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# 1961 AHSME Problems/Problem 31
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In $\triangle ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). The ratio $PA:AB$ is:
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$
## Solution
$[asy] draw((0,0)--(40,0)--(16,18)--(0,0)); draw((40,0)--(64,72)--(16,18)); draw((40,0)--(160,0)--(64,72),dotted); dot((0,0)); label("B",(0,0),SW); dot((16,18)); label("A",(16,18),NW); dot((40,0)); label("C",(40,0),S); dot((64,72)); label("P",(64,72),N); dot((160,0)); label("X",(160,0),SE); label("4n",(20,0),S); label("3n",(33,17)); label("4an-4n",(100,0),S); label("3an",(112,36),NE); [/asy]$ Let $AC = 3n$ and $BC = 4n$. Draw $X$, where $X$ is on $BC$ and $AC \parallel PX$. By AA Similarity, $\triangle ABC \sim \triangle PBX$, so $PX = 3an$, $BX = 4an$, and $CX = 4an - 4n$.
Also, let $\angle ABC = a$ and $\angle BAC = b$. Since the angles of a triangle add up to $180^{\circ}$, $\angle BCA = 180-a-b$. By Exterior Angle Theorem, $\angle ACX = a+b$, and since $CP$ bisects $\angle ACX$, $\angle PCX = \frac{a+b}{2}$. Because $AC \parallel PX$, $\angle BXP = 180 - a - b$. Thus, $\angle CPX = \frac{a+b}{2}$, making $\triangle CPX$ an isosceles triangle.
Because $\triangle CPX$ is isosceles, $PX = CX$, so $4an - 4n = 3an$. That means $a = 4$, so $PB = 4 \cdot AB$. Thus, $PA = PB - AB = 3 \cdot AB$, so $PA : AB = 3:1$. The answer is $\boxed{\textbf{(D)}}$, and it can be verified (or obtained) by making $ABC$ a 3-4-5 right triangle.
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# 4.8: Problem-Solving Strategies: Read a Graph; Make a Graph
Difficulty Level: At Grade Created by: CK-12
Graphing is a very useful tool when analyzing a situation. This lesson will focus on using graphs to help solve linear situations that occur in real life.
Remember the 4-Step Problem-Solving Plan:
1. Understand the problem and underline or highlight key information.
2. Translate the problem and devise a method to solve the problem.
3. Carry out the plan and solve the problem.
Example: A cell phone company is offering its costumers the following deal. You can buy a new cell phone for $60 and pay a monthly flat rate of$40 per month for unlimited calls. How much money will this deal cost you after 9 months?
Solution: Begin by translating the sentence into an algebraic equation.
cell phone=$60,calling plan=$40 per month\begin{align*}\text{cell phone} = \60, \text{calling plan} = \40 \ \text{per month}\end{align*}
Let m=\begin{align*}m=\end{align*}the number of months and t=\begin{align*}t=\end{align*}total cost. The equation becomes
t(m)=60+40m\begin{align*}t(m)=60+40m\end{align*}
You could use guess and check or solve this equation. However, this lesson focuses on using a graph to problem-solve. This equation is in slope-intercept form. By graphing the line of this equation, you will find all the ordered pairs that are solutions to the cell phone problem.
Finding the cost at month 9, you can see the cost is approximately 425.00. To check if this is approximately correct, substitute 9 in for the variable m\begin{align*}m\end{align*}. PhoneCalling planTotal cost=60=$40×9=$360=420.\begin{align*}\text{Phone} & = \60\\ \text{Calling plan} & = \40 \times 9 = \360\\ \text{Total cost} & = \420.\end{align*} Our answer,425.00 is approximately equal to the exact solution 420.00. Example: Christine took one hour to read 22 pages of “Harry Potter and the Order of the Phoenix.” She has 100 pages left to read in order to finish the book. Assuming that she reads at a constant rate of pages per hour, how much time should she expect to spend reading in order to finish the book? Solution: We do not have enough information to write an equation. We do not know the slope or the y\begin{align*}y-\end{align*}intercept. However, we have two points we can graph. We know that if Christine had never picked up the book, she would have read zero pages. So it takes Christine 0 hours to read 0 pages. We also know it took Christine one hour to read 22 pages. The two coordinates we can graph are (0, 0) and (1, 22). Using the graph and finding 100 pages, you can determine it will take Christine about 4.5 hours to read 100 pages. You can also think of this as a direct variation situation and solve it by writing a proportion. 22 pages1 hour=100 pagesh hours\begin{align*}\frac{22 \ pages}{1 \ hour}=\frac{100 \ pages}{h \ hours}\end{align*} By using the Cross Products Theorem, you can find out h4.55\begin{align*}h \approx 4.55\end{align*}. It will take Christine about 4.55 hours to read 100 pages, which is very close to your original estimate of 4.5 hours. ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Word Problem Solving 4 (10:05) 1. Using the following graph, determine these values: 1. The amount of earnings after 40 hours 2. How many hours it takes to earn250.00
3. The slope of the line and what it represents
4. The y\begin{align*}y-\end{align*}intercept of the line and what it represents
2. A stretched spring has a length of 12 inches when a weight of 2 lbs is attached to it. The same spring has a length of 18 inches when a weight of 5 lbs is attached to it. It is known from physics that within certain weight limits, the function that describes how much a spring stretches with different weights is a linear function. What is the length of the spring when no weights are attached?
1. A gym is offering a deal to new members. Customers can sign up by paying a registration fee of $200 and a monthly fee of$39. How much will this membership cost a member by the end of one year?
2. A candle is burning at a linear rate. The candle measures five inches two minutes after it was lit. It measures three inches eight minutes after it was lit. What was the original length of the candle?
3. Tali is trying to find the thickness of a page of his telephone book. To do this, he takes a measurement and finds out that 550 pages measure 1.25 inches. What is the thickness of one page of the phone book?
4. Bobby and Petra are running a lemonade stand and they charge 45 cents for each glass of lemonade. To break even, they must make $25. How many glasses of lemonade must they sell to break even? 5. The tip for a$78.00 restaurant bill is $9.20. What is the tip for a$21.50 meal?
6. Karen left her house and walked at a rate of 4 miles/hour for 30 minutes. She realized she was late for school and began to jog at a rate of 5.5 miles/hour for 25 minutes. Using a graph, determine how far she is from her house at the end of 45 minutes.
Mixed Review
1. Simplify 4|2111|+16\begin{align*}-4|-21-11|+16\end{align*}.
2. Give an example of a direct variation equation and label its constant of variation.
3. Identify the slope and y\begin{align*}y-\end{align*}intercept: 53x=y4\begin{align*}\frac{5}{3} x=y-4\end{align*}.
4. Suppose A=(x,y)\begin{align*}A=(x,y)\end{align*} is located in quadrant I. Write a rule that would move A\begin{align*}A\end{align*} into quadrant III.
5. Find the intercepts of 0.04x+0.06y=18\begin{align*}0.04x+0.06y=18\end{align*}.
6. Evaluate f(4)\begin{align*}f(4)\end{align*} when f(x)=3x28\begin{align*}f(x)=\frac{3x^2}{8}\end{align*}.
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# Thread: Find perimeter of figure
1. ## Find perimeter of figure
Hello!
It's really basic, but what is the perimeter of this figure? r is the radius of the circle and length of the square.
2. Originally Posted by Incendiary
Hello!
It's really basic, but what is the perimeter of this figure? r is the radius of the circle and length of the square.
Hint: Fill in two more lines on this figure. First the radius to the other corner of the square, then complete the top of the square.
You will note that the triangle formed is an equilateral triangle, with an apex angle of 60 degrees. So we are cutting a 60 degree arc off from the circumference of the circle...
-Dan
3. Hello, Incendiary!
What is the perimeter of this figure?
$\displaystyle r$ is the radius of the circle and length of the square.
Label the vertices of the square: A (upper left), B (lower left), C (lower right), D (upper right).
Label the center of the circle with O.
Then: .$\displaystyle AD \,= \,BC \,= \,r.$
Hence, $\displaystyle \Delta AOD$ is equilateral: $\displaystyle \angle AOD \,= \,60^o \,=\,\frac{\pi}{3}$
The major arc AD has length: $\displaystyle \frac{5\pi}{3}r$
The three sides of the square has length: $\displaystyle 3r$
Therefore, the perimeter is: .$\displaystyle \boxed{P \;= \;\left(\frac{5\pi}{3} + 3\right)r}$
The total area is: .$\displaystyle \text{(area of square) + (area of circle) - (area of overlap)}$
The overlap is a segment of a 60° angle.
. . The area of the sector is: .$\displaystyle \frac{1}{6}\pi r^2$
. . The area of the triangle is: .$\displaystyle \frac{\sqrt{3}}{4}r^2$
Hence, the area of the segment is: . $\displaystyle \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2$
Therefore, the area of the figure is:
. . $\displaystyle A \;=\;r^2 + \pi r^2 - \left(\frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\right) \;=\;\boxed{\frac{1}{12}\left(10\pi + 12 + 3\sqrt{3}\right)}$
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# Systems of Linear Equations in Two Variables
## Solving for two variables using elimination.
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Practice Systems of Linear Equations in Two Variables
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Systems of Two Equations and Two Unknowns
The cost of two cell phone plans can be written as a system of equations based on the number of minutes used and the base monthly rate. As a consumer, it would be useful to know when the two plans cost the same and when is one plan cheaper.
Plan A costs $40 per month plus$0.10 for each minute of talk time.
Plan B costs $25 per month plus$0.50 for each minute of talk time.
Plan B has a lower starting cost, but since it costs more per minute, it may not be the right plan for someone who likes to spend a lot of time on the phone. When do the two plans cost the same amount?
#### Watch This
http://www.youtube.com/watch?v=ova8GSmPV4o James Sousa: Solving Systems of Equations by Elimination
#### Guidance
There are many ways to solve a system that you have learned in the past including substitution and graphical intersection. Here you will focus on solving using elimination because the knowledge and skills used will transfer directly into using matrices.
When solving a system, the first thing to do is to count the number of variables that are missing and the number of equations. The number of variables needs to be the same or fewer than the number of equations. Two equations and two variables can be solved, but one equation with two variables cannot.
Get into the habit of always writing systems in standard form: \begin{align*}Ax+By=C\end{align*}. This will help variables line up, avoid +/- errors and lay the groundwork for using matrices. Once two equations with two variables are in standard form, decide which variable you want to eliminate, scale each equation as necessary by multiplying through by constants and then add the equations together. This procedure should reduce both the number of equations and the number of variables leaving one equation and one variable. Solve and substitute to determine the value of the second variable.
Example A
Solve the following system of equations: \begin{align*}5x+12y=72\end{align*} and \begin{align*}3x-2y=18\end{align*}.
Solution: Here is a system of two equations and two variables in standard form. Notice that there is an \begin{align*}x\end{align*} column and a \begin{align*}y\end{align*} column on the left hand side and a constant column on the right hand side when you rewrite the equations as shown. Also notice that if you add the system as written no variable will be eliminated.
Equation 1: \begin{align*}5x+12y=72\end{align*}
Equation 2: \begin{align*}3x-2y=18\end{align*}
Strategically choose to eliminate \begin{align*}y\end{align*} by scaling the second equation by 6 so that the coefficient of \begin{align*}y\end{align*} will match at 12 and -12.
The value for \begin{align*}x\end{align*} could be substituted into either of the original equations and the result could be solved for \begin{align*}y\end{align*}; however, since the value is a fraction it will be easier to repeat the elimination process in order to solve for \begin{align*}x\end{align*}. This time you will take the first two equations and eliminate \begin{align*}x\end{align*} by making the coefficients of \begin{align*}x\end{align*} to be 15 and -15. Scale the first equation by a factor of 3 and scale the second equation by a factor of -5.
Equation 1: \begin{align*}15x+36y=216\end{align*}
Equation 2: \begin{align*}-15x+10y=-90\end{align*}
The point \begin{align*}\left(\frac{180}{23}, \frac{63}{23} \right)\end{align*} is where these two lines intersect.
Example B
Solve the following system of equations:
Solution: Scaling the first equation by -2 will allow the \begin{align*}y\end{align*} term to be eliminated when the equations are summed.
The sum is:
You can substitute \begin{align*}x\end{align*} into the first equation to solve for \begin{align*}y\end{align*}.
The point \begin{align*}\left(\frac{5}{3}, \frac{2}{7}\right)\end{align*} is where these two lines intersect.
Example C
Solve the following system of equations:
Solution: The strategy of elimination still applies. You can eliminate the \begin{align*}\frac{1}{y}\end{align*} term if the second equation is scaled by a factor of -2.
Add the equations together and solve for \begin{align*}x\end{align*}.
Substitute into the second equation and solve for \begin{align*}y\end{align*}.
The point \begin{align*}\left(-1, \frac{1}{5}\right)\end{align*} is the point of intersection between these two curves.
Concept Problem Revisited
Plan A costs $40 per month plus$0.10 for each minute of talk time.
Plan B costs $25 per month plus$0.50 for each minute of talk time.
If you want to find out when the two plans cost the same, you can represent each plan with an equation and solve the system of equations. Let \begin{align*}y\end{align*} represent cost and \begin{align*}x\end{align*} represent number of minutes.
First you put these equations in standard form.
Then you scale the second equation by -1 and add the equations together and solve for \begin{align*}y\end{align*}.
To solve for \begin{align*}x\end{align*}, you can scale the second equation by -5, add the equations together and solve for \begin{align*}x\end{align*}.
The equivalent costs of plan A and plan B will occur at 37.5 minutes of talk time with a cost of \$43.75.
#### Vocabulary
A system of equations is two or more equations.
Standard form for the equation of a line is \begin{align*}Ax+By=C\end{align*}.
To scale an equation means to multiply every term by a constant.
#### Guided Practice
1. Solve the following system using elimination:
2. Solve the following system using elimination:
3. Solve the following system using elimination:
1. Start by scaling both of the equations by \begin{align*}\frac{1}{2}\end{align*}. Then notice that you have \begin{align*}3y\end{align*} and \begin{align*}-7y\end{align*}. Rescale the first equation by 7 and the second equation by 3 to make the coefficients of \begin{align*}y\end{align*} at 21 and -21. There are a number of possible ways to eliminate \begin{align*}y\end{align*}
Add, solve for \begin{align*}x=-1\end{align*}, substitute and solve for \begin{align*}y\end{align*}
2. Start by scaling the first equation by 7 and notice that the \begin{align*}y\end{align*} coefficient will immediately be eliminated when the equations are summed.
Add, solve for \begin{align*}x=\frac{173}{33}\end{align*}. Instead of substituting, practice eliminating \begin{align*}x\end{align*} by scaling the first equation by 2 and the second equation by 5.
Add, solve for \begin{align*}y\end{align*}
Final Answer: \begin{align*}\left(\frac{173}{33}, \frac{139}{33}\right)\end{align*}
3. To eliminate \begin{align*}\frac{1}{y}\end{align*}, scale the first equation by 2 and the second equation by 5.
To eliminate \begin{align*}\frac{1}{x}\end{align*}, scale the first equation by -9 and the second equation by 11.
Final Answer: \begin{align*}\left(-\frac{1}{3}, 1 \right)\end{align*}
#### Practice
Solve each system of equations using the elimination method.
1. \begin{align*}x+y=-4; -x+2y=13\end{align*}
2. \begin{align*}\frac{3}{2}x-\frac{1}{2}y=\frac{1}{2}; -4x+2y=4\end{align*}
3. \begin{align*}6x+15y=1; 2x-y=19\end{align*}
4. \begin{align*}x-\frac{2y}{3}=\frac{-2}{3}; 5x-2y=10\end{align*}
5. \begin{align*}-9x-24y=-243; \frac{1}{2}x+y=\frac{21}{2}\end{align*}
6. \begin{align*}5x+\frac{28}{3}y=\frac{176}{3}; y+x=10\end{align*}
7. \begin{align*}2x-3y=50; 7x+8y=-10\end{align*}
8. \begin{align*}2x+3y=1; 2y=-3x+14\end{align*}
9. \begin{align*}2x+\frac{3}{5}y=3;\frac{3}{2}x-y=-5\end{align*}
10. \begin{align*}5x=9-2y;3y=2x-3\end{align*}
11. How do you know if a system of equations has no solution?
12. If a system of equations has no solution, what does this imply about the relationship of the curves on the graph?
13. Give an example of a system of two equations with two unknowns with an infinite number of solutions. Explain how you know the system has an infinite number of solutions.
14. Solve:
15. Solve:
### Vocabulary Language: English
elimination
elimination
The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable.
Periodic Decimal
Periodic Decimal
A periodic decimal is a decimal number that has a pattern of digits that repeat. The decimal number 0.146532532532..., is a periodic decimal.
scale
scale
To scale an equation means to multiply every term in the equation by the same constant.
Standard Form
Standard Form
The standard form of a line is $Ax + By = C$, where $A, B,$ and $C$ are real numbers.
system of equations
system of equations
A system of equations is a set of two or more equations.
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Estimating Product and Quotient
The procedure of estimating product and quotient are in the following examples.
Solved examples to estimate product and quotient:
1. Estimate the product 958 × 387 by rounding off each factor to its greatest place.
Solution:
Clearly, each factor is a three digit number. So, we round off each factor to nearest hundreds.
958 rounds off as 1000
387 rounds off as 400
Therefore estimated product = 1000 × 400 = 400000
2. Estimate the product to the nearest hundreds.
(i) 42 × 37
(ii) 67 × 62
(iii) 99 × 91
(iv) 147 × 51
(v) 193 × 47
Solution:
(i) 42 × 37 = 40 × 40 = 1600
(ii) 67 × 62 = 70 × 60 = 4200
(iii) 99 × 91 = 100 × 90 = 9000
(iv) 147 × 51 = 150 × 50 = 7500
(v) 193 × 47 = 190 × 50 = 9500
2. Find the estimated quotient 2838 ÷ 125 by rounding off the numerator and denominator to the nearest hundreds.
Solution:
We find that
2838 rounds off to nearest hundreds 2800
125 rounds off to nearest hundreds as 100
Therefore, estimated quotient = 2800 ÷ 100 = 28.
3. Estimate the quotient to the nearest tens.
(i) 87 ÷ 9
(ii) 163 ÷ 11
(iii) 461 ÷ 7
(iv) 1223 ÷ 17
Solution:
(i) 87 ÷ 9 = 90 ÷ 10 = 9 = 10
(ii) 163 ÷ 11 = 160 ÷ 10 = 16 = 20
(iii) 451 ÷ 7 = 460 ÷ 10 = 46 = 50
(iv) 1223 ÷ 17 = 1220 ÷ 20 = 61 = 60
Estimate to Nearest Tens
Estimate to Nearest Hundreds
Estimate to Nearest Thousands
Estimating Sum and Difference
Numbers Page
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# APEX Calculus
## Section11.3The Dot Product
The previous section introduced vectors and described how to add them together and how to multiply them by scalars. This section introduces a multiplication on vectors called the dot product.
### Definition11.3.1.Dot Product.
1. Let $$\vec u = \la u_1,u_2\ra$$ and $$\vec v = \la v_1,v_2\ra$$ in $$\mathbb{R}^2\text{.}$$ The dot product of $$\vec u$$ and $$\vec v\text{,}$$ denoted $$\dotp uv\text{,}$$ is
\begin{equation*} \dotp uv = u_1v_1+u_2v_2\text{.} \end{equation*}
2. Let $$\vec u = \la u_1,u_2,u_3\ra$$ and $$\vec v = \la v_1,v_2,v_3\ra$$ in $$\mathbb{R}^3\text{.}$$ The dot product of $$\vec u$$ and $$\vec v\text{,}$$ denoted $$\dotp uv\text{,}$$ is
\begin{equation*} \dotp uv = u_1v_1+u_2v_2+u_3v_3\text{.} \end{equation*}
Note how this product of vectors returns a scalar, not another vector. We practice evaluating a dot product in the following example, then we will discuss why this product is useful.
### Example11.3.2.Evaluating dot products.
1. Let $$\vec u=\la 1,2\ra\text{,}$$ $$\vec v=\la 3,-1\ra$$ in $$\mathbb{R}^2\text{.}$$ Find $$\dotp uv\text{.}$$
2. Let $$\vec x = \la 2,-2,5\ra$$ and $$\vec y = \la -1, 0, 3\ra$$ in $$\mathbb{R}^3\text{.}$$ Find $$\dotp xy\text{.}$$
Solution.
1. Using Definition 11.3.1, we have
\begin{equation*} \dotp uv = 1(3)+2(-1) = 1\text{.} \end{equation*}
2. Using the definition, we have
\begin{equation*} \dotp xy = 2(-1) -2(0) + 5(3) = 13\text{.} \end{equation*}
The dot product, as shown by the preceding example, is very simple to evaluate. It is only the sum of products. While the definition gives no hint as to why we would care about this operation, there is an amazing connection between the dot product and angles formed by the vectors. Before stating this connection, we give a theorem stating some of the properties of the dot product.
The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. Our definition and theorem give properties of the dot product, but we are still likely wondering “What does the dot product mean?” It is helpful to understand that the dot product of a vector with itself is connected to its magnitude.
The next theorem extends this understanding by connecting the dot product to magnitudes and angles. Given vectors $$\vec u$$ and $$\vec v$$ in the plane, an angle $$\theta$$ is clearly formed when $$\vec u$$ and $$\vec v$$ are drawn with the same initial point as illustrated in Figure 11.3.4.(a). (We always take $$\theta$$ to be the angle in $$[0,\pi]$$ as two angles are actually created.)
The same is also true of 2 vectors in space: given $$\vec u$$ and $$\vec v$$ in $$\mathbb{R}^3$$ with the same initial point, there is a plane that contains both $$\vec u$$ and $$\vec v\text{.}$$ (When $$\vec u$$ and $$\vec v$$ are co-linear, there are infinitely many planes that contain both vectors.) In that plane, we can again find an angle $$\theta$$ between them (and again, $$0\leq \theta\leq \pi$$). This is illustrated in Figure 11.3.4.(b).
The following theorem connects this angle $$\theta$$ to the dot product of $$\vec u$$ and $$\vec v\text{.}$$
Using Theorem 11.3.3, we can rewrite this theorem as
\begin{equation*} \frac{\vec u}{\norm{\vec u}}\cdot \frac{\vec v}{\norm{\vec v}} = \cos(\theta)\text{.} \end{equation*}
Note how on the left hand side of the equation, we are computing the dot product of two unit vectors. Recalling that unit vectors essentially only provide direction information, we can informally restate Theorem 11.3.5 as saying “The dot product of two directions gives the cosine of the angle between them.”
When $$\theta$$ is an acute angle (i.e., $$0\leq \theta \lt \pi/2$$), $$\cos(\theta)$$ is positive; when $$\theta = \pi/2\text{,}$$ $$\cos(\theta) = 0\text{;}$$ when $$\theta$$ is an obtuse angle ($$\pi/2\lt \theta \leq \pi$$), $$\cos(\theta)$$ is negative. Thus the sign of the dot product gives a general indication of the angle between the vectors, illustrated in Figure 11.3.6.
We can use Theorem 11.3.5 to compute the dot product, but generally this theorem is used to find the angle between known vectors (since the dot product is generally easy to compute). To this end, we rewrite the theorem’s equation as
\begin{equation*} \cos(\theta) = \frac{\dotp uv}{\norm{\vec u}\norm{\vec v}} \Leftrightarrow \theta = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\text{.} \end{equation*}
We practice using this theorem in the following example.
### Example11.3.7.Using the dot product to find angles.
Let $$\vec u = \la 3,1\ra\text{,}$$ $$\vec v = \la -2,6\ra$$ and $$\vec w = \la -4,3\ra\text{,}$$ as shown in Figure 11.3.8. Find the angles $$\alpha\text{,}$$ $$\beta$$ and $$\theta\text{.}$$
Solution.
