = summation of all values of x2,
∑y2 = summation of all values of y2,
Now we discuss properties of linear regression coefficient:
There are following properties used in linear regression coefficient-
Property 1: The value of Linear Correlation coefficient is lie between -1 and 1 means
-1 <= r <= 1, where if r = -1, then this type of correlation is called as negative correlation and if r = 1, then this type of correlation is called as a positive correlation.
Property 2: If the values of two variables like ‘x’ and ‘y’ have positive and strong correlation, then this type of linear correlation is close to 1 means when value of ‘x’ is increased, then for value of ‘x’, value of ‘y’ also increases and if Linear Correlation Coefficient is exactly +1, then this type of linear correlation coefficient is called as a positive perfect fit correlation between two variables.
Property 3: If the values of two variables like ‘x’ and ‘y’ have negative and strong correlation, then this type of linear correlation is close to -1 means when value of ‘x’ is decreased, then for value of ‘x’, value of ‘y’ is also decreased and if linear correlation coefficient is exactly -1, then this type of linear correlation coefficient is called as a negative perfect fit correlation between two variables.
Property 4: If the values of two variables like ‘x’ and ‘y’ have no linear correlation, then this type of linear correlation is close to 0 means when value of ‘x’ does not depend upon value of ‘y’ and if linear correlation coefficient is exactly 0, then this type of linear correlation coefficient is called as a weak linear correlation between two variables.
Property 5: If linear correlation coefficient is +1 or -1, then in this situation all data points are present on a straight line and Slope of these data Point is positive, when linear correlation coefficient is exactly +1 and when linear correlation coefficient is exactly -1, then Slope of these data point is negative.
Property 6: Linear correlation coefficient is a dimensionless quantity, so this quantity is not present in unit terminology.
Property 7: When linear correlation coefficient is greater than 0.5, then this type of linear correlation coefficient is called as a strong correlation and when linear correlation coefficient is less than 0.5, then this type of linear correlation coefficient is called as a weak correlation. |
Question
If one end of a focal chord of the parabola, ${{\text{y}}^2} = 16{\text{x}}$ is at (1, 4), then length of this focal chord is?A. 25B. 24C. 20D. 22
Hint: Identify the variables in the given equation and compare it to the geometrical equation of a parabola. Upon identifying the known and required variables find the points at which both the ends of the focal chord lie, then using the formula for distance between two points determine the length of focal chord.
Given data –
Equation of parabola is ${{\text{y}}^2} = 16{\text{x}}$ and one end of the focal chord lies at (1, 4)
The geometrical equation of a parabola is in the form
${{\text{y}}^2} = 4{\text{ax}}$, whereas the ends of its focal chord are in the form $\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right){\text{ and }}\left( {{\text{at}}_1^2,2{\text{at}}_1^2} \right)$respectively.
And the relation between ${\text{t and }}{{\text{t}}_1}{\text{ is }}{{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}}$.
On Comparing the equation of parabola to given equation ${{\text{y}}^2} = 16{\text{x}}$, we get
a=4.
Comparing given point (1, 4) to$\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right)$, we get
2at = 4
⟹at=2
We know that a=2
$\Rightarrow {\text{t = }}\dfrac{1}{2}$
${{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}} \\ \Rightarrow {{\text{t}}_1} = - 2 \\$
Now the other end of the focal chord is $\left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right)$
$\Rightarrow \left( {4{{( - 2)}^2},2(4)( - 2)} \right) = \left( {16, - 16} \right)$
Now distance between the two points (1, 4) and (16, -16) is
${\text{D = }}\sqrt {{{({{\text{x}}_1} - {{\text{x}}_2})}^2} + {{({{\text{y}}_1} - {{\text{y}}_2})}^2}} \\ \Rightarrow {\text{ }}\sqrt {{{(1 - 16)}^2} + {{(4 + 16)}^2}} \\ \Rightarrow \sqrt {225 + 400} \\ \Rightarrow 25. \\$
Hence the length of the focal chord of the given parabola is 25, which makes Option A the correct answer.
Note –
In this type of question first find out and compare the equation of the given parabola. Then find you’re a, t, ${{\text{t}}_1}$ acchording to the given data in the question. Then find out the distance of the focal chord. Knowing the parabolic equation and respective properties of the focal chord is essential. |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Solutions for Session 3, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7
Problem C1 Answers will vary. One possible way of describing convex figures is that they are figures that are not dented, or that a rubber band stretched around the figure will touch the figure entirely.
Problem C2 Definitions (a), (d), (f), (g), and (h) are all equivalent descriptions of convex polygons. Statement (b) is true for all polygons, so it would apply to concave ones as well (triangle inequality!). Statement (c) is not true for all convex polygons. For example, try drawing a convex quadrilateral with one very long side; this side will probably be longer than the shortest diagonal of the quadrilateral. Statement (e) certainly does not describe convex polygons. (Consider an obtuse triangle which is convex but whose longest side is opposite to its largest angle.)
Problem C3 Answers will vary. You can find some more examples in Problem C4.
Problem C4 The answers may vary for the two figures on the right.
Problem C5 If an n-sided polygon is convex, we can pick a vertex and connect it to all other vertices, thereby creating n - 2 triangles. If the polygon is concave, we pick a vertex corresponding to an interior angle greater than 180° and connect that vertex to all the vertices so that all the resulting diagonals are in the interior of the polygon. Repeat the process if necessary. This has the effect of subdividing the original polygon into some number of convex polygons. We then divide each of the convex polygons into triangles. In the end, we will end up with n - 2 triangles.
Problem C6
Number of Sides of the Polygon
Number of Triangles Formed
Sum of the Angles in the Polygon
3 1 180° 4 2 360° 5 3 540° 6 4 720° 7 5 900° n n - 2 (n - 2) • 180°
Problem C7 Assuming that the interior of an n-sided polygon can be divided into n - 2 triangles, observe that all of the angles of the triangles actually make up the interior angles of the polygon. This is true since the vertices of all the triangles coincide with the vertices of the polygon. Therefore, since each triangle contributes 180° to the overall sum of the angles, the sum is (n - 2) • 180°. |
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# 1
## Business Mathematics MTH-367
MTH-367
Lecture Notes
Books
Text Book
Applied Mathematics for Business, Economics, and the Social
Sciences (4th Edition);
by Frank S. Budnick
Introductory Mathematical Analysis for Business, Economics,
and the Life and Social Sciences;
by Ernest F. Haeussler, Jr. Richard S. Paul, 8th Edition
2
## Chapter 2: Linear Equations
Definition: Linear equations are first degree equations. Each
variable in the equation is raised to the first power.
Definition:
A linear equation involving two variables x and y has the
standard form
ax+ by= c
where a, b and c are constants and a and b cannot both equal
zero.
Note: The presence of terms having power other than
1 or product of variables, e.g. (xy) would exclude an
equation from being linear. Name of the variables may
be different from x and y.
Examples:
1. 3 x+ 4 y=7 is linear equation, where a=3,b=4, c=7
2. x=5+ y is non-liner equation as power of x is not 1.
Solution set of an equation
Given a linear equation ax+ by= c, the solution set for the
equation (2.1) is the set of all ordered pairs (x, y) which satisfy
the equation.
S={(x , y)ax+ by=c }
## For any linear equation, S consists of an infinite number of
elements.
3
Method
1. Assume a value of one variable
2. Substitute this into the equation
3. Solve for the other variable
Example
Given 2 x + 4 y=16 , determine the pair of values which satisfy
the equation when x=-2
Solution: Put x=-2 in given equation gives us4y=16-4, i.e y=3.
So the pair (-2,3) is a pair of values satisfying the given
equation.
Linear equation with n variables
Definition
A linear equation involving n variables x1, x2, . . . , xn has the
general form
a1 x1 + a2 x2+ . . . + anxn = b
where a1 , a2 , , an are non-zero.
Definition:
The solution set S of a linear equation with n variables as
defined above is the collection of n-tuples (x 1 , x 2 , , x n )
such
that
## As in the case of two variables, there are infinitely many values
in the solution set.
4
Example
Given an equation
equation when
4 x 12 x 2+ 6 x3 =0
x 1=2
and
x 3=1
.
## Solution: Put the given values of
above equation gives
the above equation.
x 2=7
. Thus
x1
and
(2,7,1)
x3
in the
is a solution of
## Graphing two variable equations
A linear equation involving two variables graphs as a straight
line in two dimensions.
Method:
1. Set one variable equal to zero
2. Solve for the value of other variable
3. Set second variable equal to zero
4. Solve for the value of first variable
5. The ordered pairs (0, y) and (x, 0) lie on the line
6. Connect these points and extend the line in both
directions.
Example
Graph the linear equation 2 x + 4 y=16
Solution: In lectures
x-intercept
The x-intercept of an equation is the point where the graph of
the equation crosses the x-axis, i.e. y=0.
y-intercept
The y-intercept of an equation is the point where the graph of
the equation crosses the y-axis, i.e. x=0
5
## Note:Equations of the form x=k has no y-intercept and
equations of the form y=k has no x-intercept
Slope
Any straight line with the exception of vertical lines can be
characterized by its slope.Slope represents the inclination
of a line or equivalently it shows the rate at which the
lineraises and fall or how steep the line is.
Explanation:The slope of a line may be positive, negative,
zero or undefined.
The line with slope
1. Positivethenthe line rises from left to right
2. Negative then the linefalls from left to right
3. Zerothen the line is horizontal line
4. Undefined if the line is vertical line
Note: The sign of the slope represents whether the line falling
or raising. Itsmagnitude shows the steepness of the line.
Two point formula (slope)
Given any two point which lie on a (non-vertical) straight line,
the slope can be computed as the ratio of change in the value
of y to the change in the value of x.
Slope = =
y
x
## = change in the value of y
= change in the vale of x
Two point formula (mathematically)
6
## The slope m of a straight line connecting two points (x1, y 1)
and (x 2, y 2) is given by the formula
Slope=
x2 x1
y2 y1
Example
Compute the slope of the line segment connecting the
two points (2, 3) and (1, 9).
Solution: Here we have ( x 1 , y 1 )=(2,3) and ( x 2 , y 2 )=(1,9)
so using the above formula we get
93
=4
1(2 )
Slope=
## Note:Along any straight line the slope is constant.The line
connecting any two points will have the same slope.
Slope Intercept form
Consider the general form of two variable equation as
ax +by=c
We can write it as
y=
a
c
x+
b
a
## The above equation is called the slope-intercept form.
Generally, it is written as:
y=mx +k
m=
slope,
k =
y-intercept
Example
Rewrite the equation
7
x +2 y
=3 x y
4
## and find the slope and -intercept.
Solution: Rewriting the above we get
slope is
13
6
y=
13
x +0.
6
Thus
## Determining the equation of a straight line
Slope and Intercept
This is the easiest situation to find an equation of line,
if slope of a line is -5 and y-intercept is (0, 15) then we
have
m= -5, k = 15. We can write down
y=5 x+15
5 x+ y =15.
## Slope and one point
If we are given the slope of a line and some point that
lies on the line, we can substitute the know slope m
and coordinates of the given point into y=mx +k and
solve for k .
Given a non-vertical straight line with slope m and containing
the point (x1, y1), the slope of the line connecting (x1, y1) with
any other point (x, y) is given by
m=
y y 1
xx 1
8
Rearranging gives:
y y 1=m(x x1 )
Example
Find the equation of line having slope m = 2 and
passing through the point (2, 8).
Solution: Here ( x 1 , y 1 )=(2,8) and m=2 , so putting the
values in the above equation yields
y8=2(x2)
2 x + y =12.
## Parallel and perpendicular lines
Two lines are parallel if they have the same slope, i.e.
m1=m2
## Two lines are perpendicular if their slopes are equal to the
negative reciprocal of each other, i.e. m1 m2=1 .
Example
Find an equation of line through the point (2,4) and
parallel to the line 8 x4 y=20.
Solution: From the given equation we have y=2 x5 .
Let
m 1=2
of line is
m2=2
y(4 )=2 ( x2 )
2 x y=8
9
Lecture 2,3
## Chapter 3:Systems of Linear Equations
Definition
A System of Equationsis a set consisting of more than one
equation.
Dimension: One way to characterize a system of equations is by
its dimensions. If system of equations has m equations and n
variables, then the system is called an m by n system. In other
words, it has m n dimensions.In solving systems of equations,
we are interested in identifying values of the variables that satisfy
all equations in the system simultaneously.
Definition
The solution set for a system of linear equations may be a Null
set, a finite set or an infinite set.
Methods to find the solutions set
1. Graphical analysis method
2. Elimination method
3. Gaussian elimination method
Graphical Analysis (2 x 2 system)
We discuss three possible outcomes of the solution of 2 x 2
system.
1. Unique solution:
We draw the two lines and If two lines intersect at only one
10
point, say
(x 1 , y 1)
(x 1 , y 1)
## , represent the solution for the system of equations.
The system is said to have a unique solution.
2.No solution:
If two lines are parallel (recall that parallel lines have same slope
but different y-intercept) then such a system has no solution. The
equations in such a system are called inconsistent.
3. Infinitely many solutions:
If both equations graph on the same line, an infinite number of
points are common in two lines. Such a system is said to have
infinitely many solutions.
Graphical analysis by slope-Intercept relationships
Given a (2 2) system of linear equations (in slope-intercept
form)
y=m1 x + k 1
y=m 2 x+ k 2
where
m1
and
m2
k2
## denote the two y-intercepts then it has
m1 m2
I.
Unique solution if
II.
No Solution if
III.
m1=m2
but
k1 k2
m1=m2
and
k 1=k 2 .
k1
and
11
I.
II.
III.
## Good for two variable system of equations
Not good for non-integer values
Algebraic solution is preferred
## The Elimination Procedure (2 x 2 system)
Given a (2 2) system of equations
I. Eliminate one of the variable by multiplying or adding the
two equations
II. Solve the remaining equation in terms of remaining variable
III. Substitute back into one of the given equation to find the
value of the eliminated variable
Example
Solve the system of equations
2 x + 4 y=20
3 x+ y =10
## Solution: Page 90 in Book or in Lectures
Example
Solve the system of equations
3 x2 y=6
15 x +10 y=30
## Solution: Page 91 in Book/in Lectures
Gaussian Elimination Method
This method can be used to solve systems of any size.The
12
## method begins with the original system of equations.Using row
operations, it transforms the original system into an equivalent
system from which the solution may be obtained easily. Recall
that an equivalent system is one which has the same solution as
the original system.In contrast to the Elimination procedure, the
transformed system maintain m n dimension
Basic Row Operations
1. Both sides of an equation may be multiplied by a nonzero
constant.
2. Equations or non-zero multiples of equations may be added or
subtracted to another equation.
3. The order of equations may be interchanged.
Example
Determine the solution set for the given system of equations,
using Gaussian elimination method.
3 x2 y=7
2 x + 4 y=10
Solution: In Lectures
(m 2) systems, m >2
When there are more than two equations involving only two
variables then
1. We solve two equations first to get a point (x , y )
2. Put the values in the rest of the equations
3. If all equations are satisfied then system has unique solution
(x , y ).
## 4. If we get no solution at (1) then system has no solution.
5. If there are infinitely many solutions at step (1), then we select
two different equations and repeat (1).
13
Example
Determine the solution set for the given
equations,
(4 2)
system of
x+ 2 y =8
2 x 3 y=5
5 x +6 y=8
x+ y=7
## Gaussian elimination procedure for 3 x 3 system
The procedure of Gaussian elimination method for the 3 x 3
system is same as for 2 x 2 system.First we will form coefficient
transformation for 3 x 3 system, then convert it to transformed
system.
Example1
Determine the solution set for the following system of equations
5 x14 x 2 +6 x 3=24
3 x1 3 x 2 + x 3=54
2 x 1+ x25 x 3=30
14
5 4 6 24
3 3 1 54
2 1 5 30
## Applying the row operations
3 R 15 R2 ,2 R1 +5 R 3 , R2 + R3
, we get the
5 4 6
24
0 3 13 198
0 0
0
0
## As one whole row is equal to zero, hence the system of equations
has infinitely many solutions.
Example2
Determine the solution set for the following system of equations
x 1x 2+ x 3=5
3 x1 + x 2x 3=25
2 x 1 + x 2+ 3 x 3=20
1 1 1 5
3 1 1 25
2 1
3 20
## Applying the row operations
3 R 1R2 ,
1
1
1
R , 2 R 1R3 , R1 R2 ,3 R2R 3 , 4 R 2R3 , R2 , R 3
4 2
4
4
, we get the
15
1 0 0 5
0 1 0 10
0 0 1 0
## x 1=5, x 2=10, x 3=0.
Example3
Determine the solution set for the following system of equations
2 x 1+ x2 +3 x 3=10
10 x15 x 215 x 3=30
x 1+ x 23 x 3=25
2 1
3
10
10 5 15 30
1
1
3 25
## Applying the row operations 5
1
1
R1 + R2 , R1 +2 R3 , R 3 , R1R3 , R 1
3
2
, we get
1 0 2 5
0 0 0 60
0 1 1 20
## From row 2 we get 0=60, hence the above system of linear
equations has no solution.
Application
16
## Product mix problem: A variety of applications are concerned
with determining the quantitates of different products which satisfy
specific requirements
Example
A company produces three products. Each needs to be processed
through 3 different departments, with following data.
Department Products
1
A
B
C
2.5
2.5
3
2
3
4
3
3
2
2
Hours
available/wee
k
1200
1150
1400
## Determine whether there are any combination of three products
which would exhaust the weekly capacities of the three
departments?
Solution: Let
x1 , x2 , x3
## of product 1, 2 and 3. The conditions to be specified are
expressed by the following system of equations.
2 x 1 +3.5 x 2+3 x 3=1200
3 x1 +2.5 x 2+ 2 x 3=1150
4 x 1 +3 x 2+2 x 3=1400
(Department A)
(Department B)
(Department C)
17
we will get
x 1=200, x 2=100
and
x 3=150.
18
Lecture 04,05
## Chapter 4: Mathematical Functions
Definition: A function is a mathematical rule that assigns to
each input value one and only one output value.
Definition: The domain of a function is the set consisting of all
possible input values.
Definition: The range of a function is the set of all possible
output values.
Notation
The assigning of output values to corresponding input values is
often called as mapping. The notation
f ( x )= y
## represents the mapping of the set of input values of
set of output values y , using the mapping rule f .
into the
The equation
y=f ( x)
## denotes a functional relationship between the variables x
and y .Here x means the input variable and y means the
outputvariable, i.e. the value of y depends upon and uniquely
determined by the values of x .
The input variable is called the independent variable and the
output variable is called the dependent variable.
Some examples
1. The fare of taxi depends upon the distance and the day of
19
the week.
2. The fee structure depends upon the program and the type
of education (on campus/off campus) you are admitting in.
3. The house prices depend on the location of the house.
Note: The variable x is not always the independent variable,
y
is not always the dependent variable and f is not always
the rule relating x and y . Once the notation of function is clear
then, from the given notation, we can easily identify the input
variable, output variable and the rule relating them, for example
u=g ( v )
has input variable v ,output variable u and g is the
rule relating u and v .
Example (Weekly Salary Function)
A person gets a job as a salesperson and his salary depends
upon the number of units he sells each week.Then, dependency
of weekly salary on the units sold per week can be represented as
y=f ( x)
, where
## your employer has given you the following equation for
determining your weekly salary: y=100 x +5000
Given any value of x will result in the value of y with respect to
the function f . If x=5 , then y=5500. We write this as,
y=f (5 )=5500.
Example
Given the functional relationship
f (x)=5 x10 ,
Find
## f (0), f (2)f ( a+b).
20
Solution: As
f ( x )=5 x10
, so
f ( 0 )=5 ( 0 )10=10
f (2 )=5 (2 )10=20
f ( a+ b )=5 a+ 5 b10
## Domain and Range
Recall that the set of all possible input values is called the domain
of a function. Domain consists of all real values of the
independent variable for which the dependent variable is defined
and real.
Example
Determine the domain of the function
Solution:
f (x)
is undefined at
## function is not defined at
Domain(f) { x|x isrealx 2 }
f (x)=
4x =0
x= 2.
10
2
4x
Thus
## Restricted domain and range
Up to now we have solved mathematically to find the domains of
some types of functions.But for some real world problems, there
may be more restriction on the domain e.g. in the weekly salary
equation:
y=100 x +5000
Clearly, the number of units sold per week can not be negative.
21
Also, they can not be in fractions, so the domain in this case will
be all positive natural numbers {1,2,3, }. Further, the employer
can also put the condition on the maximum number of units sold
per week. In this case, the domain will be defined as:
D= {1,2, , u }
## where u is the maximum number of units sold.
Multivariate Functions
For many mathematical functions, the value of the dependent
variable depends upon more than one independent variable.
Definition: A functions which contain more than one independent
variable are called multivariate function.
Definition:A function having two independent variables is called
bivariate function.
They are denoted by z=f ( x , y ) , where x and y are the
independent variables and z is the dependent variable e.g.
z=2 x+5 y .
## In general the notation for a function f where the value of
dependent variable depends on the values of n independent
variables is
z=f ( x 1 , , x n ) .
For example,
z=2 x1 +5 x 2+4 x 34 x 4 + x 5
Types of Functions
Constant Functions
22
## A constant function has the general form
y=f ( x ) =a0
Here, domain is the set of all real numbers and range is the single
value a0 , e.g. f ( x )=20.
Linear Functions
A linear function has the general (slope-intercept) form
y=f ( x ) =a1 x + a0
where
a1
is slope and
a0
is
y=2 x+3
and y-intercept
## The weekly salary function is also an example of linear function.
A quadratic function has the general form
y=f ( x ) =a2 x2 +a 1 x +a0
provided that
a2 0
, e.g.
y=2 x 2+3.
Cubic Function
A cubic function has the general form
y=f ( x ) =a3 x3 + a2 x 2 +a1 x+ a0
23
provided that
a3 0,
e.g.
## y=f ( x ) =x3 50 x 2 +10 x1.
Polynomial Functions
n
## has the general form
y=f ( x ) =an x ++ a1 x +a 0
## Where a1 , , a n and a0 are real constants such that an 0 .
All the previous types of functions are polynomial functions.
Rational Functions
A rational function has the general form
y=f ( x ) =
g(x)
h (x )
g( x)
Where
y=f ( x ) =
and
h(x )
## are both polynomial functions, e.g.
2x
.
5 x 2 x+3
3
Composite functions
A composite function exists when one function can be viewed as
a function of the values of another function. If y=g(u) and
u=h( x)
## then composite function
y=f ( x ) =g ( h ( x ) ) .
Here
## must be in the domain of
domain of
then
. For example, if
g ( h (2 ) ) =132.
and
2
h( x )
y=g ( u )=3u + 4 u
must be in the
and
u=h ( x )=x +8
24
## Graphical Representation of Functions
The function of one or two variables (independent) can be
represented graphically.The functions of one independent variable
are graphed in two dimensions, 2-space.The functions in two
independent variables are graphed in three dimension, 3-space.
Method of graphing
1) To graph a mathematical function, one can simply assign
different values from the domain of the independent variable
and compute the values of dependent variable.
2) Locate the resulting order pairs on the co-ordinate axes, the
vertical axis ( y -axis) is used to denote the dependent
variable and the horizontal axis ( x -axis) is used to denote the
independent variable.
3) Connect all the points approximately.
Vertical Line Test
By definition of a function, to each element in the domain there
should correspond only one element in the range. This allows a
simple graphical check to determine whether a graph represents
a function or not.If a vertical line is drawn through any value in the
domain, it will intersect the graph of the function at one point
only.If the vertical line intersects at more than one point then, the
graph depicts a relation and not a function.
25
Lecture 05,06
## Chapter 5:Linear functions, Applications
Recall that a linear function f involving one independent
variable x and a dependent variable y has the general form
y=f ( x ) =a1 x + a0 ,
where
a1 0
and
a0
are constants.
Example
Consider the weekly salary function
y=f ( x)=300 x +2500
26
## where y is defined as the weekly salary and x is the number
of units sold per week.
Clearly, this is a weekly function in one independent variable x .
1. 2500 represents the base salary, i.e. when no units are sold per
week and 300 is the commission of each unit sold.
2. The change in weekly salary is directly proportional to the
change in the no. of units sold.
3. Slope of 3 indicates the increase in weekly salary associated
with each additional unit sold.
In general, a linear function having the form y=f ( x ) =a1 x + a0
a
change in the value of
the variable x .
## Linear function in two independent variables
f
A linear function
x2
x1
and
## y=f ( x1 , x2 ) =a1 x 1+ a2 x 2 +a0
a1
where
and a2
are non-zero constants and a0 is any
constant.
1. This equation tells us that the variable y depends jointly on
the values of
2. The value of
values of
x1
x1
y
and
x2.
## is directly proportional to the changes in the
and
x2.
Example
Assume that a salesperson salary depends on the number of
units sold of each of two products, i.e. the salary function is given
27
as
y=300 x 1+ 500 x 2 +2500
where
= weekly salary,
x1
## and x 2 is number of units sold of product 2. This salary function
gives a base salary of 2500, commission of 300 on each unit sold
of product 1 and 500 on each unit sold of product 2.
Linear cost function
The organizations are concerned with the costs as they reflect the
money flowing out of the organisation. The total cost usually
consists of two components: total variable cost and total fixed
cost. These two components determine the total cost of the
organisation.
Example
A firm which produces a single product is interested in
determining the functions that expresses annual total cost
as
## a function of the number of units produced x .Accountants
indicate that the fixed expenditure each year are 50,000. They
also have estimated that raw material costs for each unit
produced are 5.50,labour costs per unit are 1.50 in the assembly
department, 0.75 in the finishing room, and 1.25 in the packaging
and shipping department. Find the total cost function.
Solution
Total cost function =total variable cost + total fixed cost
Total fixed cost = 50,000
Total variable cost = total raw material cost +total labour cost
28
## y=f ( x ) =5.5 x + ( 1.5 x+ 0.75 x +1.25 x )+ 50,000=9 x +50,000.
The 9 represents the combined variable cost per unit of \$9. That
is, for each additional unit produced, total cost will increase by \$9.
Linear Revenue function
Revenue: The money which flows out into an organisation from
either selling or providing services is often referred to as revenue.
Total Revenue = Price
Suppose a firm sells product. Let
pi
Quantity sold
and
xi
## product and number of units per product respectively. Thenthe
revenue
R= p1 x 1++ pn x n .
## Linear Profit function
Profit:The profit of an organisation is the difference between total
revenue and total cost. In equation form if total revenue is
denoted by
R( x)
and
Total
## produced and sold, then profit
cost is
P( x)
C(x),
where
is quantity
is defined as
P( x)=R ( x)C( x ).
## 1. If total revenue exceeds total cost the profit is positive
2. In such case, profit is referred as net gain or net profit
3. On the other hand the negative profit is referred to as a net loss
or net deficit.
29
Example
A firm sells single product for \$65 per unit.Variable costs per unit
are \$20 for materials and \$27.5 for labour.Annual fixed costs are
\$100,000. Construct the profit function stated in terms of x ,
which is the number of units produced and sold. How much profit
is earned if annual sales are 20,000 units.
Solution:Here
R ( x ) =65 x
## material costs, labour costs, and fixed cost.
C ( x )=20 x+ 27.5 x +100,000=47.5 x+ 100,000.
Thus
P ( x )=R ( x )C ( x ) =17.5 x100,000.
As
x=20,000,
so
P (20,000 )=250,000
## Straight Line Depreciation
When organizations purchase an item, usually cost is allocated
for the item over the period the item is used.
Example
A company purchases a vehicle costing \$20,000 having a useful
life of 5 years, then accountants might allocate \$4,000 per year as
a cost of owning the vehicle.The cost allocated to any given
period is called depreciation.The value of the truck at the limp of
purchase is \$20,000 but, after 1 year the price will be\$20000\$4000 = \$ 16000and so forth.In this case, depreciation can also
be thought of as an amount by which the book value of an asset
has decreased over the period of time.
Thus, the book value declines as a linear function over time. If
30
## equals the book value of an asset and t equals time (in
years) measured from the purchase date for the previously
mentioned truck, then V =f ( t ) .
Linear demand function
A demand function is a mathematical relationship expressing the
way in which the quantity demanded of an item varies with the
price charged for it.The relationship between these two variables,
quantity demanded and price per unit, is usually inversely
proportional, i.e. a decrease in price results in increase in
demand.
Most demand functions are nonlinear, but there are situations in
which the demand relationship either is, or can be approximated
by a linear function.
Quantity demanded=q d =f ( price per unit)
## Linear Supply Function
A supply function relates market price to the quantities that
suppliers are willing to produce or sell. The supply function
implicates that what is brought to the market depends upon the
price people are willing to pay.
In contrast to the demand function, the quantity which suppliers
are willing to supply usually varies directly with the market
price.The higher the market price, the more a supplier would like
to produce and sell.The lower the price, the less a supplier would
like to produce and sell.
Quantity supplied=q s=f (market price)
Break-Even Models
31
## Break-even model is a set of planning tools which can be useful in
managing organizations. One significant indication of the
performance of a company is reflected by how much profit is
earned. Break-even analysis focuses upon the profitability of a
firm and identifies the level of operation or level of output that
would result in a zero profit.
The level of operations or output is called the break-even point.
The break-even point represents the level of operation at which
total revenue equals total cost. Any changes from the level of
operations will result in either a profit or a loss.
Break-even analysis is mostly used when:
1. Firms are offering new products or services.
2. Evaluating the pros and cons of starting a new business.
Assumption
Total cost function and total revenue function are linear.
Break-even Analysis
In break-even analysis the main goal is to determine the breakeven point.
The break-even point may be expressed in terms of
i. Volume of output (or level of activity)
ii. Total sale in dollars
iii. Percentage of production capacity
e.g. a firm will break-even at 1000 units of output, when total
sales equal 2 million dollars or when the firm is operating
60% of its plant capacity.
Method of performing break-even analysis
1. Formulate total cost as a function of x , the level of output.
2. Formulate total revenue as a function of x .
3. As break-even conditions exist when total revenue equals
32
## total cost, so we set C ( x ) equals R ( x ) and solve for x .
The resulting value of x is the break-even level of output
and denoted by
x BE
## An alternate, to step 3 is to construct the profit function
P( x)=R ( x)C( x )
, set P( x) equal to zero and solve to find
x BE .
Example
A Group of engineers is interested in forming a company to
produce smoke detectors.They have developed a design and
estimated that variable costs per unit, including materials, labor,
and marketing costs are \$22.50.Fixed costs associated with the
formation, operation, management of the company and purchase
of the machinery costs \$250,000.They estimated that the selling
price will be 30 dollars per detector.
a) Determine the number of smoke detectors which must be
sold in order for the firm to break-even on the venture.
b) Preliminary marketing data indicate that the firm can expect
to sell approximately 30,000 smoke detectors over the life of
the project, if the detectors are sold at \$30 per unit.
Determine expected profits at this level of output.
Solution:
a) If
## sold, the total revenue function
function is
x BE=33333.3
b)
C ( x )=22.5 x+ 250,000.
R ( x ) =30 x .
We put
R ( x ) =C ( x )
to get
units.
## P ( x )=R ( x )C ( x ) =7.5 x250,000.
P (30,000 )=25,000.
Now at
x=30,000,
we have
## Thus theexpected loss is \$25,000.
33
Market Equilibrium
Given supply and demand functions of a product, market
equilibrium exits if there is a price at which the quantity demanded
equals the quantity supplied.
Example
Suppose demand and supply functions have been estimated for
two competing products.
q d =1002 p1 +3 p2
(Demand, Product 1)
q d =150+ 4 p1 p2
(Demand, Product 2)
q s =2 p 14
(Supply, Product 1)
q s =3 p26
(Supply, Product 2)
## Determine the price for which the market equilibrium would
exist.
Solution:The demand and supply functions are linear. The
quantity demanded of a given product depends on the price of the
product and also on the price of the competing product and the
quantity supplied of a product depends only on the price of that
product.
Market equilibrium would exist in this two-product market place if
prices existed and were offered such that
solving we get
p1=221
equations we get
and
q d =q s =438
1
p2=260.
and
q d =q s
1
and
q d =q s
2
q d =q s =774.
2
34
Lecture 07
## Chapter 6: Quadratic and Polynomial Functions
We focused on linear and non-linear mathematics and linear
mathematics is very useful and convenient.There are many
phenomena which do not behave in a linear manner and can not
be approximated by using linear functions.We need to introduce
nonlinear functions.One of the more common nonlinear function is
Definition: A quadratic function involving the independent
variable
## has the general form
y=f ( x ) =a x2 +bx +c ,
where
a,b,c
## are constants and
a 0.
Graphical Representation
All quadratic functions have graphs as curves called parabolas.
2
Consider the function y=x then we have
35
## The graph of the function is given in the figure given below.
Properties of Parabolas
a) A parabola which opens upward is said to be concave up.
The above parabola is concave up.
b) A parabola which opens downward is said to be concave
down.
c) The point at which a parabola either concaves up or down is
called the vertex of the parabola.
Note: A quadratic function of the form
2
y=a x +bx +c
b 4 acb
,
2a
4a
).
36
## As we have discussed earlier, here are the results about
concavity.
1. If a>0 ; the function will graph as parabola which is concave
up.
2. If a<0 ; the function will graph as parabola which is concave
down.
Sketching of Parabola
Parabolas can be sketched by using the method of chapter 4.
But, there are certain things which can make the sketching
relative easy. These include,
1. Concavity of the parabola
2. Y-intercept, where graph meets y-axis.
3. X-intercept, where graph meets x-axis
4. Vertex
How to find intercepts
1) Algebraically, y -intercept is obtained when the value of
is equal to zero in the given function.
2) Algebraically, x -intercept is obtained when the value of
is set equal to zero
x
y
## Methods to find the x-intercept
The x -intercept of a quadratic equation is determined by finding
the roots of an equation.
1) Finding roots by factorisation:
If a quadratic can be factored, it is an easy way to find the roots,
2
e.g. x 3 x+2=0 can be written as ( x1 ) ( x 2 )=0, which gives
x=1,2
as its roots..
37
## 2) Finding roots by using the quadratic formula
The quadratic formula will always identify the real roots of an
equation if any exist.
The quadratic formula of an equation which has the general form
a x 2+ bx+ c
b b24 ac
x=
.
2a
will be
## Taking the same example we get
x=
(3 ) (3 ) +4 ( 1 ) ( 2 )
2
x=
3 1
2
So we get
x=1,2
## Quadratic functions; Applications
Suppose that the demand function for the product is q=f ( p) .
Here q is the quantity demanded and p is the price in dollars.
The total revenue R from selling q units of price p is stated
as the product of p and q or R= pq .
Since the demand function q is stated in terms of price
totalrevenue can be stated as a function of price,
R= p . f ( p )= p . q .
If we let
q=150050 p ,
then
R=1500 p50 p2 .
38
## Quadratic supply function
Example
Market surveys of suppliers of a particular product have resulted
in the conclusion that the supply function is approximately
quadratic in form. Suppliers were asked what quantities they
would be willing to supply at different market prices. The results of
the survey indicated that at market prices of \$25, \$30 and \$ 40,
the quantities which suppliers would be willing to offer to the
market were 112.5, 250 and 600 (thousands) units, respectively.
Determine the equation of the quadratic supply function?
Solution:We determine the solution by substituting the three
price-quantity combinations into the general equation
2
q s=a p +bp+ c .
## The resulting system of equations is
625 a+25 b+ c=112.5
900 a+30 b+ c=250
1600 a+40 b +c=600
c=200
a=0.5, b=0,
and
## Quadratic demand function
39
Example
A consumer survey was conducted to determine the demand
function for the same product as in the previous example
discussed for supply function. The researchers asked consumers
if they would purchase the product at various prices and from their
responses constructed estimates of market demand at various
market prices. After sample data points were plotted, it was
concluded that the demand relationship was estimated best by a
quadratic function. The researchers concluded that the quadratic
representation was valid for prices between \$5 and \$45.Three
data points chosen for fitting the curve were (5, 2025), (10, 1600)
and (20, 900). Just like last example, substituting these data
points into the general equation for a quadratic function and
solving the resulting system simultaneously gives the demand
function
q d= p2 100 p+2500 .
Here
qd
## equals the selling price in dollars.
Polynomial functions
A polynomial function of degree
variable
## has the general form
y=an x n ++a1 x+ a0 ,
## where an 0, and a1 , , a n are constants.
The degree of a polynomial is the exponent of the highest
powered term in the expression.
Rational Functions
40
## Rational functions are the ratios of two polynomial functions with
the general form as:
n
g ( x ) a n x ++ a1 x + a0
f ( x )=
=
.
h ( x ) bm x m ++b1 x+ b0
## Here g is n-th degree polynomial function and
degree polynomial function.
is m-th
Lecture 08
## Chapter 7:Exponential and Logarithmic Functions
Properties of exponents and radicals
If a , b are positive numbers and m , are real numbers then:
m
m +n
1.
b . b =b
2.
bm
=b mn , b 0
n
b
3.
b =b
mn
m. n
41
m
4.
a . b =( ab )
5.
b n = bm=( n b )
6.
b0 =1
7.
bm =
1
m
b
Exponential Function
x
## Definition:A function of the form b , where b>0 ,b 1 and x is
any real number is called an exponential function to the base
x
x
. e.g. f ( x )=10 .