We start by computing the magnitude of each vector.
\begin{equation*} \norm{\vec u} = \sqrt{10}; \norm{\vec v} = 2\sqrt{10}; \norm{\vec w} = 5\text{.} \end{equation*}
We now apply Theorem 11.3.5 to find the angles.
\begin{align*} \alpha \amp = \cos^{-1}\left(\frac{\dotp uv}{(\sqrt{10})(2\sqrt{10})}\right)\\ \amp = \cos^{-1}(0) = \frac{\pi}2 = 90^\circ\text{.} \end{align*}
\begin{align*} \beta \amp = \cos^{-1}\left(\frac{\dotp vw}{(2\sqrt{10})(5)}\right)\\ \amp = \cos^{-1}\left(\frac{26}{10\sqrt{10}}\right)\\ \amp \approx 0.6055 \approx 34.7^\circ.\\ \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{(\sqrt{10})(5)}\right)\\ \amp = \cos^{-1}\left(\frac{-9}{5\sqrt{10}}\right)\\ \amp \approx 2.1763 \approx 124.7^\circ \end{align*}
We see from our computation that $$\alpha + \beta = \theta\text{,}$$ as indicated by Figure 11.3.8. While we knew this should be the case, it is nice to see that this non-intuitive formula indeed returns the results we expected.
We do a similar example next in the context of vectors in space.
### Example11.3.9.Using the dot product to find angles.
Let $$\vec u = \la 1,1,1\ra\text{,}$$ $$\vec v = \la -1,3,-2\ra$$ and $$\vec w = \la -5,1,4\ra\text{,}$$ as illustrated in Figure 11.3.10. Find the angle between each pair of vectors.
Solution.
1. Between $$\vec u$$ and $$\vec v\text{:}$$
\begin{align*} \theta \amp = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\\ \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{14}}\right)\\ \amp = \frac{\pi}2\text{.} \end{align*}
2. Between $$\vec u$$ and $$\vec w\text{:}$$
\begin{align*} \theta \amp = \cos^{-1}\left(\frac{\dotp uw}{\norm{\vec u}\norm{\vec w}}\right)\\ \amp = \cos^{-1}\left(\frac{0}{\sqrt{3}\sqrt{42}}\right)\\ \amp = \frac{\pi}2\text{.} \end{align*}
3. Between $$\vec v$$ and $$\vec w\text{:}$$
\begin{align*} \theta \amp = \cos^{-1}\left(\frac{\dotp vw}{\norm{\vec v}\norm{\vec w}}\right)\\ \amp = \cos^{-1}\left(\frac{0}{\sqrt{14}\sqrt{42}}\right)\\ \amp = \frac{\pi}2\text{.} \end{align*}
While our work shows that each angle is $$\pi/2\text{,}$$ i.e., $$90^\circ\text{,}$$ none of these angles looks to be a right angle in Figure 11.3.10. Such is the case when drawing three-dimensional objects on the page.
All three angles between these vectors was $$\pi/2\text{,}$$ or $$90^\circ\text{.}$$ We know from geometry and everyday life that $$90^\circ$$ angles are “nice” for a variety of reasons, so it should seem significant that these angles are all $$\pi/2\text{.}$$ Notice the common feature in each calculation (and also the calculation of $$\alpha$$ in Example 11.3.7): the dot products of each pair of angles was 0. We use this as a basis for a definition of the term orthogonal, which is essentially synonymous to perpendicular.
### Definition11.3.11.Orthogonal.
Nonzero vectors $$\vec u$$ and $$\vec v$$ are orthogonal if their dot product is 0.
### Example11.3.12.Finding orthogonal vectors.
Let $$\vec u = \la 3,5\ra$$ and $$\vec v = \la 1,2,3\ra\text{.}$$
1. Find two vectors in $$\mathbb{R}^2$$ that are orthogonal to $$\vec u\text{.}$$
2. Find two non-parallel vectors in $$\mathbb{R}^3$$ that are orthogonal to $$\vec v\text{.}$$
Solution.
1. Recall that a line perpendicular to a line with slope $$m$$ has slope $$-1/m\text{,}$$ the “opposite reciprocal slope.” We can think of the slope of $$\vec u$$ as $$5/3\text{,}$$ its “rise over run.” A vector orthogonal to $$\vec u$$ will have slope $$-3/5\text{.}$$ There are many such choices, though all parallel:
\begin{equation*} \la -5,3\ra \text{ or } \la 5,-3\ra \text{ or } \la -10,6\ra \text{ or } \la 15,-9\ra,\text{ etc. } \end{equation*}
2. There are infinitely many directions in space orthogonal to any given direction, so there are an infinite number of non-parallel vectors orthogonal to $$\vec v\text{.}$$ Since there are so many, we have great leeway in finding some. One way is to arbitrarily pick values for the first two components, leaving the third unknown. For instance, let $$\vec v_1 = \la 2,7,z\ra\text{.}$$ If $$\vec v_1$$ is to be orthogonal to $$\vec v\text{,}$$ then $$\vec v_1\cdot\vec v = 0\text{,}$$ so
\begin{equation*} 2+14+3z=0 \Rightarrow z = \frac{-16}{3}\text{.} \end{equation*}
So $$\vec v_1 = \la 2, 7, -16/3\ra$$ is orthogonal to $$\vec v\text{.}$$ We can apply a similar technique by leaving the first or second component unknown. Another method of finding a vector orthogonal to $$\vec v$$ mirrors what we did in part 1. Let $$\vec v_2 = \la-2,1,0\ra\text{.}$$ Here we switched the first two components of $$\vec v\text{,}$$ changing the sign of one of them (similar to the “opposite reciprocal” concept before). Letting the third component be 0 effectively ignores the third component of $$\vec v\text{,}$$ and it is easy to see that
\begin{equation*} \vec v_2\cdot\vec v = \la -2,1,0\ra\cdot\la 1,2,3\ra = 0\text{.} \end{equation*}
Clearly $$\vec v_1$$ and $$\vec v_2$$ are not parallel.
An important construction is illustrated in Figure 11.3.13, where vectors $$\vec u$$ and $$\vec v$$ are sketched. In Figure 11.3.13.(a), a dotted line is drawn from the tip of $$\vec u$$ to the line containing $$\vec v\text{,}$$ where the dotted line is orthogonal to $$\vec v\text{.}$$ In Figure 11.3.13.(b), the dotted line is replaced with the vector $$\vec z$$ and $$\vec w$$ is formed, parallel to $$\vec v\text{.}$$ It is clear by the diagram that $$\vec u = \vec w+\vec z\text{.}$$ What is important about this construction is this: $$\vec u$$ is decomposed as the sum of two vectors, one of which is parallel to $$\vec v$$ and one that is perpendicular to $$\vec v\text{.}$$ It is hard to overstate the importance of this construction (as we’ll see in upcoming examples).
The vectors $$\vec w\text{,}$$ $$\vec z$$ and $$\vec u$$ as shown in Figure 11.3.13.(b) form a right triangle, where the angle between $$\vec v$$ and $$\vec u$$ is labeled $$\theta\text{.}$$ We can find $$\vec w$$ in terms of $$\vec v$$ and $$\vec u\text{.}$$
Using trigonometry, we can state that
$$\norm{\vec w} = \norm{\vec u}\cos(\theta)\text{.}\tag{11.3.1}$$
We also know that $$\vec w$$ is parallel to to $$\vec v$$ ; that is, the direction of $$\vec w$$ is the direction of $$\vec v\text{,}$$ described by the unit vector $$\vec v/\norm{\vec v}\text{.}$$ The vector $$\vec w$$ is the vector in the direction $$\vec v/\norm{\vec v}$$ with magnitude $$\norm{\vec u}\cos(\theta)\text{:}$$
\begin{align*} \vec w \amp = \Big(\norm{\vec u}\cos(\theta) \Big)\frac{1}{\norm{\vec v}}\vec v.\\ \amp = \left(\norm{\vec u}\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\frac{1}{\norm{\vec v}} \vec v \text{ (Replacing } \cos(\theta) \text{ using } \knowl{./knowl/thm_dot_product.html}{\text{Theorem 11.3.5}}\text{)}\\ \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v.\\ \amp = \frac{\dotp uv}{\dotp vv}\vec v \text{ (Applying }\knowl{./knowl/thm_dot_product_properties.html}{\text{Theorem 11.3.3}}\text{)}\text{.} \end{align*}
Since this construction is so important, it is given a special name.
### Definition11.3.14.Orthogonal Projection.
Let nonzero vectors $$\vec u$$ and $$\vec v$$ be given. The orthogonal projection of $$\vec u$$ onto $$\vec v\text{,}$$ denoted $$\proj uv\text{,}$$ is
\begin{equation*} \proj uv = \frac{\dotp uv}{\dotp vv}\vec v\text{.} \end{equation*}
### Example11.3.15.Computing the orthogonal projection.
1. Let $$\vec u= \la -2,1\ra$$ and $$\vec v=\la 3,1\ra\text{.}$$ Find $$\proj uv\text{,}$$ and sketch all three vectors with initial points at the origin.
2. Let $$\vec w = \la 2,1,3\ra$$ and $$\vec x = \la 1,1,1\ra\text{.}$$ Find $$\proj wx\text{,}$$ and sketch all three vectors with initial points at the origin.
Solution.
1. Applying Definition 11.3.14, we have
\begin{align*} \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v\\ \amp = \frac{-5}{10}\la 3,1\ra\\ \amp = \la -\frac32,-\frac12\ra\text{.} \end{align*}
Vectors $$\vec u\text{,}$$ $$\vec v$$ and $$\proj uv$$ are sketched in Figure 11.3.16. Note how the projection is parallel to $$\vec v\text{;}$$ that is, it lies on the same line through the origin as $$\vec v\text{,}$$ although it points in the opposite direction. That is because the angle between $$\vec u$$ and $$\vec v$$ is obtuse (i.e., greater than $$90^\circ$$).
2. Apply the definition:
\begin{align*} \proj wx \amp = \frac{\dotp wx}{\dotp xx}\vec x\\ \amp = \frac{6}{3}\la 1,1,1\ra\\ \amp = \la 2,2,2\ra\text{.} \end{align*}
These vectors are sketched in Figure 11.3.17.(a), and again in Figure 11.3.17.(b) from a different perspective. Because of the nature of graphing these vectors, the sketch in Figure 11.3.17.(a) makes it difficult to recognize that the drawn projection has the geometric properties it should. The graph shown in Figure 11.3.17.(b) illustrates these properties better.
We can use the properties of the dot product found in Theorem 11.3.3 to rearrange the formula found in Definition 11.3.14:
\begin{align*} \proj uv \amp = \frac{\dotp uv}{\dotp vv}\vec v\\ \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v\\ \amp = \left(\vec u \cdot \frac{\vec v}{\norm{\vec v}}\right) \frac{\vec v}{\norm{\vec v}}\text{.} \end{align*}
The above formula shows that the orthogonal projection of $$\vec u$$ onto $$\vec v$$ is only concerned with the direction of $$\vec v\text{,}$$ as both instances of $$\vec v$$ in the formula come in the form $$\vec v/\norm{\vec v}\text{,}$$ the unit vector in the direction of $$\vec v\text{.}$$
A special case of orthogonal projection occurs when $$\vec v$$ is a unit vector. In this situation, the formula for the orthogonal projection of a vector $$\vec u$$ onto $$\vec v$$ reduces to just $$\proj uv = (\vec u\cdot\vec v)\vec v\text{,}$$ as $$\vec v\cdot\vec v = 1\text{.}$$
This gives us a new understanding of the dot product. When $$\vec v$$ is a unit vector, essentially providing only direction information, the dot product of $$\vec u$$ and $$\vec v$$ gives “how much of $$\vec u$$ is in the direction of $$\vec v\text{.}$$” This use of the dot product will be very useful in future sections.
Now consider Figure 11.3.18 where the concept of the orthogonal projection is again illustrated. It is clear that
$$\vec u = \proj uv + \vec z\text{.}\tag{11.3.2}$$
As we know what $$\vec u$$ and $$\proj uv$$ are, we can solve for $$\vec z$$ and state that
\begin{equation*} \vec z = \vec u - \proj uv\text{.} \end{equation*}
This leads us to rewrite Equation (11.3.2) in a seemingly silly way:
\begin{equation*} \vec u = \proj uv + (\vec u - \proj uv)\text{.} \end{equation*}
This is not nonsense, as pointed out in the following Key Idea. (Notation note: the expression “$$\parallel \vec y$$” means “is parallel to $$\vec y\text{.}$$” We can use this notation to state “$$\vec x\parallel\vec y$$” which means “$$\vec x$$ is parallel to $$\vec y\text{.}$$” The expression “$$\perp \vec y$$” means “is orthogonal to $$\vec y\text{,}$$” and is used similarly.)
### Key Idea11.3.19.Orthogonal Decomposition of Vectors.
Let nonzero vectors $$\vec u$$ and $$\vec v$$ be given. Then $$\vec u$$ can be written as the sum of two vectors, one of which is parallel to $$\vec v\text{,}$$ and one of which is orthogonal to $$\vec v\text{:}$$
\begin{equation*} \vec u = \underbrace{\proj uv}_{\parallel\ \vec v}\ +\ (\underbrace{\vec u-\proj uv}_{\perp\ \vec v})\text{.} \end{equation*}
We illustrate the use of this equality in the following example.
### Example11.3.20.Orthogonal decomposition of vectors.
1. Let $$\vec u = \la -2,1\ra$$ and $$\vec v = \la 3,1\ra$$ as in Example 11.3.15. Decompose $$\vec u$$ as the sum of a vector parallel to $$\vec v$$ and a vector orthogonal to $$\vec v\text{.}$$
2. Let $$\vec w =\la 2,1,3\ra$$ and $$\vec x =\la 1,1,1\ra$$ as in Example 11.3.15. Decompose $$\vec w$$ as the sum of a vector parallel to $$\vec x$$ and a vector orthogonal to $$\vec x\text{.}$$
Solution.
1. In Example 11.3.15, we found that $$\proj uv = \la -1.5,-0.5\ra\text{.}$$ Let
\begin{equation*} \vec z = \vec u - \proj uv = \la -2,1\ra - \la -1.5,-0.5\ra = \la-0.5, 1.5\ra\text{.} \end{equation*}
Is $$\vec z$$ orthogonal to $$\vec v$$ ? (i.e., is $$\vec z \perp\vec v$$ ?) We check for orthogonality with the dot product:
\begin{equation*} \dotp zv = \la -0.5,1.5\ra \cdot \la 3,1\ra =0\text{.} \end{equation*}
Since the dot product is 0, we know $$\vec z \perp \vec v\text{.}$$ Thus:
\begin{align*} \vec u \amp = \proj uv\ +\ (\vec u - \proj uv)\\ \la -2,1\ra \amp = \underbrace{\la -1.5,-0.5\ra}_{\parallel\ \vec v}\ +\ \underbrace{\la -0.5,1.5\ra}_{\perp \ \vec v}\text{.} \end{align*}
2. We found in Example 11.3.15 that $$\proj wx = \la 2,2,2\ra\text{.}$$ Applying the Key Idea, we have:
\begin{equation*} \vec z = \vec w - \proj wx = \la 2,1,3\ra - \la 2,2,2\ra = \la 0,-1,1\ra\text{.} \end{equation*}
We check to see if $$\vec z \perp \vec x\text{:}$$
\begin{equation*} \dotp zx = \la 0,-1,1\ra \cdot \la 1,1,1\ra = 0\text{.} \end{equation*}
Since the dot product is 0, we know the two vectors are orthogonal. We now write $$\vec w$$ as the sum of two vectors, one parallel and one orthogonal to $$\vec x\text{:}$$
\begin{align*} \vec w \amp = \proj wx\ +\ (\vec w - \proj wx)\\ \la 2,1,3\ra \amp = \underbrace{\la 2,2,2\ra}_{\parallel\ \vec x}\ +\ \underbrace{\la 0,-1,1\ra}_{\perp \ \vec x} \end{align*}
We give an example of where this decomposition is useful.
### Example11.3.21.Orthogonally decomposing a force vector.
Consider Figure 11.3.22.(a), showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. Find the components of force, and their magnitudes, acting on the box (as sketched in Figure 11.3.22.(b)):
1. in the direction of the ramp, and
2. orthogonal to the ramp.
Solution.
As the ramp rises 5ft over a horizontal distance of 20ft, we can represent the direction of the ramp with the vector $$\vec r= \la 20,5\ra\text{.}$$ Gravity pulls down with a force of 50lb, which we represent with $$\vec g = \la 0,-50\ra\text{.}$$
1. To find the force of gravity in the direction of the ramp, we compute $$\proj gr\text{:}$$
\begin{align*} \proj gr \amp = \frac{\dotp gr}{\dotp rr}\vec r\\ \amp = \frac{-250}{425}\la 20,5\ra\\ \amp = \la -\frac{200}{17},-\frac{50}{17}\ra \approx \la -11.76,-2.94\ra\text{.} \end{align*}
The magnitude of $$\proj gr$$ is $$\norm{\proj gr} = 50/\sqrt{17} \approx 12.13\text{ lb }\text{.}$$ Though the box weighs 50lb, a force of about 12lb is enough to keep the box from sliding down the ramp.
2. To find the component $$\vec z$$ of gravity orthogonal to the ramp, we use Key Idea 11.3.19.
\begin{align*} \vec z \amp = \vec g - \proj gr\\ \amp = \la \frac{200}{17},-\frac{800}{17}\ra \approx \la 11.76,-47.06\ra\text{.} \end{align*}
The magnitude of this force is $$\norm{\vec z} \approx 48.51$$lb. In physics and engineering, knowing this force is important when computing things like static frictional force. (For instance, we could easily compute if the static frictional force alone was enough to keep the box from sliding down the ramp.)
### Subsection11.3.1Application to Work
In physics, the application of a force $$F$$ to move an object in a straight line a distance $$d$$ produces work; the amount of work $$W$$ is $$W=Fd\text{,}$$ (where $$F$$ is in the direction of travel). The orthogonal projection allows us to compute work when the force is not in the direction of travel.
Consider Figure 11.3.23, where a force $$\vec F$$ is being applied to an object moving in the direction of $$\vec d\text{.}$$ (The distance the object travels is the magnitude of $$\vec d\text{.}$$) The work done is the amount of force in the direction of $$\vec d\text{,}$$ $$\norm{\proj Fd}\text{,}$$ times $$\vnorm d\text{:}$$
\begin{align*} \norm{\proj Fd}\cdot\vnorm d \amp = \norm{\frac{\dotp Fd}{\dotp dd}\vec d}\cdot \vnorm d\\ \amp = \abs{\frac{\dotp Fd}{\norm{\vec{d}}^2}}\cdot \vnorm d\cdot\vnorm d\\ \amp = \frac{\abs{\dotp Fd}}{\norm{\vec{d}}^2}\norm{\vec{d}}^2\\ \amp = \abs{\dotp Fd}\text{.} \end{align*}
The expression $$\dotp Fd$$ will be positive if the angle between $$\vec F$$ and $$\vec d$$ is acute; when the angle is obtuse (hence $$\dotp Fd$$ is negative), the force is causing motion in the opposite direction of $$\vec d\text{,}$$ resulting in “negative work.” We want to capture this sign, so we drop the absolute value and find that $$W = \dotp Fd\text{.}$$
#### Definition11.3.24.Work.
Let $$\vec F$$ be a constant force that moves an object in a straight line from point $$P$$ to point $$Q\text{.}$$ Let $$\vec d = \overrightarrow{PQ}\text{.}$$ The work $$W$$ done by $$\vec F$$ along $$\vec d$$ is $$W = \dotp Fd\text{.}$$
#### Example11.3.25.Computing work.
A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in Figure 11.3.26. Compute the work done.
Solution.
The figure indicates that the force applied makes a $$30^\circ$$ angle with the horizontal, so $$\vec F = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \approx \la 43.3,25\ra\text{.}$$ The ramp is represented by $$\vec d = \la 15,3\ra\text{.}$$ The work done is simply
\begin{equation*} \dotp Fd = 50\la \cos(30^\circ) ,\sin(30^\circ) \ra \cdot \la 15,3\ra \approx 724.5 \text{ ft--lb }\text{.} \end{equation*}
Note how we did not actually compute the distance the object traveled, nor the magnitude of the force in the direction of travel; this is all inherently computed by the dot product!
The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. The next section explores another “product” on vectors, the cross product. Once again, angles play an important role, though in a much different way.
### Exercises11.3.2Exercises
#### Terms and Concepts
##### 1.
The dot product of two vectors is a , not a vector.
##### 2.
How are the concepts of the dot product and vector magnitude related?
##### 3.
How can one quickly tell if the angle between two vectors is acute or obtuse?
##### 4.
Give a synonym for “orthogonal.”
#### Problems
##### Exercise Group.
In the following exercises, find the dot product of the given vectors.
###### 5.
$$\vec u = \la 2,-4\ra\text{,}$$ $$\vec v = \la 3,7\ra$$
$$\vec u \cdot\vec v=$$
###### 6.
$$\vec u = \la 5,3\ra\text{,}$$ $$\vec v = \la 6,1\ra$$
$$\vec u \cdot\vec v=$$
###### 7.
$$\vec u = \la 1,-1,2\ra\text{,}$$ $$\vec v = \la 2,5,3\ra$$
$$\vec u \cdot\vec v=$$
###### 8.
$$\vec u = \la 3,5,-1\ra\text{,}$$ $$\vec v = \la 4,-1,7\ra$$
$$\vec u \cdot\vec v=$$
###### 9.
$$\vec u = \la 1,1\ra\text{,}$$ $$\vec v = \la 1,2,3\ra$$
###### 10.
$$\vec u = \la 1,2,3\ra\text{,}$$ $$\vec v = \la 0,0,0\ra$$
$$\vec u \cdot\vec v=$$
##### 11.
Create your own vectors $$\vec u\text{,}$$ $$\vec v$$ and $$\vec w$$ in $$\mathbb{R}^2$$ and show that $$\vec u\cdot (\vec v+\vec w) = \vec u\cdot \vec v + \vec u\cdot \vec w\text{.}$$
##### 12.
Create your own vectors $$\vec u$$ and $$\vec v$$ in $$\mathbb{R}^3$$ and scalar $$c$$ and show that $$c(\vec u\cdot \vec v) = \vec u\cdot (c\vec v)\text{.}$$
##### Exercise Group.
In the following exercises, find the measure of the angle between the two vectors in both radians and degrees.
###### 13.
The angle between $$\vec u = \la 1,1\ra$$ and $$\vec v = \la 1,2\ra$$ is .
###### 14.
The angle between $$\vec u = \la -2,1\ra$$ and $$\vec v = \la 3,5\ra$$ is .
###### 15.
The angle between $$\vec u = \la 8,1,-4\ra$$ and $$\vec v = \la 2,2,0\ra$$ is .
###### 16.
The angle between $$\vec u = \la 1,7,2\ra$$ and $$\vec v = \la 4,-2,5\ra$$ is .
##### Exercise Group.
In the following exercises, a vector $$\vec v$$ is given. Give two vectors that are orthogonal to $$\vec v\text{.}$$
###### 17.
Find two nonzero vectors orthogonal to $$\vec v = \la 4,7\ra\text{.}$$
and
###### 18.
Find two nonzero vectors orthogonal to $$\vec v = \la -3,5\ra\text{.}$$
and
###### 19.
Find two nonzero vectors orthogonal to $$\vec v = \la 1,1,1\ra\text{.}$$
and
###### 20.
Find two nonzero vectors orthogonal to $$\vec v = \la 1,-2,3\ra\text{.}$$
and
##### Exercise Group.
In the following exercises, vectors $$\vec u$$ and $$\vec v$$ are given. Find $$\proj uv\text{,}$$ the orthogonal projection of $$\vec u$$ onto $$\vec v\text{,}$$ and sketch all three vectors with the same initial point.
###### 21.
If $$\vec u = \la 1,2\ra$$ and $$\vec v = \la -1,3\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
###### 22.
If $$\vec u = \la 5,5\ra$$ and $$\vec v = \la 1,3\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
###### 23.
If $$\vec u = \la -3,2\ra$$ and $$\vec v = \la 1,1\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
###### 24.
If $$\vec u = \la -3,2\ra$$ and $$\vec v = \la 2,3\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
###### 25.
If $$\vec u = \la 1,5,1\ra$$ and $$\vec v = \la 1,2,3\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
###### 26.
If $$\vec u = \la 3,-1,2\ra$$ and $$\vec v = \la 2,2,1\ra\text{,}$$ then $$\proj uv=$$.
Sketch all three vectors on the same axes.
##### Exercise Group.
In the following exercises, vectors $$\vec u$$ and $$\vec v$$ are given. Write $$\vec u$$ as the sum of two vectors, one of which is parallel to $$\vec v$$ (or is zero) and one of which is orthogonal to $$\vec v\text{.}$$ Note: these are the same pairs of vectors as found in Exercises 11.3.2.21Exercise 11.3.2.26.
###### 27.
Write $$\vec u = \la 1,2\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la -1,3\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
###### 28.
Write $$\vec u = \la 5,5\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la 1,3\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
###### 29.
Write $$\vec u = \la -3,2\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la 1,1\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
###### 30.
Write $$\vec u = \la -3,2\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la 2,3\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
###### 31.
Write $$\vec u = \la 1,5,1\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la 1,2,3\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
###### 32.
Write $$\vec u = \la 3,-1,2\ra$$ as the sum of two vectors, one parallel to $$\vec v = \la 2,2,1\ra$$ (or zero) and the other perpendicular.
$$\vec u=$$$${}+{}$$
##### 33.
A 10lb box sits on a ramp that rises 4ft over a distance of 20ft. How much force is required to keep the box from sliding down the ramp?
##### 34.
A 10lb box sits on a 15ft ramp that makes a $$30^\circ$$ angle with the horizontal. How much force is required to keep the box from sliding down the ramp?
##### 35.
How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of $$45^\circ$$ to the horizontal?
##### 36.
How much work is performed in moving a box horizontally 10ft with a force of 20lb applied at an angle of $$10^\circ$$ to the horizontal?
##### 37.
How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied horizontally?
##### 38.
How much work is performed in moving a box up the length of a ramp that rises 2ft over a distance of 10ft, with a force of 50lb applied at an angle of $$45^\circ$$ to the horizontal?
##### 39.
How much work is performed in moving a box up the length of a 10ft ramp that makes a $$5^\circ$$ angle with the horizontal, with 50lb of force applied in the direction of the ramp?
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Find your Math Personality!
Trigonometric chart
Trigonometric chart
Emma is standing in the balcony of her house.
She is looking down at the flower pot placed in the balcony of another house.
You can observe a right-angled triangle in this situation.
Emma says that if she knows the height at which she is standing, she can find the distance between both the buildings.
Do you agree with her?
We can find the heights or distances using some mathematical techniques which fall under a branch of mathematics called "Trigonometry."
In this short lesson, we will learn about trigonometry, values in the trigonometric chart, and trigonometric tracer on unit circle.
Lesson Plan
1 What Does Trigonometry Formula Refer To? 2 Important Notes on Trigonometry 3 Solved Examples on Trigonometric Chart 4 Tips and Tricks on Trigonometric Chart 5 Interactive Questions on Trigonometric Chart
What Does Trigonometry Formula Refer To?
Look at the right-angled triangle $$ABC$$ shown below.
$$AC$$ is the hypotenuse.
The side $$AB$$ is the extended portion of $$\angle A$$.
Hence, we call it as the "side adjacent to $$\angle A$$".
The side $$BC$$ faces $$\angle A$$.
Hence, we call it the "side opposite to $$\angle A$$".
Trigonometric Ratios
We will now define some ratios involving the sides of the triangle $$ABC$$.
$$\sin{A}=\dfrac{\text{Side Opposite to } \angle A}{\text{Hypotenuse}}=\dfrac{BC}{AC}$$
$$\cos{A}=\dfrac{\text{Side Adjacent to } \angle A}{\text{Hypotenuse}}=\dfrac{AB}{AC}$$
$$\tan{A}=\dfrac{\text{Side Opposite to } \angle A}{\text{Side Adjacent to }\angle A}=\dfrac{BC}{AB}$$
$$\csc{A}=\dfrac{1}{\sin A}=\dfrac{AC}{BC}$$
$$\sec{A}=\dfrac{1}{\cos A}=\dfrac{AC}{AB}$$
$$\cot{A}=\dfrac{1}{\tan A}=\dfrac{AB}{BC}$$
Trigonometric Chart
You are already familiar with construction of angles like $$30^{\circ}$$, $$45^{\circ}$$, $$60^{\circ}$$, and $$90^{\circ}$$.
We will now see the values in the trigonometric chart for these angles.
$$\angle A$$ $$0^{\circ}$$ $$30^{\circ}$$ $$45^{\circ}$$ $$60^{\circ}$$ $$90^{\circ}$$
$$\sin {A}$$ 0 $$\dfrac{1}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\dfrac{\sqrt{3}}{2}$$ 1
$$\cos {A}$$ 1 $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\dfrac{1}{2}$$ 0
$$\tan {A}$$ 0 $$\dfrac{1}{\sqrt{3}}$$ 1 $$\sqrt{3}$$ Not defined
$$\csc {A}$$ Not defined 2 $$\sqrt{2}$$ $$\dfrac{2}{\sqrt{3}}$$ 1
$$\sec {A}$$ 1 $$\dfrac{2}{\sqrt{3}}$$ $$\sqrt{2}$$ 2 Not defined
$$\cot {A}$$ Not defined $$\sqrt{3}$$ 1 $$\dfrac{1}{\sqrt{3}}$$ 0
Experiment with the simulation below to determine the values of all trigonometric functions for distinct angles and observe the values being plotted on the graphs.
What Are the 8 Trigonometric Identities?
An equation becomes an identity when it is true for all the variables involved in it.
Now we will discuss trigonometric identities which are true for all angles involved in it.
$$\cos^{2}{A}+\sin^{2}{A}=1$$
$$1+\tan^{2}{A}=\sec^{2}{A}$$
$$1+\cot^{2}{A}=\csc^{2}{A}$$
$$\cos{(2A)}=2\cos^{2}{A}-1=1-2\sin^{2}{A}$$
$$\sin{(2A)}=2\cos{A}\sin{A}$$
$$\tan{(2A)}=\dfrac{2\tan{A}}{1-\tan^{2}{A}}$$
$$\cos{(2A)}=\cos^{2}{A}-\sin^{2}{A}$$
$$\sin{(2A)}=\dfrac{2\tan{A}}{1+\tan^{2}{A}}$$
Apart from these identities, we have a few more formulas which are helpful for you in solving trigonometric problems.
Example on Trigonometry
Trigonometry is a branch of mathematics that is used to solve geometric problems involving triangles.
Example
Let's assume that $$ABC$$ is a right-angled triangle with $$\tan{A}=\dfrac{4}{3}$$.
Let's find all other trigonometric ratios of $$\angle A$$.
We know that $$\tan{A}=\dfrac{BC}{AB}=\dfrac{4}{3}$$.
Hence, we can write $$BC=4k$$ and $$AB=3k$$, where $$k$$ is a positive integer.
Using Pythagoras theorem,
\begin{align}(AC)^{2}&=(AB)^{2}+(BC)^{2}\\&=9k^{2}+16k^{2}\\&=25k^{2}\end{align}
Thus, $$AC=5k$$.
Other trigonometric ratio are:
$\sin{A}=\dfrac{BC}{AC}=\dfrac{4k}{5k}=\dfrac{4}{5}$
$\cos{A}=\dfrac{AB}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}$
$\cot{A}=\dfrac{1}{\tan{A}}=\dfrac{3}{4}$
$\csc{A}=\dfrac{1}{\sin{A}}=\dfrac{5}{4}$
$\sec{A}=\dfrac{1}{\cos{A}}=\dfrac{5}{3}$
Trigonometric Tracer on Unit Circle
Here is a trigonometric tracer that shows all the trigonometric functions on a unit circle.
Choose a trigonometric function and observe its relation with the unit circle and see how its graph is traced out over a complete angle.
What Are the Rules of Trigonometry?
Signs of Trigonometric Functions
The following table shows the signs of different trigonometric functions in different quadrants.
$$\sin {A}$$ + + - -
$$\cos {A}$$ + - - +
$$\tan {A}$$ + - + -
$$\csc {A}$$ + + - -
$$\sec {A}$$ + - - +
$$\cot {A}$$ + - + -
Where is Trigonometry Used?
Do you know that trigonometry is one of the most ancient subjects which is studied by scholars all over the world?
The need for trigonometry first arose in astronomy.
In astronomy, it is used to determine the distances from the Earth to the planets and stars.
In geography and navigation, it is used to construct maps. It is also used to find the position of an island in relation to the longitudes and latitudes.
Even today, some of the technologically advanced methods which are used in engineering and physical sciences are based on the concepts of trigonometry.
Trigonometry is also used in the science of seismology to design electric circuits, describe the state of an atom, and predict the heights of tides in the ocean.
Important Notes
1. The word "Trigonometry" originated from the words, "Trigonon" which means "triangle" and "Metron" which means "to measure".
2. All trigonometric functions are periodic in nature.
3. All trigonometric functions give positive values in quadrant I.
4. We can use trigonometric ratios to find the height of an object or the distance between two objects.
Solved Examples
Example 1
A contractor wants to install a slide for children in a park.
He designs the slide such that its top is at a height of 5 feet and the angle is $$30^{\circ}$$ with the ground.
Can you determine the length of the slide?
Solution
Let's represent the length of the slide by $$AC$$.
Since the ratio involves the sides $$AB$$ and $$AC$$, we will use the trigonometric ratio $$\sin{30^{\circ}}$$.
\begin{align}\sin{30^{\circ}}&=\dfrac{AB}{AC}\\\dfrac{1}{2}&=\dfrac{5}{AC}\\AC&=10\end{align}
$$\therefore$$ The length of the slide is 10 feet.
Example 2
Look at the triangle below.
The triangle is right-angled at $$C$$ with $$AB=29\;\text{units}$$ and $$BC=21\;\text{units}$$.
Can you verify the trigonometric identity $$\cos^{2}{A}+\sin^{2}{A}=1$$ using these values?
Solution
We will find $$AC$$ using Pythagoras theorem.
\begin{align}AC&=\sqrt{(AB)^{2}-(BC)^{2}}\\&=\sqrt{29^{2}-21^{2}}\\&=\sqrt{400}\\&=20\end{align}
Let's determine the values of $$\sin{\theta}$$ and $$\cos{\theta}$$.
$\sin{\theta}=\dfrac{AC}{AB}=\dfrac{20}{29}$
$\cos{\theta}=\dfrac{BC}{AB}=\dfrac{21}{29}$
Now let's verify the identity.
\begin{align}\cos^{2}{A}+\sin^{2}{A}&=\left(\dfrac{21}{29}\right)^{2}+\left(\dfrac{20}{29}\right)^{2}\\&=\dfrac{400+441}{841}\\&=1\end{align}
$$\therefore$$ The identity is verified.
Example 3
Jack is standing on the ground and looking at the top of the tower with the angle of elevation as $$60^{\circ}$$.
If he is standing 15 feet away from the foot of the tower, can you determine the height of the tower?
Solution
Since the ratio involves the sides $$AB$$ and $$BC$$, we will use the trigonometric ratio $$\tan{60^{\circ}}$$.
\begin{align}\tan{60^{\circ}}&=\dfrac{AB}{BC}\\\sqrt{3}&=\dfrac{AB}{15}\\AB&=15\sqrt{3}\end{align}
$$\therefore$$ The height of the tower is $$15\sqrt{3}$$ feet.
Example 4
Rachel drew a triangle right-angled at Q with PQ as 3 units and PR as 6 units.
Can you determine the angles $$\angle QPR$$ and $$\angle PRQ$$?
Solution
Using sine trigonometric ratio we have,
$\sin{R}=\dfrac{PQ}{PR}=\dfrac{3}{6}=\dfrac{1}{2}$
Sine function takes the value $$\dfrac{1}{2}$$ at $$30^{\circ}$$.
So, $$\angle PRQ=30^{\circ}$$.
Using the angle sum property of triangles,
\begin{align}\angle PRQ+\angle QPR+\angle PQR&=180^{\circ}\\30^{\circ}+90^{\circ}+\angle QPR&=180^{\circ}\\\angle QPR&=180^{\circ}-120^{\circ}\\\angle QPR&=60^{\circ}\end{align}
$$\therefore$$ The values of angles are $$\angle QPR=60^{\circ}$$, $$\angle PRQ=30^{\circ}$$.
Here are some tips and tricks that will help you solve trigonometric problems.
Tips and Tricks
1. The values of trigonometric ratios do not change with the change in the side lengths of the triangle if the angle remains the same.
2. The values of $$\sin{A}$$ and $$\cos{A}$$ is always less than or equal to 1.
3. From the trigonometric chart, you can observe that as $$\angle A$$ increases from $$0^{\circ}$$ to $$90^{\circ}$$, $$\sin{A}$$ increases from 0 to 1 and $$\cos{A}$$ decreases from 1 to 0.
4. $$\sin{A}$$, $$\tan{A}$$, $$\cot{A}$$ and $$\csc{A}$$ are odd functions.
5. $$\cos{A}$$ and $$\sec{A}$$ are even functions.
Interactive Questions
Here are a few activities for you to practice. Select/type your answer and click the "Check Answer" button to see the result.
Let's Summarize
This mini-lesson targeted the fascinating concept of the Trigonometric Chart. The math journey around Trigonometric Chart starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
We hope you enjoyed trying your hand at solving a few interactive questions on the concept of Trigonometric Chart.
At Cuemath, our team of math experts are dedicated to making learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it problems, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
1. What are the basic formulas of trigonometry?
The basic formulas of trigonometry are:
• $$\cos^{2}{A}+\sin^{2}{A}=1$$
• $$1+\tan^{2}{A}=\sec^{2}{A}$$
• $$\cot^{2}{A}+1=\csc^{2}{A}$$
2. What is the easiest way to learn trigonometry?
Follow the points mentioned below to learn trigonometry in the easiest way.
• Familiarize yourself with the parts of a triangle like a hypotenuse, obtuse angle, and acute angle.
• Learn to make a unit circle which will help you relate all trigonometric functions.
• Remember the six trigonometric functions.
• Practice problems to memorize all the trigonometric identities.
3. Who is the father of Trigonometry?
The great mathematician Hipparchus was the founder of trigonometry. Therefore, he is known as the father of trigonometry.
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How do you use implicit differentiation to find dy/dx given x^2+3xy-y^2=0?
Oct 24, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$
Explanation:
$0 = {x}^{2} + 3 x y - {y}^{2} = \left(x + \frac{1}{2} \left(3 - \sqrt{13}\right) y\right) \left(x + \frac{1}{2} \left(3 + \sqrt{13}\right) y\right)$
so we have two solutions
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$
This result can be obtained the usual way
$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$ so
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / \left({f}_{y}\right) = - \frac{2 x + 3 y}{3 x - 2 y}$
now substituting
$y = \frac{1}{2} \left(3 x \pm \sqrt{13} x\right)$ we obtain the same result.
This result can be also obtained by observing that this is a homogeneous equation. Making $y = \lambda x$ we obtain
${x}^{2} + 3 \lambda {x}^{2} - {\lambda}^{2} {x}^{2} = 0$ being true for all $x$ we have
$1 + 3 \lambda - {\lambda}^{2} = 0$ with solutions
$\lambda = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$
Oct 24, 2016
See below.
Explanation:
We consider $y$ an unknown function of $x$. It may help to think of it as $\text{some stuff in parentheses}$
So we have
x^2+3xunderbrace((" "))_("this stuff is called "y) - underbrace((" "))_("this is also "y) ^2 = 0
When we differentiate $3 x$ times some stuff in parentheses, we need the product rule. When we differentiate the square of some stuff in parentheses, we use the chain rule. So when we differentiate both sides of the equation (with respect to $x$), we get
${x}^{2} + 3 x y - {y}^{2} = 0$
$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(\left(3 x\right) \left(y\right)\right) - \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(0\right)$
$2 x + \left(3\right) \left(y\right) + \left(3 x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 {y}^{1} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
Solve to get
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 3 y}{2 y - 3 x}$.
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Home » Greatest Common Factor » GCF of 65 and 24
# GCF of 65 and 24
The gcf of 65 and 24 is the largest positive integer that divides the numbers 65 and 24 without a remainder. Spelled out, it is the greatest common factor of 65 and 24. Here you can find the gcf of 65 and 24, along with a total of three methods for computing it.
1
In addition, we have a calculator you should check out. Not only can it determine the gcf of 65 and 24, but also that of three or more integers including sixty-five and twenty-four for example. Keep reading to learn everything about the gcf (65,24) and the terms related to it.
## What is the GCF of 65 and 24
If you just want to know what is the greatest common factor of 65 and 24, it is 1. Usually, this is written as
gcf(65,24) = 1
The gcf of 65 and 24 can be obtained like this:
• The factors of 65 are 65, 13, 5, 1.
• The factors of 24 are 24, 12, 8, 6, 4, 3, 2, 1.
• The common factors of 65 and 24 are 1, intersecting the two sets above.
• In the intersection factors of 65 ∩ factors of 24 the greatest element is 1.
• Therefore, the greatest common factor of 65 and 24 is 1.
Taking the above into account you also know how to find all the common factors of 65 and 24, not just the greatest. In the next section we show you how to calculate the gcf of sixty-five and twenty-four by means of two more methods.
## How to find the GCF of 65 and 24
The greatest common factor of 65 and 24 can be computed by using the least common multiple aka lcm of 65 and 24. This is the easiest approach:
gcf (65,24) = = 1
Alternatively, the gcf of 65 and 24 can be found using the prime factorization of 65 and 24:
• The prime factorization of 65 is: 5 x 13
• The prime factorization of 24 is: 2 x 2 x 2 x 3
• The prime factors and multiplicities 65 and 24 have in common are: 1
• 1 is the gcf of 65 and 24
• gcf(65,24) = 1
In any case, the easiest way to compute the gcf of two numbers like 65 and 24 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 65,24. The calculation is conducted automatically.
Similar searched terms on our site also include:
## Use of GCF of 65 and 24
What is the greatest common factor of 65 and 24 used for? Answer: It is helpful for reducing fractions like 65 / 24. Just divide the nominator as well as the denominator by the gcf (65,24) to reduce the fraction to lowest terms.
.
## Properties of GCF of 65 and 24
The most important properties of the gcf(65,24) are:
• Commutative property: gcf(65,24) = gcf(24,65)
• Associative property: gcf(65,24,n) = gcf(gcf(24,65),n)
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 65 and 24 is 1. In common notation: gcf (65,24) = 1.
If you have been searching for gcf 65 and 24 or gcf 65 24 then you have come to the correct page, too. The same is the true if you typed gcf for 65 and 24 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including sixty-five / twenty-four by using the search form in the sidebar of this page.
Questions and comments related to the gcf of 65 and 24 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the greatest common factor of 65 and 24 has been useful to you, and make sure to bookmark our site.
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An office receptionist took Rs. 8000 as a personal loan from the manager for 6 months at 11% p.a. simple interest. At the end of six months, she gave the manager an Rs. 3500 and a gold coin to settle her dues. At what price did the manager get the gold coin?
Last updated date: 22nd Feb 2024
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Hint: In the above question you were asked to find the price of the gold coin which the receptionist has paid. For solving this question, you will need to use the formula of simple interest. After finding the simple interest you will have to find the amount. So, let us see how we can solve this problem.
Complete Step by Step Solution:
In this question we have:
Principle(P) = Rs. 8000
Rate(R) = 11%
Time(T) = 6 months
In the formula of simple interest, we write time in terms of years. Here, we have 6 months, so $\dfrac{1}{2}$ years.
The formula of simple interest is: $\dfrac{{P \times R \times T}}{{100}}$ .
Putting the values of P, R and T in the formula of simple interest we get
$= \dfrac{{8000 \times 11 \times \dfrac{1}{2}}}{{100}}$
$= \dfrac{{4000 \times 11 \times 1}}{{100}}$
After solving the above expression we get,
$= 440$
Therefore, S.I = Rs. 440
Amount = Principle + S.I
Therefore, Amount = Rs. (8000+440)
= Rs. 8440
Therefore, after 6 months receptionist has to pay Rs. 8440
It is given in the question that receptionist has paid Rs. 3500, so the price of gold coins will be
= Rs. (8440 - 3500)
= Rs. 4900
Therefore, the price of gold coins which the receptionist has paid is Rs. 4900.
Note:
In the above solution we used the formula of simple interest and then we calculated the total amount which the receptionist has to pay after 6 months. After which we subtracted Rs. 3500 that the receptionist has already paid. And for the rest of the amount, she gave the gold terms, which should be equal to the remaining amount. So we get to know that the price of the gold coin should be equal to the remaining money after payment of Rs. 3500.
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# Parallel Lines and Angles Vertical Angles Vertical Angles
• Slides: 15
Parallel Lines and Angles
Vertical Angles • Vertical Angles are angles that are opposite each other at an intersection. • Vertical Angles are equal 1 2 3 4 Angles 1 and 4 are vertical angles Angles 2 and 3 are vertical angles On your notesheet: Write a word and a picture definition for vertical angles.
Supplementary Angles • Supplementary angles are angles that form lines (also called linear pairs) • Supplementary angles add up to 180 1 2 3 4 Angles 1 and 2 are supplementary, Angles 1 and 3 are supplementary Angles 2 and 4 are supplementary, Angles 3 and 4 are supplementary On your notesheet: Write a word and a picture definition for supplementary angles.
Parallel Lines and Planes You will learn to describe relationships among lines, parts of lines, and planes. In geometry, two lines in a plane that are always the same parallel lines distance apart are ______. No two parallel lines intersect, no matter how far you extend them.
Parallel Lines and Planes Definition of Parallel Lines Two lines are parallel if they are in the same plane and intersect do not ____. -This means the lines never touch -This means the lines have the same slope -This means the lines are always the same distance apart -The symbol for parallel is two vertical lines (II). For example if line m and line t are parallel you could write m II t. On your notesheet: Write a verbal and picture definition of parallel lines On your paper: #1. What are three ways to describe how lines are parallel? #2. What is the symbol for parallel?
Parallel Lines and Transversals You will learn to identify the relationships among pairs of interior and exterior angles formed by two parallel lines and a transversal.
Parallel Lines and Transversals In geometry, a line, line segment, or ray that intersects two or more lines at transversal different points is called a _____ A 2 1 4 5 8 6 7 3 l m B is an example of a transversal. It intercepts lines l and m. Note all of the different angles formed at the points of intersection.
Parallel Lines and Transversals Definition of Transversal In a plane, a line is a transversal if it intersects two or more lines, each at a different point. The lines cut by a transversal may or may not be parallel. Parallel Lines Nonparallel Lines l 1 2 4 3 m 5 6 8 7 is a transversal for l and m. c 5 6 8 7 t t b r r is a transversal for b and c.
Parallel Lines and Transversals Two lines divide the plane into three regions. The region between the lines is referred to as the interior. The two regions not between the lines is referred to as the exterior. Exterior Interior Exterior
Parallel Lines and Transversals eight angles are formed. When a transversal intersects two lines, _____ These angles are given special names. Alternate Interior angles are between the two lines on the opposite sides of the transversal. Ex. 4 and 6, 3 and 5 Consectutive Interior angles between the two lines are on the same side of the transversal. Ex. 4 and 5, 3 and 6 Alternate Exterior angles are outside the two lines on the opposite sides of the transversal. Ex. 1 and 7, 2 and 8 Corresponding angles are in the same position at each intersection. Ex. 1 and 5, 2 and 6, 4 and 8, 3 and 7 l 1 2 4 3 m 5 6 8 7 t
Parallel Lines and Transversals Alternate Interior Angles between the two lines on the opposite sides of the transversal. Ex. 4 and 6, 3 and 5 If two parallel lines are cut by a transversal, then each pair of congruent (equal) Alternate interior angles is _____. 1 2 4 3 5 6 8 7 Angles 4 and 6 are alternate interior angles so we know On your paper: #3. Write down one other pair of alternate interior angles.