Characteristics of function
f ( x)=b x , b>1
## 1. Each function f is defined for all values of x . The
domainof f is the set of real numbers.
2. The graph of f is entirely above the x-axis. The range of is
the set of positive real numbers.
3. The y -intercept occurs at (0,1) . There is no x -intercept.
4. The value of y approaches but never reaches zero as x
approaches negative infinity.
5. Function y is an increasing function of x , i.e. for
x 1 < x 2 , f ( x 1) < f ( x 2 ) .
## 6. The larger the magnitude of the base b , the greater the
rate of increase in y as x increases in value.
Characteristics of function
f ( x )=b x , b<1
42
## 1. Each function f is defined for all values of x . The
domainof f is the set of real numbers.
2. The graph of f is entirely above the x-axis. The range of is
the set of positive real numbers.
3. The y -intercept occurs at (0,1) . There is no x -intercept.
4. The value of y approaches but never reaches zero as x
approaches positive infinity.
5. Function y is adecreasing function of x , i.e. for
x 1 < x 2 , f ( x 1) > f ( x 2 ) .
## 6. The smaller the magnitude of the base b , the greater the
rate of decrease in y as x increases in value.
Base
exponential functions
where
mx
y=a e ,
## 1. Base e exponential functions also called as natural
exponential functions and are particularly appropriate in
modelling continuous growth and decay process, continuous
compounding of interest.
2. Base e exponential functions are more widely applied than
any other class of functions.
The two special exponential functions are
of
## is given by the following figure.
and
e .
The graph
43
Conversion to base e
There are instances where base e exponential functions are
preferred to those having another base. Exponential functions
having a base other than e can be transformed into equivalent
base
1.1
e =3.
## functions. For example, if we take
Thus
e =3 ,
then we have
e 1.1x =3 x =f ( x ) .
## Any positive number x can be expressed equivalently as some
power of the base e , or we can find an exponent n such that
n
e =x .
## Applications of exponential functions
1. When a growth process is characterized by a constant per
cent increase in value, it is referred as an exponential growth
process.
2. Decay processes: When a growth process is characterized
by a constant per cent decrease in value, it is referred as an
exponential decay process.
Exponential functions have particular application to growth
processes and decay processes.Growth processesinclude
population growth, appreciation in the value of assets, inflation,
growth in the rate at which some resources (like energy) are
used.Decay processes include declining value of certain assets
such as machinery, decline in the efficiency of machine, decline in
44
## the rate of incidence of certain diseases with the improvement of
medical research and technology. Both process are usually stated
in terms of time.
Example(population growth process)
This function characterized by a constant percentage of increase
in the value over time. Such process may be describe by the
kt
general function; V =f ( t )=V 0 e , where
V =
V0
t=0
## = time measured in the appropriate units (hours, days, weeks,
years, etc).
The population of a country was 100 million in 1990 and is
growing at the constant rate of 4 per cent per year. The size of
populations is
0.04 t
P=f ( t )=100 e
## The projected population for 2015, will be
P=f ( 25 )=100 e 0.04(25)=271.83
Million.
## Example(Decay functions:price of a machinery)
The general form of the exponential decay function is
kt
V =f ( t )=V 0 e
45
## The resale value V of certain type of industrial equipment has
been found to behave according to the function
V =f ( t )=100,000 e0.1 t
then
a) Find original value of the price of equipment.
b) its value after 5 years.
Solution: a)We need to find f at t=0, so
f ( 0 )=100,000 e0 =100,000.
## Thus the original price is \$100,000.
b) After 5 years the value will be
0.1 ( 5 )
f ( 5 ) =100,000 e
=\$ 60,650.
46
Lecture 09,10,11
## Chapter 8:Mathematics of Finance
Definition: The Interest is a fee which is paid for having the use
of money.
We pay interest on loans for having the use of banks money.
Similarly, the bank pay us interest on money invested in savings
accounts because the bank has temporary access to our money.
1. Interest is usually paid in proportion to the principal amount
and the period of time over which the money is used.
2. The interest rate specifies the rate at which the interest
accumulates.
3. It is typically stated as a percentage of the principal amount
per period of time, e.g. 18 % per year, 12 % quarterly or 13.5
% per month.
Definition:The amount of money that is lent or invested is called
the principal.
Simple Interest
Definition: Interest that is paid solely on the amount of the
principal is called simple interest.
Simple interest is usually associated with loans or investments
which are short-term in nature, computed as
I =Pin
47
I=
simple interest
P=
principal
i=
n=
## While calculating the interest I , it should be noted that both
and n are consistent with each other, i.e. expressed in same
duration.
Example
A credit union has issued a 3 year loan of \$ 5000. Simple interest
is charged at the rate of 10% per year. The principal plus interest
is to be repaid at the end of the third year. Compute the interest
for the 3 year period. What amount will be repaid at the end of the
third year?
Solution:Here P=5,000, i=0.1, n=3. Then by using the above
formula, we get the simple interest
I =( 5,000 ) ( 0.1 )( 3 ) =\$ 1,500.
Compound Interest
In compound interest, the interest earned each period is
reinvested, i.e. added to the principal for purposes of computing
interest for the next period. The amount of interest computed
using this procedure is called the compound interest.
Example
Assume that we have deposited \$ 8000 in a credit union which
pays interest of 8% per year compounded quarterly.
The amount of interest at the end of 1 quarter would be:
48
## I 1 =8000 ( 0.08 )( 0.25 )=\$ 160.
Here n=0.25 year, with interest left in the account, the principal on
which interest is earned in the second quarter is the original
principal plus \$160 earned during the first quarter
P2=P1 + I 1 =8000+160=\$ 8160.
## The interest earned during second quarter is
I 2 =8160 ( .25 )( .08 ) =163.2.
Continuing this way, after the year the total interest earned would
be
\$ 659.46
be
\$ 8659.49.
## The simple interest after 1 year would be:
I =800(.08)(1)=\$ 640
## The difference between simple and compound interest is
659.49640=\$ 19.46 .
Continuous compounding
The continuous compounding can be thought of as occurring
infinite number of times.It can be computed by the following
formula.
Continuous compounding
S=P e it ,
i=
Rate of interest,
t=
where
Time period,
P=
Principal.
Example
Compute the growth in a \$ 10,000 investment which earns
interest at 10 per cent per year over the period of 10 years.
49
Solution:Here
S=( 10,000 ) e
0.1 ( 10 )
i=0.1, t=10,
and
P=10,000.
Hence
=\$ 27,183.
## Note:The value of an investment increases with increased
frequency of compounding. If we compute the interest by using
different frequency, we get the following.
1. Simple interest in 10 years will raise the amount to \$ 20,000.
2. Annual compounding in 10 years will raise the amount to
\$ 25,937.
## 3. Semi-annual compounding in 10 years will raise the amount
to \$ 26,533.
4. Quarterly compounding in 10 years will raise the amount to
\$ 26,830.
Single-Payment computations
Suppose we have invested a sum of money and we wish to know
that: what will be the value of money at sometime in the future?
In knowing this, we assume that any interest is computed on
compounding basis.Recall that the amount of money invested is
called the principal and the interest earned is called (after some
period of time) the compound amount.
Given any principal invested at the beginning of a time period, the
compound amount at the end of the period is calculated as:
S=P+iP
P=
i=
principal,
interest rate in per cent,
, where
50
S=
compound amount.
## A period may be any unit of time, e.g. annually, quarterly.
Compound amount formula
To get a general formula, we let
P
= principal ,
## = number of compounding periods (number of periods in
which the principal has earned interest)
S=P ( 1+i ) .
## Compound amount after 2 periods
P ( 1+i ) +i[P (1+ i)]=P ( 1+i )2
Thus after
## number of compounding periods
Definition: The expression
factor.
( 1+i )n
P (1+i )3
S=P ( 1+i )n .
n,
so
S=f ( n ) .
## is called a compound amount
Example
Suppose that \$ 1000 is invested in a saving bank which earns
interest at the rate of 8 % per year compounded annually. If all
interest is left in the account, what will be the account balance
after 10 years?
51
Solution:Using
S=P ( 1+i ) ,
we get
## Present value computation
The compound amount formula
S=P ( 1+i )n
## is an equation involving four variables, .i.e. S , P , i, n .
Thus knowing the values of any of the 3 variables, we can easily
find the 4th one.
Rewriting the formula as:
P
P=
S
( 1+i )n
## denotes the present value of the compound amount.
Example
A person can invest money in saving account at the rate of 10 %
per year compounded quarterly.The person wishes to deposit a
lump sum at the beginning of the year and have that sum grow to
\$ 20,000 over the next 10 years.How much money should be
deposited? How much amount of money should be invested at
the rate of 10 % per year compounded quarterly, if the compound
amount is \$ 20000 after 10 years?
Solution:As
P=
S
( 1+i )n
P=
20000
=\$ 7448.62.
( 1+0.025 )40
Definition:The factor
1
( 1+i )n
52
## Other applications of compound amount formula
1. The following examples will illustrate the other applications
of the compound amount formula, when i and n are
unknown.
2. When a sum of money is invested, there may be a desire to
know that how long it will take for the principal to grow by
certain percentage.
Example
A person wishes to invest \$ 10000 and wants the investment to
grow to \$ 20000 over the next 10 years. At what annual interest
rate the required amount is obtained assuming annual
compounding?
Solution:Page number 315 in book or in Lectures
Effective Interest Rates
The stated annual interest rate is usually called nominal rate.We
know that when interest is compounded more frequently then
interest earned is greater than earned when compounded
annually.When compounding is done more frequently than
annually, then effective annual interest rates can be
determined.Two rates would be considered equivalent if both
results in the same compound amount.
Let r equals the effective annual interest rate, i is the nominal
annual interest rate and m is the number of compounding
periods per year. The equivalence between the two rates
suggests that if a principal P is invested for n years, the two
compound amounts would be the same, or
n
i
P (1+r ) =P 1+
m
nm
( )
Solving for
i m
1.
m
( )
r= 1+
r,
we get
53
## Annuities and their Future value
An annuity is a series of periodic payments, e.g. monthly car
payment, regular deposits to savings accounts, insurance
payments.
We assume that an annuity involves a series of equal
payments.All payments are made at the end of a compounding
period, e.g.
a series of payments R , each of which equals \$ 1000 at the end
of each period, earn full interest in the next period and does not
qualify for the interest in the previous period.
Example
A person plans to deposit \$ 1000 in a tax-exempt savings plan at
the end of this year and an equal sum at the end of each year
following year. If interest is expected to be earned 6 % per year
compounded annually, to what sum will the investment grow at
the time of the 4th deposit?
Solution: We can determine the value of S 4 by applying the
compound amount formula to each deposit, determining its value
at the time of the 4 -th deposit. These compound amounts may
be summed for the four deposits to determine
S4 .
## First deposit earns interest for 3 years, 4th deposit earns no
interest. The interest earned on first 3 deposits is \$ 374.62. Thus
S 4 =4374.62 .
Formula
The procedure used in the above example is not practical when
dealing with large number of payments. In general to compute the
sum S n , we use the following formula.
54
R [ ( 1+ i )n 1 ]
S n=
=R S n i
i
## The special symbol
angle
Sni
, which is pronounced as
sub
amount factor
[ (1+i )n1 ]
i
## Now we solve the last example by this formula. We have
S 4 =1000 S 4 (.06)=4374.62,
## by using the table III at the page T-10 in book.
Example
A boy plans to deposit \$ 50 in a savings account for the next 6
years. Interest is earned at the rate of 8% per year compounded
quarterly. What should her account balance be 6 years from now?
How much interest will he earn?
Solution:Here
R=50, i=
.08
=0.02, n=6 4=24.
4
We need to find
## Determining the size of an annuity
The formula S n=R Sn i has four variables. So like compound
amount formula,if any three of them are know, we can find the
fourth one.For example if the rate of interest is known, what
amount should be deposited each period in order to reach some
other specific amount? We solve for R ,
As
S n=R Sn i
, so we have
55
R=
Sn
1
=S n
Sn i
Sni
[ ]
## The expression in brackets is the reciprocal of the series
compound amount factor. This factor is called sinking fund factor.
Because the series of deposits used to accumulate some future
sum of money is often called a sinking fund. The values for the
sinking fund factor
[ ]
1
Sn i
## T17 in the book.
Example
A corporation wants to establish a sinking fund beginning at the
end of this year. Annual deposits will be made at the end of this
year and for the following 9 years. If deposits earn interest at the
rate of 8 % per year compounded annually, how much money
must be deposited each year in order to have \$ 12 million at the
time of the 10th deposit.? How much interest will be earned?
Solution:Here
Thus
R=S10
## S 10=\$ 12 million ,i=0.08,n=10.
1
=12,000,000 0.06903=\$ 828360.
S 10(0.08)
## 10 deposits of \$828360 will be made during this period, total
deposits will equal to\$8283600.
Interest earned = 12 million 8283600 = \$3716400
Example
Assume in the last example that the corporation is going to make
quarterly deposits and that the interest is earned at the rate of 8
% per year compounded quarterly. How much money should be
deposited each quarter? How much less will the company have to
56
## deposit over the 10 year period as compared with annual deposits
and annual compounding?
Solution:Here
0.08
=0.02, n=40.
4
Thus
R=S 40
1
S40 (0.02)
## Since there will be 40 deposits of \$198720, total deposits over the
10 year period will equal \$7948800.Comparing with annual
deposits and annual compounding in the last example, total
deposit required to accumulate the \$12million, will be
82836007948800=\$ 334800
less under quarterly compounding.
Annuities and their Present value
There are applications which relate an annuity to its present value
equivalent. e.g. we may be interested in knowing the size of a
deposit which will generate a series of payments (an annuity) for
college, retirement years, or given that a loan has been made, we
may be interested in knowing the series of payments (annuity)
necessary to repay the loan with interest.
The present value of an annuity is an amount of money today
which is equivalent to a series of equal payments in the future. An
assumption is that: the final withdrawal would deplete the
investment completely.
Example
A person recently won a state lottery. The terms of the lottery are
that the winner will receive annual payments of \$ 20,000 at the
end of this year and each of the following 3 years. If the winner
could invest money today at the rate of 8 % per year compounded
annually, what is the present value of the four payments?
57
## Solution: If A defines the present value of the annuity, we might
determine the value of A by computing the present value of each
20000 payment. Here
For
For
For
For
S=20000, i=0.08
then using
P=
S
( 1+i )n
A=\$ 66242.6 .
## As with the future value of an annuity, we can find the general
formula for the present value of an annuity. In case of large
number of payments the method of example is not practical.
Formula
If
R=
Amount of an annuity
i=
n=
A=
then
( 1+i )n1
A=R
=R a ni
n
i ( i+ 1 )
58
## The above equationis used to compute the present value A
of an annuity consisting of n equal payments, each made at
the end of n periods.
Definition:The expression
( 1+i )n1
n
i ( i+ 1 )
] is
## present worth factor. Its value can be found in table V in book.
Example
Parents of a teenager girl want to deposit a sum of money which
will earn interest at the rate of 9 % per year compounded semiannually. The deposit will be used to generate a series of 8 semi
annual payments of \$2500 beginning 6 months after the deposit.
These payments will be used to help finance their daughters
college education.What amount must be deposited to achieve
their goal? How much interest will be earned?
Solution:Here
R=2500, i=
0.09
=0.045, n=8.
2
Thus we find
## Since the \$16489.73 will generate eight payments totalling
\$20000, interest earned will be 20000-16489.73=\$3510.27.
Determining the size of an annuity
Given the value of A , we can find R = size of the
corresponding annuity.
The expression
[ ]
1
a ni
R=
[ ]
A
1
=A
a ni
a ni
Example
59
## For example given a loan of \$ 10000 which is received today,
what quarterly payments must be made to repay the loan in 5
years if interest is charged at the rate of 10 % per year,
compounded quarterly?How much interest will be paid on the
loan?
Solution:Here
R=10,000, i=
0.1
=0.025, n=20.
4
R=10000
1
a20 (0.025)
## There will be 20 payments totalling \$12830, thus interest will be
equal to \$2830 on the loan.
60
Lecture 12,13
## Chapter 10Linear Programming:An Introduction
Basic concepts
Linear Programming (LP) is a mathematical optimization
technique.By optimization we mean a method or procedure
which attempts to maximize or minimize some objective, e.g.,
maximize profit or minimize cost. In any LP problem certain
decisions need to be made, these are represented by decision
variables x .The basic structure of a LP problem is either to
maximize or minimize an objective functionwhile satisfying a set
of constraints.
1. Objective function is the mathematical representation of
overall goal stated as a function of decision variables x ,
examples are profit levels, total revenue, total cost, pollution
levels, market share etc.
2. The Constraints, also stated in terms of , are conditions
that must be satisfied when determining levels for the
decision variables. They can be represented by equations or
by inequalities ( and/or types).
The term Linear is due to the fact that all functions and constraints
in the problem are linear.
Example
A simple linear programming problem is
61
Maximize
z=4 x1 +2 x 2
Subject to
2 x 1 +2 x2 24
4 x 1 +3 x 2 30
## The objective is to maximize
function of two decision variables
z,
## which is stated as a linear
x1
and
x2
. In choosing values
## for x 1 and x 2 two constraints must be satisfied. The constraints
are represented by the two linear inequalities.
A Scenario (a LP example)
A firm manufactures two products, each of which must be
processed through two departments 1 and 2.
Product A
Product B
Weekly
Labor
Capacity
Departmen 3 h per
t1
unit
2 h per
unit
120 h
Departmen 4 h per
t2
unit
6 h per
unit
260 h
Profit
Margins
\$6 per unit
\$5 per unit
## If x and y are the number of units produced and sold,
respectively of product A and B, then the total profit z is
62
z=5 x+ 6 y
inequalities:
3 x+2 y 120
(Department 1)
4 x +6 y 260
(Department 1)
## We also know that x and y can not be negative.
So the LP model which represents the stated problem is:
Maximize
Subject to
z=5 x+ 6 y
3 x+2 y 120
(1)
4 y+ 6 y 260
(2)
x0
(3)
y0
(4)
Inequalities (1) and (2) are called structural constraints, while
(3) and (4) are non-negativity constraints.
Notably, the function
maximized.
## is our objective function that needs to be
Graphical Solutions
When a LP model is stated in terms of 2 decision variables, it can
be solved by graphical methods.Before discussing the graphical
solution method, we will discuss the graphics of linear
inequalities.
Graphics of Linear Inequalities
Linear inequalities which involve two variables can be
represented graphically in two dimensions by a closed half space
of a certain type of the certain plane. The half space consists of
the boundary line representing the equality part of the inequality
63
and all points on one side of the boundary line (representing the
strict inequality). Graph the boundary which represents the
equation.Determine the side that satisfy the strict inequality.
System of Linear Inequalities
In LP problems we will be dealing with system of linear
inequalities. We will be interested to determine the solution set
which satisfies all the inequalities in the system of constraints.
Region of feasible solution
The first step in graphical procedure is to identify the solution set
for the system constraints.This solution set is calledregion of
feasible solution.It includes all combinations of decision variables
which satisfy the structural and non-negativity constraints.
Corner point solution
Given a linear objective function in a linear programming problem,
the optimal solution will always include a corner point on the
region of feasible solution.This will hold, irrespective of the slope
of objective function and for either maximization or minimization
problems.We can use a method, called corner point method, to
solve the linear programming problems.
The corner point method for solving LP problems is as follows.
1. Graphically identify the region of feasible solutions.
2. Determine the coordinates of each corner point on the region
of feasible solution.
3. Substitute the coordinates of the corner points into the
objective function z to determine the corresponding value
of z .
4. The required solution occur at the corner point yielding,
(highest value of z in maximization problems)/ (lowest
value of z in minimization problems).
Alternative Optimal Solutions
There is a possibility of more than one optimal solutions in LP
64
## problem.Alternative optimal solutionexists if the following two
conditions are satisfied.
1. The objective function must be parallel to the constraint
which forms an edge or boundary on the feasible region
2. The constraint must form a boundary on the feasible region
in the direction of optimal movement of the objective function
Example (Page 437 in Book)
Use corner-point method to solve:
z=20 x 1+15 x 2
Maximize
Subject to
3 x1 + 4 x 2 60(1)
4 x 1 +3 x 2 60(2)
x 1 10
(3)
x 2 12
(4 )
x1 , x2 0
(5)
Solution: The figure shows all the corner points of the given LP
problem.
20
60 60
Corner O(0,0)
A (0,12)
B (10,0)
C( 4,12)
D(10, )
E( , )
3
7 7
point
Value
0
180
200
260
300
300
z
of
Optimal Solution occursat the point D and E.In fact, any point
along the line DE will give this optimal solution.Both the conditions
for alternative optimal solutions were satisfied in this example.The
objective function was parallel to the constraint (2) and the
constraint (2) was a boundary line in the direction of the optimal
movement of objective function.
65
No Feasible Solutions
The system of constraints in an LP problem may have no points
which satisfy all constraints.In such cases, there are no points in
the solution set and hence the LP problem is said to have no
feasible solution.
Applications of Linear Programming
Diet Mix Model
A dietician is planning the menu for the evening meal at a
university dining hall.Three main items will be served, all having
different nutritional content.The dietician is interested in providing
at least the minimum daily requirement of each of three vitamins
in this one meal. Following table summarizes the vitamin contents
per ounce of each type of food, the cost per ounce of each food,
and minimum daily requirements (MDR) for the three vitamins.
Any combination of the three foods may be selected as long as
the total serving size is at least 9 ounces.
Vitamins
Food
Cost
66
per Oz,
\$
1
50m
g
20m
g
10m
g
0.10
30m
g
10m
g
50m
g
0.15
20m
g
30m
g
20m
g
0.12
Minimum daily
requirement
(MDR)
290
mg
200
mg
210
mg
## The problem is to determine the number of ounces of each food
to be included in the meal. The objective is to minimize the cost of
each meal subject to satisfying minimum daily requirements of the
three vitamins as well as the restriction on minimum serving size.
Solution:Let xj equal the number of ounces included of food j.
The objective function should represent the total cost of the meal.
Stated in dollars, the total cost equals the sum of the costs of the
three items, or
Z = 0.10x1 + 0.15x2 + 0.12x3
The constraint for each vitamin will have the form.
Milligrams of vitamin intake MDR
Milligrams from food 1 + milligram from food 2+ milligram from
food 3 MDR
The constraints are, respectively,
50x1 + 30x2 + 20x3 290
(vitamin 1)
67
## 20x1 + 10x2 + 30x3 200 (vitamin 2)
10x1 + 50x2 + 20x3 210 (vitamin 3)
The restriction that the serving size be at least 9 ounces is stated
as
x1 + x2+ x3 9 (minimum serving size)
The complete formulation of the problem is as follows:
Minimize z = 0.10x1 + 0.15x2 + 0.12x3
subject to 50x1 + 30x2 + 20x3 290
20x1 + 10x2 + 30x3 200
10x1 + 50x2 + 20x3 210
x1 + x 2 + x 3 9
x 1 , x2 , x3 0
Transportation Model
Transportation models are possibly the most widely used linear
programming models.Oil companies commit tremendous
resources to the implementation of such models. The classic
example of a transportation problem involves the shipment of
some homogeneous commodity from m sources of supply, or
origins, to n points of demand, or destinations. By homogeneous
we mean that there are no significant differences in the quality of
the item provided by the different sources of supply. The item
characteristics are essentially the same.
Example (Highway Maintenance)
A medium-size city has two locations in the city at which salt and
sand stockpiles are maintained for use during winter icing and
snowstorms. During a storm, salt and sand are distributed from
68
## these two locations to four different city zones. Occasionally
additional salt and sand are needed. However, it is usually
impossible to get additional supplies during a storm since they are
stockpiled at a central location some distance outside the city.City
officials hope that there will not be back-to-back storms. The
director of public works is interested in determining the minimum
cost of allocating salt and sand supplies during a storm.Following
table summarizes the cost of supplying 1 ton of salt or sand from
each stockpile to each city zone.In addition, stockpile capacities
and normal levels of demand for each zone are indicated (in tons)
Solution: In formulating the linear programming model for this
problem, there are eight decisions to make-how many tons should
be shipped from each stockpile to each zone. Let x ij equal the
number of tons supplied from stockpile i to zone j. For example,
x11 equals the number of tons supplied by stockpile 1 to zone 1.
Similarly, x23 equals the number of tons supplied by stockpile 2 to
zone 3.This double-subscripted variable conveys more
information than the variables x1, x2,, x8.
Give this definition of the decision variables, the total cost of
distributing salt and sand has the form:
Total Cost = 2x11 + 3x12 + 1.5x13 + 2.5x14 +4x21 + 3.5x22 +2.5x23
+3x24
This is the objective function that we wish to minimize.For
stockpile 1, the sum of the shipments to all zones cannot exceed
900 tons, or
x11 + x12 + x13 + x14 900 (stockpile 1)
The same constraint for stockpile 2 is
x21 + x22 + x23 + x24 750 (stockpile 2)
69
## The final class of constraints should guarantee that each zone
receives the quantity demanded.
For zone 1, the sum of the shipments from stockpiles 1 and 2
should equal 300 tons, or
x11 + x21 = 300 (zone 1)
The same constraints for the other three zones are, respectively,
x12 + x22 = 450
(zone 2)
x13 + x23 = 500
(zone 3)
x14 + x24 = 350
(zone 4)
The complete formulation of the linear programming model is as
follows:
Minimize z = 2x11 + 3x12 + 1.5x13 + 2.5x14 + 4x11
+ 3.5x22 +2.5x23 +3x24
Subject to x11 + x12 + x13 + x14 900
x21 + x22 + x23 + x24 750
x11 + x21 = 300
x12 + x22 = 450
x13 + x23 = 500
x14 + x24 = 350
x11 , x12 , x13 , x14 , x11 , x22 , x23 , x24 0
70
Lecture 14,15,16,17
## Chapter 11:The Simplex and Computer Solution
Methods
Simplex Preliminaries
Graphical solution methods are applicable to LP problems
involving two variables. The geometry of three variable problems
is very complicated, and beyond threevariables, there is no
geometric frame of reference, so we need some nongraphic
Method.
The most popular non-graphical procedure is called The Simplex
Method. This is an algebraic procedure for solving system of
equations where an objective function needs to be optimized.It is
an iterative process, which identifies a feasible starting solution.
The procedure then searches to see whether there exists a better
solution.Better is measured by whether the value of the
objective function can be improved.If better solution is signaled,
the search resumes. The generation of each successive solution
requires solving a system of linear equations. The search
continues until no further improvement is possible in the objective
function
Requirements of the Simplex Method
71
## 1. All constraints must be stated as equations.
2. The right side of the constraint cannot be negative.
3. All variables are restricted to nonnegative values.
Most linear programming problems contain constraints which are
inequalities.Before we solve by the simplex method, these
inequalities must be restated as equations.The transformation
from inequalities to equations varies, depending on the nature of
the constraints.
Transformation procedure for constraints
For each less than or equal to constraint, a non-negative
variable, called a slack variable, is added to the left side of the
constraint.Note that the slack variables become additional
variables in the problem and must be treated like any other
variables. That means they are also subject to Requirement 3
that is, they cannot assume negative values.
For example 3 x1 2 x 2 4 will be transformed as
3 x1 +2 x 2+ S1 =4.
## Transformation procedure for constraints
For each greater than or equal to constraint, a non-negative
variable E, called a surplus variable, is subtracted from the left
side of the constraint. In addition, a non-negative variable A,
called an artificial variable, is added to the left of side of the
constraint.The artificial variable has no real meaning in the
problem; its only function is to provide a convenient starting point
(initial guess) for the simplex.
For example x 1+ x 2 10 will be transformed to
x 1+ x 2E 1+ A 1=10.
72
## For each equal to constraint, an artificial variable, is added to
the left of side of the constraint. For example 3 x1 + x 2=10 will be
transformed to 3 x1 + x 2 + A 1=10.
Example
Transform the following constraint set into the standard form
required by the simplex method
x1 + x2 100, 2x1 + 3x2 40, x1 x2 = 25,
x1, x2 0
Solution: The transformed constraint set is
x1 + x2 + S1
100
2x1 + 3x2 E2 + A2 40
x1 x2
+ A3 = 25
x1, x2, S1, E2, A2, A3 0
Note that each supplemental variable (slack, surplus, artificial) is
assigned a subscript which corresponds to the constraint number.
Also, the non-negativity restriction (requirement 3) applies to all
supplemental variables.
Example
An LP problem has 5 decision variables;
10 () constraints; 8 () constraints; and
2 (=) constraints.
When this problem is restated to comply with requirement 1 of the
simplex method, how many variables will there be and of what
types?
Solution: There will be 33 variables:
5 decision variables, 10 slack variables associated with the 10 ()
constraints,8 surplus variables associated with the 8 ()
constraints, and10 artificial variables associated with the () and
(=) constraints.
5 + 10 + 8 + 10 = 33
73
## Basic Feasible Solutions
Lets state some definitions which are significant to our coming
discussions. Assume the standard form of an LP problem which
has m structural constraints and a total of n decision and
supplemental variables.
Definition: A feasible solutionis any set of values for the n
variables which satisfies both the structural and non-negativity
constraints.
Definition: A basic solution is any solution obtained by setting
(nm) variables equal to 0 and solving the system of equations
for the values of the remaining m variables.
Definition: The (nm) variables which have been assigned
values of 0, are called non-basic variables. The rest of the
variables are called basic variables.
Definition:A basic feasible solutionis a basic solution which
also satisfies the non-negativity constraints. Thus optimal solution
can be found by performing a search of the basic feasible
solution.
The Simplex Method
Before we begin our discussions of the simplex method, lets
provide a generalized statement of an LP model. Given the
definitions
xj = j-th decision variable
cj = coefficient on j-th decision variable in the objective function.
aij = coefficient in the i-th constraint for the j-th variable.
bi = right-hand-side constant for the i-th constraint
The generalized LP model can be stated as follows:
Optimize (maximize or minimize)
z = c1x1 + c2x2 + ... + cnxn
subject to
74
## a11x1 + a12x2 + ... + a1nxn ( , , =) b1
a21x1 + a22x2 + ... + a2nxn ( , , =) b2
...
am1x1 + am2x2 + ... + amnxn ( , , =)bm
x1
0
x2
0
...
xn
(1)
(2)
(m)
Solution by Enumeration
Consider a problem having m () constraints and n variables.
Prior to solving by the simplex method, the m constraints would
be changed into equations by adding m slack variables. This
restatement results in a constraint set consisting of m equations
and m + n variables.
Example
Solve the following LP problem.
Maximize
z = 5x1 + 6x2
subject to
3x1 + 2x2 120
4x1 + 6x2 260
x 1 , x2 0
(1)
(2)
## Solution: The constraint set must be transformed into the
equivalent set.
3x1 + 2x2 + S1 = 120
4x1 + 6x2 + S2 = 260
x1 , x2 , S1 , S2 0
The constraint set involves two equations and four variables. Of
all the possible solutions to the constraint set, an optimal solution
occurs when two of the four variables in this problem are set
equal to zero and the system is solved for the other two variables.
The question is, which two variables should be set equal to 0
75
## (should be non-basic variables)? Lets enumerate the different
possibilities.
1. If S1 and S2 are set equal to 0, the constraint equations
become
3x1 + 2x2 = 120
4x1 + 6x2 = 260
Solving for the corresponding basic variable x 1 and x2 results in
x1= 20 and x2 = 30
2. If S1 and x1 are set equal to 0, the system becomes
2x2 = 120
6x2 + S2 = 260
Solving for the corresponding basic variables x2 and S2 results in
x2 = 60 and S2 = 100
Following table summarizes the basic solutions, that is, all the
solution possibilities given that two of the four variables are
assigned 0 values.
Notice that solutions 2 and 5 are not feasible. They each contain
a variable which has a negative value, violating the non-negativity
restriction.However, solutions 1, 3, 4 and 6 are basic feasible
solutions to the linear programming problem and are candidates
76
## for the optimal solution.
The following figure is the graphical representation of the set of
constraints.
## Specifically,solution 1 corresponds to corner point C, solution 3
corresponds to corner point B, solution 4 corresponds to corner
point D, and solution 6 corresponds to corner point A. Solution 2
and 5, which are not feasible, correspond to the points E and F
shown in the figure.
Incorporating the Objective Function
In solving by the simplex method, the objective function and
constraints are combined to form a system of equations.The
objective function is one of the equations, and z becomes an
additional variable in the system.In rearranging the variables in
the objective function so that they are all on the left side of the
equation, the problem is represented by the system of equations.
z 5x1 6x2 0S1 OS2 = 0
(0)
77
3x1 + 2x2 + S1
= 120
(1)
4x1 + 6x2 + S2
= 260
(2)
Note:The objective function is labeled as Eq.(0).The objective is
to solve this (3 x 5) system of equations so as to maximize the
value of z.Since we are particularly concerned about the value of
z and will want to know its value for any solution, z will always be
a basic variable.The standard practice, however, is not to refer to
z as a basic variable. The terms basic variable and non-basic
variable are usually reserved for other variables in the problem.
Simplex Procedure
The simplex operations are performed in a tabular format. The
initial table, or tableau, for our problem is shown in the table
below. Note that there is one row for each equation and the table
contains the coefficients of each variable in the equations and bi
column contains right hand sides of the equation.
## Summary of simplex procedure
We summaries the simplex procedure for maximization problems
having all () constraints.
First, add slack variables to each constraint and the objective
function and place the variable coefficients and right-hand-side
constants in a simplex tableau:
1- Identify the initial solution by declaring each of the slack
variables as basic variables. All other variables are non-basic
in the initial solution.
2- Determine whether the current solution is optimal by applying
78
## rule 1 [Are all row (0) coefficients 0?]. If it is not optimal,
proceed to step 3.
3-
## Determine the non-basic variable which should become a
basic variable in the next solution by applying rule 2 [most
negative row (0) coefficient].
4- Determine the basic variable which should be replaced in
the next solution by applying rule 3 (min bi/aik ratio where aik>
0)
5- Applying the Gaussian elimination operations to generate
the new solution (or new tableau). Go to step 2.
Example(Page 488 in Book)
Solve the following linear programming problem using the simplex
method.
Maximize
subject to
## z = 2x1 + 12x2 + 8x3
2x1 + 2x2 + x3 100
x1 2x2 + 5x3 80
10x1 + 5x2 + 4x3 300
x 1 , x2 , x 3 0
Solution: Detailed solution is in Book/Lectures.
Maximization Problems with Mixed Constraints
For a maximization problem having a mix of (, , and =)
constraints the simplex method itself does not change. The only
change is in transforming constraints to the standard equation
form with appropriate supplemental variables. Recall that for each
() constraint, a surplus variable is subtracted and an artificial
variable is added to the left side of the constraint. For each (=)
constraint, an artificial variable is added to the left side.An
additional column is added to the simplex tableau for each
supplemental variable.Also, surplus and artificial variables must
be assigned appropriate objective function coefficients (c, values)
Surplus variables usually re-assigned an objective function
79
## coefficient of 0.Artificial variables are assigned objective function
coefficient of M, where |M| is a very large number. This is to
make the artificial variables unattractive in the problem.
In any linear programming problem, the initial set of basic
variables will consist of all the slack variables and all the artificial
variables which appear in the problem.
Example (Page 491 in Book)
Maximize
z = 8x1 + 6x2
subject to
2x1 + x2 10
3x1 + 8x2 96
x1, x2 0
We have a maximization problem which has a mix of a ()
constraint and a () constraint.
Solution: (In Book/ Lectures)
Minimization Problems
The simplex procedure is slightly different when minimization
problems are solved.Aside from assigning artificial variables
objective function coefficients of +M, the only difference relates to
the interpretation of row (0) coefficients.The following two rules
are modifications of rule 1 and rule 2. These apply for
minimization problems.
Rule 1A: Optimality Check in Minimization Problem.
In a minimization problem, the optimal solution has been found if
all row (0) coefficients for the variables are less than or equal to
0.
If any row (0) coefficients are positive for non-basic variables, a
better solution can be found by assigning a positive quantity to
these variables.
Rule 2A: New basic variable in minimization problem.
80
## In a minimization problem, the non-basic variable which will
replace a current basic variable is the one having the largest
positive row (0) coefficient. Ties may be broken arbitrarily.
Example (page 493 in Book)
Solve the following linear programming problem using the simplex
method.
Minimize
subject to
z = 5x1 + 6x2
x1 + x2 10
2x1 + 4x2 24
x 1 , x2 0
Solution:
This problem is first rewritten with the constraints expressed as
equations, as follows:
Minimize z = 5x1 + 6x2 + 0E1 + 0E2 + MA1 + MA2
subject to
x 1 + x2 E 1
+ A1
= 10
2x1 + 4x2
E2
+ A2 = 24
x1, x2, E1, E2, A1, A2 0
x1
x2
E1
E2
A1
A2
Need to Be
Transformed
to Zero
bi
Row
Number
(0)
A1
10
(1)
A2
24
(2)
Basic
Varia
bles
The initial tableau for this problem appears in above table. Note
81
that the artificial variables are the basic variables in this initial
solution. These M coefficient in row (0) must be changed to 0
using row operations if the value of z is to be read from row (0).