Parallel Lines and Transversals Consecutive Interior Angles between the two lines are on the same side of the transversal. Ex. 4 and 5, 3 and 6 If two parallel lines are cut by a transversal, then each pair of consecutive interior angles is supplementary (add to 180) _______. 1 2 4 3 5 6 8 7 Angles 4 and 5 are consecutive interior angles so we know: On your paper: #4. Write down one other pair of consecutive interior angles.
Parallel Lines and Transversals Alternate Exterior Angles outside the two lines on the opposite sides of the transversal. Ex. 1 and 7, 2 and 8 If two parallel lines are cut by a transversal, then congruent each pair of alternate exterior angles is _____. 1 2 4 3 5 6 8 7 Angle 1 and 7 are alternate exterior angles so we know: On your paper: #5. Write down one other pair of alternate exterior angles.
Parallel Lines and Transversals Corresponding Angles are in the same position at each intersection. Ex. 1 and 5, 2 and 6, 4 and 8, 3 and 7 If two parallel lines are cut by a transversal, then each pair of corresponding angles is _____. congruent l 1 2 4 3 m 5 6 8 7 t Angle 1 and 5 are both in the upper left of each intersection so they are corresponding angles and we then know angle 1=angle 5 On your paper: #6. Write down three other pairs of corresponding angles.
Transversals and Corresponding Angles Concept Summary Types of angle pairs formed when a transversal cuts two parallel lines. Congruent Supplementary alternate interior consecutive interior alternate exterior corresponding On your notesheet: Under “Special pair of Angles” for each pair write equal or supplementary, then using the examples on the notesheet write one pair from A and B. Turn in your half piece of paper. Go to a table and complete your notesheet page on parallel lines. After you have completed this, start the next sheet in your packet. Whatever you don’t complete is homework for tonight.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Exterior Angles Theorems
## Exterior angles equal the sum of the remote interiors.
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Practice Exterior Angles Theorems
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Exterior Angles in Given Triangles
Learning Goal
By the end of the lesson I will be able to . . . describe the properties and relationships of the exterior angles of triangles
What if you knew that two of the exterior angles of a triangle measured \begin{align*}130^\circ\end{align*}? How could you find the measure of the third exterior angle? After completing this Concept, you'll be able to apply the Exterior Angle Sum Theorem to solve problems like this one.
### Watch This
Then watch this video.
Finally, watch this video.
### Guidance
An Exterior Angle is the angle formed by one side of a polygon and the extension of the adjacent side.
In all polygons, there are two sets of exterior angles, one that goes around clockwise and the other goes around counter-clockwise.
Notice that the interior angle and its adjacent exterior angle form a linear pair and add up to \begin{align*}180^\circ\end{align*}.
\begin{align*}m\angle1 + m\angle2 = 180^\circ\end{align*}
There are two important theorems to know involving exterior angles: the Exterior Angle Sum Theorem and the Exterior Angle Theorem.
The Exterior Angle Sum Theorem states that the exterior angles of any polygon will always add up to \begin{align*}360^\circ\end{align*}.
\begin{align*}m\angle{1} + m\angle{2}+m\angle{3} &= 360^\circ\\ m\angle{4} + m\angle{5} + m\angle{6} & = 360^\circ\end{align*}.
The Exterior Angle Theorem states that an exterior angle of a triangle is equal to the sum of its remote interior angles. (Remote Interior Angles are the two interior angles in a triangle that are not adjacent to the indicated exterior angle.)
\begin{align*}m\angle{A} + m\angle{B}=m\angle{ACD}\end{align*}.
#### Example A
Find the measure of \begin{align*}\angle{RQS}\end{align*}.
Notice that \begin{align*}112^\circ\end{align*} is an exterior angle of \begin{align*}\triangle{RQS}\end{align*} and is supplementary to \begin{align*}\angle{RQS}\end{align*}.
Set up an equation to solve for the missing angle.
\begin{align*}112^\circ + m\angle{RQS} & = 180^\circ\\ m\angle{RQS} &= 68^\circ\end{align*}
#### Example B
Find the measures of the numbered interior and exterior angles in the triangle.
We know that \begin{align*}m\angle{1} + 92^\circ = 180^\circ\end{align*} because they form a linear pair. So, \begin{align*}m\angle{1} = 88^\circ\end{align*}.
Similarly, \begin{align*}m\angle{2} + 123^\circ = 180^\circ\end{align*} because they form a linear pair. So, \begin{align*}m\angle{2} = 57^\circ\end{align*}.
We also know that the three interior angles must add up to \begin{align*}180^\circ\end{align*} by the Triangle Sum Theorem.
\begin{align*}m\angle{1} + m\angle{2} +m\angle{3} & = 180^\circ \qquad \text{by the Triangle Sum Theorem.}\\ 88^\circ + 57^\circ + m\angle{3} &= 180\\ m\angle{3} & = 35^\circ\end{align*}
\begin{align*}\text{Lastly}, \ m\angle{3} + m\angle{4} & = 180^\circ \qquad \text{because they form a linear pair}.\\ 35^\circ + m\angle{4} &= 180^\circ\\ m\angle{4} &= 145^\circ\end{align*}
#### Example C
What is the value of \begin{align*}p\end{align*} in the triangle below?
First, we need to find the missing exterior angle, which we will call \begin{align*}x\end{align*}. Set up an equation using the Exterior Angle Sum Theorem.
\begin{align*}130^\circ + 110^\circ + x &= 360^\circ\\ x& = 360^\circ-130^\circ-110^\circ\\ x& = 120^\circ\end{align*}
\begin{align*}x\end{align*} and \begin{align*}p\end{align*} add up to \begin{align*}180^\circ\end{align*} because they are a linear pair.
\begin{align*}x + p & = 180^\circ\\ 120^\circ + p & = 180^\circ\\ p & = 60^\circ\end{align*}
### Guided Practice
1. Find \begin{align*}m\angle{C}\end{align*}.
2. Two interior angles of a triangle are \begin{align*}40^\circ\end{align*} and \begin{align*}73^\circ\end{align*}. What are the measures of the three exterior angles of the triangle?
3. Find the value of \begin{align*}x\end{align*} and the measure of each angle.
1. Using the Exterior Angle Theorem
\begin{align*}m\angle{C} + 16^\circ & = 121^\circ\\ m\angle{C} & = 105^\circ\end{align*}
If you forget the Exterior Angle Theorem, you can do this problem just like Example C.
2. Remember that every interior angle forms a linear pair (adds up to \begin{align*}180^\circ\end{align*}) with an exterior angle. So, since one of the interior angles is \begin{align*}40^\circ\end{align*} that means that one of the exterior angles is \begin{align*}140^\circ\end{align*} (because \begin{align*}40+140=180\end{align*}). Similarly, since another one of the interior angles is \begin{align*}73^\circ\end{align*}, one of the exterior angles must be \begin{align*}107^\circ\end{align*}. The third interior angle is not given to us, but we could figure it out using the Triangle Sum Theorem. We can also use the Exterior Angle Sum Theorem. If two of the exterior angles are \begin{align*}140^\circ\end{align*} and \begin{align*}107^\circ\end{align*}, then the third Exterior Angle must be \begin{align*}113^\circ\end{align*} since \begin{align*}140+107+113=360\end{align*}.
So, the measures of the three exterior angles are \begin{align*}140\end{align*}, \begin{align*}107\end{align*} and \begin{align*}113\end{align*}.
3. Set up an equation using the Exterior Angle Theorem.
\begin{align*}&(4x+2)^\circ + (2x-9)^\circ = (5x+13)^\circ\\ & \quad \uparrow \qquad \qquad \nearrow \qquad \qquad \qquad \uparrow\\ & \text{remote interior angles} \qquad \qquad \text{exterior angle}\\ & \qquad \qquad \quad \ (6x-7)^\circ = (5x+13)^\circ\\ & \qquad \qquad \qquad \qquad \ \ x = 20\end{align*}
Substitute in \begin{align*}20\end{align*} for \begin{align*}x\end{align*} to find each angle.
\begin{align*}[4(20)+2]^\circ=82^\circ && [2(20)-9]^\circ=31^\circ && \text{Exterior angle:} \ [5(20)+13]^\circ=113^\circ\end{align*}
### Practice
Determine \begin{align*}m\angle{1}\end{align*}.
Use the following picture for the next three problems:
1. What is \begin{align*}m\angle{1}+m\angle{2}+m\angle{3}\end{align*}?
2. What is \begin{align*}m\angle{4}+m\angle{5}+m\angle{6}\end{align*}?
3. What is \begin{align*}m\angle{7}+m\angle{8}+m\angle{9}\end{align*}?
Solve for \begin{align*}x\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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# 10.8: t Distribution
Learning Objectives
• State the difference between the shape of the $$t$$ distribution and the normal distribution
• State how the difference between the shape of the $$t$$ distribution and normal distribution is affected by the degrees of freedom
• Use a $$t$$ table to find the value of $$t$$ to use in a confidence interval
• Use the $$t$$ calculator to find the value of $$t$$ to use in a confidence interval
In the introduction to normal distributions it was shown that $$95\%$$ of the area of a normal distribution is within $$1.96$$ standard deviations of the mean. Therefore, if you randomly sampled a value from a normal distribution with a mean of $$100$$, the probability it would be within $$1.96\sigma$$ of $$100$$ is $$0.95$$. Similarly, if you sample $$N$$ values from the population, the probability that the sample mean ($$M$$) will be within $$1.96\sigma _M$$ of $$100$$ is $$0.95$$.
Now consider the case in which you have a normal distribution but you do not know the standard deviation. You sample $$N$$ values and compute the sample mean ($$M$$) and estimate the standard error of the mean ($$\sigma _M$$) with $$s_M$$. What is the probability that $$M$$ will be within $$1.96 s_M$$ of the population mean ($$\mu$$)?
This is a difficult problem because there are two ways in which $$M$$ could be more than $$1.96 s_M$$ from $$\mu$$:
1. $$M$$ could, by chance, be either very high or very low and
2. $$s_M$$ could, by chance, be very low.
Intuitively, it makes sense that the probability of being within $$1.96$$ standard errors of the mean should be smaller than in the case when the standard deviation is known (and cannot be underestimated). But exactly how much smaller? Fortunately, the way to work out this type of problem was solved in the early $$20^{th}$$ century by W. S. Gosset who determined the distribution of a mean divided by an estimate of its standard error. This distribution is called the Student's $$t$$ distribution or sometimes just the $$t$$ distribution. Gosset worked out the $$t$$ distribution and associated statistical tests while working for a brewery in Ireland. Because of a contractual agreement with the brewery, he published the article under the pseudonym "Student." That is why the $$t$$ test is called the "Student's $$t$$ test."
The $$t$$ distribution is very similar to the normal distribution when the estimate of variance is based on many degrees of freedom, but has relatively more scores in its tails when there are fewer degrees of freedom. Figure $$\PageIndex{1}$$ shows $$t$$ distributions with $$2$$, $$4$$, and $$10$$ degrees of freedom and the standard normal distribution. Notice that the normal distribution has relatively more scores in the center of the distribution and the $$t$$ distribution has relatively more in the tails. The $$t$$ distribution is therefore leptokurtic. The $$t$$ distribution approaches the normal distribution as the degrees of freedom increase.
Since the $$t$$ distribution is leptokurtic, the percentage of the distribution within $$1.96$$ standard deviations of the mean is less than the $$95\%$$ for the normal distribution. Table $$\PageIndex{1}$$ shows the number of standard deviations from the mean required to contain $$95\%$$ and $$99\%$$ of the area of the $$t$$ distribution for various degrees of freedom. These are the values of $$t$$ that you use in a confidence interval. The corresponding values for the normal distribution are $$1.96$$ and $$2.58$$ respectively. Notice that with few degrees of freedom, the values of $$t$$ are much higher than the corresponding values for a normal distribution and that the difference decreases as the degrees of freedom increase. The values in Table $$\PageIndex{1}$$ can be obtained from the "Find $$t$$ for a confidence interval" calculator.
Table $$\PageIndex{1}$$: Abbreviated $$t$$ table
df 0.95 0.99
2 4.303 9.925
3 3.182 5.841
4 2.776 4.604
5 2.571 4.032
8 2.306 3.355
10 2.228 3.169
20 2.086 2.845
50 2.009 2.678
100 1.984 2.626
Returning to the problem posed at the beginning of this section, suppose you sampled $$9$$ values from a normal population and estimated the standard error of the mean ($$\sigma _M$$) with $$s_M$$. What is the probability that $$M$$ would be within $$1.96 s_M$$ of $$\mu$$? Since the sample size is $$9$$, there are $$N - 1 = 8 df$$. From Table $$\PageIndex{1}$$ you can see that with $$8 df$$ the probability is $$0.95$$ that the mean will be within $$2.306 s_M$$ of $$\mu$$. The probability that it will be within $$1.96 s_M$$ of $$\mu$$ is therefore lower than $$0.95$$.
As shown in Figure $$\PageIndex{2}$$, the "$$t$$ distribution" calculator can be used to find that $$0.086$$ of the area of a $$t$$ distribution is more than $$1.96$$ standard deviations from the mean, so the probability that $$M$$ would be less than $$1.96 s_M$$ from $$\mu$$ is $$1 - 0.086 = 0.914$$.
As expected, this probability is less than $$0.95$$ that would have been obtained if $$\sigma _M$$ had been known instead of estimated.
## Contributor
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
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×
Inscribing Triangles in Curves (Part 1)
Some of you may have heard about the unsolved geometry problem called "Inscribed Square Problem" or "Toeplitz' conjecture". It states that in any closed curve, there exists 4 points on the curve that form a square. It sounds like a really simple problem, yet mathematicians have failed to prove it.
In this note, we will attack a similar problem: finding three points on a curve that are the vertices of an equilateral triangle. Unlike the square problem, this one is solvable.
First try it on your own. Can you devise a fool-proof way to find three points on a curve that make an equilateral triangle? Don't read on until you give it a try.
The key insight we need to think of in order to find an equilateral triangle inscribed in this curve is rotations. Here's how:
First, we take our curve. Although I will use a polygon in my examples, the idea is the same.
Imgur
Now we pick any point on our polygon. It doesn't matter which; we can always find an equilateral triangle inscribed in the curve with this point as a vertex.
Imgur
Using this point as the center of revolution, turn the polygon $$60^{\circ}$$. It doesn't matter which direction.
Imgur
Now we can see that we have some points of intersection. Mark one of these points of intersection. Then, rotate it back $$60^{\circ}$$ in the opposite direction of how you rotated the polygon.
Imgur
We can get rid of our rotated polygon now, and connect these two marked points with our point of rotation:
Imgur
And we're done! We have find an equilateral triangle inscribed in the curve.
In fact, this is only one of the many inscribed equilateral triangles in this curve. We could have picked any of the intersections and created a whole new equilateral triangle.
In addition, we could have picked any point on the entire curve to begin with, and found equilateral triangles. Thus, there is an infinite number of equilateral triangles that can be inscribed in a random closed curve.
Now that we've found a way to inscribe equilateral triangles, can we do the same thing for other triangles? Yes, we can! The method of inscribing any arbitrary isosceles triangle is similar to the equilateral triangle; just change the number of degrees you rotate the curve by. For example, rotating the curve by $$90^{\circ}$$ will give you $$45-45-90$$ triangles inscribed in the curve. As an example, here is one inscribed in the same curve we used:
Imgur
This is cool and all, but here's the finale: we can actually inscribe any arbitrary triangle in any arbitrary closed curve! However, you'll have to wait until the next note in order to find out how. Try to figure it out yourself in the meantime.
Daniel $$\ddot\smile$$
Note by Daniel Liu
2 years, 9 months ago
Sort by:
This may be dumb but what if there are no intersection points after the rotation? · 2 years, 9 months ago
Don't worry, your question isn't trivial.
This isn't possible, as long as the rotation is not $$180^{\circ}$$. To see this, note that on any curve, the region very close to the point of rotation can be approximated by a straight line. Thus, after rotating, we must have that one half of this straight line is in the curve, and one half is outside. But the curve is closed, so by the Jordan Curve Theorem, the part of the rotated curve inside the original curve must cross the original curve at another place to reach the part of the rotated curve outside the original curve. Thus there must be at least one point of intersection.
Does this make sense to you? · 2 years, 9 months ago
Yup, I see. · 2 years, 9 months ago
Awesome note, @Daniel Liu . Very insightful. · 2 years, 9 months ago
If $$C$$ is a simple closed curve and $$T$$ is any triangle, we can always find a triangle inscribed in $$C$$ similar to $$T$$. · 2 years, 9 months ago
Yes, that is what I will show. In fact, I can show that there exists infinitely many triangles inscribed in $$C$$ similar to $$T$$. · 2 years, 9 months ago
There's a more general transformation called the spiral similarity which is basically a combination of rotation and homothety, we can use that to prove the general case.
Funny enough I've actually known this conjecture for quite a long time eversince I saw it in the list of unsolved problems on wiki. and tried to tackle it. Here's an even more interesting question: Out of the infinitely many triangles inscribed in $$C$$ similar to $$T$$, are there two that are congruent. This step is essential to proving(or disproving) the larger conjecture, eventhough it's a small one.
btw the only progress that I made back then when I trid to solve this was that we could use translation to obtain parallelograms :) a step closer I guess, next step: rectangles. · 2 years, 9 months ago
@Xuming Liang I think that has already been proven. It takes some knowledge about topology. If you know topology you should give it another try. · 2 years, 9 months ago
Yes, @Xuming Liang the case of rectangles has also been proven. · 2 years, 9 months ago
lol pretty legit solution m8 · 2 years, 9 months ago
And you are a troll. · 2 years, 9 months ago
|
### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Adding All Nine
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
### Have You Got It?
Can you explain the strategy for winning this game with any target?
# Counting Factors
##### Stage: 3 Challenge Level:
Charlie wants to know how many factors 360 has.
How would you work it out?
Click below to see what Alison did.
Alison divided 360 by each number in turn to see if it's a factor, and wrote down the factor pairs:
(1, 360)
(2, 180
(3, 120)
(4, 90)
(5, 72)
(6, 60)
(8, 45)
(9, 40)
(10, 36)
(12, 30)
(15, 24)
(18, 20)
"I can stop there, because the next factor would be 20 and I've already got that. So there are 24 factors."
Charlie thought about it in a different way. Click below to see what he did.
Charlie started by working out the prime factorisation of 360.
\begin{align} 360 &= 2 \times 180 \\ &= 2 \times 2 \times 90 \\ &= 2 \times 2 \times 2 \times 45 \\ &= 2 \times 2 \times 2 \times 3 \times 15 \\ &= 2 \times 2 \times 2 \times 3 \times 3 \times 5 \end{align}
So $360 = 2^3 \times 3^2 \times 5$.
Then he used a table to find all the possible combinations of the prime factors.
$2^0$
$3^0$
$5^0$ $5^1$
$3^1$
$5^0$ $5^1$
$3^2$
$5^0$ $5^1$
$2^1$
$3^0$
$5^0$ $5^1$
$3^1$
$5^0$ $5^1$
$3^2$
$5^0$ $5^1$
$2^2$
$3^0$
$5^0$ $5^1$
$3^1$
$5^0$ $5^1$
$3^2$
$5^0$ $5^1$
$2^3$
$3^0$
$5^0$ $5^1$
$3^1$
$5^0$ $5^1$
$3^2$
$5^0$ $5^1$
So the top branch gives us $2^0 \times 3^0 \times 5^0 =1$
and the eleventh branch gives us $2^1 \times 3^2 \times 5^0 = 18$
When she saw Charlie's method, Alison said "There must be lots of numbers which have exactly 24 factors!"
Charlie and Alison think all of these numbers have exactly 24 factors. Can you see why?
$25725 = 5^2 \times 3^1 \times 7^3$
$217503 = 11^1 \times 13^3 \times 3^2$
$312500 = 5^7 \times 2^2$
$690625 = 17^1 \times 13^1 \times 5^5$
$94143178827 = 3^{23}$
Here are some questions to consider:
How can I find a number with exactly 14 factors?
How can I find the smallest such number?
How can I find a number with exactly 15 factors?
How can I find the smallest such number?
How can I find a number with exactly 18 factors?
How can I find the smallest such number?
Which numbers have an odd number of factors?
Extension:
What is the smallest number with exactly 100 factors?
Which number less than 1000 has the most factors?
|
# How do you solve abs(9+7x)=30?
##### 1 Answer
Apr 30, 2017
See the entire solution process below:
#### Explanation:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Solution 1)
$9 + 7 x = - 30$
$- \textcolor{red}{9} + 9 + 7 x = - \textcolor{red}{9} - 30$
$0 + 7 x = - 39$
$7 x = - 39$
$\frac{7 x}{\textcolor{red}{7}} = - \frac{39}{\textcolor{red}{7}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} = - \frac{39}{7}$
$x = - \frac{39}{7}$
Solution 2)
$9 + 7 x = 30$
$- \textcolor{red}{9} + 9 + 7 x = - \textcolor{red}{9} + 30$
$0 + 7 x = 21$
$7 x = 21$
$\frac{7 x}{\textcolor{red}{7}} = \frac{21}{\textcolor{red}{7}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} = 3$
$x = 3$
The solution is: $x = - \frac{39}{7}$ and $x = 3$
|
# Matrices
A Matrix is an array of numbers:
A Matrix
(This one has 2 Rows and 3 Columns)
We talk about one matrix, or several matrices.
There are many things we can do with them ...
To add two matrices: add the numbers in the matching positions:
These are the calculations:
3+4=7 8+0=8 4+1=5 6-9=-3
The two matrices must be the same size, i.e. the rows must match in size, and the columns must match in size.
Example: a matrix with 3 rows and 5 columns can be added to another matrix of 3 rows and 5 columns.
But it could not be added to a matrix with 3 rows and 4 columns (the columns don't match in size)
## Negative
The negative of a matrix is also simple:
These are the calculations:
-(2)=-2 -(-4)=+4 -(7)=-7 -(10)=-10
## Subtracting
To subtract two matrices: subtract the numbers in the matching positions:
These are the calculations:
3-4=-1 8-0=8 4-1=3 6-(-9)=15
Note: subtracting is actually defined as the addition of a negative matrix: A + (-B)
## Multiply by a Constant
We can multiply a matrix by some value:
These are the calculations:
2×4=8 2×0=0 2×1=2 2×-9=-18
We call the constant a scalar, so officially this is called "scalar multiplication".
## Multiplying by Another Matrix
To multiply two matrices together is a bit more difficult ... read Multiplying Matrices to learn how.
## Dividing
And what about division? Well we don't actually divide matrices, we do it this way:
A/B = A × (1/B) = A × B-1
where B-1 means the "inverse" of B.
So we don't divide, instead we multiply by an inverse.
And there are special ways to find the Inverse ...
## Transposing
To "transpose" a matrix, swap the rows and columns. We put a "T" in the top right-hand corner to mean transpose:
## Notation
A matrix is usually shown by a capital letter (such as A, or B)
Each entry (or "element") is shown by a lower case letter with a "subscript" of row,column:
### Rows and Columns
So which is the row and which is the column?
• Rows go left-right
• Columns go up-down
To remember that rows come before columns use the word "arc":
ar,c
### Example:
B =
Here are some sample entries:
b1,1 = 6 (the entry at row 1, column 1 is 6)
b1,3 = 24 (the entry at row 1, column 3 is 24)
b2,3 = 8 (the entry at row 2, column 3 is 8)
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# Finishing Five Point Graphs
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## Objective
SWBAT sketch quick graphs of quadratic functions, identifying and labeling the key points on each: the roots, the vertex, and the y-intercept.
#### Big Idea
A focus on special cases can reveal a lot about the fascinating behavior of quadratic functions.
## Opener: Can We Make a Five Point Graph?
10 minutes
The purpose of today's opener (on the first slide of these lesson notes) is to review the idea of "Five Point Graphs" from yesterday's lesson, and to show one special case of such a graph. A big part of today's lesson will be to examine some of the special cases where the five points laid out yesterday might overlap, or not exist at all. If you're unfamiliar with the "Five Point Graphs" activity, please take a look at yesterday's lesson to get up to speed.
I give students a few minutes to get started on this problem, and it soon becomes clear that we're not going to be able to find the five unique points that we did yesterday. That's because this function has one root, and that root is also the vertex of the parabola. The best we'll be able to do, if we follow the steps laid out yesterday, is to make a "three point graph."
When we left of yesterday, students had a few choices for how to go about creating a five point graph. Depending on the route students take to complete this problem, there are different ways that the uniqueness of this function might present itself. I watch to see how each student approaches the problem, and then I ask questions accordingly:
• If a student starts by factoring to find the roots, they'll see that this function has two equal roots, which, for our purposes, means there's just one root to plot on the graph. I'll ask, "So if the roots are equal, and we know that they axis of symmetry is halfway in-between the roots, then where is the axis of symmetry on this function?"
• If they start by using the formula to find the axis of symmetry and then evaluating the function to get the vertex, they'll see that the vertex is on the x-axis. I'll ask, "Wait, doesn't that mean that the vertex is a root? So where is the other root?"
• If they determine the axis of symmetry and then the discriminant of the function, they'll see that the discriminant is zero. I'll ask, "But if the discriminant is zero, then its square root is also zero, which means that the distance from the axis of symmetry to the root is zero. What will that look like?"
I take a few laps around the classroom, watching students work, encouraging them to help each other out, and asking these questions. When it seems like the time is right, I ask for a volunteer to show us what they've got on the board, which leads to a conversation in which we synthesize some of the points that are hinted at in my questions above. This won't be the last special case we look at today. There's more of that to come as we move into work time.
## Work Time: Finishing Up Five Point Graphs
28 minutes
You can see on the second slide of today's lesson notes that I give students a few options of what to do today. I expect them to finish the Five Point Graphs assignment first, however, and for most kids, the time they get today will be exactly what they need. The primary purpose of this work list is to ensure that anyone who finishes early can quickly make plan for what to do next. The middle two work options are the same as they were at the end of last week.
If you haven't already, take a look at yesterday's lesson to see how the Five Point Graphs assignment works. Following the opener, I tell students to find their work and to pick up where they left off. I circulate to make sure that everyone has what they need, and soon everyone is cruising.
I have two roles today. One is simply to keep moving around and providing help as needed, to individual students or small groups. The other is to look for opportunities to lead a whole-class discussion about a problem or two, just like I did at the end of the opener, in order to help everyone learn something new. Such opportunities will come at different times in any given lesson. It's all about paying attention to kids, waiting for a quorum to have a similar question, and then guiding everyone to see what's going on.
Whenever it's impossible to find five unique points on the graph of a quadratic function, there's something to learn. Each of the special cases on this assignment represents an opportunity to do that. Rather than trying to record every detail of what might happen here, I've shared some pictures of the board. Take a look at these photos and try to reconstruct the kinds of conversations that would lead to notes like these:
One thing that makes the "Five Point Graph" idea particularly useful is that we can't always get five unique points on a the graph of a quadratic function. We already saw this in today's opener. We might not have time to look at functions #5 and #7 during whole-class discussions, but each of these problems allows students to develop a deeper understanding of quadratic functions, their graphs, and their key points.
Extension
For students who make short work of this assignment, it may be worthwhile to have them take another look at the first four problems. I point out that the y-coordinate of the vertex is the same for functions #1 through #4. We might also observe that the discriminant is the same for each function. Why is this happening? Is there any relationship between the discriminant and the y-coordinate of the vertex? If you have a conjecture about this, does it hold for other quadratic functions?
The relationships seen here are quite evident if we rewrite all four of these functions in vertex form. Slides #3 through #7 of the lesson notes can also be used to transition into a conversation about vertex form, and functions d, e, and f have a similar relationship to those on the front of the Five Point Graphs assignment.
## Exit Task: Sketch a Quick Graph
5 minutes
Today's exit task (on the last slide of the lesson notes) is the same as it was a week ago, at the end of the Features of a Parabola lesson. I expect students to notice their progress when they complete this task. As they submit their work and leave class, I'll check with as many students as I can, to ask if they feel like they've gotten better at quickly sketching parabolas.
No matter what kids have done the last few days, I want to see how well students can graph and label the key points on a quadratic function in a short amount of time. In addition to demonstrating to individual students how much they've learned, these exit slips will give a quick snapshot of where everyone stands, and I'll be able to plan scaffolds and extensions accordingly for the coming days. Everyone is going to master each learning target at different times, and it's a given that everyone will have a different depth of understanding at the end of this unit. This exit slip is a base level that I hope everyone reaches by now, and it provides a clear picture of whether or not we're reaching that goal.
|
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# A Bit About Vectors for AP Physics B & C
(not rated)
By McGraw-Hill Professional
Updated on Feb 10, 2011
Practice problems for these concepts can be found at: Vectors Practice Problems for AP Physics B & C
### Scalars
Scalars are numbers that have a magnitude but no direction.
For example, temperature is a scalar. On a cold winter day, you might say that it is "4 degrees" outside. The units you used were "degrees." But the temperature was not oriented in a particular way; it did not have a direction.
Another scalar quantity is speed. While traveling on a highway, your car's speedometer may read "70 miles per hour." It does not matter whether you are traveling north or south, if you are going forward or in reverse: your speed is 70 miles per hour.
### Vector Basics
p>Vectors, by comparison, have both magnitude and direction.
An example of a vector is velocity. Velocity, unlike speed, always has a direction. So, let's say you are traveling on the highway again at a speed of 70 miles per hour. First, define what direction is positive—we'll call north the positive direction. So, if you are going north, your velocity is +70 miles per hour. The magnitude of your velocity is "70 miles per hour," and the direction is "north."
If you turn around and travel south, your velocity is –70 miles per hour. The magnitude (the speed) is still the same, but the sign is reversed because you are traveling in the negative direction. The direction of your velocity is "south."
IMPORTANT: If the answer to a free-response question is a vector quantity, be sure to state both the magnitude and direction. However, don't use a negative sign if you can help it! Rather than "–70 miles per hour," state the true meaning of the negative sign: "70 miles per hour, south."
### Graphic Representation of Vectors
Vectors are drawn as arrows. The length of the arrow corresponds to the magnitude of the vector—the longer the arrow, the greater the magnitude of the vector. The direction in which the arrow points represents the direction of the vector. Figure 9.1 shows a few examples:
Vector A has a magnitude of 3 meters. Its direction is "60° degrees above the positive x-axis." Vector B also has a magnitude of 3 meters. Its direction is "β degrees above the negative x-axis." Vector C has a magnitude of 1.5 meters. Its direction is "in the negative y-direction" or "90 degrees below the x-axis."
### Vector Components
Any vector can be broken into its x- and y -components. Here's what we mean:
Place your finger at the tail of the vector in Figure 9.2 (that's the end of the vector that does not have a on it). Let's say that you want to get your finger to the head of the vector without moving diagonally. You would have to move your finger three units to the right and four units up. Therefore, the magnitude of left–right component (x -component) of the vector is "3 units" and the magnitude of up–down (y-component) of the vector is "4 units."
If your languages of choice are Greek and math, then you may prefer this explanation.
You may want to check to see that these formulas work by plugging in the values from our last example.
Vx = 5 cos 53° = 3 units
Vy = 5 sin 53° = 4 units
### Some Hints
2. Always use units. Always. We mean it. Always.
Practice problems for these concepts can be found at: Vectors Practice Problems for AP Physics
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### Related Questions
#### Q:
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How do you show greater than 3?
How do you show greater than 3?
All inequality signs give us the relationship between the first number and the second, beginning with the first number, so 4 > 3 translates to “4 is greater than 3.” This also works the other way around. If you see 5 < 8, imagine the < sign as a little alligator mouth about to chomp down on some fish.
Which is the greater number?
The greater than symbol is >. So, 9>7 is read as ‘9 is greater than 7’. The less than symbol is <. Two other comparison symbols are ≥ (greater than or equal to) and ≤ (less than or equal to).
What are some numbers greater than?
Symbol Meaning Example in Symbols
> Greater than More than Bigger than Larger than 7 > 4
< Less than Fewer than Smaller than 4 < 7
= Equal to Same as 7 = 7
How do you write greater than on a keyboard?
Creating the > symbol on a U.S. keyboard On English PC and Mac keyboards, the greater than symbol is on the same key as the period. Pressing and holding down the Shift , and then pressing > creates the greater than symbol.
What number is bigger than 26 by 20?
What order is least to greatest in numbers?
In Mathematics, the set of numbers are arranged in two ways. It means that the numbers can be arranged either in ascending order or descending order. If the numbers are arranged from the least to the greatest, then it is called ascending order.
What is meaning of greater than?
Greater than can be defined as an inequality used to compare two or more numbers, quantities or values. It is used when a quantity or number is bigger or larger than the second or rest quantities or numbers.
What are greater and smaller numbers?
Equal, Greater or Less Than
= When two values are equal we use the “equals” sign
When two values are definitely not equal we use the “not equal to” sign
< When one value is smaller than another we use a “less than” sign
> When one value is bigger than another we use a “greater than” sign
Is it greater or less than 4?
Equal, Greater or Less Than
= When two values are equal we use the “equals” sign example: 2+2 = 4
< When one value is smaller than another we use a “less than” sign example: 3 < 5
> When one value is bigger than another we use a “greater than” sign example: 9 > 6
What does ≤ mean?
≤ ≥ These symbols mean ‘less than or equal to’ and ‘greater than or equal to’ and are commonly used in algebra.
How do I get the third symbol on my keyboard?
The AltGr key is used today as an additional ‘shift’ key, to provide a third and a fourth range of graphemes for most keys – especially the accented variants of the letters on the keys but also some additional typographical symbols and punctuation marks.
Begin typing your search term above and press enter to search. Press ESC to cancel.
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# Roman Numerals to Number
## This is a helpful tool that you can use online for free. It can convert Roman numerals into regular decimal numbers.
Roman numerals may seem like a mysterious code from ancient times, but they're actually quite simple once you understand the rules. In this article, we'll break down Roman numerals step by step, so you can confidently convert them into regular numbers. By the end, you'll be able to read and interpret Roman numerals like a pro.
## What Are Roman Numerals?
Roman numerals are a numeral system that originated in ancient Rome and were widely used throughout the Roman Empire. Unlike our everyday numbers (1, 2, 3, etc.), Roman numerals use letters from the Latin alphabet to represent quantities. Each letter has a specific value, and by combining these letters, you can express a wide range of numbers.
## Roman Numeral Symbols
Here are the basic Roman numeral symbols and their corresponding values:
1. I = 1
2. V = 5
3. X = 10
4. L = 50
5. C = 100
6. D = 500
7. M = 1000
Now that you know the symbols and their values, let's dive into how to read and convert Roman numerals into regular numbers.
Reading Roman numerals is a bit like solving a puzzle. You look for patterns and follow these rules:
1. Symbols are read from left to right.
2. When a smaller numeral appears before a larger one, you subtract the smaller from the larger.
3. When a smaller numeral appears after a larger one, you add the smaller to the larger.
Let's go through some examples to illustrate these rules.
Example 1: IV = 4
Here, "I" (1) comes before "V" (5). According to our rules, we subtract "I" from "V," which gives us 4.
Example 2: VII = 7
In this case, "V" (5) comes before "II" (2). We add them together to get 7.
Example 3: XL = 40
"X" (10) appears before "L" (50). We subtract "X" from "L" to get 40.
Example 4: XC = 90
Here, "X" (10) appears before "C" (100). We subtract "X" from "C" to obtain 90.
Example 5: MCM = 1900
In this more complex example, "M" (1000) appears before "CM" (900). We add them together to get 1900.
## Converting Roman Numerals to Numbers
Now that you know how to read Roman numerals, let's learn how to convert them into regular numbers. To do this, you'll follow these steps:
1. Start from left to right.
2. If a smaller numeral appears before a larger one, subtract its value.
3. If a smaller numeral appears after a larger one, add its value.
4. Continue this process until you've gone through all the numerals.
Let's try converting a few Roman numerals into numbers.
Example 1: VII = 7
We already read this as 7. To convert it, we follow the same steps: "V" (5) comes before "II" (2), so we add them together to get 7.
Example 2: XLVII = 47
Start from the left and move to the right. "X" (10) comes before "L" (50), so we subtract 10 from 50, which gives us 40. Then, "V" (5) comes before "II" (2), so we add them together to get 7. Combine these results, and you have 40 + 7 = 47.
Example 3: CCXLIII = 243
Begin from the left. "C" (100) appears twice, so we add 100 + 100 = 200. Then, "X" (10) comes before "L" (50), so we subtract 10 from 50, which equals 40. Finally, "I" (1) comes before "II" (2), so we add 1 + 2 = 3. Combine these results: 200 + 40 + 3 = 243.
## Conclusion
Roman numerals may seem daunting at first, but with some practice and understanding of the rules, you can easily convert them into regular numbers. Remember to read Roman numerals from left to right, subtract when a smaller numeral appears before a larger one, and add when it appears after. With these basic principles, you'll be able to decode Roman numerals with confidence. So go ahead, embrace the ancient Roman system, and impress your friends with your newfound Roman numeral skills!
### Azahar Ahmed
CEO / Co-Founder
I am Azahar Ahmed, a youthful Engineer, Entrepreneur, Digital Marketer, and Motivational speaker native to Nagaon, Assam, India. Originating from a middle-class background, I am the sole son. My accomplishments are indebted to my father, a Teacher, and my mother, formerly a Teacher but now devoted to our well-being. My mother has been my closest ally, and unitedly, my parents have fostered and realized all my aspirations, epitomizing the perfect parents.
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# Difference between revisions of "2019 AMC 10B Problems/Problem 8"
## Problem
The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?
$(A)$ $4$ $(B)$ $12 - 4\sqrt{3}$ $(C)$ $3\sqrt{3}$ $(D)$ $4\sqrt{3}$ $(E)$ $16 - \sqrt{3}$
## Solution
We notice that the square can be split into $4$ congruent smaller squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it has already been split in half).
When we split a equilateral triangle in half, we get $2$ triangles with a $30-60-90$relationship. Therefore, we get that the altitude and a side length of a square is $\sqrt{3}$.
We can then compute the area of the two triangles using the base-height-area relationship and get $\frac{2 \cdot \sqrt{3}}{2} = \sqrt{3}$.
The area of the small squares is the altitude squared which is $(\sqrt{3})^2 = 3$. Therefore, the area of the shaded region in each of the four squares is $3 - \sqrt{3}$
Since there are four of these squares, we multiply this by $4$ to get $4(3 - \sqrt{3}) = 12 - 4 \sqrt{3}$ as our answer. This is choice $(B)$.
~ Awesome2.1 Edited by greersc
2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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Sum of Interior Angles of a Polygon Topic Index | Geometry Index | Regents Exam Prep Center
Sum of Interior Angles of a Polygon = 180(n - 2) (where n = number of sides)
Let's investigate why this formula is true.
Start with vertex A and connect it to all other vertices (it is already connected to B and E by the sides of the figure). Three triangles are formed. The sum of the angles in each triangle contains 180°. The total number of degrees in all three triangles will be 3 times 180. Consequently, the sum of the interior angles of a pentagon is: 3 180 = 540 Notice that a pentagon has 5 sides, and that 3 triangles were formed by connecting the vertices. The number of triangles formed will be 2 less than the number of sides. This pattern is constant for all polygons. Representing the number of sides of a polygon as n, the number of triangles formed is (n - 2). Since each triangle contains 180°, the sum of the interior angles of a polygon is 180(n - 2).
Using the Formula
There are two types of problems that arise when using this formula:
1. Questions that ask you to find the number of degrees in the sum of the interior angles of a polygon. 2. Questions that ask you to find the number of sides of a polygon.
Hint: When working with the angle formulas for polygons, be sure to read each question carefully for clues as to which formula you will need to use to solve the problem. Look for the words that describe each kind of formula, such as the words sum, interior, each, exterior and degrees.
Example 1:
Find the number of degrees in the sum of the interior angles of an octagon.
An octagon has 8 sides. So n = 8. Using the formula from above, 180(n - 2) = 180(8 - 2) = 180(6) = 1080 degrees.
Example 2: How many sides does a polygon have if the sum of its interior angles is 720°?
Since, the number of degrees is given, set the formula above equal to 720°, and solve for n. 180(n - 2) = 720 n - 2 = 4 n = 6 Set the formula = 720° Divide both sides by 180 Add 2 to both sides
Topic Index | Geometry Index | Regents Exam Prep Center Created by Michael Murray
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# Eureka Math Algebra 1 Module 1 Lesson 7 Answer Key
## Engage NY Eureka Math Algebra 1 Module 1 Lesson 7 Answer Key
### Eureka Math Algebra 1 Module 1 Lesson 7 Exercise Answer Key
Exercise 1.
Suzy draws the following picture to represent the sum 3+4:
Ben looks at this picture from the opposite side of the table and says, “You drew 4+3.”
Explain why Ben might interpret the picture this way.
Ben read the picture from his left to his right on his side of the table.
Exercise 2.
Suzy adds more to her picture and says, “The picture now represents (3+4)+2.”
How might Ben interpret this picture? Explain your reasoning.
Reading from right to left, the solution would be (2+4)+3. Make sure students have parentheses around 2+4.
Exercise 3.
Suzy then draws another picture of squares to represent the product 3×4. Ben moves to the end of the table and says, “From my new seat, your picture looks like the product 4×3.”
What picture might Suzy have drawn? Why would Ben see it differently from his viewpoint?
Squares should be arranged in a grid. If a student responds that Suzy made 3 rows of 4, then Ben’s viewpoint would be 4 rows of 3. Students should understand that Ben is seated to Suzy’s left or right now, not across from her. Some students may need scaffolding here—have them physically move to see the different viewpoint.
Exercise 4.
Draw a picture to represent the quantity (3×4)×5 that also could represent the quantity (4×5)×3 when seen from a different viewpoint.
Student solutions could vary here. Students may consider representing the problem as a 3 by 4 by 5 rectangular box. When viewed from different faces, the different expressions appear. With the 3 by 4 rectangle viewed as its base, the volume of the box might be computed as (3×4)×5. But with the 4 by 5 rectangle viewed as its base, its volume would be computed as (4×5)×3. Some students will likely repeat the 3×4 pattern 5 times in a row. This diagram viewed from the end of the table would be 4 dots repeated 5 times arranged in 3 columns.
Ask students a series of questions of the following type:
→ Could the ideas developed in Exercises 1 and 2 be modified so as to explain why 2 $$\frac{1}{2}$$+$$\frac{3}{19}$$ should equal
$$\frac{3}{19}$$ +2 $$\frac{1}{2}$$ ?
→ Or that ((-6562.65)+(-9980.77))+22 should equal (22+(-9980.77))+(-6562.65)?
→ How about that $$\sqrt{2}$$+$$\frac{1}{\pi}$$ should equal $$\frac{1}{\pi}$$+$$\sqrt{2}$$?
→ Is it possible for a rectangle or a rectangular box to have a negative side length?
→ Could the ideas developed in Exercises 3 and 4 be used to show that (-3)×($$\frac{1}{\sqrt{7}}$$) should equal ($$\frac{1}{\sqrt{7}}$$)×(-3) or that (π×17.2)×(-$$\frac{16}{5}$$ ) should equal ((-$$\frac{16}{5}$$ )×17.2)×π?
Next, have students review the four properties of arithmetic provided in the student materials and ask the following:
Four Properties of Arithmetic:
THE COMMUTATIVE PROPERTY OF ADDITION: If a and b are real numbers, then a+b=b+a.
THE ASSOCIATIVE PROPERTY OF ADDITION: If a, b, and c are real numbers, then (a+b)+c=a+(b+c).
THE COMMUTATIVE PROPERTY OF MULTIPLICATION: If a and b are real numbers, then a×b=b×a.
THE ASSOCIATIVE PROPERTY OF MULTIPLICATION: If a, b, and c are real numbers, then (ab)c=a(bc).
→ Can you relate each of these properties to one of the previous exercises?
→ Exercise 1 connects with the commutative property of addition.
→ Exercise 2 connects with the associative property of addition. (Students might mention that the commutative property of addition is also relevant to this exercise. This will be discussed fully in
Exercise 5.)
→ Exercise 3 connects with the commutative property of multiplication.
→ Exercise 4 connects the associative property of multiplication. (Students might mention that the commutative property of multiplication is also relevant to this exercise. This will be discussed fully in Exercise 6.)
Point out that the four opening exercises suggest that the commutative and associative properties of addition and multiplication are valid for whole numbers and probably apply to all real numbers as well. However, there is a weakness in the geometric models since negative side lengths and areas are not meaningful. We choose to believe these properties hold for all real numbers, including negative numbers.
Exercise 5.
Viewing the diagram below from two different perspectives illustrates that (3+4)+2 equals 2+(4+3).
Is it true for all real numbers x, y, and z that (x+y)+z should equal (z+y)+x?
(Note: The direct application of the associative property of addition only gives (x+y)+z=x+(y+z).)
To answer this exercise with the class, create a flow diagram on the board as follows. This flow diagram will show how one can apply both the commutative and associative properties to prove the equivalence of these two expressions. Have students copy this work into their handouts.
Start by showing the application of each property on the expression (x+y)+z.
Here A represents an application of the associative property and C an application of the commutative property. Be sure students understand the application of the commutative property shown here.
Point out that we can extend this diagram by applying the commutative and associative properties to the new expressions in the diagram.
Note that there are multiple branches and options for extending this diagram. Direct the students to discover options that will chart a path from (x+y)+z to (z+y)+x. Two possible paths are as follows:
Choose one of the paths in the flow diagram and show on the board how to write it as a mathematical proof of the statement that (x+y)+z and (z+y)+x are equivalent expressions. For example, for the lower of the two paths shown, write the following:
(x+y)+z =z+(x+y) commutative property
=z+(y+x) commutative property
=(z+y)+x associative property
Exercise 6.
Draw a flow diagram and use it to prove that (xy)z=(zy)x for all real numbers x, y, and z.
Here is the start of the diagram.
Students will likely realize the answer here is completely analogous to the solution to the previous exercise.
(xy)z =z(xy) commutative property
=z(yx) commutative property
=(zy)x associative property
Have students complete Exercises 7 and 8.
Exercise 7.
Use these abbreviations for the properties of real numbers, and complete the flow diagram.
C+ for the commutative property of addition
C× for the commutative property of multiplication
A+ for the associative property of addition
A× for the associative property of multiplication
Exercise 8.
Let a, b, c, and d be real numbers. Fill in the missing term of the following diagram to show that ((a+b)+c)+d is sure to equal a+(b+(c+d)).
→ This example illustrates that it is possible to prove, through repeated use of the associative property, that any two arrangements of parentheses around a given sum are equivalent expressions. For this reason it is deemed unnecessary to place parentheses among a sum of terms. (Present the following on the board.)
((a+b)+c)+d
a+(b+(c+d)) → a+b+c+d (a+b)+(c+d)
→ From now on, we will accept this as common practice. In presenting a proof, writing the following:
(x+(y+z)+(w+6))=((x+y)+(z+w)+6) by the associative property
OR
a+b+c+d=a+(b+c)+d by the associative property
for instance, is accepted.