In this initial solution the non-basic variables are x 1, x2, E1, and E2.
The basic variables are the two artificial variables with A1 = 10, A2
= 24, and
z = 34M.Applying rule 1A, we conclude that this
solution is not optimal. The x2 column becomes the new key
column.
## Applying rule 1A, we conclude that this solution is optimal.
Therefore, z is minimized at a value of 52 when x1 = 8 and x2 = 2.
Special Phenomena
In Chapter 10, certain phenomena which can arise when solving
LP problems were discussed. Specifically, the phenomena of
alternative optimal solutions, no feasible solution, and unbounded
solutions were presented. In this section we discuss the manner
in which these phenomena occur when solving by the simplex
method.
Alternative Optimal Solutions
An Alternative optimal solution results when the objective function
is parallel to a constraint which binds in the direction of
82
## optimization. In two-variable problems we are made aware of
alternative optimal solutions with the corner-point method when a
tie occurs for the optimal corner point.
When using the simplex methods, alternative optimal solutions
are indicated when
1- an optimal solution has been identified,
2- the row (0) coefficient for a non-basic variable equals zero.
The first condition confirms that there is no better solution than
the present solution. The presence of a 0 in row (0) for a nonbasic variable indicates that the non-basic variable can become a
basic variable (can become positive) and the current value of the
objective function (known to be optimal) will not change.
Example (page 491 in Book)
Maximum
z = 6x1 + 4x2
subject to
x 1 + x2 5
3x1 + 2x2 12
x1, x2 0
Solution(In Lectures/Book)
No Feasible Solutions
A problem has no feasible solution if there are no values for the
variables which satisfy all the constraints. The condition of no
feasible solution is signaled in the simplex method when an
artificial variable appears in an optimal basis at a positive value.
Lets solve the following LP problem, which, by inspection, has no
feasible solution.
Example (Page 499 in Book )
Maximize
z = 10x1 + 20x2
subject to
x + x2 5
x1 + x2 20
x1 , x 2 0
Solution:(In Lectures/Book)
Unbounded Solutions
83
## Unbounded solutions exist when there is an unbounded solution
space. Improvement in the objective function occurs with
movement in the direction of the unbounded portion of the
solution space.If at any iteration of the simplex method the
aikvalues are all 0 or negative for the variable selected to become
the new basic variable, there is an unbounded solution for the LP
problem.
Example (Page 500-Book )
Maximize
z = 2x1 + 3x2
subject to
x1
10
2x1 x2 30
x 1 , x2 0
Solution:(In Lectures/Book)
The Dual Problem
Every LP problem has a related problem called the dual problem
or, simply, the dual. Given an original LP problem, referred to as
the primal problem, or primal, the dual can be formulated from
information contained in the primal.Given an LP problem, its
solution can be determined by solving either the original problem
or its dual.
Formulation of the Dual
The parameters and structure of the primal provide all the
information necessary to formulate the dual.
The following problem illustrates the formulation of a
maximization problem and the dual of the problem.
1- The primal is a maximization problem and the dual is a
minimization problem. The sense of optimization is always
opposite for corresponding primal and dual problems.
2- The primal consists of two variables and three constraints
and the dual consists of three variables and two constraints.
The number of variables in the primal always equals the
84
## number of constraints in the dual. The number of constraints
in the primal always equals the number of variables in the
dual.
Primal Problem
maximize
z=
Dual Problem
minimize
2x1 + 4x2
subject to
## z = 800y1 + 350y2 + 125y3
subject to
5x1 + 4x2
800
(1) 5 y1 +3 y2 +4 y3
3x1 + 2x2
350
(2) 4 y1 +2 y2 +3 y3
4x1 + 3x2
125
(3)
x 1 , x2
y1 , y2 , y3
2 (1)
(2)
## 3- The objective function coefficients for x1 and x2 in the primal
equal the right-hand-side constants for constraints (1) and (2)
in the dual. The objective function coefficient for the j-th primal
variable equals the right-hand-side constant for the j-th dual
constraint.
4- The right-hand-side constants for constraints (1)(3) in the
primal equal the objective functions coefficient for the dual
variables y1, y2 and y3. The right-hand-side constant for the i-th
primal constraint equal the objective function coefficient for the ith dual variable.
5- The variable coefficients for constraint (1) of the primal equal
the column coefficient for the dual variable y1. The variable
85
coefficients for constraints (2) and (3) of the primal equal the
column coefficients of the dual variables y2 and y3. The
coefficients aij in the primal are the transpose of those in the
dual. That is, the row coefficients in the primal become
column coefficients in the dual, and vice versa.
The following table summarizes the symmetry of the two types of
problems and their relationships.
## Relationship 4 and 8 indicate that an equality constraint in one
problem corresponds to an unrestricted variable in the other
problem. An unrestricted variable can assume a value which is
positive, negative, or 0.Similarly, relationship 3 and 7 indicate that
a problem may have non-positive variables (for example xj = 0).
86
## Unrestricted and non-positive variables appear to violate the third
requirement of the simplex method, the non-negativity restriction.
There are methods which allow us to adjust the formulation to
satisfy the third requirement.
Example
Given the primal problem find the corresponding dual problem.
Maximize
z = 10x1 + 20x2 + 15x3 + 12x4
subject to
x1 + x2 + x3 + x4
2x1
100 (1)
## x3 + 3x4 140 (2)
x1 + 4x2
2x4 = 50 (3)
x1, x3, x4 0
x2 unrestricted
Solution:The corresponding dual is
Maximize
## z = 100y1 + 140y2 + 50y3
subject to
y1 + 2y2 + y3 10
y1
+ 4y3 = 20
y1 y2
15
y1 + 3y2 2y3 12
y1 0
y2 0
y3 unrestricted
Primal-Dual Solutions
It was indicated earlier that the solution to the primal problem can
be obtained from the solution to the dual problem and vice versa.
87
Maximize
z = 5x1 + 6x2
subject to
(1)
## 4x1 + 6x2 260
(2)
x 1 , x2
0
The corresponding dual is
Minimize
z = 120y1 + 260y2
Subject to
3y1 + 4y2 5
2y1 + 6y2 6
y 1 , y2 0
The following table presents the final (optimal) tableau for the
primal and dual problem.
88
## Note from this tableau that z is minimized at a value of 280 when
y1 = 3/5 and y2 = 4/5. Lets illustrate how the solution to each
problem can be read from the optimal tableau of the
corresponding dual problem.
Primal-Dual Property 1
1. If feasible solutions exist for both the primal and dual
problems, then both problems have an optimal solution for
which the objective function values are equal.
2. A peripheral relation-ship is that if one problem has an
unbounded solution, its dual has no feasible solution.
Primal-Dual Property 2
The optimal values for decision variables in one problem are read
from row (0) of the optimal tableau for the other problem.
89
1. The optimal values y1 = 3/5 and y2 = 4/5 are read from above
table as the row (0) coefficients for the slack variables S1
and S2.
2. The optimal values x1 = 20 and x2 = 30 are read from above
table as the negatives of the row (0) coefficients for the
surplus variables E1 and E2.
3. These values can be read, alternatively, under the
respective artificial variables, as the portion (term) of the row
(0) coefficient not involving M.
Lecture 18,19,20
## Chapter12: Transportation and Assignment Models
The aim of this chapter is to provide an overview of several
extensions of the basic linear programming model. It will Includea
discussion of the assumptions,distinguishing characteristics,
methods of solution, and applications of the transportation model
and the assignment model.
The transportation model
The classic transportation model involves the shipment of some
homogeneous commodity from a set of origins to a set of
destinations. Each origin represents a source of supply for the
commodity; each destination represents a point of demand for the
commodity.
90
Assumption-1
The standard model assumes a homogeneous commodity.
This first assumption implies that there are no significant
differences in the characteristics of the commodity available at
each origin. This suggests that unless other restrictions exist,
each origin can supply units to any of the destinations.
Assumption-2
The standard model assumes that total supply and total demand
are equal.
This second assumption is required by a special solution
algorithm for this type of model.
Lets state the generalized transportation model associated with
the structure shown in the figure given below.
91
If
xij = number of units distributed from origin i to destination j
cij = contribution to the objective function from distributing one unit
from origin i to destination j
Si = number of units available at origin i
dj = number of units demanded at destination j
m = number of origins
n = number of destinations
the generalized model can be stated as follows:
92
## Minimize z = c11x11 + c12x12 + . . . + c1nx1n + c21x21 + c22x22 + . . .
(or maximize)
. . . + c2nx2n + . . . + cmnxmn
Subject to
x11 + x12 + . . . +x1n = S1
x21 + x22 + . . . + x2n = S2
supply
.
.
.
Constraints
.
.
.
xm1 + xm2 + . . . + xmn = Sm
x11 + x21 . . . + xm1 = d1
x12 + x22 + . . . + xm2 = d2
.
.
.
demand
.
.
.
constraints
x1n + x2n + . . . + xmn = dn
xij 0
for all i and j
Implicit in the model is the balance between supply and demand.
S 1 + S 2 + . . . + S m = d1 + d2 + . . . + dn
The transportation model is a very flexible model which can be
applied to the problems that have nothing to do with the
distribution of commodities.
Example (Job Placement Screening)
A job placement agency works on a contract basis with
employers. A computer manufacturer is opening a new plant and
has contracted with the placement agency to process job
applications for prospective employees.Because of the uneven
demands in workload at the agency, it often uses part-time
personnel for the purpose of processing applications.For this
particular contract, five placement analysts must be hired.Each
analyst has provided an estimate of the maximum number of job
applications he or she can evaluate during the coming
month.Analysts are compensated on a piecework basis, with the
93
## rate determined by the type of application evaluated and the
experience of the analyst.
## If xij equals the number of job applications of type j assigned to
analyst i, the problem can be formulated as shown in the model
below.
Notice that the total supply (the maximum number of applications
which can be processed by five analysts) exceeds total demand
(expected number of applications). As a result, constraints (1) to
(5) cannot be stated as equalities.
According to assumption 2, total supply and demand must be
brought into balance, artificially, before solving the problem.
Maximize z = 15x11 + 10x12 + 8x13 + 7x14 + . . . 6x54
subject to
x11 + x12 + x13 + x14
90 (1)
x21 + x22 + x23 + x24
120 (2)
x31 + x32 + x33 + x34
140 (3)
x41 + x42 + x43 + x44
100 (4)
x51 + x52 + x53 + x54 110 (5)
x11 +x21 +x31 +x41 +x51
= 100 (6)
94
x12
+x22 +x32 +x42 +x52
= 150 (7)
x13 +x23 +x33 +x43 +x53
= 175 (8)
x14 +x24
+x34 + x44
+x54 = 125 (9)
xij 0 for all
i and j
Solutions to transportation models
The simplex method can be used to solve transportation models.
However, methods such as the stepping stone algorithm and a
dual-based enhancement called the MODI method prove much
more efficient.
Initial solutions
The increased efficiency can occur during two different phases
of solution:
1. Determination of the initial solution
2. Progress from the initial solution to the optimal
solution
With the simplex method the initial solution is predetermined by
the constraint structure. The initial set of basic variables will
always consist of the slack and artificial variable in the
problem.
With transportation models the stepping stone algorithm (or the
MODI method) will accept any feasible solution as a starting
point. Consequently, various approaches to finding a good
starting solution have been proposed. These include the
northwest corner method, the least cost method, and Vogels
approximation method.
Example
Consider the data contained in the table for a transportation
problem involving three origins and three destinations.
95
Assume that the elements in the body of the table represent the
costs of shipping a unit from each origin to each destination.
Also shown are the supply capacities of the three origins and the
demands at each destination.Conveniently, the total supply and
total demand are equal to one another.
The problem is to determine how many units to ship from each
origin to each destination so as to satisfy the demands at the
three destinations while not violating the capacities of the three
origins.
The objective is to make these allocations in such a way as to
minimize total transportation costs.
Solution: We will solve this problem using two special
algorithms.
We will illustrate the northwest corner method, which can be used
to determine an initial (starting) solution. Next, we illustrate the
stepping stone algorithm, which can be used to solve these types
of models.
Before we begin these examples, lets discuss come
requirements of the stepping stone algorithm.
Requirements of stepping stone algorithm
1- Total supply = Total demand
Since this is not typically the case in actual applications, the
96
## balance between supply and demand often is created artificially.
This is done by adding a dummy origin or a dummy destination
having sufficient supply (demand) to create the necessary
balance. Our example has been contrived so that balance already
exists.
2-
## Given a transportation problem with m origins and
destination (where m and n include any dummy origins or
destination added to create balance), the number of basic
variables in any given solution must equal m + n 1. In our
problem, any solution should contain 3 + 3 1 = 5 basic
variables..
## For each origin/destination combination, there is a cell which
contains the value of the corresponding decision variable x ij and
the objective function coefficient or unit transportation cost. As we
proceed on to solve a problem, we will substitute actual values for
the xijs in the table as we try different allocations.
97
## The northwest corner method is a popular technique for arriving
at an initial solution.The technique starts in the upper left-hand
cell (northwest corner) of a transportation table and assigns units
from origin 1 to destination 1. Assignments are continued in such
a way that the supply at origin 1 is completely allocated before
moving on to origin 2. The supply at origin 2 is completely
allocated before moving to origin 3, and so on.
Similarly, a sequential allocation to the destinations assures that
the demand at destination 1 is satisfied before making allocations
to destination 2; and so forth.
The following table indicates the initial solution to our problem as
derived using the northwest corner method.
Table 1
98
1-
## Starting in the northwest corner, the supply at origin 1 is 55
and the demand at destination 1 is 70. Thus, we allocate
supply at origin 1 in an attempt to satisfy the demand at
destination 1 (x11 = 55).
2-
## When the complete supply at origin 1 has been allocated,
the next allocation will be from origin 2. The allocation in cell
(1, 1) did not satisfy the demand at destination 1 completely.
Fifteen additional units are demanded. Comparing the
supply at origin 2 with the remaining demand atdestination
1, we allocate 15 units from origin 2 to destination 1 (x21 =
15). This allocation completes the needs of destination 1,
and the demand at destination 2 will be addressed next.
99
3-
## The last allocation left origin 2 with 65 units. The demand at
destination 2 is 100 units. Thus, we allocate the remaining
supply of 65 units to destination 2 (x22 = 65). The next
allocation will come from origin 3.
4-
## The allocation of 65 units from origin 2 left destination 2 with
unfulfilled demand of 35 units. Since origin 3 has a supply of
75 units, 35 units are allocated to complete the demand for
that destination (x32 = 35). Thus, the next allocation will be
destination 3.
The allocation of 35 units from origin 3 leaves that origin with
40 remaining units. The demand at destination 3 also equals
40; thus, the final allocation of 40 units is made from origin 3
to destination 3 (x33 = 40).
5-
Notice in the above table that all allocations are circled in the
appropriate cells. These represent the basic variables for this
solution. There could be five basic variables (m + n 1) to satisfy
the requirement of the stepping stone algorithm. The
recommended allocations and associated costs for this initial
solution are summarized in the following table.
100
## Stepping Stone Algorithm
Step 1:Determine the improvement index for each non-basic
variable (cell).
As we examine the effects of introducing one unit of a non-basic
variable, we focus upon two marginal effects:
1. What adjustments must be made to the values of the current
basic variables (in order to continue satisfying all supply and
demand constraints)?
2. What is the resulting change in the value of the objective
function?
101
## 1) To illustrate, focus upon the above table. Cell (1,2) has no
allocation in the initial solution and is a non-basic cell. The
question we want to ask is what trade-offs (or adjustments)
with the existing basic variables would be required if 1 unit is
shipped from origin 1 to destination 2?
2) Summarizing, to compensate for adding a unit to cell (1, 2),
shipments in cell (1, 1) must be decreased by 1 unit,
shipments in cell (2, 1) must be increased by 1 unit, and
shipments in cell (2,2) must be decreased by 1 unit.
3) The next question is, What is the marginal effect on the
value of the objective function?
4) For each cell (i, j) receiving an increased allocation of 1 unit,
costs increase by the corresponding cost coefficient (cij).
5) Similarly, costs decrease by the value of the cost coefficient
wherever allocations have been reduced by 1 unit.
6) These effects are summarized in the following table.
102
## The net (marginal) effect associated with allocating 1 unit from
origin 1 to destination 2 is to reduce total cost by \$5.00. This
marginal change in the objective function is called
theimprovement index for cell (1, 2).
Trace a closed path which begins at the unoccupied cell of
interest; moves alternately in horizontal and vertical directions,
pivoting only on occupied cells, and terminates on the unoccupied
cell.
The pluses and minuses indicate the necessary adjustments for
satisfying the row (supply) and column (demand) requirements.
Note:The direction in which the path is traced is not important.
Tracking the path clockwise or counter-clockwise will result in the
same path and identical adjustments.
Once the closed path has been identified for a non-basic cell, the
improvement index for that cell is calculated by adding all
objective function coefficient for cells in plus positions on the path
103
## and subtracting corresponding objective function coefficients for
cells in negative positions on the path.
104
We can also make the above table for cell (3,1) and cell (2,3).
(See Lectures)
## The above table summarizes the closed paths and
improvement indices for all non-basic cells in the initial
solution.
Verify these path and values for the improvement indices to
make sure you understand what we have been discussing.
Step 2: If a better solution exists, determine which variable
(cell) should enter the basis.
An examination of the improvement indices in the above table
indicates that the introduction of three of the four non-basic
variables would lead to a reduction in total costs.
For minimization problems, a better solution exists if there are
any negative improvement indices. An optimal solution has been
found when all improvement indices are non-negative.
For maximization problems a better solution has been found
when all improvement indices are non-positive.
As in the simplex method, we select the variable (cell) which
leads to the greatest marginal improvement in the objective
105
function.
Entering Variable
For minimization problems, the entering variable is identified
as the cell having the largest negative improvement index
(ties may be broken arbitrarily).
## For maximization problems, the entering variable is the cell
having the largest positive improvement index.
variable.
106
## Step 3: Determine the departing variable and the number of
units to assign the entering variable.
This step is performed by returning to the closed path
associated with the incoming cell.
The above table shows the closed path for cell (2, 3).
The stepping stone algorithm parallels the simplex exactly,
we need to determine the number of units we can assign to
cell (2, 3) such that the value of one of the current basic
variables is driven to 0.
From the above table we see that only two basic variables
decrease in values additional units are allocated to cell (2,
3): cells (2, 2) and (3, 3), both of which are in minus
positions on the closed path.
The question is which of these will go to 0 first as more units
are added to cell (2, 3).
We can reason that when the 40th unit is added to cell (2, 3),
the value for cell (2, 2) reduces to 25, the value for cell (3, 2)
increases to 75, and the value of cell (3, 3) goes to 0.
Departing variable
The departing variable is identified as the smallest basic variable
in a minus position on the closed path for the entering variable.
Number of units to assign entering variable
The number of units equals the size of the departing variable (the
smallest value in a minus position).
Step 4: Develop the new solution and return to step 1.
Again referring to the closed path for the incoming cell (2, 3), add
he quantity determined in step 3 to all cells in plus positions and
subtract this quantity from those in minus positions.
Thus, given that the entering variable x23 = 40 from step 3, the
107
## cells on the closed path are adjusted, leading to the second
solution shown in the following table.
When you determine a new solution you should check the
allocations along each row and column to make sure that they
add to the respective supply and demand values.Also, make sure
that there are m + n 1 basic variables.
It can be seen in the following table that both of the above
requirements are satisfied.
## One other piece of information which is of interest is the new
value of the objective function.We can multiply the value of
each basic variable times its corresponding objective function
coefficient and sum, as we did in the above table.
Given that the original value of the objective function was
\$4,425 and that each unit introduced to cell (2, 3) decreases
the value of the objective by \$20, introducing 40 units to cell (2,
3) results in a new value for total cost of.
108
## z = \$4,425 (40) (\$20)
= \$4,425 \$800
= \$3,625
The following tables summarize the remaining steps in solving
this problem. See whether you can verify these steps and the
final result.
109
110
## Since all improvement indices are nonnegative in the above table,
we conclude that the solution in the above table is optimal. That
111
## is, total cost will be minimized at a value of \$3,350 if
55 units are shipped from origin 1 to destination 2
40 units from origin 2 to destination 1
40 units from origin 2 to destination 3
350 units from origin 3 to destination 1
45 units from origin 3 to destination 2
Alternative optimal solution
Given that an optimal solution has been identified for a
transportation model, alternative optimal solutions exist if any
improvement indices equal 0. If the conditions for optimality exist,
allocation of units to cells having improvement indices of 0 results
is no change in the (optimal) value for the objective function. In
our optimal solution, the above table indicates that allocation of
units to cell (2, 2) would result in no change in the total cost.
## The Assignment Model and Methods of Solution
A special case of the transportation model is the assignment
model. This model is appropriate in problems which involve the
assignment of resources to tasks (e.g, assign n persons to n
different tasks or jobs).As the special structures of the
transportation model allows for solution procedures which are
more efficient than the simplex method, the structure of the
assignment model allows for solution methods more efficient than
the transportation method.
General Form and Assumptions
The general assignment problem involves the assignment of n
resources (origins) to n tasks (destinations).Typical examples of
assignment problems include the assignment of salespersons to
sales territories, airline crews to flights, ambulance units to calls
for service, referees and official to sports events, and lawyers
within a law firm to cases or clients. The objective in making
assignment can be one of minimization or maximization (e.g.,
minimization of total time required to complete n tasks or
112
## maximization of total profit from assigning salespersons to sales
territories).
The following assumptions are significant in formulating
assignment models.
Assumption 1
Each resource is assigned exclusively to one task.
Assumption 2
Each task is assigned exactly one resource.
Assumption 3
For purposes of solution, the number of resources available for
assignment must equal the number of tasks to be performed.
General Form
If
x ij=1
x ij =0
cij =
## if resource i is assigned to task j
if resource i is not assigned to task j
## n = number of resources and number of tasks.
The generalized assignment model is shown as follows.
113
Notice for this model that the variables are restricted to the two
values of 0 (non-assignment of the resource) or 1 (assignment of
the resource). This restriction on the values of the variables is
quite different from the other linear programming models we have
examined. Constraints (1) to (n) ensure that each resource is
assigned to one task only. Constraints (n + 1) to (n + n) ensure
that each task is assigned exactly to one resource.
According to assumption 3, the number of resources must equal
the number of tasks for purposes of solving the problem. This
condition might have to be artificially imposed for a given
problem.
Finally, notice that all right-hand-side constants, which are
equivalent to si and dj values in the transportation model, equal 1.
The supply of each resource is 1 unit and the demand for each
task is 1 unit.
114
x ij
=1
0
## if team i is assigned to tournament j
if team i is not assigned to tournament j
115
## Constraints (1) to (4) assure that each team of officials is
assigned to one tourney site only; constraints (5) to (8) assure
that each site is assigned exactly one team of officials.
Solution Methods
Assignment models can be solved using various procedures.
These include total enumeration of all
solutions, 0 1
programming methods, the simplex method, transportation
methods (stepping stone), and special-purpose algorithms.
These techniques are listed in an order reflecting increasing
efficiency.
One of the most popular methods is the Hungarian method.
The Hungarian Method
The Hungarian method is based on the concept of opportunity
costs. There are three steps in implementing the method.
116
## 1) An opportunity cost table is constructed from the table of
assignment costs.
2) It is determined whether an optimal assignment can be made.
3) If an optimal assignment cannot be made, the third step
involves a revision of the opportunity cost table.
Lets illustrate the algorithm with the following example.
Example(Court Scheduling)
A court administration is in the process of scheduling four court
dockets. Four judge are available to be assigned, one judge to
each docket. The court administration has information regarding
the type of cases on each of the dockets as well as data,
indicating the relative efficiently of each of the judges in
processing different types of court cases.
Estimated Days to Clear Docket
## Based upon this information, the court administrator has
compiled the data in the above table. The above table shows
117
## estimates of the number of court-days each judge would require
in order to completely process each court docket. The court
administrator would like to assign the four judges so as to
minimize the total number of court-days needed to process all
four dockets.
Solution:
Step 1:Finding opportunity cost table requires two steps.
1. Identify the least cost element in each row and subtract it
from all elements in the row. The resulting table is called
row-reduced table.
2. Identify the least cost element in each column and subtract it
from all elements in the column. The resulting table is called
opportunity cost table.
Row-Reduced Cost Table
118
## Step 2: Determine whether an optimal assignment can be
Draw the straight lines (horizontally and vertically) through
opportunity cost table in such a way as to minimize the
number of lines necessary to cover all zero entries.
If number of lines equals either the number of rows or number
of columns in the table, an optimal solution can be found.
Otherwise, opportunity cost table must be revisited.
Step 3: Revise the opportunity cost table.
Identify the smallest number in the table not covered by
straight lines.
Subtract this number from all the numbers not covered by
straight lines and add this number to all numbers lying at the
intersection of any two lines.
Revised Opportunity Cost Table
119
## Step 2: Determine whether an optimal assignment can be
As number of lines now of rows, optimal assignment is
possible.
How to make optimal assignment?
Select a row or column in which is only one zero and make
an assignment to that cell.
Cross out that row and column and repeat step 1.
Optimal Assignments
120
## Select a row/column in which there is only one zero.
Make an assignment to that cell.
There is only one zero in column 4, thus first
assignment is judge 2 to docket 4.
Since no further assignment is possible in row 2 or
column 4, we cross them off.
Again find the row or column in which there is only one zero
and repeat the steps above.
Continue until all the judges are assigned to respective
dockets.
Summary of the Hungarian Method
Step 1: Determine the opportunity cost table.
a-
## Determine the row-reduced cost table by subtracting
the least cost element in each row from all elements in
the same row.
b-
## Using the row-reduced cost table, identify the least cost
element in each column, and subtract from all elements
in that column.
121
## Step 2: Determine whether or not an optimal assignment can be
made. Draw the minimum number of straight lines
necessary to cover all zero elements in the opportunity
cost table. If the number of straight lines is less than
the number of rows (or columns) in the table, the
optimal assignment cannot be made. Go to step 3. If
the number of straight lines equals the number of rows
(columns), the optimal assignments can be identified.
Step 3: Revise the opportunity cost table. Identify the smallest
element in the opportunity cost table not covered by a straight
line.
a-
## Subtract this element from every element not covered
by a straight line.
b-
## Add this element to any element(s) found at the
intersection of two straight lines.
c-
Go to step 2.
The Hungarian method can be used when the objective
function is to be maximized. Two alternative approaches can be
used in this situation. The signs on the objective function
coefficients can be changed, and the objective function can be
minimized, or opportunity costs can be determined by subtracting
the largest element (e.g., profit) in a row or column rather than the
smallest element.
Lecture 21,22
## Chapter 15: Differentiation
This is the first of six chapters which examine the calculus and its
application to business, economics, and other areas of problem
solving. Two major areas of study within the calculus are
differential calculus and integral calculus. Differential calculus
122
## focuses on rates of change in analyzing a situation. Graphically,
differential calculus solves of following problem: Given a function
whose graph is a smooth curve and given a point in the domain of
the function, what is the slope of the line tangent to the curve at
this point.
Limits
Two concepts which are important in the theory of differential and
integral calculus are the limit of a function and continuity.
Limits of Functions
In the calculus there is often an interest in the limiting value of a
function as the independent variable approaches some specific
real number. This limiting value, when it exists, is called limit. The
notation
lim f ( x ) =L
x a
## is used to express the limiting value of a function. When
investigating a limit, one is asking whether f ( x) approaches a
specific value L as the value of x gets closer and closer to
a
.
Notation of Limits
The notation
x a f ( x )=L
lim
## represents the limit of
(left-hand limit).
Thenotation
f ( x)
as
approaches
x a f ( x )=L
lim
f (x)
as
approaches
from the
123
## right (right-hand limit).
If the value of the function approaches the same number L as
x
approaches a from either direction, then the limit is equal
to L .
Test For Existence Of Limit
If
x a f ( x)=L
lim
x a+ f (x)
lim
then
lim f ( x )=L
x a
## If the limiting values of f (x) are different when x approaches
a
from each direction, then the function does not approach a
limit as x approaches a .
Example
Determine whether the
lim x
x 2
exists or not?
## Solution: In order to determine the limit
lets construct a table of assumed value for x and corresponding
values for f(x).
The following table indicates these values.
124
Note that the value of x=2 has been approached from both the
left and the right. From either direction, f(x) is approaching the
same value, 8.
Note:For more examples, consult lectures and book.
Properties Of Limits And Continuity
1. If
f ( x)=c
2. If
f ( x)=x n
n
, where
n
lim x =a
lim f ( x ) =c
x a
x a
xa
## and c is any real number, then
c . f ( x )= c . lim f ( x )
x a
lim
xa
4. If
lim f ( x )
and
x a
lim g ( x )
x a
exist, then
[f ( x ) g ( x ) ]= lim f ( x ) lim g ( x )
xa
x a
lim
x a
5. If
lim f ( x )
x a
lim g ( x )
x a
exist, then
125
x a
lim f ( x )
6. If
x a
x a
and
xa
lim g ( x )
x a
exist, then
lim f ( x )
f ( x) xa
lim
=
x a g ( x )
lim g ( x )
x a
## Limits and Infinity
Frequently there is an interest in the behavior of a function as the
independent variable becomes large without limit (approaching
either positive or negative infinity).
Horizontal Asymptote
The line y = a is a horizontal asymptote of the graph of f if and
only if
lim f ( x )=a
x
## Note: See detailed discussion and examples in books.
Vertical Asymptote
The line x = a is a vertical asymptote of the graph of f if and only if
lim f ( x ) =
x a
Continuity
In an informal sense, a function is described as continuous if it
can be sketched without lifting your pen or pencil from the paper
(i.e., it has no gaps, no jumps, and no breaks).A function that is
not continuous is termed as discontinuous.
126
Continuity at a Point
A function f is said to be continuous at x = a if
12-
## the function is defined at x = a, and
lim f ( x ) =f ( a)
x a
Examples
Determine that
Solution:Since
f ( x)=x
is continuous at
x=2
3
lim x =f ( 2 )=8 .
x 2
Thus
is continuous at
x=2.
## Average Rate of Change and the Slope
The slope of a straight line can be determined by the two-point
formula
y y 2 y1
m
x x 2 x1
## The slope provides an exact measure of the rate of change in the
value of y with respect to a change in the value of x.
With nonlinear functions the rate of change in the value of y with
respect to a change in x is not constant.One way of describing
nonlinear functions is by the average rate of change over some
interval.
In moving from point A to point B, the change in the value of x is
(x + x) x, or x. The associated change in the value of y is
y = f(x + x) f(x)
The ratio of these changes is.
y f ( x+ x )f ( x )
=
x
x
127
## The above equation is sometimes referred to as the difference
quotient.
What Does The Difference Quotient Represent
Given any two points on a function f having coordinates
and
[x , f ( x )]
## , the difference quotient represents.
1. The average rate of change in the value of y with respect to
the change in x while moving from
[x , f (x )][( x + x ), f (x+ x)]
## 2. The slope of the secant line connecting the two points.
Examples
Find the general expression for the difference quotient of the
2
function y=f ( x)=x .Find the slope of the line connecting (-2, 4)
and (3, 9) using the two-point formula.Find the slope in part (b)
using the expression for the difference quotient found in part (a).
Solution(In Lectures)
The Derivative
DEFINITION:
Given a function of the form y = f(x), the derivative of the function
is defined as
lim f ( x+ x )f ( x )
dy
y x0
= lim
=
dx x 0 x
x
## If this limit exists.
i. The above equation is the general expression for the
derivative of the function f.
ii. The derivative represents the instantaneous rate of change
in the dependent variable given a change in the independent
128
iii.
iv.
## variable. The notation dy /dx is used to represent the
instantaneous rate of change in y with respect to a change in
x
. This notation is distinguished from y / x which
represents the average rate of change.
The derivative is a general expression for the slope of the
graph of f at any point x in the domain.
If the limit in the above figure does not exist, the derivative
does not exist.
Step 1
given function.
Step 2
## Find the limit of the difference quotient as x
0.
Example
2
Find the derivative of f ( x )=x .
Solution:(In Lectures)
## Using And Interpreting The Derivative
To determine the instantaneous rate of change (or equivalently,
the slope) at any point on the graph of a function f, substitute the
dy
.
dx
## value of the independent variable into the expression for
The derivative, evaluated at
x=c
, can be denoted by
dy
c,
dx x
at x=c .
Differentiation
evaluated
129
## The process of finding the derivative is called the differentiation.
A set of rules of differentiation exists for finding the derivative of
dy
dx
## many common functions.An alternate to
f'
notation is to let
## represent the derivative of the function
at
Rules of Differentiation
Rule 1: Constant Function
If f (x)=c , where c is any constant then
f ( x)=0.
## Rule 2: Power Rule
n
If f (x)=x , where n is a real number
f (x)=n xn1
Algebra Flashback
m
Recall that
1
n
=xn , x m=x n .
n
x
## Rule 3: Constant Times a Function
If f ( x)=c . g ( x) , where c is a constant and
function then
is a differentiable
## Rule 4: Sum or Difference of Functions
If f (x)=u ( x) v ( x), where u and v are differentiable,
f ' ( x)=u ' ( x) v ' ( x)
130
## Rule 5: Product Rule
If f ( x)=u ( x). v (x) , where
and
are differentiable,
If
f (x)=
u(x)
,
v (x)
## where u and v are differentiable and
f ' ( x)
Rule 7: If
f ( x )=[ u ( x ) ]
v ( x ) 0
[v( x)]2
then
## RULE 8: Base-e Exponential Functions
u ( x)
If f ( x )=e , where u is differentiable, then
'
'
u(x)
f (x )=u ( x ) e
## Rule 9: Natural Logarithm Functions
If f ( x )=lnu (x) , where u is differentiable, then
f ' (x )=
u' (x)
u( x)
## Rule 10: Chain Rule
If y=f (u) , is a differentiable function and
u=g ( x)
is a
, then
131
## differentiable function, then
dy dy du
=
dx du dx
Example:
An object is dropped from a cliff which is, 1,296 feet above the
ground. The height of the object is described as a function of
time. The function is
h=f (t )=16 t 2 +1,296
## where h equals the height in feet and t equals the time
measured in seconds from the time the object is dropped.
a) How far will the object drop in 2 seconds?
b) What is the instantaneous velocity of the object at t = 2?
c) What is the velocity of the object at the instant it hits the
ground?
Solution:
a) h=f ( 2 )f ( 0 )
[ 16 ( 2 ) 2+1,296 ] [ 16 ( 0 ) 2+1,296 ]
(64 +1,296 ) 1,296=64
## Thus, the object drops 64 feet during the first 2 seconds.
'
b) Since f ( x )=32t , the object will have a velocity equal to
f ' ( x )=32 ( 2 )
64 feet /second
## at t=2. the minus sign indicates the direction of velocity
(down)
c) In order to determine the velocity of the object when it hits
the ground, we must know when it will hit the ground. The
object will hit the ground when h=0 , or when
132
16 t 2 +1,296=0
and
t= 9 .
## Since a negative root is meaningless, we can conclude that the
object will hit the ground after 9 seconds. The velocity at this
time will be
'
f ( 9 )=32 ( 9 )=288
feet/second.
## Exponential Growth Processes
The exponential growth processes were introduced in Chap. 7.
The generalized exponential growth function was presented as,
kt
V =f (t)=V 0 e
## We indicated that these processes are characterized by a
constant percentage rate of growth.
To verify this, lets find the derivative.
'
kt
f ( t ) =V 0 ( k ) e
## which can be written as
'
f ( t ) =k V 0 e kt
## The derivative represents the instantaneous rate of change in
the value V with respect to a change in t .
The percentage rate of change would be found by the ratio.
Instantaneous rate of change f ' (t)
=
Value of the function
f (t)
133
kt
f '(t ) k V 0 e
=
f (t )
V o e kt
## This confirms that for an exponential growth function of the form
of Eq. (7.9), k represents the percentage rate of growth. Given
that is a constant, the percentage rate of growth is the same for
all values of t .
The Second Derivative:
The second derivative f of a function is the derivative of the first
2
d y
d x2
or
f (x) .
## The Second Derivative:
As the first derivative is a measure of the instantaneous rate of
change in the value of y with respect to a change in x , the
second derivative is a measure of the instantaneous rate of
change in the value of the first derivative with respect to the
change in x .Described differently, the second derivative is a
measure of the instantaneous rate of change in the slope with
respect to a change in x .
Example
Consider the function
f ( x )=x 2 .