The same holds for a product of symbols. Repeated application of the associative property of multiplication establishes the equivalency of ((xy)z)w and x((yz)w), for example, and these can both be written simply as xyzw.
### Eureka Math Algebra 1 Module 1 Lesson 7 Problem Set Answer Key
Question 1.
The following portion of a flow diagram shows that the expression ab+cd is equivalent to the expression dc+ba.
Fill in each circle with the appropriate symbol: Either C+ (for the commutative property of addition) or C× (for the commutative property of multiplication).
Question 2.
Fill in the blanks of this proof showing that (w+5)(w+2) is equivalent to w2 +7w+10. Write either commutative property, associative property, or distributive property in each blank.
(w+5)(w+2) =(w+5)w+(w+5)×2 _____________
distributive property
=w(w+5)+(w+5)×2 _____
commutative property
=w(w+5)+2(w+5) _____________
commutative property
=w2 +w×5+2(w+5) _____
distributive property
=w2 +5w+2(w+5) _____
commutative property
=w2 +5w+2w+10 _____
distributive property
=w2 +(5w+2w)+10 _____
associative property
=w2 +7w+10
Question 3.
Fill in each circle of the following flow diagram with one of the letters: C for commutative property (for either addition or multiplication), A for associative property (for either addition or multiplication), or D for distributive property.
Question 4.
What is a quick way to see that the value of the sum 53+18+47+82 is 200?
53+18+47+82=(53+47)+(18+82)=100+100
Question 5.
a. If ab=37 and y=$$\frac{1}{37}$$, what is the value of the product x×b×y×a?
x×b×y×a=(xy)(ab)=1
b. Give some indication as to how you used the commutative and associative properties of multiplication to evaluate x×b×y×a in part (a).
x×b×y×a=x×y×a×b by two applications of the commutative property of multiplication and
x×y×a×b=(xy)(ab) by the associative property of multiplication.
c. Did you use the associative and commutative properties of addition to answer Problem 4?
Yes, they were used in an analogous manner.
Question 6.
The following is a proof of the algebraic equivalency of (2x)3 and 8x3 . Fill in each of the blanks with either the statement commutative property or associative property.
(2x)3 = 2x∙2x∙2x
=2(x×2)(x×2)x ____
associative property
=2(2x)(2x)x ___
commutative property
=2∙2(x×2)x∙x ___
associative property
=2∙2(2x)x∙x ___
commutative property
=(2∙2∙2)(x∙x∙x) ____
associative property
=8x3
Question 7.
Write a mathematical proof of the algebraic equivalency of (ab)2 and a2 b2 .
(ab)2 =(ab)(ab)
=a(ba)b associative property
=a(ab)b commutative property
=(aa)(bb) associative property
=a2 b2
Question 8.
Suppose we are to play the 4-number game with the symbols a, b, c, and d to represent numbers, each used at most once, combined by the operation of addition ONLY. If we acknowledge that parentheses are unneeded, show there are essentially only 15 expressions one can write.
By also making use of the commutative property of addition, we have the expressions:
a, b, c, d, a+b, a+c, a+d, b+c, b+d, c+d,
a+b+c, a+b+d, a+c+d, b+c+d, a+b+c+d
b. How many answers are there for the multiplication ONLY version of this game?
By analogous reasoning, there are only 15 expressions here too.
Question 9.
Write a mathematical proof to show that (x+a)(x+b) is equivalent to x2 +ax+bx+ab.
(x+a)(x+b) =(x+a)x+(x+a)b (D)
=x(x+a)+b(x+a) (C)
=x2 +xa+bx+ba (D)
=x2 +ax+bx+ab (C)
Question 10.
Recall the following rules of exponents:
xa ∙xb =xa+b
$$\frac{x^{a}}{x^{b}}$$ = xa-b
(xa)b =xab
(xy)a =xaya
($$\frac{x}{y}$$)a =$$\frac{x^{a}}{y^{a}}$$
Here x, y, a, and b are real numbers with x and y nonzero.
Replace each of the following expressions with an equivalent expression in which the variable of the expression appears only once with a positive number for its exponent. (For example, $$\frac{7}{b^{2}}$$∙b-4 is equivalent to $$\frac{7}{b^{6}}$$.)
a. (16x2 )÷(16x5 )
$$\frac{1}{x^{3}}$$
b. (2x)4 (2x)3
128x7
c. (9z-2 ) ((3z)-1)-3
$$\frac{z}{3}$$
d. ((25w4 )÷(5w3 ))÷(5w-7)
w8
e. (25w4 )÷((5w3 )÷(5w-7 ))
$$\frac{25}{w^{6}}$$
Optional Challenge:
Question 11.
Grizelda has invented a new operation that she calls the average operator. For any two real numbers a and b, she declares a b to be the average of a and b:
a. Does the average operator satisfy a commutative property? That is, does a b=b a for all real numbers a and b?
Yes, use the fact that $$\frac{x}{2}$$=$$\frac{1}{2}$$ ∙x for any real number x and the commutative property.
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# Catenary Arch Math for an 11 cubic foot kiln
Lets waste no time in digging into the math for a Catenary arch. First I will explain what we are building and then do the brick math and simplify the formulas needed to calculate whats needed.
This kiln is tall enough and wide enough to work with single shelves in the middle and stacked vertically. The shelves measure 12" x 24" which gives you 288 square inches per shelf. Picture this:
Our interior opening is 32" wide and 32" high and the kiln will be 28" deep.
Our formulas for determining catenary volume is "L * ((4/3*H) * (1/2 * W)) The best way to use the formula is to input the numbers in inches. With our numbers above we find that our arch volume is equal to 19114.66. To covert to cubic feet we divide this number by 1728 which is how many cubic inches are in a cubic foot. We now get 11.06 cubic feet. You can adjust the dimensions as needed to get the cubic footage you want
The next question I wanted to answer was how many bricks do I need for the front and back faces and how many bricks do i need for the arch? We now need to do two calculations for surface area. The faces are easy since we already did that calculation when finding the volume. (4/3*H) * (1/2 * W) gave us 682 square inches.
We should quickly identify the face edges of a standard firebrick. Standard bricks are 9" long by 4.5" wide and 2.5" thick. If we do the quick math for face edges we get 11.25" for the end, 22.5" for the side edge and 40.5" for the top edge
Now to identify how many bricks for the front brick wall we simply take the arch area which is 682 square inches and divide by the face area for the side edge 22.5 square inches. We get 30.31 bricks. Now an important note is that this brick wall will be inset into the arch so that its flush. Like the picture below.
To get the surface area of the arch I hung a string 32" wide and 32" low. The string is what i copied to make the arch for the form. I was then able to measure the string length and got 80". **THIS IS IMPORTANT** Because we want the arch to cover the front and back walls we need to add a brick width to each end. We know that we want the interior to be 28" so we add 4.5" to each end. We now know that we want the arch surface to be 37" long. Now its a simple area formula 80" * 37" and we get 2,960" divide this by the brick side face edge and we get a 131.55. So it will take approximately 132 bricks mounted on its side edge to cover the entire arch form.
Now something to keep in mind is that the arch is not straight and the brick edges are. So we will still need to either cut brick or use refractory mortar with the bricks to seal the arch. You can see in the images that there will need to be some cutting so that the edges are flush
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# How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left.
## Full text
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Pre-Algebra (page 1/9) The verbal answers to all of the following questions should be memorized before completion of
pre-algebra. Answers that are not memorized will hinder your ability to succeed in algebra 1.
### Number Basics
What are the first six place values to the left of the decimal point? Ones, tens, hundreds, one thousands, ten thousands, hundred thousands. What are the main place value groupings to the left of the decimal point? Ones, thousands, millions, billions, trillions, etc.
What are the first three place values to the right of the decimal point? Tenths, hundredths, thousandths.
How do you round numbers?
Look one digit to the right of the place you are rounding to. If that digit is a 0-4, round down. If that digit is a 5-9, round up.
examples: Round 87 to the nearest ten. 90
Round 354,918 to the nearest ten thousand. 350,000 Round 45.0836 to the nearest tenth. 45.1
How do you compare numbers?
On a number line, larger numbers are to the right and smaller numbers are to the left. What is the meaning of an exponent?
The base needs to be multiplied by itself the number of times of the exponent. example: 25 = ⋅ ⋅ ⋅ ⋅ =2 2 2 2 2 32
What is the rule for order of operations?
PEMDAS. Parenthesis first, exponents next, then multiplication and division left to right, finally addition and subtraction left to right.
example: 24 − + − =5
8 2
### )
24 − + = − + = + =5 6 16 5 6 11 6 17 How do you approximate a square root of a number?
The square root of a given number will be a number between the square roots of the two closest perfect square numbers.
example: The 12 will be a number between the 9 =3 and the 16 =4.
### Decimals
Align the decimal points. Add right to left, carrying as necessary. example: 12.8 9 12.8 21.8
9.
+ = =
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Pre-Algebra (page 2/9) How do you subtract decimals?
Align the decimal points. Subtract right to left, borrowing as necessary. example:
3 13
4 3.8 4 3 .8
43.8 9 34.8
9. 9.
− = = =
− −
How do you multiply decimals?
Multiply normally. The sum of the number of decimal places gives the new number of decimal places. example: .03 1.002× =0.03006
How do you multiply decimals by powers of ten?
Shift the decimal place to the right by the numbers of zeros in the power of ten. example: 12.405 10,000 124,050.× =
Where are the dividend, divisor, and quotient located in a division problem? quotient
divisor dividend
How do you divide decimals?
Eliminate decimals in the divisor. Shift the decimal point the same number of places in the dividend. Divide normally.
example:
7.9 .02 .158=2 15.8 =2 15.8
How do you divide decimals by powers of ten?
Shift the decimal place to the left by the numbers of zeros in the power of ten. example: 12.405 10,000÷ =0.0012405
How do you write a number in scientific notation?
Starting at the left, put a decimal point to the right of the 1st nonzero number. Count the number places the decimal point moved. If the number got smaller, the exponent on the ten will be positive. If the number got larger, the exponent on the ten will be negative.
examples: a) 48,000=4.8 10 b) 0.0000048× 4 =4.8 10× −6
### Signed Numbers
What does absolute value do to a number? Absolute value makes numbers positive. examples: 5 =5, 8− =8
How do you add signed numbers?
If the numbers are the same sign, add the numbers and keep the sign. If the numbers are different signs, subtract the numbers and keep the sign of the larger magnitude number.
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Pre-Algebra (page 3/9) How do you subtract signed numbers?
(Subtracting a negative number is the same as adding a positive number. Subtracting a positive number is the same as adding a negative number.) If the numbers are the same sign, add the numbers and keep the sign. If the numbers are different signs, subtract the numbers and keep the sign of the larger magnitude number.
examples: 3 8− = + − = −3 8 5, − − = − + − = −8 2 8 2 10, 6− − = + =7 6 7 13 How do you multiply signed numbers?
Multiply normally. If the numbers have the same sign, the answer is positive. If the numbers have different signs, the answer is negative.
examples: 2⋅ − = −4 8, − ⋅ − =3 6 18, − ⋅ = −5 4 20 How do you divide signed numbers?
Divide normally. If the numbers have the same sign, the answer is positive. If the numbers have different signs, the answer is negative.
examples: 32− ÷ − =4 8, −30÷ = −2 15, 12÷ − = −2 6
### Factors and Multiples
What is a factor?
A factor is a number that divides evenly into another number. example: 4 is a factor of 20.
What is the greatest common factor of two numbers? The largest number that divides evenly into both numbers. example: 12 is the greatest common factor of 24 and 36. What is a prime number?
A number having exactly two factors.
example: 11 is a prime number since 11 is divisible evenly by only 1 and 11. What is a composite number?
A number having more than two factors.
example: 10 is a composite number since its factors are 1, 2, 5, and 10. What is the prime factorization of a number?
A number written as the product of only prime numbers. example: The prime factorization of 80 is 2 2 2 2 5⋅ ⋅ ⋅ ⋅ = ⋅24 5 What is a multiple?
A number obtained by multiplying a given number by an integer. example: 5, 10, and 15 are multiples of 5.
What is the least common multiple of two numbers?
The smallest positive number that is evenly divisible by both numbers. example: 30 is the least common multiple of 10 and 15.
(4)
Pre-Algebra (page 4/9)
### Fractions and Mixed Numbers
When is the value of a fraction equivalent to one? When the numerator and denominator are the same.
examples: All of the following fractions are equivalent to one. 3 10 17 3, ,10 17
How do you simplify fractions?
Divide the numerator and denominator by all common factors. example: 36 36 2 18 18 3 6
78 78 2 39 39 3 13
÷ ÷
= = = =
÷ ÷
What is an improper fraction?
A fraction in which the value of the numerator is greater than or equal to the value of the denominator. example: The following two fractions are both improper. 8 and 6
5 6
How do you compare fractions?
Find a common denominator and compare the numerators.
example: 3 and 1 3 3 and 1 7 9 and 7 3 1
7 3 → ⋅7 3 3 7⋅ → 21 21→ 7> 3
Find a common denominator. Add the numerators. Leave the denominator the same. Simplify. example: 1 3 1 5 3 3 5 9 14 7
6+10 = ⋅ +6 5 10 3⋅ =30+30=30 =15
How do you subtract fractions?
Find a common denominator. Subtract the numerators. Leave the denominator the same. Simplify. example: 5 1 5 5 1 4 25 4 21
8−10 = ⋅ −8 5 10 4⋅ =40−40 =40
How do you add a whole number to a fraction? Find a common denominator. Add the fractions example: 3 5 3 20 23
4+ = +4 4 = 4
How do you multiply fractions?
Cancel common factors. Multiply the numerators. Multiply the denominators. example: 15 4 3 1 3
32 35⋅ = ⋅ =8 7 56
How do you divide fractions? Multiply by the reciprocal fraction. example: 1 3 1 4 1 2 2
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Pre-Algebra (page 5/9) How do you convert a mixed number into an improper fraction?
Multiply the denominator by the whole number and add the numerator. Leave the denominator the same.
example: 37 3 8 7 31
8 8 8
⋅ +
= =
How do you convert an improper fraction into a mixed number?
Divide the numerator by the denominator. The result is the whole number and the remainder is the numerator of the fraction. The denominator stays the same.
example:
5 1
21 1
4 21 5
4 4
R
= =
How do you convert a fraction into a decimal? Divide the numerator by the denominator.
example:
0.6 3
5 3.0 0.6
5= =
How do you convert a mixed number into a decimal?
The whole number stays the same. Convert the fraction into a decimal and combine it with the whole number.
example:
0.25
1 1
3 3 3 4 1.00 3 0.25 3.25
4 = + = +4 = + =
How do you convert a decimal into a mixed number?
The part to the left of the decimal becomes the whole number. The part to the right of the decimal is written as a fraction and then simplified.
example: 12.85 12 85 12 85 5 1217
100 100 5 20
÷
= + = + =
÷
How do you add mixed numbers?
Add the fraction parts (carrying if necessary). Add the whole number parts.
example: 13 57 1 5 3 7 1 5 6 7 6 13 6 15 6 1 5 7 5 75
4+ 8= + + + = + + + = +4 8 8 8 8 = + 8 = + + = + =8 8 8
How do you subtract mixed numbers?
Subtract the fraction parts (borrowing if necessary). Subtract the whole number parts. example: 43 2 9 4 6 2 9 316 2 9 1 7
5− 10 = 10− 10= 10− 10= 10
How do you multiply mixed numbers?
Convert to improper fractions. Multiply the improper fractions. Convert back to mixed numbers. example:
6 1
2 4 5 19 1 19 19 1
1 3 3 19 6
3 5 3 5 3 1 3 3
R
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Pre-Algebra (page 6/9) How do you divide mixed numbers?
Convert to improper fractions. Divide the improper fractions. Convert back to mixed numbers. example:
1 11
1 3 25 7 25 4 25 1 25 11
3 1 14 25 1
8 4 8 4 8 7 2 7 14 14
R
÷ = ÷ = ⋅ = ⋅ = = =
### Ratio, Proportion, and Percentage
What is a ratio?
A comparison of two numbers. example: 3:8 is a ratio. 3 8 is a ratio.
What is a proportion? Two ratios that are equal. example: 3 6
4=8.
How do you solve a proportion? Cross multiply and divide.
example: 3 3 20 3 5 15
4 20 4 1
x
x ⋅ ⋅
= = = =
What is a percentage?
A number representing a part out of 100. example: 23% means 23 parts out of 100.
How do you convert a percentage to a decimal? Divide the percentage by 100.
example: 45% is the same as 45 100÷ which is 0.45 as a decimal. How do you convert a percentage to a fraction?
Write the number with a denominator of 100. example: 60% 60 6 3
100 10 5
= = =
How do you convert a decimal into a percentage? Multiply by 100 and add the percentage symbol. example: 0.4 written as a percentage is 40%.
What is the formula for percentage increase and percentage decrease? The change divided by the original and then multiplied by 100%.
example: Last week 10 students said they loved math. This week 13 students say they love math. What is the percentage increase in math loving students? The change is 3 students so the percentage increase
is 3 100% 30% increase.
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Pre-Algebra (page 7/9) How do you set up a proportion equation for solving a percent problem?
is %
of =100
example: What percentage of 30 is 12? 12 , 12 100 12 10 4 10 40%
30 100 30 3 1
x
x ⋅ ⋅ ⋅
= = = = =
How are the following units related?
a) feet and inches? b) feet and yards? c) feet and miles? d) millimeters and centimeters? e) centimeters and meters? f) millimeters and meters? g) kilometers and meters?
a) 1 foot equals 12 inches. b) 3 feet equals 1 yard. c) 5280 feet equals 1 mile. d) 10 millimeters equals 1 centimeter. e) 100 centimeters equals 1 meter. f) 1000 millimeters equals 1 meter. g) 1 kilometer equals 1000 meters.
### Statistics and Probability
How is theoretical probability calculated?
Theoretical probability is the ratio of the number of successful outcomes to the number of possible outcomes.
example: What is the probability of picking a prime number given the numbers {1, 2, 3, 4, 5, 6, 7, 8, 9}? The prime numbers are {2, 3, 5, 7} so the probability would be 4 9 .
How do you find the mean (average) of a set of numbers? Add up all the numbers and then divide by the number of numbers. example: The mean of 2, 6, and 13 is 2 6 13 7
3
+ + =
How do you find the median of a set of numbers?
Order the numbers from small to large. The median is the middle number. If there is no middle number, the median is the average of the two numbers in the middle.
example: The median of 2, 8, and 1 is 2. (2 is the middle number for 1, 2, 8) What is the mode of a set of numbers?
The most frequently occurring number. If there is a tie, all the tied numbers are the mode. example: For the numbers 2, 4, 3, 4, 4, 4, 8, 1, the mode is 4.
What is the minimum of a set of numbers? What is the maximum of a set of numbers? The minimum is the smallest number in the set and the maximum is the largest number in the set. example: The minimum of 2, 5, 10, 1, 8 is 1. The maximum of 2, 5, 10, 1, 8 is 10.
What is the range of a set of numbers?
The difference between the maximum number and the minimum number. example: The range of 2, 5, 10, 1, 8 is 10 1− =9.
### Geometry
What is perimeter? (Give some example units for perimeter.)
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Pre-Algebra (page 8/9) How do you find the perimeter of a two dimensional shape?
example: The perimeter of the shape below is 14 meters.
What is area? (Give some example units for area.)
The amount of surface covered by a two dimensional shape. (ft2, in2, m2, km2, cm2, …) What is the formula for the area of a rectangle?
Length times width. (Area of a rectangle is also written as base times height.) example: The area of the rectangle below is 10 in2.
What is the formula for the area of a triangle? One half base times height.
1
### )
2
A= bh
example: The area of the triangle below is 12 ft2.
What is the formula for the area of a circle? Pi times radius squared.
A=
r2
## )
example: The area of the circle below is 9π cm2.
What is the formula for the circumference of a circle? Two times pi times radius.
A=2πr
### )
example: The circumference of the circle below is 6π yards. 5 m
2 m 2 m 5 m
5 in
2 in 2 in 5 in
3 ft 8 ft
3 cm
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Pre-Algebra (page 9/9) 6 8 10 12 14
6 8 10 12 14 What is volume? (Give some example units for volume.)
The amount of three dimensional space enclosed by an object. (ft3, in3, m3, km3, cm3, …) What is the formula for the volume of a box?
Length times width times height.
example: The volume of the box below is 60 mm3.
What is the Pythagorean Theorem and when can it be used?
a squared plus b squared equals c squared
a2+b2 =c2
## )
. It can only be used on right triangles to find a missing side when two sides are already known.
example: The missing side in the following triangle is 12.
2 2 2 2 2
## )
5 +b =13 → 25+b =169 → b =144 → b=12
### Graphical
How do you graph inequalities on a number line?
Less than " "< and greater than " "> are open circles. Less than or equal " "≤ and greater than or equal
" "≥ are closed circles. Greater "> or ≥" is shaded to the right. Less "< or ≤" is shaded to the left. examples: x≥10
x<10
How is a point (x, y) plotted in a coordinate plane?
The x coordinate is plotted left and right. Positive numbers are to the right and negative numbers are to the left. The y coordinate is
plotted up and down. Positive numbers are up and negative numbers are down.
examples: Plot the points A (2, 4)− and B ( 1,3)− . 4 mm
3 mm 5 mm
5 13
b
A B
-5 -3 -1 1 3 5
Updating...
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## Elementary Linear Algebra 7th Edition
$$\operatorname{adj}(A)=\left[ \begin {array}{cccc} -1&-1&-1&2\\ -1&-1&2&-1 \\ -1&2&-1&-1\\ 2&-1&-1&-1 \end {array} \right] .$$ $$A^{-1}= -\frac{1}{3}\left[ \begin {array}{cccc} -1&-1&-1&2\\ -1&-1&2&-1 \\ -1&2&-1&-1\\ 2&-1&-1&-1 \end {array} \right].$$
The matrix is given by $A=\left[ \begin {array}{cccc} 1&1&1&0\\ 1&1&0&1 \\ 1&0&1&1\\ 0&1&1&1\end {array} \right] .$ To find $\operatorname{adj}(A)$, we calculate first the cofactor matrix of $A$ as follows $$\left[ \begin {array}{cccc} -1&-1&-1&2\\ -1&-1&2&-1 \\ -1&2&-1&-1\\ 2&-1&-1&-1 \end {array} \right] .$$ Now, the adjoint of $A$ is $$\operatorname{adj}(A)=\left[ \begin {array}{cccc} -1&-1&-1&2\\ -1&-1&2&-1 \\ -1&2&-1&-1\\ 2&-1&-1&-1 \end {array} \right] .$$ To find $A^{-1}$, we have to calculate $\det(A)$ which is given by $$\det(A)=-3.$$ Finally, we have $$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)=-\frac{1}{3}\left[ \begin {array}{cccc} -1&-1&-1&2\\ -1&-1&2&-1 \\ -1&2&-1&-1\\ 2&-1&-1&-1 \end {array} \right].$$
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Paul's Online Math Notes
[Notes]
Calculus I - Notes
Derivatives Previous Chapter Next Chapter Integrals Optimization Previous Section Next Section L'Hospital's Rule and Indeterminate Forms
## More Optimization Problems
Because these notes are also being presented on the web we’ve broken the optimization examples up into several sections to keep the load times to a minimum. Do not forget the various methods for verifying that we have the optimal value that we looked at in the previous section. In this section we’ll just use them without acknowledging so make sure you understand them and can use them. So let’s get going on some more examples.