## derivatives of this function are
Definition: nth-order Derivative
## The n-th order derivative of , denoted by
differentiating the derivative of order
n 1.
f n( x)
is found by
That is, at ,
134
f ( n) ( x ) =
d (n1)
[f
(x )]
dx
Lecture 23,24,25
## Chapter 16:Optimization Methodology
(The First Derivative)
Increasing Function
The function f is said to be an increasing function on an
interval I if for any x1 and x2 within the interval, x1<x2 implies
that f(x1) <f(x2).
Increasing functions can be identified by slope conditions.If the
first derivative off is positive throughout an interval, then the slope
is positive and f is an increasing function on the interval.Which
mean that at any point within the interval, a slight increase in the
value of x will be accompanied by an increase in the value of f(x).
Decreasing Function
The function f is said to be a decreasing function on an interval I if
for any x1 and x2, within the interval, x1<x2 implies that f(x1) >f(x2).
As with increasing functions, decreasing functions can be
identified by tangent slope conditions.
If the first derivative of f is negative throughout an interval, then
the slope is negative and f is a decreasing function on the interval.
This means that, at any point within the interval a slight increase
in the value of x will be accompanied by a decrease in the value
of f(x).
135
Note
If a function is increasing (decreasing) on an interval, the function
is increasing (decreasing) at every point within the interval.
Example
2
Given f ( x )=5 x 2 x+3 , determine the intervals over which f can be
described as (a) an increasing function, (b) a decreasing
function, and (c) neither increasing nor decreasing.
Solution:(In Lectures)
Note:
a) If f(x) is negative on an interval Iof f, the first derivative is
decreasing on I.
b) Graphically, the slope is decreasing in value, on the interval.
c) If f(x) is positive on an interval Iof f, the first derivative is
increasing on I.
d) Graphically, the slope is increasing in value, on the interval.
Concavity
The graph of a function f is concave up (down) on an interval if f
increases (decreases) on the entire interval.
Inflection Point
A point at which the concavity changes is called an inflection
point.
Relationships Between The Second Derivative And Concavity
IIf f(x) < 0 on an interval a x b, the graph of f is concave
down over that interval. For any point x = c within the
interval, f is said to be concave down at [c, f(c)].
II- If f(x) > 0 on any interval a x b, the graph of f is concave
up over that interval. For any point x = c within the interval, f
is said to be concave up at [c, f(c)].
136
III-
## If f(x) = 0 at any point x = c in the domain of f, no conclusion
can be drawn about the concavity at [c, f(c)]
## Note:Please find examples in the lectures and book.
Locating Inflection Points
I-
II-
## If f(x) change sign when passing through
an inflection point at x = a.
x = a, there is
Relative Extrema
Relative Maximum
f
If
is defined on an interval
(b , c )
which contains
x
all
## within the interval
(b , c )
close to
x=a
x=a
if
x=a
f (a) f (x)
is
for
Relative Minimum
f
If
is defined on an interval
(b , c )
which contains
all
(b , c )
close to
x=a
x=a
if
x=a
f ( a ) f ( x)
is
for
## Both definitions focus upon the value of () within an interval.
A relative maximum refers to a point where the value of f (x) is
greater than the values for any points which are nearby.
A relative minimum refers to a point where the value of f (x) is
lower than the values for any points which are nearby.
If we use these definitions and examine the above figure, f
137
## has relative maxima at x = aand x = c.
Similarly, f has relative minima at x = band x = d.
Collectively, relative maxima and minima are called relative
extrema.
Absolute Maximum
A function f is said to reach an absolute maximum at x = a if f (a)
>f (x) for any other x in the domain of f.
Absolute Minimum
A function f is said to reach an absolute minimum at x = a if f (a)
<f (x) for any other x in the domain of f.
Critical Points
Necessary Conditions For Relative Maxima (Minima)
Give the function f, necessary conditions for the existence of a
relative maximum or minimum at x = a (a contained in the domain
of f.
1.
f ' ( a )=0
2.
f ' (a )
, or
is undefined
## Points which satisfy either of the conditions in this definition
are candidates for relative maxima (minima).
Such points are often referred to as critical points.
Any critical point where f(x) =0 will be a relative maximum, a
relative minimum, or an inflection point.
Points which satisfy condition 1 are those on the graph of f
where the slope equals 0.
Points satisfying condition 2 are exemplified by discontinuities
on f or points where f (x) cannot be evaluated.
138
## Values of x in the domain of f which satisfy either condition 1
or condition 2 are called critical values.
These are denoted with an asterisk (x*) in order to distinguish
them from other values of x.
Given a critical value for f, the corresponding critical point is
(x*, f (x*)].
The First Derivative Test
After the locations of critical points are identified, the firstderivative test requires an examination of slope conditions to the
left and right of the critical point.
The First-Derivative Test
1. Locate all critical values x*.
2. For any critical value x*, determine the value of f (x) to the
left (xl) and right (xr) of x*.
a. If f (xl) > 0 and f (xr) < 0, there is a relative maximum
for f at [x*,f(x*)].
b. If f(xl) < 0 and f (xr) > 0, there is a relative minimum
for f at [x*,f(x*)].
c. If f(x) has the same sign of both x l and xr, an inflection
point exists at [x*,f (x*)].
Example
Determine the location(s) of any critical points on the graph of
f ( x)=2 x 2 12 x 10 , and determine their nature.
Solution:(In Lectures)
The Second-Derivative Test
For critical points, where f(x)=0, the most expedient test in the
second-derivative test.
The Second-Derivative Test
1. Find all critical values x*, such that f(x) = 0.
139
## 2. For any critical value x*, determine the value of f(x*).
(a) If f(x*) > 0, the function is concave up at x* and there
is a relative minimum for f at [x*, f(x*)]
(b) If f(x*) < 0, the function is concave down at x* and
there is a relative maximum for f at [x*, f(x*)].
(c) If f(x*) = 0, no conclusions can be drawn about the
concavity at x* nor the nature of the critical point.
Another test such as the first-derivative test is
necessary.
Note:See detailed examples in Lectures.
When the Second-Derivative Test Fails
If f(x*)=0, the second derivative does not allow for any
conclusion about the behavior of f at x*.
Curve Sketching
Sketching functions is facilitated with the information we have
acquired in this chapter.One can get a feeling for the general
shape of the graph of a function without determining and plotting
a large number of ordered pairs.
Key Date Points
In determining the general shapes of the graph of a function, the
following attributes are the most significant.
1. Relative maxima and minima.
2. Inflection points.
3. x and y intercepts.
4. Ultimate direction
Note:See detailed example in lectures.
Restricted-domain Considerations
In this section we will examine procedures for identifying absolute
maxima and minima when the domain of a function is restricted.
When the Domain is Restricted
140
## Very often in applied problems the domain is restricted. For
example, if profit P is stated as a function of the number of units
produced x, it is likely that x will be restricted to values such that
0 . In this case x is restricted to nonnegative values (there is
no production of negative quantities) which are less than or equal
to some upper limit . The value of may reflect production
capacity, as defined by limited labor, limited raw materials, or by
the physical capacity of the plant itself.
In searching for the absolute maximum or absolute minimum of a
function, consideration must be given not only to the relative
maxima and minima of the function but also to the endpoints of
the domain of the function.
Identifying Absolute Maximum and Minimum Points
1. Locate all critical points [x*, f (x*)] which lie within the
domain of the function and exclude from consideration any
critical values x* which lie outside the domain.
2. Compute the values of f(x) at the two endpoints of the
domain [f(x1) and f(xu).
3. Compare the values of f(x*) for all relevant critical points
with (x1) and (xu). The absolute maximum in the largest of
these values and the absolute minimum is the smallest of
these values.
Note: See detailed example in lectures.
Lecture 26,27
## Chapter 17:Optimization Applications
Revenue Applications
141
## The following applications focus on revenue maximization. The
money which flows into an organization from, either selling
products or providing services, isoften referred to as revenue.The
most fundamental way of computing total revenue from selling a
product (or service) is
Total revenue = (price per unit)(quantity sold)
An assumption in this relationship is that the selling price is the
same for all units sold.
Example
The demand for the product of a firm varies with the price that the
firm charges for the product. The firm estimates that annual total
revenue R (stated in \$1,000s) is a function of the price p (stated in
dollars). Specifically
R=f ( p )=50 p 2+500 p
(a)
(b)
## Determine the price which should be charged in order to
maximize total revenue.
What is the maximum value of annual total revenue?
Solution:(In Lectures)
Cost Applications
A common problem in organizations is determining how much of a
needed item should be kept on hand. For retailers, the problem
may relate to how many units of each product should be kept in
stock. For producers, the problem may involve how much of each
raw material should be kept available.This problem is identified
with an area called inventory control, or inventory management.
Concerning the question of how much inventory to keep on
hand, there may be costs associated with having too little or too
much inventory on hand.
142
## Example (Inventory Management)
A retailer of motorized bicycles has examined cost data and has
determined a cost function which expresses the annual cost of
purchasing, owning, and maintaining inventory as a function of
the size (number of units) of each other it places for the bicycles.
The cost function is
C=f ( q )=
4,860
+15 q+750,000
q
## where C equals annual inventory cost, stated in dollars, and q
equals the number of cycles ordered each time the retailer
replenishes the supply.
Determine the order size which minimizes annual inventory cost.
What is minimum annual inventory cost expected to equal?
Solution: (In Lectures)
Profit Applications
Example(Solar Energy)
A manufacturer has developed a new design for solar collection
panels.Marketing studies have indicated that annual demand for
the panels will depend on the price charged.The demand function
for the panels has been estimated as
q=1000,000200 p ,
## where q equals the number of units demanded each year and
p
equals the price in dollars.
Engineering studies indicate that the total cost of producing q
panels is estimated well by the function
C=150,000+100 q+0.003 q
P=f (q)
143
## profit P as a function of the number of units
produced and sold.
which are
## Solution: We have been asked to develop a function which
states profit P as a function of q . We must construct the profit
function. The total cost function the above figure is stated in terms
of q . However, we need to formulate a total revenue function
stated in terms of q . The basic structure for computing total
revenue is
R= pq .
As
q=10000220 p ,
so
p=500.005 q .
Thus
R=500 q0.005 q .
P=RC
2
## Example (Restricted Domain)
Assume in the last example that the manufacturers annual
production capacity is 20,000 units. Re-solve last example with
Solution: (In Lectures)
Marginal Approach to Profit Maximization
An alternative approach to finding the profit maximization point
involves marginal analysis.Popular among economists, marginal
analysis examines incremental effects on profitability.Given that a
firm is producing a certain number of units each year, marginal
analysis would be concerned with the effect on profit if one
additional unit is produced and sold.To utilize the marginal
approach to profit maximization, the following conditions must
hold.
144
## Requirements for Using the Marginal Approach
I
It must be possible to identify the total revenue function and
the total cost function, separately.
II
The revenue and cost functions must be stated in terms of
the level of output or number of units produced and sold.
Marginal Revenue
One of the two important concepts in marginal analysis is
marginal revenue.Marginal revenue is the additional revenue
derived from selling one more unit of a product or service.For a
total revenue function R(q), the derivative R(q) represents the
instantaneous rate of change in total revenue given a change in
the number of units sold. R also represents a general expression
for the slope of the graph of the total revenue function.
For purposes of marginal analysis, the derivative is used to
represent the marginal revenue, or
MR=R (q)
Marginal Cost
The other important concept in marginal analysis is marginal cost.
Marginal cost is the additional cost incurred as a result of
producing and selling one more unit of a product or service.For a
total cost function C(q), the derivative C(q) represents the
instantaneous rate of change in total cost given a change in the
number of units produced. C(q) also represents a general
expression for the slope of the graph of the total cost function.
For purposes of marginal analysis, the derivative is used to
represent the marginal cost, or
MC=C (q)
Marginal Profit
145
## As indicated earlier, marginal profit analysis is concerned with the
effect of profit if one additional unit of a product is processes and
sold. As long as additional revenue brought in by the next unit
exceeds the cost of producing and selling that unit, there is a net
profit from producing and selling that unit and total profit
increases. If however the additional revenue from selling the next
unit is exceeded by the cost of producing and selling the
additional unit, there is a net loss from that next unit and total
profit decreases.
1. If
2. If
MR> MC ,
MR< MC ,
## produce the next unit
do not produce the next unit.
A large multinational conglomerate is interested in purchasing
some prime boardwalk real estate at a major ocean resort.The
conglomerate is interested in acquiring a rectangular lot which is
located on the boardwalk. The only restriction is that the lot have
an area of 100,000 square feet.Figure 17.11(book) presents a
sketch of the layout with x equaling the boardwalk frontage for the
lot and y equaling the depth of the lot (both measured in feet).
The seller of the property is pricing the lots at \$5,000 per foot of
frontage along the boardwalk and \$2,000 per foot of depth away
from the boardwalk.The conglomerate is interested in determining
the dimensions of the lot which will minimize the total purchase
cost.
Solution: Total purchase cost for a lot having dimensions of
feet by y feet is
C=5,000 x +2,000 y
where
is cost in dollars.
146
## The problem is to determine the values of x and y which
minimize C .However, C is stated as a function of two
variables, and we are unable, as yet, to handle functions which
have two independent variables.
Since the conglomerate has specified that the area of the lot must
equal 100,000 square feet, a relationship which must exist
between x and y is
xy=100,000
## Given this relationship, we can solve for either variable in terms of
the other.
For instance
y=
100,000
x
.
We can substitute the right side of this equation into the cost
function wherever the variable y appears, or
C=f ( x )=5,000 x+ 2,000
Now
As
100,000
200,000,000
=5,000 x+
x
x
, put
C' ( x ) =0,
to get
x= 200.
## Thus the minimum value of C , occur at x=200, which gives
y=500.
So ( x , y )=( 200,500 ) , i.e. if the lot is 200 feet by 500 feet,
total cost will be minimized at a value of
C=5000 ( 200 )+ 2000 (500 )=\$ 2000,000.
## Emergency Response: Location Model
Three resort cities had agreed jointly to build and support an
147
## emergency response facility which would house rescue trucks
and trained paramedics.The key question is to deal with the
location of the facility. The criterion selected was to choose the
location so as to minimize S. S is the sum of the products of the
summer populations of each town and the square of the distance
between the town and the facility.
## The criterion function to be minimized was determined to be
S=f ( x ) =450 x 219,600 x+241,600,
## where x is the location of the facility relative to the zero point in
Fig17.12(Lectures/book).
Given the criterion function, the first derivative is
'
f ( x )=900 x19,600
## If f is set equal to 0, we get the critical value
the nature of the critical point, we find
x=21.77
. Checking
## f (x)=900 for x>0
In particular, f (21.77)=900>0 .
Thus, f is minimized when x = 21.77.The criterion S is minimized
at x = 21.77.
Example (Welfare Management)
A newly created state welfare agency is attempting to determine
the number of analysts to hire to process welfare applications.
Efficiency experts estimate that the average cost C of processing
an application is a function of the number of analysts x.
Specially, the cost function is
C=f ( x )=0.001 x 25x +60
148
## minimize the average cost per application.
Solution: The derivative of f is
'
f ( x )=0.002 x5
0.002 x
We put
1
x
5
x
f ' ( x )=0,
to get
The value of
f ( x)
x= 50.
## at the critical point is
f ( 50 ) =0.001 ( 50 ) 2550+60
0.001 ( 2,500 )5 ( 3.912 )+ 60
2.519.56+ 60=\$ 42.94
## We can check the nature of the critical point, as
f ' ' (50 )=0.004> 0,
We conclude
is minimized at
x=50.
## Example (Elasticity of Demand)
An important concept in economics and price theory is the price
elasticity of demand, or more simply, the elasticity of demand.
Given the demand function for a product q=f ( p) and a particular
point
( p , q)
## Percentange changequantitiy demanded
Percentage change price
## This ratio is a measure of the relative response of demand to
changes in price. The above equation can be expressed
symbolically as
149
q
q
p
p
## We use the Greek letter (eta) to denote the point elasticity of
demand. The point elasticity is the limit of the above ratio as
0, at a point (p, q).
Note:Detailed example is in lectures.
Economists classify point elasticity values into three categories.
Case 1 (||> 1) : The percentage change in demand is greater than
the percentage change in price
(e.g., a 1 percent change in price results in a greater than 1
percent change in demand).In these regions of a demand
function, demand is said to be elastic.
Case 2 (||< 1) :
The percentage change in demand is less than the percentage
change in price.In these regions of a demand function, demand is
said to be inelastic.
Case 3 (||=1) :
The percentage change in demand is equals the percentage
change in price.In these regions of a demand function, demand is
said to be unit elastic.
150
Lecture 28
## Chapter 18- Integral Calculus:an Introduction
Antiderivatives
Given a function f , we are acquainted with how to find the
derivative f .There may be occasions in which we are given the
derivative f and wish to determine the original function f .
Since the process of finding the original function is the reverse of
differentiation, f is said to be an anti-derivative of f .
Example
'
Consider the derivative f ( x )=4
By using a trial-error approach, it is not very difficult to conclude
that the function
f ( x )=4 x
151
## has a derivative of the form above. Another function having the
same derivative is
f ( x )=4 x+ 1
## In fact, any function having the form
f ( x )=4 x+ c
Example
Find the antiderivative of
f ' ( x )=2 x 5.
## Solution:Using a trial-and-error approach and working with each
term separately, you should conclude that the antiderivative is
f ( x )=x 25 x+ c
## Revenue and Cost Functions
If we have an expression for either marginal revenue or marginal
cost, the respective antiderivatives will be the total revenue and
total cost functions.
Example (Marginal Cost)
The function describing the marginal cost of producing a product
is
MC=x +100
## where x equals the number of units produced. It is also known
that total cost equals \$ 40,000 when x=100 . Determine the total
cost function.
Solution:
152
## To determine the total cost function, we must first find the
antiderivative of the marginal cost function,
C ( x )=
x2
=100 x+ c
2
## Given that C(100)=40,000 , we can solve for the value of
happens to represent the fixed cost.
, which
## Example (Marginal Revenue)
The marginal revenue function for a companys product is
M R=50,000x
## Where x equals the number of units produced and sold. If total
revenue equals 0 when no units are sold, determine the total
revenue function for the product.
Solution: (In Lectures)
Rules Of Integration
Integration
The process of finding antiderivatives is more frequently called
integration. The family of functions obtained through this process
is called the indefinite integral.
The notation
f ( x ) dx
is often used to indicate the indefinite integral of the function f.
The symbol is the integral sign; f is the integrand, and dx, as we
will deal with it, indicates the variable with respect to which the
integration process is performed.
Two verbal descriptions of the process are: integrate the function
f with respect to the variable x or find the indefinite integral of f
with respect to x.
153
Indefinite Integral
Given that f is a continuous function,
f ( x ) dx=F ( x ) +c
if
F' ( x ) =f ( x )
## In this definition c is termed the constant of integration.
Rule 1: Constant Functions
k dx =kx +c
where k is any constant.
Rule 2: Power Rule
xn +1
x dx= n+1 + c , n 1
n
For example
x3
x dx = + c
3
2
Rule 3
kf ( x ) dx=k f ( x ) dx
where k is any constant. For example
2 xdx=2 xdx
Rule 4
If f ( x ) dx g ( x ) dx exist, then
[f ( x ) g ( x ) ]dx= f ( x ) dx g ( x ) dx
154
## Rule 5: Power Rule, Exception
x1 dx =1n x +c
Rule 6
e x dx=e x+ c
Rule 7
n +1
[ f ( x )]
[ f ( x ) ] f ( x ) dx= n+1 +c n 1
n '
For example,
3
( 3 x 2+ 2 ) ( 6 x ) dx=
( 3 x2 +2 )
4
+c
Rule 8
## f ' ( x) e f (x) dx=e f (x) + c
2
2 x e x dx=e x +c
For example,
Rule 9
f '( x)
f ( x ) dx=ln f ( x )+ c
For example
x
( 3 x610
) dx=ln (3 x 10 )+c .
2
Differential Equations
A differential equation is an equation which involves derivatives
and/or differentials. If a differential equation involves derivatives of
a function of one independent variable, it is called an ordinary
differential equation.
The following equation is an example, where the independent
variable is x .
155
dy
=5 x2
dx
## Differential equations are also classified by their order, which is
the order of the highest-order derivative appearing in the
equation.
Another way of classifying differential equations is by their
degree. The degree is the power of the highest-order derivative in
the differential equation. For example,
2
dy
10= y
dx
( )
## is an ordinary differential equation of first order and second
degree.
Solutions of Ordinary Differential Equations
Solution to differential equations can be classified into general
solutions and particular solutions.
A general solution is one which contains arbitrary constants of
integration.
A particular solution is one which is obtained from the general
solution.
For particular solutions, specific values are assigned to the
constants of integration based on initial conditions or boundary
conditions.
Consider the differential equation
dy
=3 x 22 x +5
dx
## The general solution to this differential equation is found by
integrating the equation,
156
3
y=f ( x ) =
3x 2x
+5 x+ c
3
2
x x +5 x +c
## If we are given the initial condition that f ( 0 )=15 , the particular
solution is derived by substituting these values into the general
solution and solving for c .
15=0302 +5 ( 0 ) +C
15=C
## The particular solution to the differential equation is
f ( x )=x 3x 2 +5 x+15
Example
Given the differential equation
'
## and the boundary conditions f ( 2 )=4 and f ( 0 )=10 , determine the
general solution and particular solution.
Solution:(In Lectures)
Application of Differential Equations
In Chaps. 7 and 15 exponential growth and exponential decay
functions were discussed.
Exponential growth functions have the general form
V =V 0 e
kt
## where V0 equals the value of the function when t = 0 and
positive constant.
is a
157
dV
kt
=k V 0 e
dt
or
dV
=kV
dt
## Example (Species Growth)
The population of a rare species of fish is believed to be growing
exponentially.
When first identified and classified, the population was estimated
at 50,000.
Five years later the population was estimated to equal 75,000.
If P equals the population of this species at time t , where t is
measured in years, the population growth occurs at a rate
described by the differential equation
dP
=kP=k P0 ekt
dt
P=50,000
At
t=0,
at
P0
t=0 .
we get
P0=50,000.
## Now we will find the value of
75000=50000 e 5 k
## which gives k =0.0811 .
Thus the particular function describing the growth of thepopulation
is
P=50,000 e 0.0811t
158
Lecture 29,30
## Chapter 19- Integral Calculus:Applications
The Definite Integral
If f is a bounded function on the interval (a, b), then
b
f ( x) dx
a
## is called definite integral of f .
The values a and b which appear, respectively, below and
above the integral sign are called the limits of integration. The
lower limit of integration is a , and the upper limit of integration is
b
.
Thenotation
b
f ( x) dx
a
## can be described as the definite integer of between a lower limit
x=a
and an upper limit x=b , or more simply the integral of
f
between a and b .
Evaluating Definite Integrals
The evaluation of definite integrals is facilitated by the following
important theorem.
Fundamental Theorem of Integral Calculus
If a function f is continuous over an interval and F is any antix=a
derivative of f , then for any points
and x=b on the
interval, where a b ,
159
b
f (x)dx=F ( b )F (b)
a
## According to the fundamental theorem of integral calculus, the
definite integral can be evaluated by determining the indefinite
integral F(x )+c and Computing F( b) F (a) , sometimes denoted
b
## by [ F( x )] a . As you will see in the following example, there is no
need to include the constant of integration in evaluating definite
integrals.
Example
3
3
Evaluate 0 x dx
Solution:
F ( x )= x 2 dx
x3
+c
3
## Applying limits, we obtain
3
3
3
3
x dx= x3 + c = 33 + c 03 +c =9.
0
0
2
( )
## This implies that the constant of integration always drop out.
Properties of Definite Integrals
Property-1
If
## is defined and continuous on the interval
(a , b)
, then
160
a
f ( x ) dx= f ( x ) dx
b
For example
1
4 x 3 dx= x 4|2=116=15
2
And
2
4 x 3 dx= x 4|1
=161=15
Property-2
a
f ( x ) dx=0
a
For example
10
e x dx=e10e 10=0
10
Property-3
If f is continuous on the interval
b
f ( x ) dx + f ( x ) dx= f ( x ) dx
a
For example
4
3 x2 dx + 3 x 2 dx
2
3 x dx=
2
(a , c )
and
a<b <c
, then
161
4
## x3|0= x 3|0 + x3|2
4 3=23 + 4323
L. H . S .=R . H . S .
Property-4
b
cf (x) dx=c f (x ) dx
a
where
is constant.
Property-5
b
If
f ( x)dx
a
exists and
## g( x )dx exists, then
a
f ( x)
b
[ g( x )]dx= f ( x ) dx g (x) dx
a
## Define Integrals and Areas
Areas between a Function and x-axis
Definite integrals can be used to compute the area between the
curve representing a function and the x axis. Several different
situations may occur. The treatment of each varies and is now
discussed.
Case 1: (f ( x)>0)
When the value of a continuous function f is positive over the
interval x b , that is, the graph of f lies above the x -axis,
the area which is bounded by f , the x -axis, x=a , and
162
x=b
is determined by
f (x)dx .
a
Example
Determine the area beneath
between
x=1
and
x=3
f ( x)=x 2
## and above the
-axis
Solution:
3
x3
33 1 271 26
A= x dx=
= =
=
3 1 3 3
3
3
1
2
Case 2: (f ( x)<0)
When the value of a continuous function f is negative over the
interval a x b , that is, the graph of f lies below the x -axis,
the area which is bounded by f , the x -axis, x=a , and x=b
is determined by
b
f (x)dx .
a
## However, the definite integral evaluates the area as negative
when it lies below the x axis, because area is absolute (or
positive), the area will be computed as
b
f ( x ) dx .
a
## Case 3: (f ( x)<0f ( x)>0)
When the value of a continuous function f is positive over part
of the interval a x b and negative over the remainder of the
interval, part of the area between f and the x -axis is above
b
the
-axis then
f (x) dx
a
163
## calculates the net area.
That is, area above the x -axis is evaluated as positive, and
those below are evaluated as negative. The two are combined
algebraically to yield the net value.
Finding Areas between Curves
The following examples illustrate procedures for determining
areas between curves.
Example
Find the area between the curves
between
f ( x )=x 2
and
g ( x ) =2 x 2+ 27
x=0, x=3.
## Solution: The net area is
3
A= g ( x) dx f (x) dx
0
## Area between Two Curves
If the function y=f ( x) lies above the function y=g( x ) over the
interval a x b , the area between the two functions over the
interval is
b
[f ( x ) g (x)]dx
a
Example
Consider the functions f (x)=x and g( x)=x . Suppose we want
to determine the area between functions over the interval 0 x 4 ,
given in the figure.
Solution: Using the above property
164
4
0
## Applications of Integral Calculus
Revenue: The total revenue function can be determined by
integrating the marginal revenue function. As a simple extension
of this concept, assume that the price of a product is constant at a
value of \$10 per unit, or the marginal revenue function is
MR=f ( x )
10
## where x equals the number of units sold. Total revenue from
selling x units can be determined by integrating the marginal
revenue function between 0 and x. For example, the total revenue
from selling 1,500 units can be computed as
1500
1500
10 dx=10 x|0
## This is a rather elaborate procedure for calculating total revenue
since we simply could have multiplied price by quantity sold to
determine the same result. However, it does illustrate how the
area beneath the marginal revenue function (see figure in
lectures) can be interpreted as total revenue or incremental
revenue. The additional revenue associated with increasing sales
from 1,500 to 1,800 units can be computed as
1800
1800
1500
## Example (Maintenance Expenditures)
An automobile manufacturer estimates that the annual rate of
expenditure r(t) for maintenance on one of its models is
represented by the function
r ( t ) =100+10 t 2
165
## where t is the age of the automobile stated in years and r(t) is
measured in dollars per year.
This function suggests that when the car is 1 year old,
maintenance expenses are being incurred at a rate of.
r ( 1 )=100+10 ( 1 )2=110 per year
## When the car is 3 year old, , maintenance expenses are being
incurred at a rate of.
r ( 3 )=100+10 ( 3 )2=190 per year
## As would be anticipated, the older the automobile, the more
maintenance is required. The expected maintenance cost during
the automobiles first 5 years are computed as
5
3
10 ( 5 )
10 t 3
(
)
(100+10 t )=100 t+ 3 =100 5 3 =916.67
0
0
2
Example (Fund-Raising)
A state civic organization is conducting its annual fund-raising
campaign for its summer camp program for the disadvantaged.
Campaign expenditures will be incurred at a rate of \$10,000 per
day. From past experience it is known that contributions will be
high during the early stages of the campaign and will tend to fall
off as the campaign continues. The function describing the rate at
which contributions are received is
c (t )=100 t+ 20000
where t represents the day of the campaign, and c(t) equals the
rate of which contribution are received, measured in dollars per
day. The organization wishes to maximize the net proceeds from
the campaign. Determine how long the campaign should be
conducted in order to maximize the net profit.
a) What are total campaign expenditures expected to equal?
b) What are total contributions expected to equal?
c) What are the net proceeds (total contributions less total
166
## expenditures) expected to equal?
Solution:
(a)
The function which describes the rate at which
expenditures e (t) are incurred is
e ( t )=10000
## As long as the rate at which contributions are made exceeds
the rate of expenditures for the campaign, net proceeds are
positive. Net proceeds will be positive up until the time when
the graphs of the two functions intersect. Beyond this point,
the rate of expenditure exceeds the rate of contribution. That
is, contributions would be coming in at a rate of less than
\$10,000 per day.
The graphs of the two functions intersect when
c (t )=e(t )
2
100 t + 20000=10000
This implies
t=10 d
ays.
(b)
Total campaign expenditures are represented by the
area under e(t) between t = 0 and t = 10. This could be
found by integrating e (t) between these limits or more
simply by multiplying
E= (10000 )( 10 days )=\$ 100,000
(c)
Total contributions during the 10 days are represented
by the area under c (t) between t=0 and t=10 , or
10
10
t3
c= (100 t + 20000)dt =100 +20000 t =\$ 166666.67
3
0
0
2
167
(d)
## Net proceeds are expected to equal
CE=166666.67100000=\$ 66666.67
## Example (Blood Bank Management)
A hospital blood bank is conducting a blood drive to replenish its
inventory of blood. The hospital estimates that blood will be
donated at a rate of d(t) pints per day, where
d ( t )=500 e0.4 t
and t equals the length of the blood drive in days. If the goal of the
blood drive is 1,000 pints, when will the hospital reach its goal?
Solution:In this problem the area between the graph of d and the
t axis represents total donations of blood in pints. Unlike previous
applications, the desired area is known; the unknown is the upper
limit of integration. The hospital will reach its goal when
t
0
This implies
t
500 0.4 t
e
=1000
0.4
0
500 0.4 t
[ e e0 ]=1000
0.4
500 0.4t
[ e 1 ]=1000
0.4
Thus, we obtain,
e0.4t =0.2
0.4 t =1.6094
168
t =4.0235
## Thus the hospital will reach its goal in approximately 4 days.
Example (Consumers Surplus)
Consumer surplus is the difference between the maximum price a
consumer is willing to pay and the actual price they do pay. If a
consumer would be willing to pay more than the current asking
price, then they are getting more benefit from the purchased
product than they spent to buy it.
One way of measuring the value or utility that a product holds for
a consumer is the price that he or she is willing to pay for it.
Economists contend that consumers actually receive bonus or
surplus value from the products they purchase according to the
way in which the marketplace operates.
The figure (in lectures) portrays the supply and demand functions
for a product. Equilibrium occurs when a price of \$10 is charged
and demand equals 100 units. If dollars are used to represent the
value of this product to consumers, accounting practices would
suggest that the total revenue (\$10 . 100units = \$1,000) is a
measure of the economic value of this product.
The area of rectangle ABCE represents this measure of value.
However, if you consider the nature of the demand function, there
would have been a demand for the product at prices higher than
\$10.
That is, there would have been consumers willing to pay almost
\$20 for the product. Additional consumers would have been
drawn into the market at prices between \$10 and \$20. If we
assume that the price these people would be willing to pay is a
measure of the utility the product holds for them, they actually
receive a bonus when the market price is \$10.
Economists would claim that a measure of the actual utility of the
169
## product is the area ABCDE.
When the market is in equilibrium, the extra utility received by
consumers, referred to as the consumers surplus, is represented
by the shaded area CDE. This area can be found as
p
20
p3 40 p2
( 240 p+ 400)dx=
+ 400 p
3
2
10
20
10
20 3
103
20 ( 20 2 )+ 400 ( 20 )
20 ( 102 ) + 400 (10 )
3
3
333.34
## Our accounting methods would value the utility of the product at
\$1,000. Economists would contend that the actual utility is
\$1,333.34, or that the consumers surplus equals \$333.34. This
measure of added, or bonus, utility applies particularly to those
consumers who would have been willing to pay more than \$10.
170
Lecture 31,32
## Chapter 20 Optimization: Functions of Several Variables
Partial Derivatives
The calculus of bivariate functions is very similar to that of singlevariable functions. In this section we will discuss derivatives of
bivariate functions and their interpretation.
Derivatives of Bivariate Functions
With single-variable functions, the derivative represents the
instantaneous rate of change in the dependent variable with
respect to a change in the independent variable. For bivariate
functions, there are two partial derivatives. These derivatives
represent the instantaneous rate of change in the dependent
variable with respect to changes in the two independent variables,
separately.
Given a function f (x , y) , a partial derivative can be found with
respect to each independent variable. The partial derivative taken
with respect to x is denoted by
171
z
f
x x
z
f y
y
## Although both notational forms are used to denote the partial
derivative, we will use the subscripted notation f x and f y .
Partial Derivative
Given the function z = f (x, y), the partial derivative of z with
respect to x at (x, y)is
f x = lim
x 0
f ( x + x , y )f ( x , y)
x
## provided the limit exists. The partial derivative of z with respect to
y at (x, y)is
f y = lim
y 0
f ( x , y + y )f (x , y)
x
## provide the limit exists.
Example
2
3
Consider the function f ( x , y ) =3 x +5 y . To find the partial derivative
with respect to x we consider
f x = lim
x 0
f ( x + x , y )f ( x , y)
x
x+ x
3( 2+5 y )( 3 x 2+5 y 3 )
lim
x 0
172
lim
( 3( x 2+ x 2+ 2 x x)+5 y 3 )( 3 x 2 +5 y 3 )
x
x 0
lim
( 3 x 2 +3 x 2 +6 x x +5 y 3 )( 3 x 2 +5 y 3 )
x 0
3 x2 +6 x x
= lim 3 x+ 6 x=6 x
x
x 0
x 0
lim
## Fortunately, partial derivatives are found more easily using the
same differentiation rules we used in Chap. 15 17. The only
exception is that when a partial derivative is found with respect to
one independent variable, the other independent variable is
assumed to be held constant. For instance, in finding the partial
derivative with respect to x, y is assumed to be constant. A very
important point is that the variable which is assumed constant
must be treated as a constant in applying the rules of
differentiation.
EXAMPLE
Find the partial derivatives
fx
and
fy
## for the function
f ( x , y ) =5 x 2 +6 y 2
f x =10 x
## The partial derivative with respect to y is
f y =12 y .
Example
Find fx and fy for the function
f ( x , y ) =4 xy .
173
f x =4 y
f y =4 x .
Example
Find f x and
fy
## for the function
f ( x , y ) =e x + y .
2
f x =2 x e x + y
## The partial derivative with respect to y is
2
f y =2 y e x + y .
## Second Order Derivatives
Type-1 (Pure Second Order Partial Derivatives)
Pure second order partial derivatives are denoted by
found by first finding
fx
f yy .
## Type-2 (Mixed Partial Derivatives)
fx
f xx , f yy . f xx
is
with respect
174
determining
fx
f xy f yx
## and then differentiating
f xy
.
fx
is found by
with respect to
y.
Example
Determine all first and second derivatives for the function
f ( x , y ) =8 x3 4 x 2 y+ 10 y 3 .
Solution:
f x =24 x 28 xy ( y is constant ) ,
f y =4 x 2 +30 y 2( x is constant )
Type I
As f x =24 x 8 xy f xx=48 x 8 y .
As f y =4 x 2+30 y 2 f yy=60 y .
Type II
As f x =24 x 28 xy f xy=8 x .
As f y =4 x 2+30 y 2 f yx=8 x .
## Optimization of Bivariate Functions
The process of finding optimum values of bivariate functions is
very similar to that used for single variable functions.
As with single-variable functions, we will have a particular interest
in identifying relative maximum and minimum points on the
surface representing a function f(x,y).
Relative maximum and minimum points have the same meaning
in three dimensions as in two dimensions.
175
Relative Maximum
A function z = f (x, y)is said to have a relative maximum at x = a
and y = b if for all points (x, y) sufficiently close to (a, b)
f (a, b) f (x, y)
Relative Minimum
A function z = f (x, y)is said to have a relative minimum at x = a
and y = b if for all points
(x, y) sufficiently close (a, b)
f (a, b) f (x, y)
Necessary Condition For Relative Extrema
A necessary condition for the existence of a relative maximum or
a relative minimum of a function f whose partial derivatives fx and
fy both exist is that:
f x =0f y =0
Critical Points
The values x* and y* at which fx = 0 and fy = 0 are critical values.
The corresponding point (x*, y*, f (x*, y*))is a candidate for a
relative maximum or minimum on f and is called a critical point.
Example
Locate any critical points of the function
f ( x , y ) =4 x 212 x+ y 2+2 y10.
Solution: Since
set
f x =0
and
f x =8 x12,
f y =0.
f x =0 8 x12=0 x=
3
2
f y =0 2 y+ 2=0 y =1
and
f y =2 y+ 2
176
Now
f
3
3 2
3 ( )2 ( )
,1 =4
12
+ 1 +2 1 10=20
2
2
2
) () ()
( 32 ,1,20) .
## Distinguishing Nature of Critical Points
Once a critical point has been identified, it is necessary to
determine its nature. Aside from relative maximum and minimum
points, there is one other situation in which fx and fy both equal
0,which is referred to as a saddle point. A saddle pointp is a
portion of a surface which has the shape of a saddle , at point
where you sit on the horse the value of fx and fy both equal 0.
However, the function does not reach either a relative maximum
or a relative minimum at p.
Test of Critical Point
## Given that a critical point of f is located at ( x , y , z ) where all
second partial derivatives are continuous, determine the value of
D ( x , y , z ) , where
D ( x , y )=
I-
II-
If
2
f xx f xy
=f xx ( x , y ) f yy ( x , y ) [ f xy ( x , y ) ]
f yx f yy
D ( x , y )> 0
## a) A relative maximum if both
f xx ( x , y )
negative.
b) A relative minimum if both
f xx ( x , y )
and
and
f yy ( x , y )
f yy ( x , y )
positive.
are
are
177
III-
If
D ( x , y )=0
## , no conclusion can be drawn.
Example
2
3
2
Given the function f ( x , y ) =x y +12 y . Determine the location and
nature of the critical points.
Solution:
Here
f x =2 x , f y =3 y (8 y)
Put
f x =0
x=0,
which gives us
and for
f y =0
gives us
3 y ( 8 y )=0 y=0, 8.
(0, 0, 0)
and
(0,8, 256)
. Second
## order derivatives are
f xx =2, f xy=0=f yx
Taking
and
f yy=6 y +24
(0, 0, 0)
Taking
( 0, 0,0 ) .
(0, 0, 0)
## This implies a critical point is either relative maximum or
minimum. As
f xx <0
maximum occurs on
and
f
f yy <0
at
( 0, 8,256 ) .
178
## Applications of Bivariate Optimization
This section presents some applications of the optimization of
bivariate functions.
A manufacturers estimated annual sales (in units) to is a function
of the expenditures made for radio and TV advertising. The
function specifying this relationship was stated as
z=50000 x+ 40000 y10 x 220 y 210 xy
where z equals the number of units sold each year, x equals the
amount spent for TV advertising and y equals the amount spent
for radio advertising (x and y both in \$1,000s). Determine how
much money should be spent on TV and radio advertising in order
to sell maximum number of units.
Solution:
Here f x =5000020 x 10 y and
f y =0
f y =4000040 y10 x .
Put
f x =0
and
we obtain
20 x+10 y =50000
10 x+ 40 y=40000
x=2285.72
f yy=40
and
y=428.57
## Testing critical point
D (2285.72, 428.57 )=(20 ) (40 )(10 )2=700> 0
. Now
and
179
Since
D>0
, both
f xx , f yy
## are negative, hence annual sales are
maximum at
f ( x , y ) =f ( 2285.72, 428.57 ) =65714296units .
## Example (Pricing Model)
A manufacturer sells two related products, the demands for which
are estimated by the following two demand functions:
q1 =1502 p1 p2
q 2=200 p13 p2
where pj equals the price (in dollars) of product j and qj equals the
demand (in thousands of units) for product j. Examination of these
demand functions indicates that the two products are related. The
demand for one product depends not only on the price charged
for the product itself but also on the price charged for the other
product. The firm wants to determine the price it should charge for
each product in order to maximize total revenue from the sale of
the two products.
Solution:This revenue-maximizing problem is exactly like the
single-product problems discussed in Chap. 17. The only
difference is that there are two products and two pricing decisions
to be made. Total revenue from selling the two products is
determined by the function
R= p1 q1 + p2 q2
180
R p1 (150 2 p1 p2 ) p2 (200 p1 3 p2 )
150 p1 2 p12 p 1 p2 3 p22
150 p1 2 p12 2 p 1 p2 200 p2 3 p22
## This function is stated in
terms of four independent variables. As with the single-product
problems, we can substitute the right side of equations
## We can now proceed to examine the revenue surface for relative
maximum points.
The first partial derivatives are
f p =1504 p12 p2 , f p =2 p1 +2006 p2
1
1
p1=25, p 2=25.
## Testing for critical points
f p p =4, f p p =2=f p p , f p
1
Therefore
p2=6
D (25,25 )=20>0.
maximum occurs at
As
f p p ,f p
1
p2
p1= p2=25.
f ( 25,25 ) =
\$ 4,375
181
p1
and
p2
## into the demand equations, or
q1 =1502 ( 25 )25=75
q 2=200253(25)=100
(Thousand units)
(Thousand units)
## Example (Satellite Clinic Location)
A large health maintenance organization (HMO) is planning to
locate a satellite clinic in a location which is convenient to three
suburban townships. The HMO wants to select a preliminary site
by using the following criterion: Determine the location (x, y)
which minimize the sum of the squares of the distances from each
township to the satellite clinic.
Solution:The unknown in this problem are x and y, the
coordinates of the satellite clinic location.
We need to determine an expression for the square of the
distance separating the clinic and each of the towns. Given the
point (x 1 , y 1) and ( x 2 , y 2) , the distance formula gives
2
d = ( x 2x 1 ) + ( y 2 y 1 )
We obtain
S f ( x, y )
[( x 40) 2 ( y 20) 2 ] [( x 10) 2 ( y 10) 2 ]
[( x 20) 2 ( y 10) 2 ]
182
2
## 3 x +3 y +60 x40 y +2700
Taking derivatives
f x =6 x60, f y =6 y40 .
Now we put
x=10, y=
f x =0=f y
which gives us
20
.
3
## Critical point testing
f xx =6, f yy=6, f xy =f yx =0
D 10,
20
=36> 0
3
x=10, y=
20
.
3 |
## How fast should multiplication facts?
The minimum correct rate for basic facts should be set at 30 to 40 problems per minute, since this rate has been shown to be an indicator of success with more complex tasks (Miller & Heward, 1992, p. 100).” Rates of 40 problems per minute seem more likely to continue to accelerate than the lower end at 30.
## How fast is fluent for math facts?
When students achieve automaticity of their facts, a common benchmark is that they can recall them in two seconds or less (30 problems in 60 seconds). This skill is essential for students to master and tremendously impacts their future in mathematics.
How many multiplication facts are in 3 minutes?
A student should be able to work out the 100 problems correctly in 5 minutes, 60 problems in 3 minutes, or 20 problems in 1 minute. This multiplication math drill worksheet is appropriate for Kindergarten, 1st Grade, 2nd Grade, and 3rd Grade.
### How do you create a multiplication fact fluency?
What are the Foundation Strategies for Multiplication Fact Fluency?
2. finding a pattern (related to skip-counting)
3. knowing the Identity and Zero Properties of Multiplication.
4. and learning by memory the all-important products square numbers (6×6 = 36, etc.).
### What is the fastest way to memorize multiplication facts?
There’s 5 steps to mastering the multiplication facts:
1. Step 1: Break up the facts into manageable chunks.
2. Step 2: Make the facts concrete with a simple visual.
3. Step 3: Teach your child to use easier facts as stepping stones to the harder facts.
4. Step 4: Practice each times table on its own until it’s mastered.
How long is a timed multiplication test?
three to four seconds
Often teachers ask us how long we suggest giving students to complete a timed test. We consistently see that it is recommended students are given three to four seconds per math fact. You can find research to support both.
#### Is a fact family?
The definition of “Fact Family is a collection of related addition and subtraction facts, or multiplication and division facts, made from the same numbers. For example, for the numbers 7, 8, and 15, the addition/subtraction fact family consists of 7 + 8 = 15, 8 + 7 = 15, 15 – 8 = 7, 15 – 7 = 8.
#### How many multiplication facts are there?
There is an infinite number of multiplication facts because numbers continue indefinitely! Students typically learn 144 of these multiplication facts from 1 to 12. Often multiplication facts are presented in a multiplication or times table.
How many multiplication facts should a third grader know?
By the end of third grade, here are the 10 math skills your child should learn (Four of them have to do with fractions!): Knowing the multiplication tables from 1 to 10 by heart. Multiplying and dividing with numbers up to 100. Understanding fractions as numbers that represent part of a whole.
## What is multiplication fact fluency?
“According to CCSSM, fluency is “skill in carrying out procedures flexibly, accurately, efficiently and appropriately” (CCSSO 2010, p. 6). Thus, far from just being a measure of speed, fluency with multiplication facts involves flexibly and accurately using an appropriate strategy to find the answer efficiently.”
## What is the best multiplication app?
Best Multiplication Apps for Kids
• Math Bingo \$2.99 *RECOMMENDED APP. Android.
• Monster Math FREE + In App Purchases. Android.
• Math Ninja Times Table FREE. Android.
• MathTango FREE + In App Purchases.
• Love to Count 2.
• Tic Tac Math \$1.99.
• Multiplication Kids: Math Game FREE.
• Factor Samurai \$2.99. |
Home » Math Formulas » Area of Rectangle Formula
# Area of Rectangle Formula, Definition, Derivation, Examples
Area of Rectangle Formula: Imagine you want to fill a rectangular park with artificial grass. But Do you know how much grass you need to cover the whole park? To determine the quantity of the grass first, you must know the area of that rectangular park. So, understanding the Area of the Rectangle Formula is necessary not only in maths but also in daily life. In this article, we are going to learn about the Area of the Rectangle Formula, its derivation with some solved Examples,
## What is Area of Rectangle
In geometry, the Area of rectangle defines as the region or area covered by the rectangle in a two-dimensional plane. Before understanding the area of the Rectangle formula in Detail, we must get a general overall view of an Rectangle. A rectangle is a sort of quadrilateral, a two-dimensional object with four sides and four vertices. The rectangle’s four angles are all right angles or equal to 90 degrees. The rectangle’s opposite sides are equal and parallel to one other.
## What is Area of Rectangle Formula
The area of rectangle os the is the space enclosed by the rectangle’s boundaries. The area of rectangle can also be defined as the number of unit squares that can fit inside it and is expressed in terms of the ‘number of’ square units. So, We can easily determine the area of a rectangle which is the product of its length multiplied by its breadth. This can be determined using the area of rectangle formula and several approaches based on the dimensions provided.
## Area of Rectangle Formula
The area of rectangle can be calculated by using the area of rectangle formula to determine how much space these things take up. Once the length and breadth of a rectangle are determined, the area is calculated. The area of a rectangle can be calculated by multiplying its length and width. As a result, the formula for the area of a rectangle whose length and width are ‘l’ and ‘w’ are as follows.
The area of rectangle Formula = (Length × Width)
## How to Calculate Area of Rectangle?
Go through below mentioned few easy steps to calculate the area of a rectangle using the area of rectangle formula:
Step 1: Take note of the length and width (breadth) dimensions of the given rectangle.
Step 2: Multiply the length and width values of the rectangle to obtain the area of rectangle.
Step 3: Provide the solution in square units.
## Unit of Area of Rectangle
The area of a rectangle is calculated by multiplying its length by its width (breadth). Square units are used to express the area of a rectangle. We all know that length is always measured and stated in units such as centimeters, inches, and so on. Because the area of a rectangle is calculated by multiplying its length and width, the area of the rectangle will be in square units. The units used to express the area of rectangles is always square units such as square centimeters, square inches, square feet, etc.
## Area of Rectangle Using Diagonal
In some geometric problems, the length of the rectangle is not given. If the diagonal and one side of a rectangle are specified, the area can be computed. A rectangle’s diagonal is the straight line connecting the rectangle’s opposite vertices. A rectangle has two diagonals that are both the same length. Lets’s understand how to find the area of rectangle using the diagonal.
Using Pythagoras’ theorem, we can find the value of the missing side and then the area. The diagonal of a rectangle is determined using the following formula:
(Diagonal)² = (Length)² + (Width)²
If the Length of the given rectanagle is not given,
(Length)² = (Diagonal)²– (Width)²
(Length) =√ (Diagonal²– Width)²
Or,
If the Widh of the given rectangle is not given,
(Width)² = (Diagonal)² – (Length)²
(Width) =√ (Diagonal²– Length))²
Now, using the area of rectangle formula we can easily find out the area of the given rectangle.
Area of rectangle = Length × Width
## Area of Rectangle Formula Derivation
Now, The question arises of how the area of rectangle formula derives, Let us understand the derivatization of the area of rectangle formula. suppose ABCD is a rectangle with a diagonal AC. The diagonal AC divides the rectangle ABCD into two congruent triangles as (ABC and ADC) , As per the below diagram, We can conclude that the area of rectangle is the sum of the areas of these two triangles.
Derivation of the area of a rectangle formula
Area of Rectangle ABCD = Sum of the area of two congruent triangle
Area of rectangle ABCD = Area of ∆ABC + Area of ∆ADC
Or, Rectangle ABCD = 2 × Area of Triangle ABC
Or, Area of Rectangle = 2 × (1/2 × Base × Height)
or, Area of ABCD = 2 × 1/2 × AB × BC
or, Area of Rectangle ABCD = AB × BC
or, As a result, Area of Rectangle = Length x Width. [proved]
## Area of Rectangle Formula Examples
Example 1: The length and width of a rectangular park are 10 km yards and 6 yards. Find the total area of the parks
Solution:
As per the Given data, the Length of the park, l = 10 Km, and the width of the Park, w = 8 Km/.
Earlier, We know that the area of the rectangle formula is,
Area of rectangle = length × width
Or, The Area of the given rectangular park is = 10 × 8 Square km.= 80 km²
Therefore, the area rectangular park is 80 square km.
Example 2: Suppose a rectangle has a diagonal of 25 units, and a width of 3 units, Find the area of rectangle.
Solution: Earlier we came to know that To find the Area of Rectangle we need its Length and Width.
We can find out the Length of the given rectangle using the area of rectangle formula,
(Length)² = (Diagonal)²– (Width)²
(Length) =√ (Diagonal²– Width)²
(Length) =⎷[(5)2 – (3)2]
(Length) = ⎷[25- 9]
(Length) = = ⎷(16) = 4 units.
Area of the given rectangle = Length × Width = 4× 3 =12 square units.
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## FAQs
### What is the Area of rectangle?
In geometry, the Area of rectangle defines as the region or area covered by the rectangle in a two-dimensional plane. The area of rectangle os the is the space enclosed by the rectangle's boundaries.
### What is the Area of Rectangle formula?
The area of rectangle Formula is = (Length × Width).The area of a rectangle is calculated by multiplying its length by its width (breadth). Square units are used to express the area of a rectangle. |
# Video: Multiplying Two Fractions Using Models
Multiply 2/3 by 4/5 using models.
05:00
### Video Transcript
Multiply two-thirds by four-fifths using models.
We can see that the two models that we’re shown underneath this question represent the two fractions that are in the question. This shape has two out of three parts shaded. This is the same as two-thirds. And this shape has four out of five parts shaded. This is the same as four-fifths. And we can use these models to give us a clue as to how to solve this problem. The problem asks us to multiply these two fractions together, two-thirds multiplied by four-fifths.
Now, the idea of multiplying two fractions by each other can be tricky to understand. So another way we can think of this calculation is as trying to find two-thirds of four-fifths. In other words, we can start with four-fifths and then try to find what two-thirds of this is worth. And it’s this second idea that we’re going to go on to use. So to find two-thirds of four-fifths, we need to start with four-fifths. And we can use our model that shows four-fifths to help us.
Let’s begin by drawing a rectangle. Now, it doesn’t need to be the same size of the rectangle that we’re shown in the diagram. But it does need to be divided into five equal parts, in other words, divided into fifths. And to show four-fifths, we need to shade four out of our five parts. So this shaded area four-fifths is the part of the whole shape that we need to work with. And we need to multiply it by two-thirds. As we’ve said already, this is the same as finding two-thirds of this amount. And to show what two-thirds of this amount is, we need to use our second model to help us. It’s a reminder of what two-thirds is like. It’s two out of a possible three parts.
And so, to show two-thirds of our four-fifths, we’re going to firstly need to split four-fifths into three equal parts. And because we already have lines going across the shape, let’s do this from top to bottom, something like this. And as we’ve seen already in the diagram to show two-thirds, we need to shade two out of these three parts. Here’s one-third of four-fifths and here are two-thirds of four-fifths. We know that this new shaded area is worth two-thirds of four-fifths. But what fraction is it of the whole amount?
To find the answer to this, we need to divide the whole shape into equal parts. And at the moment, it’s not divided into equal parts. So we need to extend our two vertical lines downwards. And by doing this, we’ve split the whole shape into equal parts. We can see now that there are five rows and each row contains three equal parts. There must be 15 equal parts in the whole shape. We’ve divided the shape into 15ths. And so, our answer is going to be a number of 15ths.
In fact, if you look at our original fractions, can you see where the denominator 15th comes from? We’ve multiplied our two denominators together. One of our fractions was a number of thirds and the other was a number of fifths. And three multiplied by five equals 15. Let’s come back to our model. If we think about our shaded area that represented two-thirds of four-fifths, how many 15ths can we see? Let’s count in twos: two, four, six, eight. The fraction that shown in our model is eight 15ths.
If we look back at our original fractions, can you see where the numerator eight comes from? We had two-thirds and four-fifths. And if we multiply these two numerators together, two times four equals eight. Although we could have found the answer here just by multiplying the numerators and the denominators together, we were told in the question to use models. And so, that’s what we’ve done. First, we drew a rectangle that represented one whole. And then, we shaded four-fifths of it. And to find two-thirds of four-fifths, we needed to divide our four-fifths into three equal parts and only select two of them. We could see that the answer was going to be eight equal parts. And by extending our two lines downwards, we could see that each part was one 15th of the whole amount.
And so, we can say two-thirds multiplied by four-fifths equals eight 15ths. |
# How Do You Construct A Line Segment
## Construction Of A Line Segment
### To draw a line segment of a given length
We can draw a line segment of a given length in two ways:
(a) By using a scale
(b) By using a pair of compasses
Construction: Draw a line segment of length 5.5 cm.
(a) By using a scale
• Step 1: Mark a point in your notebook and label it as A.
• Step 2: Place the scale in such a way that the zero (0) mark on the scale coincides with A.
• Step 3: How, move the sharp end of the pencil along the edge of the scale till it reaches the point 5.5 cm of the scale,
• Step 4: Label the second end as B.
AB is the required line segment.
(b) By using a pair of compasses
• Steps 1: Draw a line l and mark a point A on it.
• Step 2: Open the arms of the compass so that the end points of the two open arms equal to 5.5 cm.
• Step 3: Without disturbing the opening of the compasses, place its needle at point A and draw an arc to cut the line l at point B.
• Step 4: AB is the required line segment of length 5.5 cm.
### To draw a line segment equal to a given line segment
Construction: Draw a line segment (AB) equal to PQ.
Given: A line segment PQ.
• Step 1: Draw a line l. Mark a point A on it.
• Step 2: Take the compass and measure PQ.
• Step 3: Without disturbing the opening, place the needle of the compass at point A on line l and draw an arc, which cuts the line at point B.
• Step 4: AB is the required line segment which is equal to the length of PQ, i.e., AB = PQ.
Construction: Draw a line segment whose length is the sum of the lengths of two given line segments.
Given: Two line segments AB and BC.
Construct a line segment, say AC, such that
AC = AB + BC.
• Step 1: Draw a line l and mark point A on it.
• Step 2: Take the compasses and measure AB.
• Step 3: Without disturbing the opening, place its needle at A and draw an arc cutting line l at B.
• Step 4: Again adjust the compasses and measure the line segment BC.
• Step 5: Without disturbing the opening, place the pointer at point B on the line l and draw an arc cutting the line l at C.
• Step 6: AC is the required line segment whose length is equal to the sum of the lengths of line segments AB and BC, i. e., AC = AB + BC. |
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# 3.2
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## 3.2Presentation Transcript
• CHAPTER 3 LINEAR SYSTEMS3.2 Solving Systems Algebraically Part 1: Substitution
• SYSTEM OF EQUATIONS A system of equations is a set of two or more equations A linear system consists of linear equations A solution of a system is a set of values for the variables that makes all the equations true. (usually an ordered pair) Systems can be solved be various methods: graphing, substitution, and elimination
• SOLVING SYSTEMS BY SUBSTITUTION Use when it is easy to isolate one of the variables At least one variable has a coefficient that is 1
• SOLVING SYSTEMSBY SUBSTITUTION1. Solve one equation for one of the variables2. Substitute the expression into the other equation and solve3. Substitute the solution into one of the original equations and solve for the remaining variable4. Check the solution
• SOLVING SYSTEMS BY SUBSTITUTION
• SOLVING SYSTEMS BY SUBSTITUTION
• SOLVING SYSTEMS BY SUBSTITUTION
• ASSIGNMENT 3.2 Substitution WS
• CHAPTER 3 LINEAR SYSTEMS3.2 Solving Systems Algebraically Part 2: Elimination
• SYSTEM OF EQUATIONS A system of equations is a set of two or more equations A linear system consists of linear equations A solution of a system is a set of values for the variables that makes all the equations true. (usually an ordered pair) Systems can be solved be various methods: graphing, substitution, and elimination
• SOLVING BYELIMINATION1. Rewrite both equations in standard form2. Multiply one or both systems by an appropriate non-zero number (note you want one variable to drop out in the next step)3. Add the equations4. Solve for the variable5. Substitute the value into one of the original equation and solve for the remaining variable6. Check the solution
• SOLVE BY ELIMINATION
• SOLVE BY ELIMINATION
• SOLVE BY ELIMINATION
• SOLVING SYSTEMS WITHOUT UNIQUESOLUTIONS Solving a system algebraically can sometimes lead to infinitely many solutions and/or no solution If you get a true result: infinitely many solutions If you get a false result: no solution
• EXAMPLE: SOLVE THE SYSTEM
• EXAMPLE: SOLVE THE SYSTEM
• ASSIGNMENT 3.2 Elimination WS |
Question
# In the figure below, the rectangle at the corner measures $$10 cm \times 20 cm.$$ The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?
A
10 cm
B
40 cm
C
50 cm
D
30 cm
Solution
## The correct option is C 50 cm$$We\quad take\quad the\quad corner\quad O\quad of\quad the\quad given\quad square\quad OPQR,\quad containing\quad the\quad 10cm\times 20cm\\ rectangle,\quad as\quad the\quad origine\quad and\quad the\quad sides\quad OP\quad \& \quad OR\quad as\quad the\quad coordinate\quad axes.\\ Let\quad the\quad sides\quad of\quad OPQR\quad be\quad 2a.\\ So\quad OPQR\quad lies\quad in\quad the\quad third\quad quadrant.\quad \quad The\quad inscribe\quad circle's\quad radius\quad is\quad r=a\\ \& \quad centre\quad is\quad C(a,a).\\ The\quad circle\quad passes\quad through\quad A(-10,-20).\\ \therefore \quad The\quad equation\quad of\quad the\quad circle\quad is\\ { \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }..........(i)\\ Here\quad \left( h,k \right) =\left( a,a \right) \quad and\quad r=a\\ \therefore \quad equation\quad (i)\quad becomes\\ { \left( x-a \right) }^{ 2 }+{ \left( y-a \right) }^{ 2 }={ a }^{ 2 }..........(ii)\\ The\quad circle\quad passes\quad through\quad A(-10,-20).\\ i.e\quad \left( x,y \right) =\left( -10,-20 \right) \\ So\quad equation\quad (ii)\quad reduces\quad to\\ { \left( -10-a \right) }^{ 2 }+{ \left( -20-a \right) }^{ 2 }={ a }^{ 2 }.\\ On\quad solving\quad for\quad a\quad we\quad get\\ |a|=50\quad units\quad and\quad 10\quad units.\quad \\ (a\quad is\quad a\quad length.\quad So\quad we\quad do\quad not\quad consider\quad the\quad negative\quad sign).\\ Also\quad A(-10,-20)\quad lie\quad on\quad the\quad circle.midway\quad between\quad (a,0)\quad \& \quad (0,a).\\ So\quad a>10\quad units\quad \& \quad a>20\quad units.\\ \therefore \quad we\quad reject\quad a=10.\\ so\quad a=50units\\ Ans-\quad Option\quad C\\ \\ \\$$Mathematics
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Chapter 4 Congruent Triangles
# Chapter 4 Congruent Triangles
## Chapter 4 Congruent Triangles
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Chapter 4 Congruent Triangles 4-1 Congruent Figures
2. Objectives • Students will be able to: • Recognize congruent figures and their corresponding parts • Define and apply the third angles theorem • Use congruent parts to prove triangles congruent
3. Essential Understanding • You can determine whether two figures are congruent by comparing their corresponding parts.
4. Congruent Figures • What do you know about congruent figures? • Does the orientation of congruent figures matter? • No, the congruent figures can be slid, flipped, or turned and they will still be congruent
5. Congruent Figures • Congruent Polygons: • Have congruent corresponding parts (equal sides and angles) • When naming congruent figures, you must list corresponding parts (angles/vertices) in the same order
6. Corresponding Parts • If figure HIJK ≅ LMNO, what are the congruent corresponding parts? • If it helps, draw a picture • Then name the corresponding segments • HI ≅ LM IJ ≅ MN JK ≅ NO KH ≅ OL • Then name the corresponding angles • <H ≅ <L <I ≅ <M <J ≅ <N <K ≅ <O • If ΔABC ≅ ΔXYZ, what are the congruent corresponding parts?
7. Congruent Parts • Suppose ΔABC ≅ ΔFED, <B = 30 and <D = 75, what is the measure of <A? • Draw a picture of the situation • Suppose ΔWYS ≅ ΔMKV. If <W = 62 and <Y = 35, what is the measure of <V? How do you know?
8. Third Angles Theorem • If you know two angles of a triangle are congruent, then the third angles will be congruent.
9. Are the triangles congruent? • Write a two column proof • To prove two triangle congruent you need to get all the corresponding angles and segments to be congruent. This allows you to use the Def. of Congruent Triangles reason as your final statement • Statements | Reasons . • 1. AB ≅ AD, BC ≅ DC, 1. Given<B ≅ <D • 2. AC ≅ AC 2. Reflexive Property of ≅ • 3.<BCA ≅ <DCA 3. All right angles are congruent • 4. <BAC ≅ <DAC 4. Third <s Thm. • 5. ΔABC ≅ ΔADC 5. Def of Congruent Triangles
10. Are the Triangles Congruent? • No because we cannot show that all the corresponding parts are congruent.
11. Proving Triangles Congruent • Given: <A ≅ <D, AE ≅ DC, EB ≅ CB, BA ≅ BD • Prove: ΔAEB ≅ΔDCB
12. Homework • Pg. 222 -223 • 10 – 19, 30 – 32, 44 • 14 problems |
# Google Earth Math Problems using Area and proportions
Document Sample
``` Google Earth Math Problems: Using Area and Proportion
Corn Problem
Part 1 – Find the Area
1. Corn Problem
1. Find the area of the marked field of corn. Use the ruler tool and measure lengths to the
nearest hundredth of a mile. Write your work on another sheet of paper. Include a labeled
sketch of the field and work down from formulas.
2. Click on the house icon and solve the given problem.
3. You may use a calculator for assistance but write your work in steps on paper.
Part Two – Find the Value
1. Find the value. Farmer Green is trying to estimate how much money she will make from her
corn field this year. You will need three proportions to solve this problem.
1. First find how many acres of corn Farmer Green has. Write a proportion using the unit rate of
640 acres per 1 square mile with your previous answer to find how many acres she has.
2. Next, use a proportion to find how many bushels of corn her field will produce. Use 40
bushels per acre as your unit rate.
3. Finally, write a proportion to find the value of the bushels using a unit rate of \$4.655 for 1
bushel.
A=lw+lw+lw
A=(.50x.24)+(.22x.24)+(.50x.25)
A= .1200+.0528+.1250
A= .2978 or .30 mi²
.30mi²x640= 192 acres
192acresx40bushels = 7680 bushels
Wheat Problem
Part 1
1. Find the area of the marked field of wheat. Use the ruler tool and measure lengths to the nearest
hundredth of a mile. Write your work on another sheet of paper. Include a labeled sketch of the field and
work down from formulas.
2. Click on the house icon and solve the given problem.
You may use a calculator for assistance but write your work in steps on paper.
A=lxw
A=.97x.48
A=.4656mi²
A=½bh
A=.5x.41x.19
A=0.03895mi²
(A+A) =.5046mi² -(.08x.08)-(.11x.11)
.5046mi² - .0064 - .0121= .4861mi²
Part Two
Farmer Ingqvist is trying to decide whether he should plant soybeans next year. Find how much he will
make with his wheat field this year and estimate how much he will make if it was soybean.
1. First find how many acres of wheat Farmer Ingqvist has. Write a proportion using the unit rate of 640
acres per 1 square mile with your previous answer to find how many acres he has.
2. Next use a proportion to find how many bushels of wheat his field will produce. Use 40 bushels per
3. Now write a proportion to find the value of the bushels using a unit rate of \$9.45 for 1 bushel of wheat.
4. Repeat the step 3 above using a unit rate of \$12.685 for 1 bushel of soybean.
5. Report the difference of the two crops.
Round off to .49mi²
.49x640= 313.6 acres of land
313.6acresx 40 bushels= 12544 bushels of wheat
12544x\$9.45= \$118,540.80 for wheat.
12544x12.685 = \$159,120.64 for soybean
the difference = \$40,579.84
```
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# Introduction to Projective Geometry Solutions 5.12
## Equiareal Transformations
A 4 minute read, posted on 25 Sep 2019
Tags computer vision, projective geometry, problem solution
Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.
#### 1. Prove Theorem 2.
Composing two general equiareal transformations, we get a transformation of the form below. $$\begin{pmatrix} a_{12}b_{21} + a_{11}b_{11} & a_{12}b_{22} + a_{11}b_{12} & a_{12}b_{23} + a_{11}b_{13} + a_{13} \\ a_{22}b_{21} + a_{21}b_{11} & a_{22}b_{22} + a_{21}b_{12} & a_{22}b_{23} + a_{21}b_{13} + a_{23} \\ 0 & 0 & a_{33}b_{33} \end{pmatrix}$$
Let’s consider the fraction relevant to equiareal transformations $$\frac{(a_{12}b_{21} + a_{11}b_{11})(a_{22}b_{22} + a_{21}b_{12}) - (a_{12}b_{22} + a_{11}b_{12})(a_{22}b_{21} + a_{21}b_{11})}{a_{33}^2b_{33}^2}$$
Expanding this we get $$\frac{(a_{12}a_{21} - a_{11}a_{22})(b_{11}b_{22} - b_{12}b_{21})}{a^2_{33}b^2_{33}}$$ $$\implies \frac{(a_{12}a_{21} - a_{11}a_{22})}{a^2_{33}}\frac{(b_{11}b_{22} - b_{12}b_{21})}{b^2_{33}}$$
As the absolute value of each of these terms is 1, the absolute value of this fraction will also be 1. Thus, two equiareal transformations are closed under the operation of composition.
The inverse of a general equiareal transformation has the form $$\frac{1}{a_{33}(a_{11}a_{22} - a_{12}a_{21})}\begin{pmatrix} a_{22}a_{33} & -a_{12}a_{33} & a_{12}a_{33} - a_{13}a_{22} \\ -a_{21}a_{33} & a_{11}a_{33} & a_{13}a_{21} - a_{11}a_{23} \\ 0 & 0 & a_{11}a_{22} - a_{12}a_{21} \end{pmatrix}$$
Let’s consider the fraction relevant to equiareal transformations $$\frac{a_{33}^2(a_{22}a_{11} - a_{12}a_{21})}{(a_{22}a_{11} - a_{12}a_{21})^2}$$ $$= \frac{a_{33}^2}{(a_{22}a_{11} - a_{12}a_{21})}$$
The absolute value of this is clearly 1. Hence the inverse is also an equiareal transformation.
Finally, it is obvious, from the rules of matrix multiplication, that the composition of equiareal transformations is associative and that the identity is included among the equiareal transformations.
Hence the set of all equiareal transformations is a group under the operation of composition.
#### 2. Are the equiareal transformations of all six types? Give an example of each possible type.
##### Type I
$a_{33} \ne \pm1$
$\begin{pmatrix} a_{33}^2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & a_{33} \end{pmatrix}$
##### Type II
$a_{23} \ne 0$
$\begin{pmatrix} a_{11} & 0 & a_{13} \\ 0 & -a_{11} & a_{23} \\ 0 & 0 & -a_{11} \end{pmatrix}$
##### Type III
$\begin{pmatrix} a_{11} & 0 & 0 \\ 0 & a_{11} & 0 \\ 0 & 0 & -a_{11} \end{pmatrix}$
##### Type IV
$\begin{pmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{11} & 0 \\ a_{31} & a_{32} & a_{11} \end{pmatrix}$
##### Type V
$a_{13} \ne 0, a_{23} \ne 0$
$\begin{pmatrix} a_{11} & 0 & a_{13} \\ 0 & a_{11} & a_{23} \\ 0 & 0 & a_{11} \end{pmatrix}$
##### Type VI
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
#### 3. Determine which, if any, of the following mappings can be accomplished by an equiareal transformation, and when it exists, find the equations of the transformation:(a) $(0,0,1) \rightarrow (0,0,1)$, $(1,0,1) \rightarrow (1,0,1)$, $(2,1,1) \rightarrow (-1,1,1)$(b) $(0,0,1) \rightarrow (0,0,1)$, $(2,0,1) \rightarrow (0,3,1)$, $(2,1,1) \rightarrow (1,0,3)$(c) $(0,0,1) \rightarrow (0,0,1)$, $(2,0,1) \rightarrow (3,1,1)$, $(1,1,1) \rightarrow (1,2,2)$
(a) The two triangles share the same base and their heights are the same. So an equiareal transformation between the two is possible. Using the invariance of $(0, 0, 1)$, we get $a_{13} = a_{23} = 0$ and $a_{33} = 1$. Further, using the invariance of $(1, 0, 1)$, we get $a_{11} = 1$ and $a_{21} = a_{31} = 0$. Finally, using the mapping $(2, 1, 1) \rightarrow (-1, 1, 1)$ we get $a_{12} = -3$ and $a_{22} = 1$. Hence, the matrix of transformation is $$\begin{pmatrix} 1 & -3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
(b) Areas of the two triangles are $\frac{1}{2} \times 2 \times 1 = 1$ and $\frac{1}{2} \times 3 \times \frac{1}{3} = \frac{1}{2}$ respectively. Clearly, the area of the two triangles do not match; hence an equiareal transformation is not possible.
(c) Areas of the two triangles are $\frac{1}{2} \times 2 \times 1 = 1$ and $\frac{1}{2} \times 1 \times \frac{5}{2} = \frac{5}{4}$ respectively. Clearly, the area of the two triangles do not match; hence an equiareal transformation is not possible.
#### 4. If $A$ and $B$ are distinct finite points, and if $A’$ and $B’$ are distinct finite points, show that there is an equiareal transformation which maps $A$ into $A’$ and $B$ into $B’$. Is this transformation unique?
Given two pairs of distint points, it is possible to choose another pair of points C and C’ such that the areas of $\bigtriangleup ABC$ and $\bigtriangleup A’B’C’$ are the same. Hence, it is possible to define an equiareal transformation that achieves the given mapping.
As the choice of $C$ and $C’$ are arbitrary, the equiareal mapping is not unique.
#### 5. Show that for a given equiareal transformation there are in general two directions with the property that the lengths of segments in these directions are unaltered by the transformation.
By corollary 1, every equiareal transformation is an affine transformation. Hence the result in Exercise #8, Sec 5.10 holds for equiareal transformations too. |
Subject: Calculus
# Logarithmic Derivative
Logarithms (log) are exponent or power to which a number or function, can be raised to yield a specific number. We call the number being raised as the "base". Logarithms can also be referred to as "mathematical shortcut" for exponents. When you have a number say, 1000, then the logarithm of that number with respect to the base, say 10, is 3, which is written as
\log_{10}{1000} = 3
Why do I say shortcut? Because as you see , the number 1000 can be written as the multiplication of three (3) 10's (1000=10^3), which in this case is the base. So imagine if you're working with large numbers (say, trillionth degree). Then it would be laborious to oneself writing down such big amounts of numbers. That's where the logarithm takes place. So instead of writing down the whole thing with the actual large numbers, you can use the logarithmic scale and take down the logarithms of all these large numbers you're working with. Then, life would be easier.
On the other hand, when a function f is to take its logarithm, then it is written as \log{f}.
## Basic Properties of Logarithm
Properties of logarithm are also called as "logarithmic identities". This covers the basic mathematical operations for logarithms.