Example 1 A window is being built and the bottom is a rectangle and the top is a semicircle. If there is 12 meters of framing materials what must the dimensions of the window be to let in the most light? Solution Okay, let’s ask this question again in slightly easier to understand terms. We want a window in the shape described above to have a maximum area (and hence let in the most light) and have a perimeter of 12 m (because we have 12 m of framing material). Little bit easier to understand in those terms. Here’s a sketch of the window. The height of the rectangular portion is h and because the semicircle is on top we can think of the width of the rectangular portion at 2r. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius r. The area (what we want to maximize) is the area of the rectangle plus half the area of a circle of radius r. Here are the equations we’ll be working with in this example. In this case we’ll solve the constraint for h and plug that into the area equation. The first and second derivatives are, We can see that the only critical point is, We can also see that the second derivative is always negative (in fact it’s a constant) and so we can see that the maximum area must occur at this point. So, for the maximum area the semicircle on top must have a radius of 1.6803 and the rectangle must have the dimensions 3.3606 x 1.6803 (h x 2r).
Example 2 Determine the area of the largest rectangle that can be inscribed in a circle of radius 4. Solution Huh? This problem is best described with a sketch. Here is what we’re looking for. We want the area of the largest rectangle that we can fit inside a circle and have all of its corners touching the circle. To do this problem it’s easiest to assume that the circle (and hence the rectangle) is centered at the origin. Doing this we know that the equation of the circle will be and that the right upper corner of the rectangle will have the coordinates . This means that the width of the rectangle will be 2x and the height of the rectangle will be 2y. The area of the rectangle will then be, So, we’ve got the function we want to maximize (the area), but what is the constraint? Well since the coordinates of the upper right corner must be on the circle we know that x and y must satisfy the equation of the circle. In other words, the equation of the circle is the constraint. The first thing to do then is to solve the constraint for one of the variables. Since the point that we’re looking at is in the first quadrant we know that y must be positive and so we can take the “+” part of this. Plugging this into the area and computing the first derivative gives, Before getting the critical points let’s notice that we can limit x to the range since we are assuming that x is in the first quadrant and must stay inside the circle. Now the four critical points we get (two from the numerator and two from the denominator) are, We only want critical points that are in the range of possible optimal values so that means that we have two critical points to deal with : and . Notice however that the second critical point is also one of the endpoints of our interval. Now, area function is continuous and we have an interval of possible solution with finite endpoints so, So, we can see that we’ll get the maximum area if and the corresponding value of y is, It looks like the maximum area will be found if the inscribed rectangle is in fact a square.
We need to again make a point that was made several times in the previous section. We excluded several critical points in the work above. Do not always expect to do that. There will often be physical reasons to exclude zero and/or negative critical points, however, there will be problems where these are perfectly acceptable values. You should always write down every possible critical point and then exclude any that can’t be possible solutions. This keeps you in the habit of finding all the critical points and then deciding which ones you actually need and that in turn will make it less likely that you’ll miss one when it is actually needed.
Example 3 Determine the point(s) on that are closest to (0,2). Solution Here’s a quick sketch of the situation. So, we’re looking for the shortest length of the dashed line. Notice as well that if the shortest distance isn’t at there will be two points on the graph, as we’ve shown above, that will give the shortest distance. This is because the parabola is symmetric to the y-axis and the point in question is on the y-axis. This won’t always be the case of course so don’t always expect two points in these kinds of problems. In this case we need to minimize the distance between the point (0,2) and any point that is one the graph (x,y). Or, If you think about the situation here it makes sense that the point that minimizes the distance will also minimize the square of the distance and so since it will be easier to work with we will use the square of the distance and minimize that. If you aren’t convinced of this we’ll take a closer look at this after this problem. So, the function that we’re going to minimize is, The constraint in this case is the function itself since the point must lie on the graph of the function. At this point there are two methods for proceeding. One of which will require significantly more work than the other. Let’s take a look at both of them. Solution 1 In this case we will use the constraint in probably the most obvious way. We already have the constraint solved for y so let’s plug that into the square of the distance and get the derivatives. So, it looks like there are three critical points for the square of the distance and notice that this time, unlike pretty much every previous example we’ve worked, we can’t exclude zero or negative numbers. They are perfectly valid possible optimal values this time. Before going any farther, let’s check these in the second derivative to see if they are all relative minimums. So, is a relative maximum and so can’t possibly be the minimum distance. That means that we’ve got two critical points. The question is how do we verify that these give the minimum distance and yes we did mean to say that both will give the minimum distance. Recall from our sketch above that if x gives the minimum distance then so will x and so if gives the minimum distance then the other should as well. None of the methods we discussed in the previous section will really work here. We don’t have an interval of possible solutions with finite endpoints and both the first and second derivative change sign. In this case however, we can still verify that they are the points that give the minimum distance. First, notice that if we are working on the interval then the endpoints of this interval (which are also the critical points) are in fact where the absolute minimum of the function occurs in this interval. Next we can see that if then . Or in other words, if the function is decreasing until it hits and so must always be larger than the function at . Similarly, then and so the function is always increasing to the right of and so must be larger than the function at . So, putting all of this together tells us that we do in fact have an absolute minimum at . All that we need to do is to find the value of y for these points. So, the points on the graph that are closest to (0,2) are, Solution 2 The first solution that we worked was actually the long solution. There is a much shorter solution to this problem. Instead of plugging y into the square of the distance let’s plug in x. From the constraint we get, and notice that the only place x show up in the square of the distance it shows up as x2 and let’s just plug this into the square of the distance. Doing this gives, There is now a single critical point, , and since the second derivative is always positive we know that this point must give the absolute minimum. So all that we need to do at this point is find the value(s) of x that go with this value of y. The points are then, So, for significantly less work we got exactly the same answer.
This previous example had a couple of nice points. First, as pointed out in the problem, we couldn’t exclude zero or negative critical points this time as we’ve done in all the previous examples. Again, be careful to not get into the habit of always excluding them as we do many of the examples we’ll work.
Next, some of these problems will have multiple solution methods and sometimes one will be significantly easier than the other. The method you use is up to you and often the difficulty of any particular method is dependent upon the person doing the problem. One person may find one way easier and other person may find a different method easier.
Finally, as we saw in the first solution method sometimes we’ll need to use a combination of the optimal value verification methods we discussed in the previous section.
Now, before we move onto the next example let’s take a look at the claim above that we could find the location of the point that minimizes the distance by finding the point that minimizes the square of the distance. We’ll generalize things a little bit,
Fact
Suppose that we have a positive function, , that exists everywhere then and will have the same critical points and the relative extrema will occur at the same points.
This is simple enough to prove so let’s do that here. First let’s take the derivative of and see what we can determine about the critical points of .
Let’s plug into this to get,
By assumption we know that exists and and therefore the denominator of this will always exist and will never be zero. We’ll need this in several places so we can’t forget this.
If then because we know that the denominator will not be zero here we must also have . Likewise, if then we must have . So, and will have the same critical points in which the derivatives will be zero.
Next, if doesn’t exist then will also not exist and likewise if doesn’t exist then because we know that the denominator will not be zero then this means that will also not exist. Therefore, and will have the same critical points in which the derivatives does not exist.
So, the upshot of all this is that and will have the same critical points.
Next, let’s notice that because we know that then and will have the same sign and so if we apply the first derivative test (and recalling that they have the same critical points) to each of these functions we can see that the results will be the same and so the relative extrema will occur at the same points.
Note that we could also use the second derivative test to verify that the critical points will have the same classification if we wanted to. The second derivative is (and you should see if you can use the quotient rule to verify this),
Then if is a critical point such that (and so we can use the second derivative test) we get,
Now, because we know that and by assumption we can see that and will have the same sign and so will have the same conclusion from the second derivative test.
So, now that we have that out of the way let’s work some more examples.
Example 4 A 2 feet piece of wire is cut into two pieces and one piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire cut so that the total area enclosed by both is minimum and maximum? Solution Before starting the solution recall that an equilateral triangle is a triangle with three equal sides and each of the interior angles are (or ). Now, this is another problem where the constraint isn’t really going to be given by an equation, it is simply that there is 2 ft of wire to work with and this will be taken into account in our work. So, let’s cut the wire into two pieces. The first piece will have length x which we’ll bend into a square and each side will have length . The second piece will then have length (we just used the constraint here…) and we’ll bend this into an equilateral triangle and each side will have length . Here is a sketch of all this. As noted in the sketch above we also will need the height of the triangle. This is easy to get if you realize that the dashed line divides the equilateral triangle into two other triangles. Let’s look at the right one. The hypotenuse is while the lower right angle is . Finally the height is then the opposite side to the lower right angle so using basic right triangle trig we arrive at the height of the triangle as follows. So, the total area of both objects is then, Here’s the first derivative of the area. Setting this equal to zero and solving gives the single critical point of, Now, let’s notice that the problem statement asked for both the minimum and maximum enclosed area and we got a single critical point. This clearly can’t be the answer to both, but this is not the problem that it might seem to be. Let’s notice that x must be in the range and since the area function is continuous we use the basic process for finding absolute extrema of a function. So, it looks like the minimum area will arise if we take while the maximum area will arise if we take the whole piece of wire and bend it into a square.
As the previous problem illustrated we can’t get too locked into the answers always occurring at the critical points as they have to this point. That will often happen, but one of the extrema in the previous problem was at an endpoint and that will happen on occasion.
Example 5 A piece of pipe is being carried down a hallway that is 10 feet wide. At the end of the hallway the there is a right-angled turn and the hallway narrows down to 8 feet wide. What is the longest pipe that can be carried (always keeping it horizontal) around the turn in the hallway? Solution Let’s start off with a sketch of the situation so we can get a grip on what’s going on and how we’re going to have to go about solving this. The largest pipe that can go around the turn will do so in the position shown above. One end will be touching the outer wall of the hall way at A and C and the pipe will touch the inner corner at B. Let’s assume that the length of the pipe in the small hallway is while is the length of the pipe in the large hallway. The pipe then has a length of . Now, if then the pipe is completely in the wider hallway and we can see that as then . Likewise, if the pipe is completely in the narrow hallway and as we also have . So, somewhere in the interval is an angle that will minimize L and oddly enough that is the length that we’re after. The largest pipe that will fit around the turn will in fact be the minimum value of L. The constraint for this problem is not so obvious and there are actually two of them. The constraints for this problem are the widths of the hallways. We’ll use these to get an equation for L in terms of and then we’ll minimize this new equation. So, using basic right triangle trig we can see that, So, differentiating L gives, Setting this equal to zero and solving gives, Solving for gives, So, if radians then the pipe will have a minimum length and will just fit around the turn. Anything larger will not fit around the turn and so the largest pipe that can be carried around the turn is,
Example 6 Two poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of each pole and it is also staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used? Solution As always let’s start off with a sketch of this situation. The total length of the wire is and we need to determine the value of x that will minimize this. The constraint in this problem is that the poles must be 20 meters apart and that x must be in the range . The first thing that we’ll need to do here is to get the length of wire in terms of x, which is fairly simple to do using the Pythagorean Theorem. Not the nicest function we’ve had to work with but there it is. Note however, that it is a continuous function and we’ve got an interval with finite endpoints and so finding the absolute minimum won’t require much more work than just getting the critical points of this function. So, let’s do that. Here’s the derivative. Setting this equal to zero gives, It’s probably been quite a while since you’ve been asked to solve something like this. To solve this we’ll need to square both sides to get rid of the roots, but this will cause problems as well soon see. Let’s first just square both sides and solve that equation. Note that if you can’t do that factoring don't worry, you can always just use the quadratic formula and you’ll get the same answers. Okay two issues that we need to discuss briefly here. The first solution above (note that I didn’t call it a critical point…) doesn’t make any sense because it is negative and outside of the range of possible solutions and so we can ignore it. Secondly, and maybe more importantly, if you were to plug into the derivative you would not get zero and so is not even a critical point. How is this possible? It is a solution after all. We’ll recall that we squared both sides of the equation above and it was mentioned at the time that this would cause problems. We’ll we’ve hit those problems. In squaring both sides we’ve inadvertently introduced a new solution to the equation. When you do something like this you should ALWAYS go back and verify that the solutions that you get are in fact solutions to the original equation. In this case we were lucky and the “bad” solution also happened to be outside the interval of solutions we were interested in but that won’t always be the case. So, if we go back and do a quick verification we can in fact see that the only critical point is and this is nicely in our range of acceptable solutions. Now all that we need to do is plug this critical point and the endpoints of the wire into the length formula and identify the one that gives the minimum value. So, we will get the minimum length of wire if we stake it to the ground feet from the smaller pole.
Let’s do a modification of the above problem that asks a completely different question.
Example 7 Two poles, one 6 meters tall and one 15 meters tall, are 20 meters apart. A length of wire is attached to the top of each pole and it is also staked to the ground somewhere between the two poles. Where should the wire be staked so that the angle formed by the two pieces of wire at the stake is a maximum? Solution Here’s a sketch for this example. The equation that we’re going to need to work with here is not obvious. Let’s start with the following fact. Note that we need to make sure that the equation is equal to because of how we’re going to work this problem. Now, basic right triangle trig tells us the following, Plugging these into the equation above and solving for gives, Note that this is the reason for the in our equation. The inverse tangents give angles that are in radians and so can’t use the 180 that we’re used to in this kind of equation. Next we’ll need the derivative so hopefully you’ll recall how to differentiate inverse tangents. Setting this equal to zero and solving give the following two critical points. The first critical point is not in the interval of possible solutions and so we can exclude it. It’s not difficult to show that if that and if that and so when we will get the maximum value of .
Example 8 A trough for holding water is be formed by taking a piece of sheet metal 60 cm wide and folding the 20 cm on either end up as shown below. Determine the angle that will maximize the amount of water that the trough can hold. Solution Now, in this case we are being asked to maximize the volume that a trough can hold, but if you think about it the volume of a trough in this shape is nothing more than the cross-sectional area times the length of the trough. So for a given length in order to maximize the volume all you really need to do is maximize the cross-sectional area. To get a formula for the cross-sectional area let’s redo the sketch above a little. We can think of the cross-sectional area as a rectangle in the middle with width 20 and height h and two identical triangles on either end with height h, base b and hypotenuse 20. Also note that basic geometry tells us that the angle between the hypotenuse and the base must also be the same angle that we had in our original sketch. Also, basic right triangle trig tells us that the base and height can be written as, The cross-sectional area for the whole trough, in terms of , is then, The derivative of the area is, So, we have either, However, we can see that must be in the interval or we won’t get a trough in the proper shape. Therefore, the second critical point makes no sense and also note that we don’t need to add on the standard “ ” for the same reason. Finally, since the equation for the area is continuous all we need to do is plug in the critical point and the end points to find the one that gives the maximum area. So, we will get a maximum cross-sectional area, and hence a maximum volume, when .
Optimization Previous Section Next Section L'Hospital's Rule and Indeterminate Forms Derivatives Previous Chapter Next Chapter Integrals
[Notes]
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# Logarithmic Functions
A population of 50 flies is expected to double every week, leading to a function of the form f(x) = 50(2)x , where x represents the number of weeks that have passed. When will this population reach 500? Trying to solve this problem leads to:
500 = 50(2)x Dividing both sides by 50 to isolate the exponential
10 = 2x
While we have set up exponential models and used them to make predictions, you may have noticed that solving exponential equations has not yet been mentioned. The reason is simple: none of the algebraic tools discussed so far are sufficient to solve exponential equations. We must introduce a new function, named log, as the function that “undoes” an exponential function, like how a square root “undoes” a square. Since exponential functions have different bases, we will define corresponding logarithms of different bases as well.
Logarithm: The logarithm (base b) function, written logb(x) , “undoes” exponential function bx .
The statement ba = c is equivalent to the statement logb(c) = a .
Since the logarithm and exponential “undo” each other (in technical terms, they are inverses), it follows that:
Properties of Logs: Inverse Properties
logb(bx) = x
blogbx = x
Since log is a function, it is most correctly written as logb(c) , using parentheses to denote function evaluation, just as we would with f(c). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written as logbc .
Example 4.2.1
Write these exponential equations as logarithmic equations:
23 = 8 52 = 25 10-4 =
23 = 8 is equivalent to log2(8) = 3
52 = 25 is equivalent to log5(25) = 2
10-4 = is equivalent to log10() = -4
Example 4.2.2
Write these logarithmic equations as exponential equations:
is equivalent to
is equivalent to
By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic and exponential equations by rewriting.
Example 4.2.3
Solve log4(x) = 2 for x.
By rewriting this expression as an exponential, 42 = x , so x = 16
Most calculators and computers will only evaluate logarithms of two bases.
Common and Natural Logarithms:
The common log is the logarithm with base 10, and is typically written log(x) .
The natural log is the logarithm with base e, and is typically written ln(x) .
Example 4.2.4
Evaluate log(1000) using the definition of the common log.
To evaluate log(1000) , we can say x = log(1000) , then rewrite into exponential form using the common log base of 10.
10x = 1000
From this, we might recognize that 1000 is the cube of 10, so x = 3.
We also can use the inverse property of logs to write log10(103) = 3
Values of the Common Log Number Number as exponential Log(number) 1000 103 3 100 102 2 10 101 1 1 100 0 0.1 10-1 -1 0.01 10-2 -2 0.001 10-3 -3
Example 4.2.5
Evaluate ln() .
We can rewrite ln() as ln() . Since ln is a log base e, we can use the inverse property for logs: ln() = = .
# Graphs of Logarithms
Recall that the exponential function f(x) = 2x produces this table of values:
x -3 -2 -1 0 1 2 3 f(x) 1 2 4 8
Since the logarithmic function “undoes” the exponential, g(x) = log2(x) produces the table of values:
x 1 2 4 8 f(x) -3 -2 -1 0 1 2 3
In this second table, notice that:
1. As the input increases, the output increases.
2. As input increases, the output increases more slowly.
3. Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is (0, ∞) .
4. Since the exponential function can accept all real numbers as inputs, the logarithm can output any real number, so the range is all real numbers or (-∞, ∞) .
Sketching the graph, notice that as the input approaches zero from the right, the output of the function grows very large in the negative direction, indicating a vertical asymptote at x = 0. A vertical asymptote is a vertical line x = a where the graph tends towards positive or negative infinity as the inputs approach a.
In symbolic notation we write as x→0+ , f(x)→-∞ , and as x→∞ , f(x)→∞
Graphical Features of the Logarithm
Graphically, in the function g(x) = logb(x):
• The graph has a horizontal intercept at (1, 0)
• The graph has a vertical asymptote at x = 0
• The graph is increasing and concave down
• The domain of the function is x > 0, or (0, ∞)
• The range of the function is all real numbers, or (-∞, ∞)
When sketching a general logarithm with base b, it can be helpful to remember that the graph will pass through the points (1, 0) and (b, 1). To get a feeling for how the base affects the shape of the graph, examine the graphs below.
Notice that the larger the base, the slower the graph grows. For example, the common log graph, while it grows without bound, it does so very slowly. For example, to reach an output of 8, the input must be 100,000,000.
# Logarithm Properties
To utilize the common or natural logarithm functions to evaluate expressions like log2(10), we need to establish some additional properties.
Properties of Logs: Exponent Property
logb(Ar) = rlogb(A)
Example 4.2.6
Rewrite log3(25) using the exponent property for logs.
Since 25 = 52,
log3(25) = log3(52) = 2log35
Example 4.2.7
Rewrite 4ln(x) using the exponent property for logs.
Using the property in reverse, 4ln(x) = ln(x4)
The second important property allows us to change the base of a logarithmic expression.
Properties of Logs: Change of Base
Example 4.2.8
Evaluate log2(10) using the change of base formula.
According to the change of base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e:
Using our calculators to evaluate this:
This finally allows us to answer our original question – the population of flies we discussed at the beginning of the section will take 3.32 weeks to grow to 500.
Example 4.2.9
Evaluate log5(100) using the change of base formula.
We can rewrite this expression using any other base. We can rewrite using the common log, base 10:
An alternative approach to solving exponential equations is described below:
Solving exponential equations:
1. Isolate the exponential expressions when possible
2. Take the logarithm of both sides
3. Utilize the exponent property for logarithms to pull the variable out of the exponent
4. Use algebra to solve for the variable.
Example 4.2.10
Solve 2x = 10 for x.
Using this alternative approach, rather than rewrite this exponential into logarithmic form, we will take the logarithm of both sides of the equation. Since we often wish to evaluate the result to a decimal answer, we will usually utilize either the common log or natural log. For this example, we’ll use the natural log:
ln(2x) = ln(10) Utilizing the exponent property for logs,
xln(2) = ln(10) Now dividing by ln(2),
x = ≈ 2.861
Notice that this result matches the result we found using the change of base formula.
Example 4.2.11
Previously, we predicted the population (in billions) of India t years after 2008 by using the function f(t) = 1.14(1 + 0.0134)t . If the population continues following this trend, when will the population reach 2 billion?
We need to solve for the t so that f(t) = 2:
Divide by 1.14 to isolate the exponential expression Take the logarithm of both sides of the equation Apply the exponent property on the right side Divide both sides by ln(1.0134)
years.
If this growth rate continues, the model predicts the population of India will reach 2 billion about 42 years after 2008, or approximately in the year 2050.
Some situations cannot be addressed using the properties already discussed. For these, we need some additional properties:
Sum of Logs Property:
Difference of Logs Property:
With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.
Example 4.2.12
Write log3(5) + log3(8) – log3(2) as a single logarithm.
Using the sum of logs property on the first two terms:
log3(5) + log3(8) = log3(5 · 8) = log3(40)
This reduces our original expression to log3(40) – log3(2)
Then using the difference of logs property,
log3(40) – log3(2) = log3() = log3(20)
Example 4.2.13
Evaluate 2log(5) + log(4) without a calculator by first rewriting as a single logarithm.
On the first term, we can use the exponent property of logs to write:
2log(5) = log(52) = log(25)
With the expression reduced to a sum of two logs, log(25) + log(4) , we can utilize the sum of logs property:
log(25) + log(4) = log(4 · 25) = log(100)
Since 100 = 102, we can evaluate this log without a calculator:
log(100) = log(102) = 2
Example 4.2.14
Rewrite as a sum or difference of logs.
First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write:
Then seeing the product in the first term, we use the sum property:
Finally, we could use the exponent property on the first term:
# Log Properties in Solving Equations
The logarithm properties often arise when solving problems involving logarithms.
Example 4.2.15
Solve .
In order to rewrite in exponential form, we need a single logarithmic expression on the left side of the equation. Using the difference property of logs, we can rewrite the left side:
Rewriting in exponential form reduces this to an algebraic equation:
Solving:
Checking this answer in the original equation, we can verify there are no domain issues, and this answer is correct.
# Practice questions
1. Write the exponential equation 42 = 16 as a logarithmic equation.
2. Evaluate log(1,000,000).
3. Rewrite using the exponent property for logs: ln.
4. Solve 5(0.93)x = 10.
5. Without a calculator evaluate by first rewriting as a single logarithm: log2(8) + log2(4)
6. Solve log(2x – 4) = 1 + log(x + 2).
7. Suppose the population of rats in a city is estimated to be 1.1 million. If the population size increases exponentially at a rate of 15% per year, how many years would it take for the population size to reach 5 million rats?
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You need to rent a video camera for your school project. Murphy’s Camera charges a \$30 rental fee plus \$35 per day. Allied Rental charges a
Question
You need to rent a video camera for your school project. Murphy’s Camera charges a \$30 rental fee plus \$35 per day. Allied Rental charges a \$45 rental fee plus \$30 per day. Which company should you choose?
Independent variable (x):
Dependent variable (y):
System of equations:
Graph:
A. When does it make sense to rent a camera from Murphy’s? How do you know?
B. When does it make sense to rent a camera from Allied Rental? How do you know?
C. Is there ever a time where it wouldn’t matter which store to choose? Explain what this means in the context of a “system of equations.”
D. If you were renting a camera for 3 days, how much money would you pay for each store? Show all work.
in progress 0
3 months 2022-02-10T05:06:17+00:00 1 Answer 0 views 0
1. A. If a camera is rented for 1 or 2 days, it would be logical to rent it from Murphy’s.
B. If a camera is needed for more than 3 days, it would make sense to rent it from Allied rental.
C. Yes, if a camera is rented for 3 days both rentals charge the same money.
C. If you were need a camera for 3 days, you would have to pay \$135 for both stores.
Step-by-step explanation:
Step 1; Murphy’s camera costs \$30 to rent and an additional \$35 for each day of rental. So the equations are
y = \$30 + (x × \$35), where y is total cost and x is the number of days rented.