### Product
The logarithm of a product is equal to the sum of the two logarithms:
\log_{x}{ab}=\log_{x}{b}+\log_{x}{a}
### Quotient
The logarithm of a quotient is equal to the difference of the two logarithms:
\log_{x}\frac{a}{b}=\log_{x}{b} - \log_{x}{a}
### Power
The logarithm of the n-th power of a number is n times the logarithm of that number:
\log_{x}{a^n}=n\log_{x}{a}
## Differentiation of Logarithmic Function
The derivative of the logarithm of a function (\log{f} is defined as the logarithmic derivative of a function. The formula of the logarithmic derivative is given by;
D_{x}[\log{f}] = \frac{f'}{f}
So basically, logarithmic derivative is just as easy as ordinary derivatives. The only difference is that in logarithmic derivative, we have to divide the derivative of a function by the function itself. For example, we have a function f(x)=x^2, then its logarithmic derivative is given by;
f(x)=x^2
f'(x)=2x
Thus,
\frac{f'(x)}{f}=\frac{2x}{x^2}=\frac{2}{x}
## Properties of Logarithmic Derivative
We have discussed above the properties of logarithm which in turn, has its analog properties in logarithmic derivative. They are as follows;
### Product
The logarithmic derivative of a product of two functions is equal to the sum of each individual logarithmic derivatives.:
\frac{\left (ab\right )'}{ab}= \frac{a'}{a}+\frac{b'}{b}
### Quotient
The logarithmic derivative of a ratio of two functions is the difference of their individual logarithmic derivatives.:
\frac{\left (\frac{a}{b}\right )'}{\frac{a}{b}}= \frac{a'}{a}-\frac{b'}{b}
### Power
The logarithmic derivative of the n-th power of a function is n-times the logarithmic derivative of that function:
\frac{\left (f^{p}\right )'}{f}=p \frac{f'}{f}
## Standard Notations
When the base of a logarithm is 10, then the subscript that indicates the base is conventionally omitted. Thus, expressions such as \log{x}, \log{f}, \log{a} means that the logarithm has the base 10.
Furthermore, another form of logarithm is the natural logarithm written as \ln{x}. Instead of the word log, natural logarithm uses ln. The only difference between logarithm and natural logarithm are its base. While logarithms cover all values for the base, natural logarithms indicates a base of e only where e is a dimensionless natural number whose value is 2.71828183. Additionally, natural logarithms are assigned only to a range of non-zero and below numbers. In short, there is no natural logarithm of negative numbers and zero.
### Example #1
What is the logarithmic derivative of the function f=x+1?
D_x\log{f}=\frac{1}{x+1}
### Example #2
Find the logarithmic derivative when the function is f(u)=\cos{u}.
D_u\log{f(u)}=\frac{-\sin{u}}{\cos{u}} = -\tan{u}
### Example #3
Show that the logarithmic derivative of the function f(x)=e^x is one.
First, we have to get the derivative of f,
f'(x)=e^x.
Then, we divide the derivative of f by f itself, that is
\frac{f'(x)}{f(x)} = \frac{e^x}{e^x}=1
Thus,
D_x\log{e^x}=1
### Example #4
Using the product property of logarithmic derivative, find the logarithmic derivative of f(s)=e^x\sin{x}.
Product property of logarithmic derivative states that
\frac{\left (ab\right )'}{ab}= \frac{a'}{a}+\frac{b'}{b}
In this case, we can assign e^x=a and \sin{x}=b. Then we find the individual derivatives of a and b,
a' = e^x
b'=\cos{x}
Then,
\frac{a'}{a}+\frac{b'}{b}= \frac{e^x}{e^x}+\frac{\cos{x}}{\sin{x}}
Therefore,
D_s\log{e^x\sin{x}} = 1+\cot{x}
### Example #5
Find the logarithmic derivative of a logarithmic function. That is D[\log{\log{f}}]
D[\log{\log{f}}] = \frac{f'}{(\log{f})^2}
NEX TOPIC: Related rates |
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# NCERT Exemplar Class 9 Maths Chapter 14 Solutions: Statistics and Probability
NCERT Exemplar Class 9 Solutions Maths Chapter 14 will give you information on statistics and probability – the two chapters of NCERT Exemplar Class 9 Maths that help in analyzing data. You will learn about data collection. With probability, you will be able to find out an outcome of any event happening. You will also learn about the collection and presentation of data in graphical form, measuring the central tendency, or the basic approach to probability.
The chapter contains 4 exercises 72 exemplar questions. The NCERT Exemplar Class 9 Chapter 14 Problems have been designed to check your understanding level for all types of questions. From multiple-choice questions to long questions, you will have to solve all using your knowledge that you possess about the chapter. Being a Class 9 student, you will be facing issues in understanding the complex chapters but Instasolv aims at making the chapter easy for you. We provide you with the best guidance on the subject maths so that you can prepare well for your exams.
## Important Topics for NCERT Exemplar Class 9 Solutions Maths Chapter 14 – Statistics and Probability
Collection of Data
You will be understanding how to collect the data and how to present it in a proper format. You will understand the range of data that refers to the difference between high and low values. The chapter guides you about class intervals, and the upper and lower class limit.
After understanding all these concepts, you will be able to present data.
Then you will learn how to graphically present a collection of data. Under this section, you will know about bar graphs that help in making the pictorial representation of data in the form of bars that are drawn with uniform width and equal spacing between them. Also, you will learn about the histogram that helps in presenting a grouped frequency data using class intervals. Also, you will know how to find the mean, median and mode.
Mean, Median and Mode
1. Meanfor n observations.
2. Median refers to the measure of central tendency.
3. The mode is the value that occurs most frequently in a set of data.
Probability
• In this section, you will learn about the chances of a particular outcome that occur in an experiment in probability.
• You will also know about ‘trial’. When you get one or several outcomes in an event or an experiment.
• The event refers to the collection of outcomes of an experiment.
• You will also know how to find the empirical probability based on the outcome of trials. If n is the total of trials for an Event E, then the probability is given as,
• You will understand all the other concepts of probability in the chapter in detail with the different activities.
### Exercise Discussion of NCERT Exemplar Class 9 Solutions Maths Chapter 14 – Statistics and Probability
• Exercise 14.1 has 30 questions that are multiple-choice questions. These questions are based on statistics and probability in which you need to find the correct answer from the given four options.
• Exercise 14.2 has 10 questions in which you need to justify the statement if it is true or false giving a proper reason to your answer.
• Exercise 14.3 has 20 questions in which you need to find out the mean, median and mode from the given collection of data.
• Exercise 14.4 contains 12 questions in which you need to draw histograms from the given data.
• The exercises are a bit complex and you need proper knowledge of all the topics of the chapter.
## Benefits of NCERT Exemplar Class 9 Solutions Maths Chapter 14 – Statistics and Probability by Instasolv
The Instasolv team provides the best guidance in Maths. If you are facing issues in understanding the NCERT Exemplar Class 9 Maths Problems Chapter 14, then your worries totally end with us. We make use of the best technology to provide you will easy to understand solutions to the Exemplar problems.
We aim at making the concepts easy for you by providing stepwise solutions for the topics of statistics and probability. From preparing for your CBSE board exams to preparing for the other competitive exams like IIT JEE, NEET, our NCERT Exemplar solutions will be of great help.
More Chapters from Class 9 |
# Perpendicular Lines
The theory behind perpendicular lines is similar to what we saw with parallel lines. Remember that, in geometry, two lines are perpendicular if they cross each other at a right angle. The algebra way of expressing this isn't quite as straightforward.
In Math . . .
Two lines are perpendicular if they have their slopes are negative reciprocals of each other.
In English . . .
That "negative reciprocal" thing is a little confusing. What it means is that if you have a line then to get the slope of the line perpendicular to it you take the original line's slope, flip it over (that's the reciprocal part) and make it negative. Here's an easier way to think about it: If you multiply the slopes together and you get -1 then the lines are perpendicular.
# Example 1
Are the lines $y = 2x + 4$ and $x = -3y - 2$ perpendicular?
To answer this question, we need to know the slopes of the two lines. The first equation is in slope-intercept form so we can immediately see that its slope is going to be 3. Be careful with the second equation - it may look like it's in the slope-intercept form but, if you look closely, you'll see that it's solved for x not y. If we solve it for y we get:
$$x = -3y - 2$$$$-3y = x + 2$$$$y = \frac{-1}{3}x - \frac{2}{3}$$
So the slope of the second line is -1/3. If we multiply the two slopes together we get:
$$2 \cdot \frac{-1}{3}=\frac{-2}{3}$$
That isn't equal to -1 so the lines aren't perpendicular.
# Example 3
Are the lines $y=\frac{2}{3}x - 6$ and $y=-\frac{3}{2}x - 2$ perpendicular?
Both of these lines are already in the slope-intercept form so we can see right away that there slopes are 2/3 and -3/2. If we multiply those numbers together, we get:
$$\frac{2}{3}\cdot\frac{-3}{2}=\frac{-6}{6}=-1$$
Because the product equals -1, the lines must be perpendicular.
# Example 2
What's the slope of the line perpendicalar to the line y = -4x + 3?
To find the slope of the perpendicular line we have to do two things: flip the orignal slope over and reverse its sign. The reciprocal of -4 is -1/4 and if we reverse its sign we get +1/4 so that's the slope of the perpendicular line.
# Example 4
Find the equation of the line that's perpendicular to y = 2x + 4 that passes through the point (3, 5).
The procedure behind this kind of question goes like this: We know a point on the line so, if we know the line's slope, we can use the point-slope form of the equation of a line to find its equation. We can get the slope by taking the negative reciprocal of the slope of the that we're given.
Here's how it works in practice. We know that the slope of the original line is 2. That means that the slope of the perpendicular line is going to be -1/2. (Take a look at Example 2 to see how I worked that out.) Because the line passes through the point (3, 5), we can use the point-slope form of the equation of a line to get its equation.
$$y=-\frac{1}{2}(x-5)$$ $$y-6=-\frac{x}{2}+\frac{5}{2}$$ $$y=-\frac{x}{2}+\frac{5}{2} + 6$$ $$y=-\frac{x}{2}+\frac{5}{2} + \frac{12}{2}$$ $$y=-\frac{x}{2}+\frac{17}{2}$$ |
What Is 47/80 as a Decimal + Solution With Free Steps
The fraction 47/80 as a decimal is equal to 0.587.
One of the most basic operations in mathematics is division. Long Division is one of the techniques used to perform division. The division is commonly expressed in fraction form x/y or decimal form.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 47/80.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 47
Divisor = 80
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 47 $\div$ 80
This is when we go through the Long Division solution to our problem. The Long Division process can be seen by the figure 1 below:
Figure 1
47/80 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 47 and 80, we can see how 47 is Smaller than 80, and to solve this division, we require that 47 be Bigger than 80.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 47, which after getting multiplied by 10 becomes 470.
We take this 470 and divide it by 80; this can be done as follows:
470 $\div$ 80 $\approx$ 5
Where:
80 x 5 = 400
This will lead to the generation of a Remainder equal to 470 – 400 = 70. Now this means we have to repeat the process by Converting the 70 into 700 and solving for that:
700 $\div$ 80 $\approx$ 8
Where:
80 x 8 = 640
This, therefore, produces another Remainder which is equal to 700 – 640 = 60. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 600.
600 $\div$ 80 $\approx$ 7
Where:
80 x 7 = 560
Finally, we have a Quotient generated after combining the three pieces of it as 0.587=z, with a Remainder equal to 40.
Images/mathematical drawings are created with GeoGebra. |
# Math Expressions Grade 3 Unit 6 Lesson 11 Answer Key Focus on Mathematical Practices
## Math Expressions Common Core Grade 3 Unit 6 Lesson 11 Answer Key Focus on Mathematical Practices
Math Expressions Grade 3 Unit 6 Lesson 11 Homework
Use the drawing above to solve the problems.
Lesson 11 Answer Key Focus on Mathematical Practices Grade 3 Unit 6 Question 1.
Elton wants to plant grass seed in his backyard. The two parts of his yard are shaped like rectangles. He will use 1 cup of grass seed for every square foot in his yard. How many cups will he use?
38 cups
Explanation:
Elton uses 1 cup of grass seed for every square foot in his yard. By adding the total perimeter of the given rectangle now Elton will use 38 cups.
Math Expressions Grade 3 Unit 6 Lesson 11 Answer Key Question 2.
Elton wants to put a picket fence around the outside of his backyard. How many feet of fencing will he need?
30 feet
Explanation
Elton needs 30 feet of fencing around the outside of his backyard i.e 6 feet+5 feet +4 feet + 4 feet +2 feet + 9 feet = 30 feet
To the right of the garden, draw a different shape that has the same perimeter as part A of Elton’s garden. What is the area of your shape?
24 square feet
Explanation:
The area of the shape given is 8×3= 24 square feet.
Math Expressions Grade 3 Unit 6 Lesson 11 Remembering
Solve.
Question 1.
The concert hall has an audience of 831 people. Special passes to meet the musicians are given to 162 people. How many people do not receive the special passes?
669 people
Explanation:
The concert hall has a total of 831 audience and special passes are given to 162 people so people who did not receive the special passes are
831-162 = 669 people.
Question 2.
The art teacher orders 297 sheets of construction paper, 104 sheets of glitter paper, and 185 sheets of metallic paper. How many sheets of paper does she order?
586 sheets.
Explanation:
The sheets of paper she ordered are 297+104+185 = 586 sheets.
Write an equation and solve the problem.
Question 3.
Dexter buys a package of 38 plates. He already has 4 plates. He puts an equal number on each of 6 tables. How many plates are on each table?
7 plates
Explanation:
(38+4) ÷ 6 =n
n= 7
The number of plates on each table are 7 plates.
Question 4.
Use the tangram pieces to make shapes. Choose one shape and copy it on a separate sheet of paper. Find the area of the shape you made. Remember, the square is one square inch.
Question 5.
Stretch Your Thinking Anna has 30 feet of fence to go around her garden. The lettuce will be in a 8 ft × 5 ft section and the carrots will be in a 2 ft × 7 ft section. Will her layout work? If not, how can she change it to work?
No her layout wouldn’t work.
Explanation:
No, she would need 40 ft with this layout. To make it work, she can put the 2 ft×7 f section along the 8 ft side and use only 30 ft of fence. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Conversion of Systems of Measure
## Convert between measures in word problems
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Practice Conversion of Systems of Measure
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Use Metric and Customary Units of Measurement in Problem Solving
Have you ever worked with customary and metric units of measurement together? This involves conversion. Take a look at this dilemma.
Omar has a dog that weighs 30 pounds. About how many kilograms does Omar’s dog weigh?
### Guidance
Did you know that you can convert between customary and metric units of measurement? These measurement conversions will be estimates, because you cannot make an exact measurement when converting between systems of measurement.
There are some common benchmarks you can use.
Length
An inch is about 2.5 centimeters.
A meter is slightly longer than a yard.
A kilometer is about 0.6 of a mile.
Capacity
A liter is about the same as a quart.
Mass
A kilogram is a little more than 2 pounds.
Write these benchmarks down in your notebook.
Now look at this situation.
Randy ran a 20 kilometer race. How many miles did Randy run?
To figure this out, we have to use a comparison between kilometers and miles. A kilometer is about .6 of a mile. That is our first ratio.
1 km.6 mile
Next, we have to look at how many kilometers, Randy actually ran. He ran 20 kilometers. We are looking for his miles. That forms our second ratio, and we can now write a proportion.
1 km.6 mile=20 kmx miles
Lastly, we cross multiply and solve for x\begin{align*}x\end{align*}.
20×.6=x miles\begin{align*}20 \times .6 = x \ miles\end{align*}
Randy ran 12 miles in his race.
Notice that we solved this problems by using ratios and proportions. As long as you keep in mind what you are comparing, you can solve any measurement conversion problem in this way!
Solve each by using benchmarks.
#### Example A
John ran 5 kilometers. How many miles did he run?
Solution: 3.1 miles
#### Example B
Kary measured out 12 inches on a ruler. About how many centimeters would that be?
Solution: 30 cm
#### Example C
Sandy ran 15 meters. About how many feet is that?
Solution: 45 feet
Now let's go back to the dilemma from the beginning of the Concept.
Use an estimated conversion factor to write a ratio. We will compare kilograms to pounds. One kilogram is about 2 pounds, or 1 kilogram2 pounds\begin{align*}\frac{1 \ kilogram}{2 \ pounds}\end{align*}.
Next, write a proportion and solve.
1 kilogram2 poundsx(2)2xx=x kilograms30 pounds=1(30)=30=15
Omar’s dog weighs about 15 kilograms.
### Vocabulary
Metric System
a system of measurement commonly used outside of the United States. It contains units such as meters, milliliters and grams.
### Guided Practice
Here is one for you to try on your own.
How many meters are in 67 feet?
Solution
20.421\begin{align*}20.421\end{align*}
### Practice
Directions: Answer each question by using benchmarks.
1. About how many centimeters is in one inch?
2. About how many centimeters is in three inches?
3. About how many inches is in 5 centimeters?
4. About how many centimeters is in 1 foot?
5. About how many centimeters is in 3 feet?
6. About how many centimeters is in 1 yard?
7. About how many meters are there in one yard?
8. About how many meters are there in three yards?
9. About how many yards are there in twenty -four meters?
10. About how many feet are there in eighteen meters?
11. About how many kilometers are there in 1.8 miles?
12. About how many miles are there in 10 kilometers?
13. About how many miles are there in 30 kilometers?
14. About how many miles are there in 15 kilometers?
15. About how many kilometers are there in 10 miles?
### Vocabulary Language: English
Customary System
Customary System
The customary system is the measurement system commonly used in the United States, including: feet, inches, pounds, cups, gallons, etc.
Metric System
Metric System
The metric system is a system of measurement commonly used outside of the United States. It contains units such as meters, liters, and grams, all in multiples of ten.
Proportion
Proportion
A proportion is an equation that shows two equivalent ratios.
Ratio
Ratio
A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”. |
# Solution
An "even" number is a number of the form $2n$, where $n$ is an integer. An "odd" number is one of the form $2n+1$, where $n$ is an integer. Show that the sum of two even or two odd numbers is even, the sum of an odd and an even integer is odd, and the product of two odd numbers is odd.
Part I.
Let $a$ and $b$ be even numbers. As such, $a=2n_1$ and $b=2n_2$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= 2n_1 + 2n_2 \\&= 2 (n_1 + n_2)\end{align} We know that $(n_1 + n_2)$ must be an integer due to the closure of the integers under addition. So, defining $n_3 = n_1 + n_2$, we see that the sum $a+b$ can be written as $2n_3$ for some integer $n_3$, making it an even number. Hence, the sum of two even numbers is even.
Part II.
Now let $a$ and $b$ be odd numbers. As such, $a=n_1+1$ and $b=2n_2+1$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= (2n_1 + 1) + (2n_2 + 1) \\ &= 2n_1 + 2n_2 + 2\\ &= 2(n_1 + n_2 + 1)\end{align} We know that $(n_1 + n_2 + 1)$ must be an integer due to the closure of the integers under addition. So defining $n_3 = n_1 + n_2 + 1$, we see that the sum $a+b$ can be written as $2n_3$ for some integer $n_3$, making it an even number. Hence, the sum of two odd numbers is even.
Part III.
Now let $a$ be odd and $b$ be even. As such, $a=2n_1+1$ and $b=2n_2$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= (2n_1 + 1)+ 2n_2 \\ &= 2n_1 + 2n_2 + 1\\ &= 2(n_1 + n_2) + 1\end{align} We know that $(n_1 + n_2)$ must be an integer due to the closure of the integers under addition. So defining $n_3 = n_1 + n_2$, we see that the sum $a+b$ can be written as $2n_3+1$ for some integer $n_3$, making it an odd number. Hence, the sum of an even and an odd integer is odd.
Part IV.
Now let $a$ and $b$ be odd numbers. As such, $a=2n_1 + 1$ and $b=2n_2+1$ for some integers $n_1$ and $n_2$. Now consider their product: \begin{align} ab &= (2n_1 + 1)(2n_2 + 1) \\ &= 4n_1 n_2 + 2n_1 + 2n_2 + 1 \\ &= 2(2n_1 n_2 + n_1 + n_2) + 1\end{align} We know that $(2n_1 n_2 + n_1 + n_2)$ must be an integer due to the closure of the integers under addition and multiplication. So defining $n_3 = (2n_1 + n_1 + n_2)$, we see that the product $ab$ can be written as $2n_3+1$ for some integer $n_3$, making it an odd number. Hence, the product of two odd numbers is odd. |
# Integers: Prime Numbers
If $$z$$ is a positive integer smaller than integer $$n$$, is $$z$$ a factor of $$n$$? >(1) $$n$$ is divisible by all the positive integers less than or equal to 9. >(2) $$z$$ is not divisible by any prime number.
Incorrect. [[snippet]] Stat. (2) is devious, but what it really means is that $$z=1$$. The only positive number that isn't divisible by any prime number is 1. In order to understand this concept, first recall that any positive integer greater than 1 is either prime (option A) or can be broken down into prime numbers (option B): * Option A: A prime number is always divisible by itself (for example, 7 is divisible by 7). So if $$z$$ is a prime number, it is—by definition—divisible by a prime number, which contradicts the statement. Hence, $$z$$ cannot be prime. * Option B: Any nonprime integer greater than 1 can be broken down into prime numbers. For example, 15 can be broken down into $$3 \times 5$$ and is therefore divisible by 5 and 3. Hence, $$z$$ cannot be a nonprime integer greater than 1 or else it would be divisible by a prime number, contradicting the statement. Thus, the only positive value for $$z$$ (that is not divisible by any prime number) is 1. Since 1 is a factor of any integer (any number is divisible by 1), $$z$$ _must_ be a factor of $$n$$, and the answer to the question is a definite "__Yes__." Therefore **Stat.(2) → Yes → S → BD**.
Correct. [[snippet]] According to Stat. (1), $$n$$ contains all the factors from 1 through 9, inclusive. __Plug In__ for $$z$$. If $$z=5$$, then $$z$$ is indeed a factor of $$n$$ (since $$n$$ must be divisible by 5), and the answer is "__Yes__." However, is the answer always "__Yes__"? The variable $$z$$ could also be (for example) a prime number greater than 9, in which case it isn't a factor of $$n$$. For example, if $$n=9!$$ (which satisfies the question stem) and $$z=13$$, then $$z$$ is _not_ a factor of $$n$$, and the answer is "__No__." Therefore **Stat.(1) → Maybe → IS → BCE**. Stat. (2) is devious, but what it really means is that $$z=1$$. The only positive number that isn't divisible by any prime number is 1. In order to understand this concept, first recall that any positive integer greater than 1 is either prime (option A) or can be broken down into prime numbers (option B): * Option A: A prime number is always divisible by itself (for example, 7 is divisible by 7). So if $$z$$ is a prime number, it is—by definition—divisible by a prime number, which contradicts the statement. Hence, $$z$$ cannot be prime. * Option B: Any nonprime integer greater than 1 can be broken down into prime numbers. For example, 15 can be broken down into $$3 \times 5$$ and is therefore divisible by 5 and 3. Hence, $$z$$ cannot be a nonprime integer greater than 1 or else it would be divisible by a prime number, contradicting the statement. Thus, the only positive value for $$z$$ (that is not divisible by any prime number) is 1. Since 1 is a factor of any integer (any number is divisible by 1), $$z$$ _must_ be a factor of $$n$$, and the answer to the question is a definite "__Yes__." Therefore **Stat.(2) → Yes → S → B**.
Incorrect. [[snippet]] According to Stat. (1), $$n$$ contains all the factors from 1 through 9, inclusive. __Plug In__ for $$z$$. If $$z=5$$, then $$z$$ is indeed a factor of $$n$$ (since $$n$$ must be divisible by 5), and the answer is "__Yes__." However, is the answer always "__Yes__"? The variable $$z$$ could also be (for example) a prime number greater than 9, in which case it isn't a factor of $$n$$. For example, if $$n=9!$$ (which satisfies the question stem) and $$z=13$$, then $$z$$ is _not_ a factor of $$n$$, and the answer is "__No__." Therefore **Stat.(1) → Maybe → IS → BCE**.
Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked.
Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked.
BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. |
# Multiplication Lesson
Document Sample
``` Mathematics Lesson: Year 3 - Multiplication
Year group: 3
Resources: digit cards (0 to 10) or dice, multilink cubes, Multiplication Tables grid,
number fans
Vocabulary: multiply, multiplication, row (across), column (down)
Oral and mental objectives: To know by heart and use all x10 tables.
Framework ref: Section 5 pages 52-53; Section 6 pages 58-59
Oral and mental activities:
Pairs place all digit cards face down on table. Mix these up. One pupil turns over a
card and multiplies by 10 saying “number times 10 equals…..” and then says the
After 5 minutes give out 2 dice to each pair. Roll both together. Add the total and
multiply by 10 saying the answer “total of both dice times 10 equals….” and then says
the answer. Score a point if correct. Repeat.
Possible extensions could include
1. asking the pair to turn over 2 of the digit cards, adding the numbers together
and then multiplying by 10
2. asking the pair to roll one die and turning over one digit card, adding the
numbers together and then multiplying by 10.
3. asking the pair to multiply by a number other than 10.
Main teaching objectives: To understand the relationship between multiplication and
division
Framework ref: Section 5, page 57
Main teaching activities:
Teacher to show the multiplication grid (with the x10 row and column empty) on a
large monitor/ screen/ laminated sheet/ acetate. Explain the difference between a row
and a column. Explain to children that they are going to use this square to write
multiplication and related division sums. Selected pupil to click on a cell. As a class,
explore what two numbers need to be multiplied to fill in this cell. How do we know?
Complete as a class the x10 row and column.
Pick any cell and highlight red. What multiplication do I do to get this answer? What
is the related division? Write on board. Repeat.
(I found that having different coloured rows helped pupils with poor visual tracking
skills).
Tasks:- (encourage discussion of methods. Pupils may benefit by referring to 100
square, or even number lines)
Lower Ability In Pairs. Either on computer or as print out, look at multiplication grid.
Change the colour of any number in the x10 row or column to red. What
multiplication goes with this cell? Can we make a related division? Write these in
books or on the sheet.
(example: child picks 40 in x10 column. 4x10= 40. Related division 40 / 10= 4)
Middle Ability In Pairs. Either on computer or as print out, look at multiplication
grid. Change the colour of any number in the grid to red. What multiplication goes
with this cell? Can we make a related division? Write these in books or on the sheet.
Higher Ability In Pairs. Either on computer or as print out, look at adapted
multiplication grid with empty x2, x5, x10 rows. Complete the grid. Next change the
colour of any number in the grid to red. What multiplication goes with this cell? Can
we make a related division? Write these in books or on the sheet.
Plenary:
Display the multiplication grid on the board. Cover one of the answers in a cell. What
number should be in the cell? Write the number on whiteboards. Now try to write a
multiplication and matching division on whiteboards. Compare and explain methods.
Repeat, covering a different number.
Further uses for the ICT resource:
Pupils could, in future lessons, use the grid to investigate number patterns such as
odd/ even numbers. Pupils could highlight rows/ columns of even numbers. What
patterns are there? Can we explain these patterns? |
# What's 15 Of 50?
## Introduction
If you’re someone who loves numbers, then you must have come across the term “15 of 50” at some point. This term is used to represent a percentage, but what exactly does it mean? In this article, we’ll explore what 15 of 50 is and how it works.
## What is 15 of 50?
To put it simply, 15 of 50 is a fraction that represents a percentage. To calculate it, you divide 15 by 50 and then multiply the result by 100. So, 15/50 x 100 = 30%. Therefore, 15 of 50 is equal to 30%.
### Example
Let’s say you have 50 marbles, and you want to know what percentage 15 of those marbles represent. You would divide 15 by 50 and then multiply the result by 100. So, 15/50 x 100 = 30%. Therefore, 15 of 50 represents 30% of the marbles.
## Why is 15 of 50 important?
Understanding percentages is an essential skill in many areas of life, from finance to cooking. Knowing what 15 of 50 means can be helpful when you’re trying to calculate discounts, markups, or proportions.
### Discounts
Suppose you’re shopping and you see a shirt that originally cost \$50, but it’s on sale for 15% off. To calculate the new price, you would multiply the original price by the percentage of the discount. So, \$50 x 15% = \$7.50. Therefore, the new price would be \$42.50.
### Markups
Now, let’s say you’re a business owner, and you want to increase the price of your product by 15%. To do this, you would multiply the original price by the percentage increase. So, if the original price was \$50, 15% of that would be \$7.50. Therefore, the new price would be \$57.50.
### Proportions
Knowing what 15 of 50 means can also be helpful when you’re trying to compare two quantities. For example, if you have 15 red apples and 50 green apples, you would know that the ratio of red to green apples is 15:50, which reduces to 3:10.
## Conclusion
In conclusion, 15 of 50 represents 30%, and it’s a fraction that can be used to calculate percentages, discounts, markups, and proportions. Understanding percentages is an essential skill in many areas of life, and knowing what 15 of 50 means can be helpful in various situations. |
# Lesson video
In progress...
Hi year six, welcome to our fourth lesson in the decimals and measures topic.
Today, we will be solving problems involving the conversion of length.
All you'll need is a pencil and a piece of paper.
So pause the video and get your equipment, if you haven't done so already.
In today's lesson, we'll be solving problems involving the conversion of length, starting with a quiz to test your knowledge from our previous lesson.
We'll then look at calculating perimeter before moving on to calculating area and then some independent work.
So we're starting off by looking at calculating perimeter.
Now, I want you to think about how do we calculate perimeter? What is the procedure for calculating it? So the perimeter is the length of the outline of the shape.
So it's the distance all the way around the shape.
And we calculate it by adding the lengths of the sides together.
So we could either, do it by adding, nine and nine, because we know the bottom one is nine and four and four, which is equal to 26 centimetres.
The distance around the shape is 26 centimetres or we could use our knowledge of multiplication.
So nine multiplied by two plus four multiplied by two is equal to 26 or we could use a nine plus four and then multiply that by two, which will give us 26 centimetre.
So whichever approach you use, these are different ways of calculating the perimeter.
So now, if I wanted to know, what is the perimeter of the shape in millimetres? How would I go about working that out? Well, I know that one centimetre is equivalent to 10 millimetres.
So if we think back to our previous lesson, centimetres to millimetres, the conversion is to multiply by 10.
So I would multiply my 26 centimetres by 10.
So I know that the perimeter is 260 millimetres.
Let's have a look at another one together.
So I would like to calculate the perimeter of this shape in centimetres.
Now, what you will notice is that my units are not the same, so it makes sense for me to convert them, see that they're in the same unit.
So if I'm being asked for my perimeter in centimetres, it makes sense to convert these both to centimetres.
So one centimetre was 10 millimetres.
So to convert from millimetres to centimetres, I need to divide by 10, 35 divided by 10 is 3.
5 centimetres.
So 35 millimetres is equivalent to 3.
5 centimetres.
So now I need to calculate my perimeter.
So I'm either using repeated addition or I can use multiplication as we showed on the previous slide.
Either way I know that these add up together to give me a perimeter of 22 centimetres.
Now, what if I wanted to convert that to millimetres? Well, I know centimetres to millimetres I'm multiplied by 10, so that would be, 220 millimetres as the perimeter.
So now I'd like you to pause the video and calculate the perimeter of each shape in the given unit.
So you may have to convert the units given so that you can give your perimeter in the correct unit.
Pause the video now and calculate the perimeter.
So for the first one, which is a square, you were asked to calculate the perimeter in millimetres.
So I'm converting this by multiplying it by 10 to 34 millimetres.
And I know that as it's a square, I'm adding 34, four times, or I'm doing 34 multiplied by four, which is equal to 136.
So the perimeter is 136 millimetres.
For the second one, I'm asked for the perimeter in centimetres.
So I can convert these both to centimetres first, or I could do it afterwards, I'll do it first, so I know dividing by 10, this is 5.
6 centimetres, and this is 10.
7 centimetres.
And to calculate the perimeter, I'm going to do 5.
6 multiply by two plus 10.
7 multiplied by two to give me the perimeter, which will be 11.
2 plus 21.
4, which is equal to 32.
6 centimetres.
Now, that takes us on to compound rectilinear shapes.
This is where we have shapes made up of multiple rectangles or squares.
Now this shape represents the floor plan of the kitchen in Simone's new house and it's a compound rectilinear shape, as I just said.
Now, we want to work out the perimeter of the kitchen in centimetres.
Now, as the measurements are all already in metres, it would be most efficient to keep them as metres, find the perimeter in metres and then convert my final answer into centimetres.
Now there's some things we need to add on before we can calculate the perimeter.
So the first one is this left hand side.
So this is five metres tall and this is three metres tall.
So the left hand measurement is eight metres.
And then we have this one here, well, I can see that this one is 8.
5 metres wide, and 3.
2 metres is taking up some of that.
So I need to find the difference between 8.
5 and 3.
2, which is 5.
3 metres.
So now I can calculate the perimeter by adding all of the sides together.
So I've done this already eight plus 8.
5 plus 3.
5.
3 plus five plus 3.
2 is equal to 33 metres.
So the perimeter is 33 metres, but I'm being asked for it in centimetres so I need to convert it.
Now, if you think back to our previous lesson, one metre is equivalent to 100 centimetres.
So to convert from metres, to centimetres, I multiply my number by 100.
So 33 multiplied by 100 is 3,300 centimetres.
So you need to make the decision about what is most efficient.
Is it most efficient for me to convert all of my measures into my ultimate desired unit? Or is it more efficient for me to do it afterwards? Like our model there.
So use that thought process to calculate the perimeter of this compound rectilinear shape.
I've put a hint up there.
You've got three missing length that you need to find first.
And then I'd like your perimeter in metres.
Pause the video now and calculate the perimeter.
So, first of all, you needed to add in your three missing length.
So we had one here, which we have from the other side is nine centimetres.
So actually you may have combined them both to be 11 centimetres.
We had one here, which is the same as this one, two centimetres.
And then we had this gap here, which was the difference between 7.
5 and five, which was 2.
5 centimetres.
It's going to be much quicker for us to calculate it in centimetres and then convert rather than convert each of these individual measurements into metres first.
So if we add them all together, we have a perimeter in centimetres of 37 centimetres.
So we need to convert this to metres.
We know that 100 centimetres is equal to one metre.
So to convert from centimetres to metres, we divide by 100.
So 37 divided by 100, is equal to 0.
37 metres.
Now, before we move on to area, we need to have a look at practising a scale, which will help us in our subsequent questions.
So we're looking at how do we efficiently multiply decimals? So here we've got eight times 0.
4.
Now we need to think about how do we efficiently do this? Now 0.
4 is the same as four divided by 10.
So I can do this in a more efficient way.
I can do eight times four and then divide my answer by 10.
Because 0.
4 is 10 times smaller than four.
It's easier for me to multiply integers than it is for me to multiply an integer by a decimal.
So another way of thinking about it, is like this.
If I have eight times 0.
4 and I multiply the 0.
4 by 10 to make it easier to work with eight times four, then what I need to do at the end of it is to actually then convert it back by dividing my answer by 10.
So you need to convert integers, but then don't forget to use your knowledge of decimals to convert back.
So eight times four is 32, and then we divide our answer by 10, which gives us 3.
2.
So you're going to pause the video and multiply the decimal by the integer.
We'll do this one together first and then you'll do the other three.
So six times 0.
2, 0.
2 is 10 times smaller than two.
So I will do six times two and then I'll divide my answer by 10.
So six times two is 12 and divide my answer by 10 is equal to 1.
2.
So use that as a guide for working out the other three.
Pause the video now and multiply the decimal by the integer.
So for the next one, what you could have here was four multiplied by 11 and then divide your answer by 10, which gives you 4.
4.
This one, this is the equivalent to seven divided by 10.
So we can do seven multiplied by 12 and then divide the answer by 10, which is 8.
4.
And then this one was exactly the same, but slightly greater to number.
So this was six multiplied by 45.
And then the answer divided by 10, which is equal to 27.
Now we'll look at multiplying a decimal by a decimal.
So 0.
8 is the same as eight divided by 10.
And 0.
4 is the same as four divided by 10.
So this time, if I do my eight multiplied by four, which is equal to 32, then I need to make it 10 times smaller and another 10 times smaller, which is 100 times smaller.
So that will be 32 divided by 100, which is 0.
32.
It's divided by 100 because we're dividing by 10 twice.
So I can write that a bit nicer eight times four divided by 100.
So we'll do the same again, okay.
I'm going to do the first one with you and then you can do the other three independently.
So 0.
6 is the same as six divided by 10 0.
2 is the same as two divided by 10.
So to figure out 0.
6 times 0.
2, I do six times two and then I divide my answer by 100, which is 0.
12.
So use that logic to calculate the other three.
Pause the video now and multiply the decimals.
So for this one, you are working on four multiplied by 11 divided by 100, which is equal to 0.
44.
This one, 17 multiplied by 12 divided by 100, which is equal to 2.
04 and finally 61 multiplied by 45 divided by 100, which is equal to 27.
45.
Now, we will use this multiplying decimals and integers when we're calculating the area.
So how do we calculate the area? The area is calculated by multiplying length times height.
And the answer is given in unit squared.
So if this was 12 centimetres and this was six centimetres.
It would be 12 times six which is 72 and the units are centimetres squared.
So in today's lesson, when we're calculating the area and our final units are different to our initial units, what I would like you to do is to always convert the initial units into the final units first.
And we're going to go over this in more detail in a couple of lessons time.
So I need the centimetres and millimetres.
If one centimetre is equal to 10 millimetres, then to convert, I multiply it by 10.
So 11 centimetres is 110 millimetres.
23 centimetres is 230 millimetres.
To calculate the area, I'm going to multiply 230 by 110.
And that is equal to 25,300 millimetres squared.
I'll pop that in down here.
So the most important thing is that today we are going to convert our measures into the final desired unit first.
So for this question, we're asked for our final area in millimetres squared.
So again, I need to convert these initial measurements into millimetres.
One centimetre is equivalent to 10 millimetres.
So I'm going to multiply each of my measures by 10 to convert them to millimetres.
1.
4 times 10 is 14 millimetres 2.
3 times 10 is 23 millimetres.
So then I'll use long multiplication to multiply 23 by 14, Three times four is 12, two times four is eight, plus one is nine place hold zero.
One times three is three, one times two is two and then add those together.
So my answer is 322 millimetres squared.
Now it's your turn to calculate the area of each shape.
Look at the given unit and convert your measures first.
Pause the video now and calculate the area.
So for your first one, you will have converted 3.
4 centimetres into 34 millimetres and 34 millimetres multiplied by 34 millimetres as it's a square is equal to 1,156 millimetres squared For your second one, you are converting them into centimetres.
So that's 5.
6 multiplied by 10.
7.
So using our multiplying decimals approach before that would have been 56 times 107 divided by 100, which is equal to 59.
92 centimetres squared.
Now let's look at area of compound rectilinear shapes.
So it's the same rules applying here.
We need to think of it as two separate shapes.
So we find the area of each and then we add them together.
So we find the area of A then the area of B and then add to them.
We're being asked to calculate it in metre squared, but they are in centimetres.
So remember we need to convert 100 centimetres is equal to one metre.
So to convert for centimetres to metres, I divide by 100.
So this would be one metre by 0.
4 metres.
And this would be 0.
5 metres by 0.
2 metres.
To calculate the area of A, I'm doing one metre times 0.
4 metres, which is equal to 0.
4 metres squared, that's the area of that.
And for B I'm multiplying 0.
5 metres by 0.
2 metres.
So remember that's the same as two multiplied by five divided by 100, which is 0.
1 metre squared and then I add the two of them together to get my total area of 0.
5 metres squared.
So use the same approach.
this time you're calculating the area in metre squared, so you need to address these two numbers first.
Pause the video and calculate the area.
So converting from centimetres to metres you're dividing by 100.
So 700 centimetres is equivalent to seven metres and 200 centimetres is two metres.
So if we call this A, the area of A, is seven multiplied by two, which is 14 metres squared.
B was already in metres for us so that was nine metres multiply by five metres is 45 metres squared and then we were just adding the two shapes together, which is equal to 59 metres squared.
Now it's time for some independent learning.
So pause the video and complete the task and click restart once you're finished.
So for question one, we have a floor plan of a museum.
In the museum the manager needed to close off the Math Zone so that it could be updated.
She needed to put a rope around the perimeter.
So we needed to figure out how much rope she would need by calculating the perimeter of the shape.
This one here, would be the total 12, subtract this part eight.
So this is four metres.
This part here is 6.
25 metres.
And then this one here is 9.
5 metres.
So you were adding all of these measures together to calculate the perimeter.
And the perimeter was 55.
5 metres.
That's those numbers added together, but you were asked for that in centimetres.
So you know that you have to multiply it by 100 to convert metres to centimetres.
Which is equal to 5,550 centimetres as the perimeter.
Now here you were asked to calculate the missing length.
So you know that 14 multiplied by W is equal to 294.
So using your knowledge of the inverse, you will have done 294 divided by 14 is equal to W so, therefore, W is equal to 21 centimetres.
21 multiplied by 14 is equal to 294 centimetre.
Here we were looking at the perimeter and you were asked to calculate the missing length.
So first the perimeter needed to be converted into centimetres, as all of the others were in centimetres, divided by 10 is 31.
2 centimetres.
So we know that this measurement 8.
2 plus this one, plus two lots of something is equal to 31.
2.
So using the inverse, you subtract the known numbers, and that leaves us a 14.
8.
So this side and this side are 14.
8 divided by two means that each of these sides were 7.
4 centimetres.
And finally, you were asked to calculate the area and perimeter of the shape in the given units.
So your first one was asking you for centimetres, which meant that you needed to convert your millimetres to centimetres first.
So 15 centimetres and four centimetres, 15 multiplied by four is 60 centimetre squared.
And then add those sides together gives you 38 centimetres.
In your next one, you're given units with millimetres, so you needed to initially convert to a 100 millimetres, 40 millimetres, 50 millimetres and 20 millimetres, and then find the area of each shape and then add them together.
So you were working on, 100 multiplied by 40 is 4,000 millimetres squared.
And 50 multiplied by 20 is 1,000 millimetres squared.
And then you were just adding those two together.
So that gave you 5,000 millimetres squared.
And then adding all of these known sides together, is 380 millimetres.
So you needed to convert these measurements into metres by dividing by 100.
So 0.
1 metres and 0.
03 metres.
0.
1 multiplied by 0.
03, is equal to 0.
003 metres.
And this one was 0.
06 , 0.
05, multiply those two together you also get 0.
03 point metres.
And that means that when we add those two together, we get an area of 0.
006 metres squared.
And then if you add all of the sides together in metres, we reach 0.
38 metres as our perimeter.
Great work today.
That was a really tricky lesson.
But well done for persevering with it.
In our next lesson, we'll be learning to calculate the area of parallelograms and triangles.
I'll see then. |
Students can download Maths Chapter 6 Trigonometry Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2
Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = $$\frac{√3}{2}$$; cos 60° = $$\frac{1}{2}$$
L.H.S = sin² 60° + cos² 60°
L.H.S = R.H.S
Hence it is proved.
(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = $$\frac{1}{√3}$$; sec 30° = $$\frac{2}{√3}$$
L.H.S = 1 + tan² 30°
∴ L.H.S = R.H.S
Hence it is proved.
(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = $$\frac{1}{√2}$$, cos 45° = $$\frac{1}{√2}$$
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 ($$\frac{1}{√2}$$)²
= 1 – 2 × $$\frac{1}{2}$$
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2($$\frac{1}{√2}$$)² – 1
= $$\frac{2}{2}$$ – 1
= $$\frac{2 – 2}{2}$$ = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.
(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = $$\frac{1}{2}$$; cos 60° = $$\frac{1}{2}$$; cos 30° = $$\frac{√3}{2}$$; sin 60° = $$\frac{√3}{2}$$; sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°
= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.
Question 2.
Find the value of the following:
(i) $$\frac{tan 45°}{cosec 30°}$$ + $$\frac{sec 60°}{cot 45°}$$ – $$\frac{5 sin 90°}{2 cos 0°}$$
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan4 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1
= 0
(ii) sin 90° = 1; cos 60° = $$\frac{1}{2}$$; cos 45° = $$\frac{1}{√2}$$; sin 30° = $$\frac{1}{2}$$; cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
= $$\frac{7}{4}$$
(iii) sin 30° = $$\frac{1}{2}$$; cos 60° = $$\frac{1}{2}$$; tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan4 45°
= 3
Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°
= 0
∴ L.H.S = R.H.S
Hence it is proved.
Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × $$\frac{1}{2}$$ × $$\frac{1}{2}$$ × 1
= 2 |
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# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 10 –
Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ Solution 14: Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is 97 92 5 1 1995 1985 10 2 Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y). Slope of AB = Slope of BC 1 y 97 2 2010 1995 1 y 97 2 15 15 y 97 2 y 97 7.5 y 7.5 97 104.5 Thus, the slope of line AB is 1 , while in the year 2010, the population will be 104.5 crores. 2 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 10 – Straight Lines Maths ______________________________________________________________________________ Exercise 10.2 Question 1: Write the equation for the x and y-axes. Solution 1: The y-coordinate of every point on the x-axis is 0. Therefore, the equation of the x-axis is y = 0. The x-coordinate of every point on the y-axis is 0. Therefore, the equation of the y-axis is y = 0. Question 2: Find the equation of the line which passes through the point (-4, 3) with slope 1 2 . Solution 2: We know that the equation of the line passing through point x y y y mx x . 0 0 0, 0 Thus, the equation of the line passing through point (-4, 3), whose slope is 1 2 , is 1 y3 x 4 2 2 (y-3) = x + 4 2y – 6 = x + 4 i.e., x – 2y + 10 = 0 , whose slope is m, is Question 3: Find the equation of the line which passes through (0, 0) with slope m. Solution 3: We know that the equation of the line passing through point x y y y mx x 0 0 0, 0 Thus, the equation of the line passing through point (0, 0), whose slope is m, is (y – 0) = m (x – 0) i.e., y = mx , whose slope is m, is Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Question 29
The members of a consulting firm rent cars from three rental agencies: 50% from agency X, 30% from agency Y and 20% from agency Z. From past experience it is known that 9% of the cars from agency X need a service and tuning before renting, 12% of the cars from agency Y need a service and tuning before renting and 10% of the cars from agency Z need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency Z is not to be blamed.
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### Transcript
Question 29 The members of a consulting firm rent cars from three rental agencies: 50% from agency X, 30% from agency Y and 20% from agency Z. From past experience it is known that 9% of the cars from agency X need a service and tuning before renting, 12% of the cars from agency Y need a service and tuning before renting and 10% of the cars from agency Z need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency Z is not to be blamed. Let us define events as A: Car needs service E1: Car is rented from agency X E2: Car is rented from agency Y E3: Car is rented from agency Z We need to find the probability that Car is not chosen from agency Z, if the car needs service Here, we find probability that Car is chosen from agency Z, if the car needs service i.e. P(E3|A) Here, P(E1) = Probability that agency X is chosen = 50% = 50/100 P(E2) = Probability that agency Y is chosen = 30% = 30/100 P(E3) = Probability that agency Z is chosen = 20% = 20/100 P(A|E1) = Probability that car needs service if agency X is chosen = 9% = 9/100 P(A|E2) = Probability that car needs service if agency Y is chosen = 12% = 12/100 P(A|E3) = Probability that car needs service if agency Z is chosen = 10% = 10/100 Now, P(E3|A) = (𝑃(𝐸_3 ).𝑃(𝐴|𝐸_3))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2)+𝑃(𝐸_3 ).𝑃(𝐴|𝐸_3)) = (20/100 × 10/100 )/(50/100 × 9/100 + 30/100 × 12/100 + 20/100 × 10/100) = (20 × 10 )/(50 × 9 + 30 × 12 + 20 × 10) = 200/(450 + 360 + 200) = 200/1010 = 20/101 Thus, P(E3|A) = 20/101 Now, P(Car is not chosen from agency Z, if the car needs service) = 1 – P(E3|A) = 1 – 20/101 = 81/101 Thus, required probability is 𝟖𝟏/𝟏𝟎𝟏 |
Parallelogram Formula
View Notes
Introduction to Parallelogram Formula
A parallelogram is one of the types of quadrilaterals. A quadrilateral is a closed geometric shape which has 4 vertices, 4 sides and hence 4 angles that lie on the same plane. Sum of the interior angles of a quadrilateral measures 3600. A quadrilateral is a type of a polygon. There are various kinds of quadrilateral embracing trapezoids, parallelograms and kites. A parallelogram is one of the types of quadrilateral in which opposite sides are equal and parallel and opposite angles are equal.
Properties of Parallelogram
Let us consider the parallelogram ABCD represented in the above figure to understand the properties of a parallelogram in a better manner. The properties of a parallelogram are listed below.
• Opposites sides of a parallelogram are parallel and congruent.
i.e. AB || CD and AC || BD. Also AB = CD and AC = BD
• Opposite angles of a parallelogram are congruent.
• Diagonals of a parallelogram bisect each other.
i.e. AC bisects BD and BD bisects AC.
• Sum of any two interior adjacent angles of a parallelogram is a straight angle (i.e. it measures 1800).
• If one of the interior angles of a parallelogram is a right angle, then all the interior angles are right angles.
• The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
i.e. If the diagonal AC is drawn for the parallelogram ABCD shown in the above figure, the diagonal divides the parallelogram into two triangles: ΔABC and ΔADC such that ΔABC ⩭ ΔADC
Important Parallelogram Formula
1. The Perimeter of a Parallelogram
The perimeter of a parallelogram is the measure of all sides of a parallelogram. A parallelogram is a two dimensional geometric shape. The two measurable dimensions are length and width. Since opposite sides of a parallelogram are congruent, its perimeter can be written as the sum of all the four sides in terms of length and width as:
The Perimeter of a Parallelogram = 2 (L + B)
In the above equation, ‘L’ is the length and ‘B’ is the breadth or width of the parallelogram.
1. Area of Parallelogram Formula
A deeper analysis of the parallelogram properties reveal that the parallelograms are made of two congruent triangles. (i.e. the diagonal of a parallelogram divides it into two congruent triangles.) Since the triangles are congruent, they have the same area. The area of a triangle is measured as half of the product of its base and height. Since the parallelogram has 2 triangles, its area is twice the area of the triangle. Therefore, the area of a parallelogram formula is equal to the product of its base and height.
Area of Parallelogram = Base x Height
Parallelogram Formula Example Problems
1. Find the area of a parallelogram whose base is 5 cm and height is 3 cm.
Solution:
Given: Length of the parallelogram / Base of the parallelogram (B) = 5 cm
Height of the parallelogram (H) = 3 cm
Area of parallelogram formula is given as:
Area = B x H
Area = 5 x 3
Area = 15 cm2
2. Determine the perimeter of a quadrilateral whose sides measure 5 cm, 4 cm, 5 cm and 4 cm taken in an order. Identify whether the given quadrilateral is a parallelogram or not. Justify your answer.
Solution:
Perimeter of a quadrilateral is the sum of all the 4 sides of the quadrilateral. So,
Perimeter = 5 + 4 + 5 + 4 = 18 cm.
The given quadrilateral is a parallelogram because its opposite sides are found to be equal. If the opposite sides are equal, obviously they will be parallel too.
Fun Quiz:
One of the most important learning outcomes of understanding the concept of parallelogram properties is that the student should be able to identify whether the given quadrilateral is a parallelogram or not.
• All rectangles and squares will have the properties of parallelograms whereas all the parallelograms may not be squares or rectangles.
• Square is the only example of a regular quadrilateral.
• Parallelogram is a two dimensional shape and hence its volume cannot be determined.
1. What are Quadrilaterals? Mention its Types and Properties.
A quadrilateral is a four sided closed geometric shape. It has only two dimensions namely length and width. Quadrilaterals are plane geometric shapes. The different types of quadrilaterals include parallelograms, trapezoid and kite. The different types of parallelograms include rhombus and rectangles. Square is a special type of rectangle. The properties of quadrilaterals are:
• All the quadrilaterals have 4 vertices, 4 sides and 4 angles.
• Sum of all the four interior angles of a quadrilateral is 3600.
2. What are different Kinds of Parallelograms?
Parallelograms are the four sided closed plane geometric shapes in which the opposite sides are equal and parallel. The two special kinds of parallelogram are:
1. Rhombus:
Rhombus is a parallelogram in which all the sides are congruent and opposite angles are equal.
1. Rectangle:
A parallelogram can be considered as a rectangle if and only if one of the interior angles of the parallelogram is a right angle. (It is very important to note that if one of the interior angles is 900, then all the 4 interior angles also measure 900)
A square is a special type of a rectangle in which all the sides are congruent. |
# Math Expressions Grade 4 Student Activity Book Unit 2 Lesson 17 Answer Key Use the Shortcut Method
This handy Math Expressions Grade 4 Student Activity Book Answer Key Unit 2 Lesson 17 Use the Shortcut Method provides detailed solutions for the textbook questions.
## Math Expressions Grade 4 Student Activity Book Unit 2 Lesson 17 Use the Shortcut Method Answer Key
Compare Methods of Multiplication
Look at the drawing and the five numerical solutions for 4 × 2,237.
Question 1.
How are the solutions similar? List at least two ways.
Method A and Method C,
Explanation:
Method A and Method C has same result.
Question 2.
How are the solutions different? List at least three comparisons between methods.
Solutions are same comparisons betweeen methods are diiferent Method B, Method C and Method F,
Explanation:
Solutions are same comparisons betweeen methods are diiferent Method B, Method C and Method F,
Question 3.
How do Methods A-D relate to the drawing? List at least two ways.
Methods B and C,
Explanation:
Methods B – C relate to same solutions as Method A – D,
Analyze the Shortcut Method.
Look at this breakdown of solution steps for Method E and Method F.
Question 4.
Describe what happens in Step 1.
Multiplication of 4 with units place 7 gets 8,
Explanation:
If we see in step 1, 4 is multiplied by 7 of units place.
Question 5.
Describe what happens in Step 2.
Multiplication of 4 with tens place 3 and added 2 gets 4,
Explanation:
If we see in step 2, 4 is multiplied by 3 of tens place and result 12 we add 2 we get 14,
we give 4 in result at tens place and take 1 to hundreds.
Question 6.
Describe what happens in Step 3.
Multiplication of 4 with hundreds place 2 and added 1 gets 9,
Explanation:
If we see in step 3, 4 is multiplied by 2 of hundreds place and result 8 we add 1 we write 9 hundreds.
Question 7.
Describe what happens in Step 4.
Multiplication of 4 with thousands place we get 8,
Explanation:
If we see in step 4, 4 is multiplied by 2 of thousands place and result 8 we write at thousands place.
Round and Estimate With Thousands and Hundreds
You can use what you know about rounding and multiplication with thousands to estimate the product of 4 × 3,692.
Question 8.
Find the product if you round up: 4 × 4,000 = _____
16,000,
Explanation:
The product if I round up 4 X 4,000 is 4 X 4 X 1,000 = 16,000.
Question 9.
Find the product if you round down: 4 × 3,000 = _____
12,000,
Explanation:
The product if I round down is 4 X 3,000 is 4 X 3 X 1,000 = 12,000.
Question 10.
Which one of the two estimates will be closer to the actual solution? Why?
4 X 4,000 =16,000 more close,
Explanation:
One of the two estimates will be closer to the actual solution is 12,000 < 14,768 < 16,000 difference is
2,768 < 14,768 < 1,232 so 4 X 4,000 = 16,000 is much closer to 14,768.
Question 11.
Calculate the actual solution. _____
14,768,
Explanation:
The actual solution of 4 X 3,692 is 4 X (3,000 + 600 + 90 + 2) = 4 X 3,000 + 4 X 600 + 4 X 90 + 4 X 2 = 12,000 + 2,400 + 360 + 8 = 14,768.
Question 12.
Explain why neither estimate is very close to the actual solution.
More difference to the actual solution,
Explanation:
There is lot of difference to the both estimates so there is lot of difference to the actual solution two estimates will be closer to the actual solution is 12,000 < 14,768 < 16,000 difference is 2,768 < 14,768 < 1,232 so 4 X 4,000 = 16,000 is much closer to 14,768.
Question 13.
What would be the estimate if you added 4 × 600 to 4 × 3,000; (4 × 3,000) + (4 × 600)? _______
14,400,
Explanation:
The estimate if we add 4 X 600 to 4 X 3,000 is (4 X 3,000) + (4 X 600) = ( 4 X 3 X 1,000) + (4 X 6 X 100) =
12,000 + 24,00 = 14,400.
Question 14.
What would be the estimate if you added 4 × 700 to 4 × 3,000; (4 × 3,000) + (4 × 700)? _____
14,800,
Explanation:
The estimate if we add 4 X 700 to 4 X 3,000 is (4 X 3,000) + (4 X 700) = ( 4 X 3 X 1,000) + (4 X 7 X 100) =
12,000 + 2,800 = 14,800.
Question 15.
Estimate 4 × 7,821 by rounding 7,821 to the nearest thousand.
32,000,
Explanation:
The estimate 4 X 7,821 to the nearest thousand is (4 X 8,000) = 4 X 8 X 1,000 = 32,000.
Question 16.
Find the actual product. _____
31,284,
Explanation:
The actual product 4 X 7,821 is 4 X 7,000 + 4 X 800 + 4 X 20 + 4 X 1 = 28,000 + 3,200 + 80 + 4 = 31,284.
Question 17.
Find a better estimate for 4 × 7,821. Show your work.
4 X 7800 = 31,200,
Explanation:
A better estimate for 4 X 7821 is rounding it to the nearest hundred is 4 X 7800 = 4 X 7000 + 4 X 800 = 2,8000 + 3,200 = 31,200.
Round, estimate, and fix the estimate as needed.
Question 18.
6 × 3,095
Actual :18,570,
Estimate : 18,600,
Explanation:
Given to estimate 6 X 3,095 so 6 X 3,000 + 6 X 90 + 6 X 5 = 18,000 + 540 + 30 = 18,570 rounding to nearest hundred we get 18,600.
Question 19.
7 × 2,784
Actual : 19,488,
Estimate : 19,500,
Explanation:
Given to estimate 7 X 2,784 so 7 X 2,000 + 7 X 700 + 7 X 80 + 7 X 4 = 14,000 + 4,900 + 560 + 28 = 19,488 rounding to nearest hundred we get 19,500.
Estimate Products
You can use estimation to decide if an answer is reasonable.
Question 20.
5 × 3,487 = ____
Actual: 17,435,
Estimate : 17,500,
Explanation:
Given to estimate 5 X 3,487 so 5 X 3,000 + 5 X 400 + 5 X 80 + 5 X 7 = 15,000 + 2,000 + 400 + 35 = 17,435 rounding to nearest hundred we get 17,500.
Question 21.
7 × 8,894 = ____
Actual : 62,258,
Estimate : 62,000.
Explanation:
Given to estimate 7 X 8,894 so 7 X 8,000 + 7 X 800 + 7 X 90 + 7 X 4 = 56,000 + 5,600 + 630 + 28 = 62,258 rounding to nearest hundred we get 62,000.
Question 22.
4 × 7,812 = ____
Actual : 31,248,
Estimate : 31,000,
Explanation:
Given to estimate 4 X 7,812 so 4 X 7,000 + 4 X 800 + 4 X 10 + 4 X 2 = 28,000 + 3,200 + 40 + 8 = 31,248 rounding to nearest hundred we get 31,000.
Question 23.
3 × 4,109 = ____
Actual : 12,327,
Estimate : 12,000,
Explanation:
Given to estimate 3 X 4,109 so 3 X 4,000 + 3 X 100 + 3 X 0 + 3 X 9 = 12,000 + 300 + 0 + 27 = 12,327 rounding to nearest hundred we get 12,000.
What’s the Error?
Dear Math Students,
My school collected 2.468 empty cartons of milk during the day today. If the school collects about the same number of cartons each day for 5 days, I estimated that the school will collect 17,500 empty cartons of milk. I wrote this estimate.
(5 × 3,000) + (5 × 500) = 17500
I am not sure if this is a reasonable estimate.
Can you help me? |
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# 3.E: Integers (Exercises)
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
### 3.1 - Introduction to Integers
#### Locate Positive and Negative Numbers on the Number Line
In the following exercises, locate and label the integer on the number line.
1. 5
2. −5
3. −3
4. 3
5. −8
6. −7
#### Order Positive and Negative Numbers
In the following exercises, order each of the following pairs of numbers, using < or >.
1. 4__8
2. −6__3
3. −5__−10
4. −9__−4
5. 2__−7
6. −3__1
#### Find Opposites
In the following exercises, find the opposite of each number.
1. 6
2. −2
3. −4
4. 3
In the following exercises, simplify.
1. (a) −(8) (b) −(−8)
2. (a) −(9) (b) −(−9)
In the following exercises, evaluate.
1. −x, when (a) x = 32 (b) x = −32
2. −n, when (a) n = 20 (b) n = −20
#### Simplify Absolute Values
In the following exercises, simplify.
1. |−21|
2. |−42|
3. |36|
4. −|15|
5. |0|
6. −|−75|
In the following exercises, evaluate.
1. |x| when x = −14
2. −|r| when r = 27
3. −|−y| when y = 33
4. |−n| when n = −4
In the following exercises, fill in <, >, or = for each of the following pairs of numbers.
1. −|−4|__4
2. −2__|−2|
3. −|−6|__−6
4. −|−9|__|−9|
In the following exercises, simplify.
1. −(−55) and − |−55|
2. −(−48) and − |−48|
3. |12 − 5|
4. |9 + 7|
5. 6|−9|
6. |14−8| − |−2|
7. |9 − 3| − |5 − 12|
8. 5 + 4|15 − 3|
#### Translate Phrases to Expressions with Integers
In the following exercises, translate each of the following phrases into expressions with positive or negative numbers.
1. the opposite of 16
2. the opposite of −8
3. negative 3
4. 19 minus negative 12
5. a temperature of 10 below zero
6. an elevation of 85 feet below sea level
In the following exercises, model the following to find the sum.
1. 3 + 7
2. −2 + 6
3. 5 + (−4)
4. −3 + (−6)
#### Simplify Expressions with Integers
In the following exercises, simplify each expression.
1. 14 + 82
2. −33 + (−67)
3. −75 + 25
4. 54 + (−28)
5. 11 + (−15) + 3
6. −19 + (−42) + 12
7. −3 + 6(−1 + 5)
8. 10 + 4(−3 + 7)
#### Evaluate Variable Expressions with Integers
In the following exercises, evaluate each expression.
1. n + 4 when (a) n = −1 (b) n = −20
2. x + (−9) when (a) x = 3 (b) x = −3
3. (x + y)3 when x = −4, y = 1
4. (u + v)2 when u = −4, v = 11
#### Translate Word Phrases to Algebraic Expressions
In the following exercises, translate each phrase into an algebraic expression and then simplify.
1. the sum of −8 and 2
2. 4 more than −12
3. 10 more than the sum of −5 and −6
4. the sum of 3 and −5, increased by 18
In the following exercises, solve.
1. Temperature On Monday, the high temperature in Denver was −4 degrees. Tuesday’s high temperature was 20 degrees more. What was the high temperature on Tuesday?
2. Credit Frida owed $75 on her credit card. Then she charged$21 more. What was her new balance?
### 3.3 - Subtract Integers
#### Model Subtraction of Integers
In the following exercises, model the following.
1. 6 − 1
2. −4 − (−3)
3. 2 − (−5)
4. −1 − 4
#### Simplify Expressions with Integers
In the following exercises, simplify each expression.
1. 24 − 16
2. 19 − (−9)
3. −31 − 7
4. −40 − (−11)
5. −52 − (−17) − 23
6. 25 − (−3 − 9)
7. (1 − 7) − (3 − 8)
8. 32 − 72
#### Evaluate Variable Expressions with Integers
In the following exercises, evaluate each expression.
1. x − 7 when (a) x = 5 (b) x = −4
2. 10 − y when (a) y = 15 (b) y = −16
3. 2n2 − n + 5 when n = −4
4. −15 − 3u2 when u = −5
#### Translate Phrases to Algebraic Expressions
In the following exercises, translate each phrase into an algebraic expression and then simplify.
1. the difference of −12 and 5
2. subtract 23 from −50
#### Subtract Integers in Applications
In the following exercises, solve the given applications.
1. Temperature One morning the temperature in Bangor, Maine was 18 degrees. By afternoon, it had dropped 20 degrees. What was the afternoon temperature?
2. Temperature On January 4, the high temperature in Laredo, Texas was 78 degrees, and the high in Houlton, Maine was −28 degrees. What was the difference in temperature of Laredo and Houlton?
### 3.4 - Multiply and Divide Integers
#### Multiply Integers
In the following exercises, multiply.
1. −9 • 4
2. 5(−7)
3. (−11)(−11)
4. −1 • 6
#### Divide Integers
In the following exercises, divide.
1. 56 ÷ (−8)
2. −120 ÷ (−6)
3. −96 ÷ 12
4. 96 ÷ (−16)
5. 45 ÷ (−1)
6. −162 ÷ (−1)
#### Simplify Expressions with Integers
In the following exercises, simplify each expression.
1. 5(−9) − 3(−12)
2. (−2)5
3. −34
4. (−3)(4)(−5)(−6)
5. 42 − 4(6 − 9)
6. (8 − 15)(9 − 3)
7. −2(−18) ÷ 9
8. 45 ÷ (−3) − 12
#### Evaluate Variable Expressions with Integers
In the following exercises, evaluate each expression.
1. 7x − 3 when x = −9
2. 16 − 2n when n = −8
3. 5a + 8b when a = −2, b = −6
4. x2 + 5x + 4 when x = −3
#### Translate Word Phrases to Algebraic Expressions
In the following exercises, translate to an algebraic expression and simplify if possible.
1. the product of −12 and 6
2. the quotient of 3 and the sum of −7 and s
### 3.5 - Solve Equations using Integers; The Division Property of Equality
#### Determine Whether a Number is a Solution of an Equation
In the following exercises, determine whether each number is a solution of the given equation.
1. 5x − 10 = −35
1. x = −9
2. x = −5
3. x = 5
2. 8u + 24 = −32
1. u = −7
2. u = −1
3. u = 7
#### Using the Addition and Subtraction Properties of Equality
In the following exercises, solve.
1. a + 14 = 2
2. b − 9 = −15
3. c + (−10) = −17
4. d − (−6) = −26
#### Model the Division Property of Equality
In the following exercises, write the equation modeled by the envelopes and counters. Then solve it.
#### Solve Equations Using the Division Property of Equality
In the following exercises, solve each equation using the division property of equality and check the solution.
1. 8p = 72
2. −12q = 48
3. −16r = −64
4. −5s = −100
#### Translate to an Equation and Solve.
In the following exercises, translate and solve.
1. The product of −6 and y is −42
2. The difference of z and −13 is −18.
3. Four more than m is −48.
4. The product of −21 and n is 63.
#### Everyday Math
1. Describe how you have used two topics from this chapter in your life outside of your math class during the past month.
### PRACTICE TEST
1. Locate and label 0, 2, −4, and −1 on a number line.
In the following exercises, compare the numbers, using < or > or =.
1. (a) −6__3 (b) −1__−4
2. (a) −5__|−5| (b) −|−2|__−2
In the following exercises, find the opposite of each number.
1. (a) −7 (b) 8
In the following exercises, simplify.
1. −(−22)
2. |4 − 9|
3. −8 + 6
4. −15 + (−12)
5. −7 − (−3)
6. 10 − (5 − 6)
7. −3 • 8
8. −6(−9)
9. 70 ÷ (−7)
10. (−2)3
11. −42
12. 16−3(5−7)
13. |21 − 6| − |−8|
In the following exercises, evaluate.
1. 35 − a when a = −4
2. (−2r)2 when r = 3
3. 3m − 2n when m = 6, n = −8
4. −|−y| when y = 17
In the following exercises, translate each phrase into an algebraic expression and then simplify, if possible.
1. the difference of −7 and −4
2. the quotient of 25 and the sum of m and n.
In the following exercises, solve.
1. Early one morning, the temperature in Syracuse was −8°F. By noon, it had risen 12°. What was the temperature at noon?
2. Collette owed $128 on her credit card. Then she charged$65. What was her new balance?
In the following exercises, solve.
1. n + 6 = 5
2. p − 11 = −4
3. −9r = −54
In the following exercises, translate and solve.
1. The product of 15 and x is 75.
2. Eight less than y is −32. |
A multiple is the product of one number with another number. Also, we can define a multiple as the result that is obtained by multiplying a number by an integer. But it is not a function. The multiples of the whole numbers are found by doing the product of the counting numbers and that of whole numbers. For example, multiples of 5 can be obtained when we multiply 5 by 1, 5 by 2, 5 by 3, and so on.
Example 1: Find the multiples of whole number 4?
Firstly, do the multiplication of 4 with other numbers to get multiples of 4.
Multiplication: 4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6, 4 x 7, 4 x 8
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32
Solution: The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32,……
Example 2: Find the multiples of whole number 6?
Firstly, do the multiplication of 6 with other numbers to get multiples of 6.
Multiplication: 6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6, 6 x 7, 6 x 8
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48.
Solution: The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48,……
Also, Check: Common Multiples
Properties of Multiples
Here we have given some important Properties of Multiples. Check out the properties and get a grip on them to make your learning easy.
(i) Every number is a multiple of itself.
For example, the first multiple of 5 is 5 because 5 × 1 = 5.
(ii) The multiples of a number are infinite.
We know that numbers are infinite. Therefore, the multiples of a number also infinite. If you take the example of multiples of 2, we begin with 2, 4, 6, 8, 10, 12, 14,…. and so on.
(iii) The multiple of a number is greater than or equal to the number itself.
For example, if we take the multiples of 3: 3, 6, 9, 12, 15, .… and so on. We can see that: The 1st multiple of 3 is equal to 3: 3 × 1 = 3. The 2nd multiple, the 3rd multiple, and the following multiples of 3 are all greater than 3 (6 > 3, 9 > 3, ….)
(iv) 0 is a multiple of every number.
Common Multiples
Multiples that are common to any given two numbers are known as common multiples of those numbers. Check out the example for better understanding.
Consider two numbers– 2 and 3. Multiples of 2 and 3 are –
Multiples of 2 = 2, 4, 6, 8, 10, 12, ….
Multiples of 3 = 3, 6, 9, 12, 15, 18,……….
We observe that 6 and 12 are the first two common multiples of 2 and 3. But what can be the real-life use of common multiples?
Suppose Arun and Anil are cycling on a circular track. They start from the same point but Arun takes 30 seconds to cover a lap while Anil takes 45 seconds to cover the lap. So when will be the first time they meet again at the starting point?
This can be deduced from the list of common multiples. Arun and Anil will meet again after 90 minutes.
First Ten Multiples of the Numbers
Find out the first ten multiples of the numbers from the below figure.
Multiples of Different Numbers
When two numbers are multiplied the result is called the product of the multiple of given numbers. If the number 6 is multiplied with other numbers, then you get different multiples. Also, if the number 7 is multiplied with other numbers, then you get different multiples.
Solved Examples on Multiples
1. Find the first three multiples of 7.
Solution:
The given number is 7.
To find the first three multiples of 7, you need to multiply 7 with 1, 2, 3.
7 × 1 = 7
7 × 2 = 14
7 × 3 = 21
So, 7, 14, 21 are the first 3 multiples of 7.
2. Four friends Alex, Ram, Vijay, and Venu decided to pluck flowers from the garden in the order of the first four multiples of 5. Can you list the number of flowers that each of them plucked as a series of the first four multiples of 5?
Solution:
Given that four friends Alex, Ram, Vijay, and Venu decided to pluck flowers from the garden in the order of the first four multiples of 5.
To find the first four multiples of 5, you need to multiply 5 with 1, 2, 3, and 4.
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
The first four multiples of 5 are (5 × 1) = 5, (5 × 2) =10, (5 × 3) = 15, and (5 × 4) = 20.
Hence, Alex plucked 5 flowers, Ram plucked 10 flowers, Vijay plucked 15 flowers and Venu plucked 20 flowers.
3. Sam loves watering plants. Her mom asked her to water the pots which were marked in the order of the multiples of 8. However, she missed a few pots. Can you help her identify the pots that she missed in the following list: 8, 16, __, 32, __, 48, 56, 64, __?
Solution:
Given that Sam’s mom asked her to water the pots which were marked in the order of the multiples of 8.
Let us start counting the multiplication table of 8: 8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40, 8 × 6 = 48, 8 × 7 = 56, 8 × 8 = 64, 8 × 9 = 72.
The missed pots are 24, 40, and 72. |
## Engage NY Eureka Math Precalculus Module 3 Lesson 8 Answer Key
### Eureka Math Precalculus Module 3 Lesson 8 Exercise Answer Key
Exercises
Exercise 1.
Let F(0, 5) and G(0, – 5) be the foci of a hyperbola. Let the points P(x,y) on the hyperbola satisfy either PF – PG = 6 or PG – PF = 6. Use the distance formula to derive an equation for this hyperbola, writing your answer in the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1.
PG – PF = 6
$$\sqrt{x^{2} + (y + 5)^{2}}$$ – $$\sqrt{x^{2} + (y – 5)^{2}}$$ = 6
$$\sqrt{x^{2} + (y + 5)^{2}}$$ = 6 + $$\sqrt{x^{2} + (y – 5)^{2}}$$
x2 + (y2 + 10y + 25) = 36 + 12$$\sqrt{x^{2} + (y – 5)^{2}}$$ + (x2 + y2 – 10y + 25)
20y – 36 = 12$$\sqrt{x^{2} + (y – 5)^{2}}$$
5y – 9 = 3$$\sqrt{x^{2} + (y – 5)^{2}}$$
25y2 – 90y + 81 = 9(x2 + y2 – 10y + 25)
25y2 – 90y + 81 = 9x2 + 9y2 – 90y + 225
16y2 – 9x2 = 144
$$\frac{y^{2}}{9}$$ – $$\frac{x^{2}}{16}$$ = 1
Exercise 2.
Where does the hyperbola described above intersect the y – axis?
The curve intersects the y – axis at (0,3) and (0, – 3).
Exercise 3.
Find an equation for the line that acts as a boundary for the portion of the curve that lies in the first quadrant.
For large values of x, $$\frac{y}{3}$$≈$$\frac{x}{4}$$, so the line y = $$\frac{3}{4}$$ x is the boundary for the curve in the first quadrant.
Exercise 4.
Sketch the graph of the hyperbola described above.
### Eureka Math Precalculus Module 3 Lesson 8 Problem Set Answer Key
Question 1.
For each hyperbola described below: (1) Derive an equation of the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1 or $$\frac{y^{2}}{b^{2}}$$ – $$\frac{x^{2}}{a^{2}}$$ = 1. (2) State any x – or y – intercepts. (3) Find the equations for the asymptotes of the hyperbola.
a. Let the foci be A( – 2,0) and B(2,0), and let P be a point for which either PA – PB = 2 or PB – PA = 2.
i. x2 – $$\frac{y^{2}}{3}$$ = 1
ii. ( – 1,0), (1,0); no y – intercepts
iii. y≈$$\sqrt{3}$$ x, so y = ±$$\sqrt{3}$$ x
b. Let the foci be A( – 5,0) and B(5,0), and let P be a point for which either PA – PB = 5 or PB – PA = 5.
i. $$\frac{x^{2}}{\frac{25}{4}}$$ – $$\frac{y^{2}}{\frac{75}{4}}$$ = 1
ii. ( – 2.5,0), (2.5,0); no y – intercepts
iii. y≈$$\frac{\sqrt{75}}{2}$$⋅$$\frac{2}{5}$$ x = $$\sqrt{3}$$ x, so y = ±$$\sqrt{3}$$ x
c. Consider A(0, – 3) and B(0,3), and let P be a point for which either PA – PB = 2.5 or PB – PA = 2.5.
i. $$\frac{y^{2}}{\frac{9}{4}}$$ – $$\frac{x^{2}}{\frac{27}{4}}$$ = 1
ii. (0,1.5), (0, – 1.5); no x – intercepts
iii. y≈$$\frac{3}{2}$$⋅$$\frac{2}{\sqrt{27}}$$ x = $$\frac{1}{\sqrt{3}}$$x, so y = ±$$\frac{\sqrt{3}}{3}$$x
d. Consider A(0, – $$\sqrt{2}$$) and B(0,$$\sqrt{2}$$), and let P be a point for which either PA – PB = 2 or PB – PA = 2.
i. y2 – x2 = 1
ii. (0,1), (0, – 1); no x – intercepts.
iii. y = ±x
Question 2.
Graph the hyperbolas in parts (a)–(d) in Problem 1.
a.
b.
c.
d.
Question 3.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
x2 – y2 = k.
a. k = 4
b. k = 1
c. k = $$\frac{1}{4}$$
d. k = 0
e. k = – $$\frac{1}{4}$$
f. k = – 1
g. k = – 4
h. Describe the hyperbolas x2 – y2 = k for different values of k. Consider both positive and negative values of k, and consider values of k close to zero and far from zero.
If k is close to zero, then the hyperbola is very close to the asymptotes y = x and y = – x, appearing almost to have corners as the graph crosses the x – axis. If k is far from zero, the hyperbola gets farther from the asymptotes near the center. If k > 0, then the hyperbola crosses the x – axis, opening to the right and left, and if k<0, then the hyperbola crosses the y – axis, opening up and down.
i. Are there any values of k so that the equation x2 – y2 = k has no solution?
No. The equation x2 – y2 = k always has solutions. The solution points lie on either two intersecting lines or on a hyperbola.
Question 4.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
$$\frac{x^{2}}{k}$$ – y2 = 1.
a. k = – 1
There is no solution to the equation $$\frac{x^{2}}{ – 1}$$ – y2 = 1, because there are no real numbers x and y so that x2 + y2 = – 1.
b. k = 1
c. k = 2
d. k = 4
e. k = 10
f. k = 25
g. Describe what happens to the graph of $$\frac{x^{2}}{k}$$ – y2 = 1 as k → ∞.
As k → ∞, it appears that the hyperbolas with equation $$\frac{x^{2}}{k}$$ – y2 = 1 get flatter; the x – intercepts get farther from the center at the origin, and the asymptotes get less steep.
Question 5.
For each value of k specified in parts (a)–(e), plot the set of points in the plane that satisfy the equation
x2 – $$\frac{y^{2}}{k}$$ = 1.
a. k = – 1
b. k = 1
c. k = 2
d. k = 4
e. k = 10
f. Describe what happens to the graph x2 – $$\frac{y^{2}}{k}$$ = 1 as k → ∞.
As k → ∞, the hyperbola with equation x2 – $$\frac{y^{2}}{k}$$ = 1 is increasingly stretched vertically. The center and intercepts do not change, but the steepness of the asymptotes increases.
Question 6.
An equation of the form ax2 + bx + cy2 + dy + e = 0 where a and c have opposite signs might represent a hyperbola.
a. Apply the process of completing the square in both x and y to convert the equation
9x2 – 36x – 4y2 – 8y – 4 = 0 to one of the standard forms for a hyperbola: $$\frac{(x – h)^{2}}{a^{2}}$$ – $$\frac{(y – k)^{2}}{b^{2}}$$ = 1 or $$\frac{(y – k)^{2}}{b^{2}}$$ – $$\frac{(x – h)^{2}}{a^{2}}$$ = 1.
9x2 – 36x – 4y2 – 8y = 4
9(x2 – 4x) – 4(y2 + 2y) = 4
9(x2 – 4x + 4) – 4(y2 + 2y + 1) = 4 + 36 – 4
9(x – 2)2 – 4(y + 1)2 = 36
$$\frac{(x – 2)^{2}}{4}$$ – $$\frac{(y + 1)^{2}}{9}$$ = 1
b. Find the center of this hyperbola.
The center is (2, – 1).
c. Find the asymptotes of this hyperbola.
$$\frac{x – 2}{2}$$ = ±$$\frac{y + 1}{3}$$
y = $$\frac{3}{2}$$ x – 4 or y = – $$\frac{3}{2}$$ x + 2
d. Graph the hyperbola.
Question 7.
For each equation below, identify the graph as either an ellipse, a hyperbola, two lines, or a single point. If possible, write the equation in the standard form for either an ellipse or a hyperbola.
a. 4x2 – 8x + 25y2 – 100y + 4 = 0
In standard form, this is the equation of an ellipse: $$\frac{(x – 1)^{2}}{25}$$ + $$\frac{(y – 2)^{2}}{4}$$ = 1.
b. 4x2 – 16x – 9y2 – 54y – 65 = 0
When we try to put this equation in standard form, we find $$\frac{(x – 2)^{2}}{9}$$ – $$\frac{(y + 3)^{2}}{4}$$ = 0, which gives
($$\frac{x – 2}{3}$$ = ±$$\frac{y + 3}{2}$$. These are the lines with equation y = $$\frac{1}{3}$$ (2x – 13) and y = $$\frac{1}{3}$$ ( – 2x – 5).
c. 4x2 + 8x + y2 + 2y + 5 = 0
When we try to put this equation in standard form, we find (x + 1)2 + $$\frac{(y + 1)^{2}}{4}$$ = 0. The graph of this equation is the single point ( – 1, – 1).
d. – 49x2 + 98x + 4y2 – 245 = 0
In standard form, this is the equation of a hyperbola: $$\frac{y^{2}}{49}$$ – $$\frac{(x – 1)^{2}}{4}$$ = 1.
e. What can you tell about a graph of an equation of the form ax2 + bx + cy2 + dy + e = 0 by looking at the coefficients?
There are two categories; if the coefficients a and c have the same sign, then the graph is either an ellipse, a point, or an empty set. If the coefficients a and c have opposite signs, then the graph is a hyperbola or two intersecting lines. We cannot tell just by looking at the coefficients which of these sub – cases hold.
### Eureka Math Precalculus Module 3 Lesson 8 Exit Ticket Answer Key
Question 1.
Let F( – 4,0) and B(4,0) be the foci of a hyperbola. Let the points P(x,y) on the hyperbola satisfy either PF – PG = 4 or PG – PF = 4. Derive an equation for this hyperbola, writing your answer in the form $$\frac{x^{2}}{a^{2}}$$ – $$\frac{y^{2}}{b^{2}}$$ = 1.
PF = $$\sqrt{(x + 4)^{2} + y^{2}}$$
PB = $$\sqrt{(x – 4)^{2} + y^{2}}$$
$$\sqrt{(x + 4)^{2} + y^{2}}$$ = 4 + $$\sqrt{(x – 4)^{2} + y^{2}}$$
(x + 4)2 + y2 = 16 + 8$$\sqrt{(x – 4)^{2} + y^{2}}$$ + (x – 4)2 + y2
x2 + 8x + 16 + y2 = 16 + 8$$\sqrt{(x – 4)^{2} + y^{2}}$$ + x2 – 8x + 16 + y2
16x – 16 = 8$$\sqrt{(x – 4)^{2} + y^{2}}$$
2x – 2 = $$\sqrt{(x – 4)^{2} + y^{2}}$$
$$\frac{x^{2}}{4}$$ – $$\frac{y^{2}}{12}$$ = 1 |
# Class 12 RD Sharma Solutions- Chapter 33 Binomial Distribution – Exercise 33.2 | Set 1
• Last Updated : 03 Mar, 2021
### Question 1. Can the mean of a binomial distribution be less than its variance?
Solution:
Let np be the mean and npq be the variance of a binomial distribution.
So,
Mean – Variance = np – npq
Mean – Variance = np (1 – q)
Mean – Variance = np.p
Mean – Variance = np2
Since n can never be a negative number and
p2 will always be a positive number, thus np2 > 0
Then,
Mean – Variance > 0
Mean > Variance
Hence, a mean of a binomial distribution can never be less than its variance.
### Question 2. Determine the binomial distribution whose mean is 9 and variance 9/4.
Solution:
We are given mean(np) = 9 and variance(npq) = 9/4.
Solving for the value of q, we will get
q =
We know, the relation p + q = 1
So, p = 1 – (1/4) = 3/4 -(1)
Since, np = 9
So, put the value of p from equation(1), we get
n.(3/4) = 9
n = 12
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 12Cr(3/4)r(1/4)12-r for r = 0,1,2,3,4,….,12
### Question 3. If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Solution:
We are given mean, np = 9 and variance npq = 6.
Solving for the value of q, we will get
q = 6/9 = 2/3
We know, the relation p + q = 1
p = 1 – (2/3) = 1/3 -(1)
Since, np = 9
So, put the value of p from equation(1), we get
n.(1/3) = 9
n = 27
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 27Cr(1/3)r(2/3)27-r for r = 0,1,2,3,4,…,27
### Question 4. Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.
Solution:
Given n = 5 and np + npq = 4.8
np (1 + q) = 4.8
5p (1 + 1 – p) = 4.8
10p -5p2 = 4.8
50p2 – 100p + 48 = 0
Solving for the value of p we will get
p = 6/5 or p = 4/5
Since, the value of p cannot exceed 1, we will consider p = 4/5.
Therefore, q = 1 – 4/5 = 1/5
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 5Cr(4/5)r(1/5)5-r for r = 0,1,2,….,5
### Question 5. Determine the binomial distribution whose mean is 20 and variance 16.
Solution:
We are given mean, np = 20 and variance npq = 16.
Solving for the value of q, we will get
q = 16/20 = 4/5
We know, the relation p + q = 1
p = 1 – 4/5 = 1/5 -(1)
Since, np = 20
So, put the value of p from equation(1), we get
n.(1/5) = 20
n = 100
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100
### Question 6. In a binomial distribution, the sum and product of the mean and the variance are 25/3 and 50/3 respectively. Find the distribution.
Solution:
We are given sum, np + npq = 25/3
np(1 + q) = 25/3 -(1)
Product, np x npq = 50/3 -(2)
Dividing equation(2) by equation(1), we get
= (50/3) × 3/25
npq = 2 (1 + q)
np(1 – p) = 2(2 – p)
np = -(3)
On substituting the value of equation(3) in the relation np + npq = 25/3, we get
= 25/3
. (1 – p) = 25/3
(1 + 1 – p) = 25/3
(2 – p) = 25/3
6p2+ p – 1 = 0
On solving for the value of p, we will get p = 1/3, therefore q = 2/3
Now, putting value of p and q in the relation np + npq = 25/3
n.(1/3)(1 + (2/3)) = 25/3
n = 15
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15
### Question 7. The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.
Solution:
We are given mean, np = 20 -(1)
Standard deviation, √npq = 4
npq = 16 -(2)
On dividing the equation (ii) by equation (i), we get
q = 4/5
Therefore, p = 1 – q = 1 – 4/5 = 1/5
Now, since np = 20
n = 20 x 5 = 100
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 100Cr(1/5)r(4/5)100-r for r = 0,1,2,3,4,…,100
### Question 8. If the probability of a defective bolt is 0.1, find the (i) mean and (ii) standard deviation for the distribution of bolts in a total of 400 bolts.
Solution:
We are given n = 400 and q = 0.1, therefore p = 0.9
(i) Mean = np = 400 × 0.9 = 360
(ii) Standard Deviation = √npq =√(400 × 0.9 ×0.1) = 6
### Question 9. Find the binomial distribution whose mean is 5 and variance 10/3.
Solution:
We are given mean, np = 5 and variance npq = 10/3.
Solving for the value of q, we will get
q = = 2/3
We know, the relation p + q = 1
p = 1 – (2/3) = 1/3
Since, np = 5
n.(1/3) = 5
n = 15
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P(x = r) = 15Cr(1/3)r(2/3)15-r for r = 0,1,2,3,4,…,15
### Question 10. If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.
Solution:
We are given n = 500,
p = 9/10 and thus q = 1/10
Therefore, mean = np = 500 × 0.9 = 450
Standard deviation = √npq = √(450 × 0.1) = 6.71
### Question 11. The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P(X = 1), and P(X ≥ 2).
Solution:
We are given, mean (np) = 16 and variance (npq) = 8
q = 8/16 = 1/2
Therefore, p = 1 – 1/2 = 1/2
Putting the value of p in the relation, np = 16
n = 16 x 2 = 32
Now, a binomial distribution is given by the relation: nCr pr(q)n-r
P (x = r) = 32Cr(1/2)r(1/2)32-r for r = 0,1,2,3,4,…,32
Now, P(X = 0) = 32C0(1/2)32 = (1/2)32
Similarly, P(X = 1) = 32C1(1/2)1(1/2)31 = 32 × (1/2)31
Also, P(X ≥ 2) = 1 – P(X = 0) – P(X = 1)
1 – (1/2)32 – 32 × (1/2)32
1 –
### Question 12. In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.
Solution:
We are given n = 8 and p = 2/6 = 1/3 therefore q = 2/3
Now, mean = np = 8 ×(1/3) = 8/3 and
Standard deviation √npq = = 4/3
### Question 13. Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.
Solution:
We are given n = 8 and the probability of having a boy or girl is equal, so p = 1/2 and q = 1/2
Therefore, the expected number of boys in a family = np = 8 × 0.5 = 4
### Question 14. The probability that an item produced by a factory is defective is 0.02. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.
Solution:
We are given n = 10,000, also p = 0.02 therefore, q = 1 – 0.02 = 0.98
Now, the expected number of defective items = np = 10000 × 0.02 = 200
The standard deviation = √npq = = √196 = 14
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# How do you solve 2 step equations?
## How do you solve 2 step equations?
Solving Two-Step Equations
1. 1) First, add or subtract both sides of the linear equation by the same number.
2. 2) Secondly, multiply or divide both sides of the linear equation by the same number.
3. 3)* Instead of step #2, always multiply both sides of the equation by the reciprocal of the coefficient of the variable.
## What is an example of a two step equation?
Sal solves the equation -16 = x/4 + 2. It takes two steps because he first has to subtract 2 from both sides and then multiply both sides by 4.
What is the formula for integers?
Integers formulas are formulas for addition/subtraction and multiplication/division of integers. These formulas are mentioned below. Addition/Subtraction Formulas: (+) + (+) = +
### How do you solve multi step algebraic equations?
Solving Multi-Step Equations
1. (Optional) Multiply to clear any fractions or decimals.
2. Simplify each side by clearing parentheses and combining like terms.
3. Add or subtract to isolate the variable term—you may have to move a term with the variable.
4. Multiply or divide to isolate the variable.
5. Check the solution.
### What is a multi step equation?
Multi-step equations are algebraic expressions that require more than one operation, such as subtraction, addition, multiplication, division, or exponentiation, to solve.
How do you solve two step equations with integers?
Answer. To do two-step equations with integers, simplify the whole numbers in the equation. Isolate x in theequation by adding or subtracting the opposite whole number to both sides of the equation. Divide both sides by the number next to x in the equation to solve for x.
#### What is the definition of two step equations?
A two-step equation is an algebraic equation that takes you two steps to solve. You’ve solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign. 2-step estimation word problems.
#### What is a two step linear equation?
Two-step linear equation is an equation, that requires two actions (operations) in order to be solved. Basically, there are 2 kinds of such equations: Above 2 kinds of linear equations are general. Addition/subtraction, multiplication/division can come in any order.
How do you solve two step equations practice?
A two-step equation is as straightforward as it sounds. You will need to perform two steps in order to solve the equation. One goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. The other goal is to have the number in front of the variable equal to one. |
# Singapore Math Challenge - Grade 3 and Up
Product Number: TB25797
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Sorry, this product is no longer available. For your Common Core curriculum.
Get ready to take the math challenge! These problems, puzzles, and brainteasers strengthen mathematical thinking and provide students with skill-building practice based on the leading math program in the world. Step-by-step strategies are clearly explained for solving problems at varied levels of difficulty. A complete, worked solution is also provided for each problem. Correlated to the Common Core State Standards. 352 pages.
CCSS Product Alignment
3.OA.1 Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 x 7.
3.OA.2 Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 ÷ 8.
3.OA.3 Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
3.OA.5 Apply properties of operations as strategies to multiply and divide. Examples: If 6 x 4 = 24 is known, then 4 x 6 = 24 is also known. (Commutative property of multiplication.) 3 x 5 x 2 can be found by 3 x 5 = 15, then 15 x 2 = 30, or by 5 x 2 = 10, then 3 x 10 = 30. (Associative property of multiplication.) Knowing that 8 x 5 = 40 and 8 x 2 = 16, one can find 8 x 7 as 8 x (5 + 2) = (8 x 5) + (8 x 2) = 40 + 16 = 56. (Distributive property.)
3.OA.6 Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.
3.OA.7 Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 x 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
3.OA.8 Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
3.OA.9 Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends.
3.NBT.2 Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
3.MD.7c Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a x b and a x c/. Use area models to represent the distributive property in mathematical reasoning.
3.MD.8 Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.
4.OA.1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 x 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
4.OA.2 Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.
4.OA.3 Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
4.OA.4 Find all factor pairs for a whole number in the range 1-100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1-100 is prime or composite.
4.OA.5 Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule “Add 3” and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way.
4.NBT.1 Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division.
4.NBT.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm.
4.NBT.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
4.NBT.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
4.MD.2 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.
5.NBT.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
5.NBT.5 Fluently multiply multi-digit whole numbers using the standard algorithm.
5.NF.5a Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
5.G.4 Classify two-dimensional figures in a hierarchy based on properties.
Math Standards for Mathematical Practice
MP1 Make sense of problems and persevere in solving them.
MP2 Reason abstractly and quantitatively.
MP4 Model with mathematics.
MP5 Use appropriate tools strategically.
MP6 Attend to precision.
MP7 Look for and make use of structure.
MP8 Look for and express regularity in repeated reasoning.
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# Rules of Surd – Basic Forms of Surd, Similar Surds, Conjugate Surds
Surds are irrational numbers. They are the root of rational numbers whose value can not be expressed as exact fractions. Examples of surds are: √2, √7, √12, √18, etc.
1. √(a X b ) = √a X √ b
2. √(a / b ) = √a / √b
3. √(a + b ) ≠ √a + √b
4. √(a – b ) ≠ √a – √b
Basic Forms of Surds
√a is said to be in its basic form if A does not have a factor that is a perfect square. E.g. √6, √5, √3, √2 etc. √18 is not in its basic form because it can be broken into √ (9×2) = 3√2. Hence 3√2 is now in its basic form.
Similar Surds
Surds are similar if their irrational part contains the same numerals e.g.
1. 3√n and 5√ n
2. 6√2 and 7√2
Conjugate Surds
Conjugate surds are two surds whose product result is a rational number.
(i)The conjugate of √3 – √5 is √3 + √5
The conjugate of -2√7 + √3 is -2√7 – √3
In general, the conjugate of √x + √y is √x – √y
The conjugate of √x – √y = √x + √y
Simplification of Surds
Surds can be simplified either in the basic form or as a single surd.
Examples 1. Simplify the following in its basic form (a) √45 (b) √98
Solution
(a) √45 = √ (9 x 5) = √9 x √5 = 3√5
(b) √98 = √ (49 x 2) = √49 x √2 = 7√2
Example 2. Simplify the following as a single surd (a) 2√5 (b) 17√2
Solution
(a) 2√5 = √4 x √5 = √ (4 x 5) = √20
(b) 17√2 = √289 x √2 = √ (289 x 2) = √578 |
Lesson Explainer: Special Segments in a Circle | Nagwa Lesson Explainer: Special Segments in a Circle | Nagwa
# Lesson Explainer: Special Segments in a Circle Mathematics
In this explainer, we will learn how to use the theorems of intersecting chords, secants, or tangents and secants to find missing lengths in a circle.
Let’s begin by recalling the names of the various parts of a circle.
We can then focus on some specific parts. If a line segment intersects the circumference of a circle exactly once such that it is perpendicular to the radius at that point and it has an endpoint on the circumference of the circle, it is called a tangent segment. If a line segment has one endpoint outside the circle, one endpoint on the circle, and a point between these points that intersects the circle, it is called a secant segment.
Having recapped, previously, the names of different line segments in a circle and demonstrated how properties of these line segments can help us to solve problems, we will consider two different theorems that will help us to solve further problems involving circles.
### Theorem: The Intersecting Chords Theorem
When two chords intersect inside a circle, each chord is divided into two segments. These are called chord segments. In the circle given, these segments are defined by , , , and .
If chord and chord intersect at point ,
Alternatively,
This means that if we know any three of these values, we can find the fourth. Let’s demonstrate a simple application of this theorem.
### Example 1: Finding the Length of a Chord in a Circle
Given that , , and , find the length of .
Recall that the intersecting chords theorem tell us that if chord and chord of the same circle intersect at point ,
We are given , , and , so we can substitute these values into this formula, where and , to obtain
Hence, the length of is 10 units.
In our next example, we will demonstrate how to apply this theorem when the ratio of the lengths of two chord segments is given.
### Example 2: Finding the Length of Two Segments Drawn in a Circle Using the Ratio between Them
If , , and , find the lengths of and .
The first thing we can do is take the information we are given and enter it onto our diagram.
And then, we recall what we know about intersecting chords:
We can use this to form an equation in terms of and , where and :
.
At this point, it does not seem as though we have enough information to solve the problem.
However, we do know that
So,
We can then substitute this into to give
Note: We do not need to include the negative root of 49 since is a length.
So, we can therefore say that
Next, we will consider two further theorems: the intersecting secants theorem and the tangent–secant theorem.
### Theorem: Intersecting Secants Theorem
Given secant segments and ,
Alternatively,
### Theorem: The Tangent–Secant Theorem
This is a special case of the intersecting secants theorem and applies when the lines are tangent segments.
In the diagram, , , and . In the case where one line is a secant segment and the other is a tangent segment,
In our next example, we will use one of these theorems to solve a problem involving two secants that intersect outside the circle.
### Example 3: Finding an Unknown Length of a Proportion Resulting from Two Circle Secants Drawn from the Same External Point
If , , and , find the length of .
When we look at our figure, we see that we have two secant segments that intersect outside the circle at point .
We can add the dimensions that we have been given to our diagram.
To enable us to find , we recall the intersecting secants theorem:
Applying this to our question, we can say that
Now, if we substitute in the values we know, we get
Hence, the length of is 12 cm
In the next example, to find a missing length, we will have to use not only information we know about secants and tangents, but also information we know about triangles.
### Example 4: Finding the Length of a Tangent to a Circle Using the Application of Similarity in Circles
In the figure shown, the circle has a radius of 12 cm, , and . Determine the distance from to the center of the circle, , and the length of , rounding your answers to the nearest tenth.
The first thing we will do is take the information we are given and add it to our diagram.
The two lengths that we are trying to find are the perpendicular distance from to the center of the circle, , and .
To solve the first part of the question, we will calculate the distance from to .
Let’s recall some facts about triangles.
We know the length of as this is the radius of the circle, which means the distance from to will also be 12 cm.
What this now gives us is an isosceles triangle for which we can calculate the height; the height of an isosceles triangle is the length of its median, which is the line segment that joins the vertex to the midpoint of the opposite side. This means it divides the base into two equally sized segments.
Next, we can calculate the length of the base of each of the right triangles:
From here, we can use the Pythagorean theorem to find the length we are looking for:
Then, if we round this to the nearest tenth, we get 3.4 cm.
Next, we will calculate the length of .
Since is a tangent and it intersects the secant at the point , we can say that
When finding , we were only interested in the positive result as we were finding a distance, which cannot be negative.
Therefore, the distance from to the center of the circle, , is 3.4 cm (to the nearest tenth).
The length of is 20.5 cm (to the nearest tenth).
We will now solve a problem that combines algebraic manipulation with the skills we have explored in this explainer.
### Example 5: Finding the Length of the Chords in a Circle Using the Properties of Chords
In the following figure, find the value of .
Inspecting the diagram, we see it consists of a circle with two chords: and . The two chords intersect at a point inside of the circle. In the question, we are asked to find , which has been used in expressions for the lengths of the segments of the two chords.
Therefore, to solve this problem, we need to recall the intersecting chord theorem.
If chord and chord intersect at point , then
We can use this to find an equation for by substituting in the expressions we have been given for the dimensions:
This equation can then be solved to find the value of . Distributing the parentheses, then rearranging the equation to put all terms on the left-hand side, we obtain
In the final example, we will determine whether four points that define two intersecting line segments can be points on a circle given the lengths of their parts.
### Example 6: Understanding the Chords Theorem
Given that , , , and , do the points , , , and lie on a circle?
Firstly, we are going to label the diagram with the lengths that we have been given.
In order for these four points to lie on a circle, they would have to satisfy the intersecting chord theorem.
Therefore, to solve this problem, we need to recall the intersecting chord theorem.
If chord and chord intersect at point , then
Let’s now see if this is satisfied by the lengths of the line segments in our diagram: and
From our calculations, we can see that as both and are equal to 39. Therefore, we can say that yes, the points , , , and lie on a circle.
Let us finish by recapping some key points.
### Key Points
• The Intersecting Chords Theorem
• The Intersecting Secants Theorem
• The Tangent–Secant Theorem |
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# How to find the values of $p$ for which demand is elastic and the values for which demand is inelastic, with the price demand equation as $f\left( p \right) = 455 - 35p$?
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Hint: Here, we will find the demand at which the price vanishes and the price at which the demand vanishes which are the $x$-intercepts and $y$-intercepts. We will plot the demand curve by using the extreme points. We will then use the midpoint formula to find the midpoint of the demand curve. By simplifying the equation, we will find the values for which demand is elastic and is inelastic.
Formula Used:
Midpoint Formula $= \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Complete Step by Step Solution:
We are given a function $f\left( p \right) = 455 - 35p$.
The given equation is a linear equation.
Let $p$ be the Price and $f\left( p \right)$ be the Price Demand Equation. Let the Price Demand equation $f\left( p \right)$ be$x$
Thus, we get $x = 455 - 35p$.
Now, we will find the Demand at which the Price vanishes by substituting $p = 0$ in the above equation.
$\Rightarrow x = 455 - 35\left( 0 \right)$
Now, by simplifying, we get
$\Rightarrow x = 455$
At the Demand $x = 455$, the price vanishes.
Now, we will find the Price at which the Demand vanishes by substituting $x = 0$ in the equation $x = 455 - 35p$. Therefore, we get
$0 = 455 - 35p$
Subtracting 455 from both the sides, we get
$\Rightarrow - 35p = - 455$
Dividing both sides by $- 35$, we get
$\Rightarrow p = \dfrac{{ - 455}}{{ - 35}}$
$\Rightarrow p = 13$
At the Price $p = 13$, the demand vanishes.
Thus, the two extreme points are$\left( {0,13} \right)$ and $\left( {455,0} \right)$.
Now, we will plot the graph for the two extreme points which is a demand curve for the price demand equation
Now, we will find the midpoint of the demand equation, by using the midpoint formula
Midpoint Formula $= \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Now, by using the two extreme points $\left( {0,13} \right)$ and $\left( {455,0} \right)$ of the demand curve, we get
$\Rightarrow$ Midpoint of the demand $= \left( {\dfrac{{0 + 455}}{2},\dfrac{{13 + 0}}{2}} \right)$
$\Rightarrow$ Midpoint of the demand $= \left( {\dfrac{{455}}{2},\dfrac{{13}}{2}} \right)$
Dividing the terms, we get
$\Rightarrow$ Midpoint of the demand $= \left( {227.5,6.5} \right)$
Thus, at price $6.5$ , the quantity demanded is $227.5$.
Exactly at the middle of the demand equation, the length of the lower segment is equal to the length of the upper segment in the demand curve. Hence the elasticity at the midpoint of the demand curve is unitary.
We know that the Demand is relatively inelastic in the lower segment of the demand curve and Demand is relatively elastic in the upper segment of the demand curve.
Therefore, Demand is relatively elastic for any price greater than $6.5$ and Demand is relatively inelastic for any price lesser than $6.5$.
Note:
We know that Price elasticity is defined as the ratio of change in quantity demanded or supplied to the change in price. Price Elasticity is of three types: Elastic, Inelastic or Unitary. Elastic Demand is defined as the ratio of change in quantity demanded to the change in price which is greater than the proportion. Inelastic Demand is defined as the ratio of change in quantity demanded to the change in price which is less than the proportion. Demand curve gives the curve for the demand and the price. The given demand equation forms a linear demand curve. |
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# Rational Function Limits
## Value of rational functions, and cases when denominator equals zero.
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Rational Function Limits
Finding the limit of a rational function is actually much less complex than it may seem, in fact many of the limits you have already evaluated have been rational functions.
In this lesson, you will gain more experience working with rational function limits, and will use another theorem which simplifies the process of finding the limit of some rational functions.
### Watch This
James Sousa: Ex 3: Determine Limits Analytically by Factoring
### Guidance
Sometimes finding the limit of a rational function at a point a is difficult because evaluating the function at the point a leads to a denominator equal to zero. The box below describes finding the limit of a rational function.
Theorem: The Limit of a Rational Function
For the rational function and any real number a,
.
However, if then the function may or may not have any outputs that exist.
#### Example A
Find .
Solution:
Using the theorem above, we simply substitute x = 3:
#### Example B
Find .
Solution:
Notice that the domain of the function is continuous (defined) at all real numbers except at x = 3. If we check the one-sided limits we see that and . Because the one-sided limits are not equal, the limit does not exist.
#### Example C
Find .
Solution:
Notice that the function here is discontinuous at x = 2, that is, the denominator is zero at x = 2. However, it is possible to remove this discontinuity by canceling the factor x - 2 from both the numerator and the denominator and then taking the limit:
This is a common technique used to find the limits of rational functions that are discontinuous at some points. When finding the limit of a rational function, always check to see if the function can be simplified. -->
### Guided Practice
1) Find .
2) Find .
3) Find .
1) The numerator and the denominator are both equal to zero at x = 3, but there is a common factor x - 3 that can be removed (that is, we can simplify the rational function):
2) To find
..... Start by factoring the numerator
Since we have (x - 1) in both numerator and denominator, we know that the original function is equal to just except where it is undefined (1).
Therefore the closer we get to inputing 1, the closer we get to the same value, whether from the + or - side.
To find the value, just solve for .
3) To find
..... Start by factoring the numerator
Since we have (x + 2) in both numerator and denominator, we know that the original function is equal to just except where it is undefined (-2).
Therefore the closer we get to substituting -2, the closer we get to the same output value, whether from the + or - side.
To find the value, just solve for .
### Explore More
Solve the following rational function limits.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 8.5.
### Vocabulary Language: English
Conjugates
Conjugates
Conjugates are pairs of binomials that are equal aside from inverse operations between them, e.g. $(3 + 2x)$ and $(3 - 2x)$.
Continuous
Continuous
Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.
discontinuous
discontinuous
A function is discontinuous if the function exhibits breaks or holes when graphed.
limit
limit
A limit is the value that the output of a function approaches as the input of the function approaches a given value.
Rational Function
Rational Function
A rational function is any function that can be written as the ratio of two polynomial functions.
rationalization
rationalization
Rationalization generally means to multiply a rational function by a clever form of one in order to eliminate radical symbols or imaginary numbers in the denominator. Rationalization is also a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.
theorem
theorem
A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven. |
# Basic Maths - Grand Test 1 -Q9
Let -1,3 be the eigenvalues and $\begin{bmatrix} 1 \\-1 \end{bmatrix}$$\begin{bmatrix} 1 \\1 \end{bmatrix}$ be the corresponding eigenvectors? of the matrix A, then which one of the following choices is TRUE ?
$(A).A =\begin{bmatrix} 1 & 1 \\-1& 1\end{bmatrix}\begin{bmatrix} -1 & 0 \\0& 3\end{bmatrix} \begin{bmatrix} 1 & -1 \\1& 1\end{bmatrix}$ $(B).A = \begin{bmatrix} 1 & 1 \\-1& 1\end{bmatrix}\begin{bmatrix} 3 & 0 \\0& -1\end{bmatrix} \begin{bmatrix} 1 & -1 \\1& 1\end{bmatrix}$
$(C).A = \begin{bmatrix} 1 & 1 \\-1& 1\end{bmatrix}\begin{bmatrix} 3 & 0 \\0& -1\end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\\frac{1}{2}& \frac{1}{2}\end{bmatrix}$ $(D).A = \begin{bmatrix} 1 & 1 \\-1& 1\end{bmatrix}\begin{bmatrix} -1 & 0 \\0& 3\end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\\frac{1}{2}& \frac{1}{2}\end{bmatrix}$
reshown Jun 23
+1 vote
We have ;$A = P DP^{-1}$ where P is a matrix of order 2 constructed by the linearly dependent eigenvectors of matrix A as column vectors and D is a diagonal matrix such that the diagonal elements as the eigenvalues of matrix A
so
$A = \begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}$
answered Jun 24 by (4,380 points) |
# What do the fraction strips tell you?
## What do the fraction strips tell you?
Fraction strips (or fraction bars or tiles) help students see that the same “whole” can be broken up into different equal-size parts. When students move the strips and put them side by side, they can visualize the fractional amounts.
How can you use fraction strips to determine which fraction is the biggest?
Compare the two sets of strips. If they are the same size, then the fractions are equal. If one set is larger than the other, then the fractions are not equal.
### What is the purpose of using fraction strips and a 1 strip?
Answer Expert Verified 1 strip or a single strip often confuses the child. So fraction strips like ½, 1/3, 2/3 etc. are used for giving better explanation. When the strips are cut into such sizes and placed beside each other, then it becomes visually easy to see.
Which point is at 1 4 on the number line?
Answer Expert Verified 1/4 we have to mark on the right side of number line because it is a positive number having no negative number . That is why we will mark it on the right side of the number line . As we know the denominator is 4 so between every number we will keep four points distance.
## What is the purpose of using fraction?
Fractions are important because they tell you what portion of a whole you need, have, or want. Fractions are used in baking to tell how much of an ingredient to use. Fractions are used in telling time; each minute is a fraction of the hour.
How do you use fraction strips in math?
Use Fraction Strips to represent, compare, order, and operate on fractions. Represent fractions by dragging pieces from the fraction tower into the workspace. Pieces can be placed in a line to form a train. Manipulate the pieces and trains to compare and order fractions or to model fraction operations.
### Can you plot fractions on a number line?
Fraction strips are also not the only fraction manipulative that can be used, but they are the focus of this list. Being able to plot fractions on a number line is an extremely important skill! However, many students may struggle with drawing a useable number line because they just can’t get the spacing right.
What is a comparison bar in fraction strips?
A comparison bar is a grey, vertical line that is useful to compare fraction strips, especially when the left edges are aligned. Multiple comparison bars can be created. The length of a bar, its colour and its thickness can be modified. Switch between English and French. Use the zoom in button to make the pieces look bigger.
## What do the ticks on a fraction strip mean?
The ticks are illuminated in red if a piece ends exactly at that position. A comparison bar is a grey, vertical line that is useful to compare fraction strips, especially when the left edges are aligned. Multiple comparison bars can be created. |
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