So for x = 1, y = \$30 + (1 × \$35) = \$30 + \$35 = \$65
So for x = 2, y = \$30 + (2 × \$35) = \$30 + \$70 = \$100
So for x = 3, y = \$30 + (3 × \$35) = \$30 + \$105 = \$135
So for x = 4, y = \$30 + (4 × \$35) = \$30 + \$140 = \$170
So for x = 5, y = \$30 + (5 × \$35) = \$30 + \$175 = \$ 205.
Step 2; Allied rental costs \$45 to rent and a \$30 for each day of rental. So the equations are as follows
y = \$45 + (x × \$30), where y is total cost and x is the days rented
So for x = 1, \$45 + (1 × \$30) = \$45 + \$30 = \$75
So for x = 2, \$45 + (2 × \$30) = \$45 + \$60 = \$105
So for x = 3, \$45 + (3 × \$30) = \$45 + \$90 = \$135
So for x = 4, \$45 + (4 × \$30) = \$45 + \$120 = \$165
So for x = 5, \$45 + (5 × \$30) = \$45 + \$150 = \$195
Step 3; Comparing both prices over 5 days. Murphy’s is cheaper for until 2 days. On day 3 both the rentals cost the same but after day 4, allied rentals is the better option.
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# How do you simplify 21 - 1 times 2 ÷ 4 using order of operations?
Nov 23, 2015
20.5
#### Explanation:
What you need to remember when solving this is the following:
you start by looking if there are any parenthesis, because you need to solve whatever is between the parenthesis before doing anything else. Inside the parenthesis the same rules apply as discussed here.
Then you look if there are any exponents, they need to be solved after you solved the parenthesis.
When you have done that you multiply or divide if necessary. It doesn't matter in which order so you can go from left to right or from right to left with multiplying or dividing, whatever suits you best.
Finally you add or subtract. Just like with multiplying or dividing, it doesn't matter which order you choose.
So with the equation 21 - 1 * 2 ÷ 4 you see that there are no parenthesis or exponents, but you can divide and multiply.
Which leaves you with $21 - 2 : 4 = 21 - 0.5$
Then you are left with a subtraction, which you can solve as $21 - 0.5 = 20.5$
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# Operations with Numbers and Properties of Operations
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In addition to the revision notes for Operations with Numbers and Properties of Operations on this page, you can also access the following Arithmetic learning resources for Operations with Numbers and Properties of Operations
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1.3Operations with Numbers and Properties of Operations
In these revision notes for Operations with Numbers and Properties of Operations, we cover the following key points:
• What is a mathematical operation? What are the four basic operations used in math?
• Why we sometimes split an addend in two parts.
• How to add two numbers in a column and in a number line.
• What is subtraction? What is/are the property (ies) of subtraction?
• How is subtraction related to addition?
• How to subtract two numbers in a column and in a number line.
• What is multiplication? What does it have in common with addition?
• What are the properties of multiplication?
• How to do a multiplication in a column?
• What is division? How is it related to multiplication?
• What is/are the property (ies) of division?
• How to make a division operation easier?
• What is the remainder of a division?
• What is "power of numbers"? How is it expressed in exercises?
## Operations with Numbers and Properties of Operations Revision Notes
A mathematical operation is a process in which a number or quantity is changed in order to give another number or quantity. There are four basic mathematical operations: addition, subtraction, multiplication and division.
Addition is a mathematical operation that shows two or more numbers added together. The result of addition is called the sum and each element (number) participating in an addition operation is called an addend.
The operation of addition is expressed through the plus (+) symbol.
An addition expressed on a number line means a shift due right.
Addition as a mathematical operation has the following properties:
1. Closure property - If two (or more) addends belong to a given number set, the sum also belongs to this set.
2. Commutative property - If we switch the place of addends the sum does not change.
3. Associative property - if we dont start the operations from the two leftmost addends but from two other addends more in the right, the sum does not change.
4. Additive identity property - Any number added with zero gives the same number as the sum. Zero is called the identity element of addition.
Sometimes, it is more appropriate to split an addend in two parts to make the calculations easier. Then we can apply any of the aforementioned properties of addition to find the sum.
We can add numbers in columns for simplicity. In this method, we place the like placeholders in the same column (units below units, tens below tens and so on).
Subtraction is a mathematical operation in which we remove a quantity of items from a collection. Subtraction is represented mathematically through the symbol minus (-).
The participants in a subtraction are as follows: the first number (usually the biggest) from which the quantity is subtracted is called the minuend; the quantity subtracted is called the subtrahend, while the result of subtraction is known as the difference.
Subtraction on a number line is done by moving due left.
Subtraction has only one property in common with addition - the Subtractive Identity Property of Zero, where zero is the subtractive identity element. Based on this property, when we subtract zero from a number, the value of the number remains the same.
Subtraction is commonly considered as the inverse operation of addition. Hence, instead of writing a - b = c, we can write a + (-b) = c as the result is the same in both cases.
We can subtract numbers in a column in the same way as we do in addition. In this case, when the upper digit is smaller than the lower digit, we borrow one "ten" from the next placeholder on its left. We say a "ten" is borrowed from the left placeholder to make the subtraction Possible.
Multiplication is a shorter representation of the repeated addition of equal numbers. It is represented in expressions through the symbols ( × ) or ( · ).
The numbers participating in a multiplication operation are known as factors and the result of multiplication is called the product.
Properties of multiplication include:
1. Closure property of multiplication - If two (or more) factors belong to a given number set, the product also belongs to this set.
2. Commutative property of multiplication - When we switch the place of factors the product does not change.
3. Associative property of multiplication - We can start the operations from anywhere for convenience as the result does no change.
4. Distributive property - When an expression inside brackets containing addition or subtraction is multiplied by a number, the expression can be written without brackets where the given number multiplies every element of the expression separately.
5. Multiplicative identity property - In multiplication, this identity element is the number 1. This means that if we multiply a number by 1, the product is the same as the number itself.
We can multiply two numbers in column for an easier solution. We multiply each number of the upper factor to each number of the lower factor and the products obtained are written in separate rows below each other by starting (due right) from the position of the digit of the lower number involved in the process.
Division is the inverse operation of multiplication. We apply division when we want to cut a quantity into equal parts.
Division is the inverse operation of multiplication. We use the symbol (÷) to represent division in a mathematical expression. The original quantity is known as the dividend, the number that divides the original quantity in equal parts is called the divisor while the result of division is called the quotient.
Division has only one property in common with multiplication: the divisive identity property. According to this property, the number 1 is the identity element for division because any number divided by 1 gives the same number as a result.
When we divide two natural numbers, the result is not always a natural number. However, if we want to keep writing the result as a natural number, we write the closest value as quotient and also the remainder of division inside brackets, i.e. the number left from the division operation.
If we have to multiply the same factor a number of times by itself, we use a shorter notation known as power to represent this recurring multiplication by the same number. In other words, the power of a number says how many times to use that number in a multiplication. The recurring factor is called the base, the number that shows how many times this factor appears in a recurring multiplication is called the exponent and the result of this operation is called the power.
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Proof that the Square Root of 2 is Irrational
# Proof that the Square Root of 2 is Irrational
We've looked at the operations of addition and multiplication over the field of real numbers. One next might question is how $\sqrt{2} \approx 1.4142...$ might arise with only these two operations since at best we can only form rational numbers, that is numbers in the set $\mathbb{Q} := \{ \frac{a}{b} : a, b \in \mathbb{Z} \: \mathrm{and} \: b \neq 0 \}$. The following proof will illustrate that $\sqrt{2} \not \in \mathbb{Q}$.
Theorem 1: There exists no rational number $r = \frac{a}{b}$ ($a, b \in \mathbb{Z}$ and $b \neq 0$) such that $r^2 = 2$.
• Proof of Theorem: We will do this proof by contradiction. Suppose that $r$ is an irrational number in lowest terms, that is the integers $a$ and $b$ have 1 as their greatest common divisors commonly abbreviated as $\mathrm{gcd}(a, b)= 1$, and $r$ is such that $r^2 = 2$. Therefore:
(1)
\begin{align} r^2 = 2 \\ \left ( \frac{a}{b} \right )^2 = 2 \\ \frac{a^2}{b^2} = 2 \\ a^2 = 2b^2 \end{align}
• Therefore it follows that $a^2$ is even which implies that $a$ is even. Since $a$ is even it can be written as $a = 2m$ for some $m \in \mathbb{Z}$. Making this substitution we get that:
(2)
\begin{align} a^2 = 2b^2 \\ (2m)^2 = 2b^2 \\ 4m^2 = 2b^2 \\ 2m^2 = b^2 \end{align}
• And we see that $b^2$ is also even which implies $b$ is even so $b$ can be written as $b = 2n$ for some $n \in \mathbb{Z}$.
• Since $a = 2m$ and $b = 2n$ for some $m, n \in \mathbb{Z}$, it follows that $\mathrm{gcd}(a, b) = \mathrm{gcd}(2m, 2n) ≥ 2$ which contradicts our original assumption that $r$ was rational. $\blacksquare$
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8.8: Triangle Classification
Difficulty Level: Basic Created by: CK-12
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Kevin and Jake began examining a sculpture while the girls were examining the painting with the lines. This sculpture is full of triangles. The boys remember how Mrs. Gilson explained that a triangle is one of the strongest figures that there is and that is why we see triangles in construction.
“Think of a bridge,” Kevin said to Jake. “A bridge has many triangles within it. That is how the whole thing stays together. If we did not have the triangles, the structure could collapse.”
“What about here? Do you think it matters what kind of triangle is used?” Jake asked.
“I don’t know. Let’s look at what they used here.”
The two boys walked around the sculpture and looked at it from all sides. There was a lot to notice. After a little while, Jake was the first one to speak.
“I don’t think it matters which triangle you use,” he said.
“Oh, I do. The isosceles makes the most sense because it is balanced,” Kevin said smiling.
Jake is confused. He can’t remember why an isosceles triangle “is balanced” in Kevin’s words. Jake stops to think about this as Kevin looks at the next sculpture.
Do you know what Kevin means? What is an isosceles triangle and how does it “balance?” This Concept will teach you all about triangles and how to classify them. When you finish with this Concept, you will have a chance to revisit this problem. Then you may understand a little better what Kevin means by his words.
Guidance
As we have seen, the angles in a triangle can vary a lot in size and shape, but they always total .
We can identify kinds of triangles by the sizes of their angles. A triangle can either be acute, obtuse, or right.
Let’s look more closely.
Acute triangles have three acute angles. In other words, all of their angles measure less than . Below are some examples of acute triangles.
Notice that each angle in the triangles above is less than , but the total for each triangle is still .
We classify triangles that have an obtuse angle as an obtuse triangle. This means that one angle in the triangle measures more than . Here are some obtuse triangles.
You can see that obtuse triangles have one wide angle that is greater than . Still, the three angles in obtuse triangles always add up to . Only one angle must be obtuse to make it an obtuse triangle.
The third kind of triangle is a right triangle. As their name implies, right triangles have one right angle that measures exactly . Often, a small box in the corner tells you when an angle is a right angle. Let’s examine a few right triangles.
Once again, even with a right angle, the three angles still total !
Now let’s practice identifying each kind of triangle.
Label each triangle as acute, obtuse, or right.
In order to classify the triangles, we must examine the three angles in each. If there is an obtuse angle, it is an obtuse triangle. If there is a right angle, it is a right triangle. If all three angles are less than , it is an acute triangle.
One short cut we can use is to compare the angles to . If an angle is exactly , we know the triangle must be a right triangle. If any angle is more than , the triangle must be an obtuse triangle. If there are no right or obtuse angles, the triangle must be an acute triangle. Check to make sure each angle is less than .
There are no right angles in Figure 1. There are no angles measuring more than . This is probably an acute triangle. Let’s check each angle to be sure: and are all less than so this is definitely an acute triangle. Figure 1 is an acute triangle.
Is there any right or obtuse angles in the second triangle? The small box in the corner tells us that the angle is a right angle. Therefore this is a right triangle. Figure 2 is a right triangle.
What about Figure 3? It does not have any right angles. It does, however, have an extremely wide angle. Wide angles usually are obtuse. Let’s check the measure to make sure it is more than . It is , so it is definitely an obtuse angle. Therefore this is an obtuse triangle. Figure 3 is an obtuse triangle.
Figure 4 doesn’t have any right angles. It doesn’t have any wide angles either. Obtuse angles are not always wide, however. Check the angle measures to be sure. The angle measuring is greater than so it is obtuse. That makes this an obtuse triangle. Figure 4 is an obtuse triangle.
Yes. Make a few notes about each type of triangle so that you can remember how to classify them according to their angles.
You have seen that we can classify triangles by their angles. We can also classify triangles by the lengths of their sides. As you know, every triangle has three sides. Sometimes all three sides are the same length, or congruent. In some triangles, only two sides are congruent. And still other triangles have sides that are all different lengths. By comparing the lengths of the sides, we can determine the kind of triangle it is.
Let’s see how this works.
A triangle with three equal sides is an equilateral triangle. It doesn’t matter how long the sides are, as long as they are all congruent, or equal. Here are some equilateral triangles.
Just remember, equal sides means it’s an equilateral triangle.
An isosceles triangle has two congruent sides. It doesn’t matter which two sides, any two will do. Let’s look at some isosceles triangles.
The third type of triangle is a scalene triangle. In a scalene triangle, none of the sides are congruent.
Now let’s practice identifying each kind of triangle.
Classify each triangle as equilateral, isosceles, or scalene.
We need to examine the lengths of the sides in each triangle to see if any sides are congruent. In the first triangle, two sides are 7 meters long, but the third side is shorter. Which kind of triangle has two congruent sides? The first triangle is an isosceles triangle.
Now let’s look at the second triangle. All three sides are the same length, so this must be an equilateral triangle. The second triangle is an equilateral triangle.
The last triangle has sides of 5 cm, 4 cm, and 8 cm. None of the sides are congruent, so this is a scalene triangle. The last triangle is a scalene triangle.
Determine the type of triangles described in each example.
Example A
One angle is and the other two angles are acute angles.
Solution: Obtuse triangle
Example B
All three angles have the same measure.
Solution: Equiangular triangle
Example C
Two out of three angles measure .
Solution: Acute Triangle
Here is the original problem once again.
Kevin and Jake began examining a sculpture while the girls were examining the painting with the lines. This sculpture is full of triangles. The boys remember how Mrs. Gilson explained that a triangle is one of the strongest figures that there is and that is why we see triangles in construction.
“Think of a bridge,” Kevin said to Jake. “A bridge has many triangles within it. That is how the whole thing stays together. If we did not have the triangles, the structure could collapse.”
“What about here? Do you think it matters what kind of triangle is used?” Jake asked.
“I don’t know. Let’s look at what they used here.”
The two boys walked around the sculpture and looked at it from all sides. There was a lot to notice. After a little while, Jake was the first one to speak.
“I don’t think it matters which triangle you use,” he said.
“Oh, I do. The isosceles makes the most sense because it is balanced,” Kevin said smiling.
Jake is confused. He can’t remember why an isosceles triangle “is balanced” in Kevin’s words. Jake stops to think about this as Kevin looks at the next sculpture.
Kevin’s comment is a little tricky. You can think of an isosceles triangle as being balanced because it has two equal sides. Therefore, if you look at an isosceles triangle, it will be even whereas a scalene triangle would not be. In Kevin’s thinking, this type of triangle makes sense because it would be firm, solid and “balanced.”
If you think about Kevin’s statement, you can grasp the math by thinking about the properties of an isosceles triangle. Look at the sculpture again. How are triangles being used in the sculpture? Can you see any other types of triangles in this sculpture? Make a few notes in your notebook.
Vocabulary
Here are the vocabulary words in this Concept.
Triangle
a figure with three sides and three angles.
Interior angles
the three inside angles of a triangle.
Exterior angles
the angles outside of a triangle formed by intersecting lines
Acute Triangles
triangles with three angles less than
Obtuse Triangles
triangle with one angle greater than
Right Triangle
a triangle with one angle.
Congruent
exactly the same
Equilateral Triangle
all three side lengths are the same
Isosceles Triangle
two side lengths are the same and one is different.
Scalene Triangle
all three side lengths of a triangle are different lengths.
Guided Practice
Here is one for you to try on your own.
Can you identify both the sides and angles of a triangle at the same time?
As you can see, the first name identifies the triangle by its angles, and the second name groups it by its sides. Equilateral triangles do not quite fit this pattern. They are always acute. This is because the three angles in an equilateral triangle always measure .
There is one more thing to know about classifying triangles by their angles and sides. We can also tell whether a triangle is isosceles, scalene, or equilateral by its angles. Every angle is related to the side opposite it. Imagine a book opening. The wider you open it, the greater the distance between the two flaps. In other words the wider an angle is, the longer the opposite side.
Therefore we can say that if a triangle has two congruent angles, it must have two congruent sides, and must be isosceles. If it has three angles of different measures, then its sides are also all of different lengths, so it is scalene. Finally, an equilateral triangle, as we have seen, always has angles of and these angles are opposite congruent sides.
Video Review
Here is a video review.
Practice
Directions: Find the measure of angle in each figure below.
Directions: Identify each triangle as right, acute, or obtuse.
Directions: Identify each triangle as equilateral, isosceles, or scalene.
Directions: Use what you have learned to answer each question.
13. True or false. An acute triangle has three sides that are all different lengths.
14. True or false. A scalene triangle can be an acute triangle as well.
15. True or false. An isosceles triangle can also be a right triangle.
16. True or false. An equilateral triangle has three equal sides.
17. True or false. An obtuse triangle can have multiple obtuse angles.
18. True or false. A scalene triangle has three angles less than 90 degrees.
19. True or false. A triangle with a angle must be an obtuse triangle.
20. True or false. The angles of an equilateral triangle are also equal in measure.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English Spanish
TermDefinition
Acute Triangle An acute triangle has three angles that each measure less than 90 degrees.
equiangular triangle A triangle with all congruent angles.
equilateral triangle A triangle with three congruent sides.
Isosceles Triangle An isosceles triangle is a triangle in which exactly two sides are the same length.
Obtuse Triangle An obtuse triangle is a triangle with one angle that is greater than 90 degrees.
Scalene Triangle A scalene triangle is a triangle in which all three sides are different lengths.
Triangle A triangle is a polygon with three sides and three angles.
Interior angles Interior angles are the angles inside a figure.
Right Angle A right angle is an angle equal to 90 degrees.
Equilateral A polygon is equilateral if all of its sides are the same length.
Equiangular A polygon is equiangular if all angles are the same measure.
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# tan(pi-x) Formula | Simplify tan(x-π)
The tan(π-x) formula is given by tan(π-x)= -tanx. The formula of tan(pi-θ) is equal to tan(π-θ)= -tanθ. In this post, we will learn how to compute tan(pi-x) and tan(θ-pi).
Note that
• tan(π-x)= -tanx or, tan(π-θ)= -tanθ
• tan(x-π)= tanx or, tan(θ-π)= tan θ.
## Proof of tan(pi-x) Formula
Let us use the formula:
tan(a-b) = $\dfrac{\tan a -\tan b}{1+\tan a \tan b}$ …(∗)
Put a=π, b=x. Thus, we get that
tan(π-x) = $\dfrac{\tan \pi -\tan x}{1+\tan \pi \tan x}$
= $\dfrac{0 -\tan x}{1+0 \cdot \tan x}$ as the value of tanπ = 0.
= $\dfrac{-\tan x}{1}$
= -tanx
So the formula of tan(π-x) is equal to -tanx, that is,
tan(π-θ) Formula: In the above formula, if we replace x by θ, we will get the formula of tan(π-θ) which is as follows:
## Proof of tan(x-pi) Formula
In the above formula () of tan(a-b), let us put
a=x, b=π.
So, tan(x-π)
= $\dfrac{\tan x- \tan \pi}{1+\tan x \tan \pi}$
= $\dfrac{\tan x- 0}{1+\tan x \cdot 0}$ as tan(π) =0.
= tanx
So the formula of tan(x-π) is equal to tan(x-π)= tanx.
tan(θ-π) Formula: In the above formula, replacing x by θ, we get the following: the formula of tan(theta-π) is given by tan(θ-π)= tanθ.
More Formulas:
Formula of sin(π-x)
Value of sin(5π/2)
If x+y=π/4, then prove (1+tanx)(1+tany)=2
If x+y+z=π, then prove tanx + tany + tanz = tanx tany tanz
## FAQs
Q1: What is the Formula of tan(π-x)?
Answer: The formula of tan(π-x) is given by tan(π-x)= -tanx.
Q2: What is the Formula of tan(x-π)?
Answer: The formula of tan(x-π) is given by tan(x-π)= tanx.
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# Mathematics Review
#### Chapter index in this window — — Chapter index in separate window
This web page provides a review of the mathematical concepts you will apply in this course.
### Fractions and Percentages %
Fractions are a part of every-day life and they are also used in science. A fraction with the same numerator (number on top) as another but with a smaller denominator (number on bottom) is a larger number. For example 1/2 is bigger than 1/3 which is bigger than 1/4, etc. That is because if you divide 1 by 2, you get 0.5 but 1 divided by 3 is only 0.333 and 1 divided by 4 is only 0.25. The fraction 3/5 (=0.6) is bigger than 3/6 (=0.5) which is bigger than 3/7 (=0.4286).
Fractions are sometimes expressed as a percent. To express a fraction as a percentage, find the decimal form and multiply by 100. The percent symbol ``%'' simply means ``divide by 100.'' For example 1/2 = 0.5 = 0.5× 100% = 50%; 3/5 = 0.6 = 0.6× 100% = 60%. Some examples on converting percentages to fractions or decimals: 5.8% = 5.8/100 = 0.058; 0.02% = 0.02/100 = 0.0002; the Sun is 90% hydrogen means that 90 out of every 100 atoms in the Sun is hydrogen.
### ``Times'' and ``Factor of''
Several homework questions ask you to find some quantity and find out how many ``times'' smaller or larger it is than something else, e.g., star A is ____times larger than star B. This means star A = a × star B, and you must find the number a. Another example: A gallon is equivalent to 4 quarts. This means that 1 gallon is 4 times bigger than 1 quart since 1 gallon = 4 × 1 quart, or (1 gallon)/(1 quart) = 4/1. This also means that 1 quart is 4 times smaller than 1 gallon since 1 quart = 1/4 × (1 gallon), or (1 quart)/(1 gallon) = 1/4. Notice when ``times bigger'' is used and when ``times smaller'' is used.
The use of the phrase ``factor of'' is very similar to the use of ``times''. For example, 1 quart is a factor of 4 smaller than 1 gallon, or 1 gallon is a factor of 4 bigger than 1 quart.
#### Go to Astronomy Notes home
last updated: 08 August 2001
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# What is 10 raise to the power 20?
## What is 10 raise to the power 20?
10 to the 20th power is 1020 which is 1 followed by 20 zeroes.
## How do you calculate 10 to the power of?
Understand that the exponent is the amount of times that you multiply 10 by itself.
1. For example: 103 = 1000 because 10 x 10 x 10 = 1000.
2. 104 = 10 x 10 x 10 x 10 or 10,000.
3. 105 = 10 x 10 x 10 x 10 x 10 = 100,000.
4. 106 = 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000.
5. 107 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10,000,000.
What is 2 raise to the power 9?
512
Answer: The value of 2 raised to 9th power i.e., 29 is 512.
### What number is 10 to the power of 23?
100,000,000,000,000,000,000,000
Answer: 10 to the power of 23 is 100,000,000,000,000,000,000,000 or equals 1 followed by 23 zeros.
### What is 1E 30?
The “Allowable Increase” for this constraint is show as 1E+30. This is Excel’s way of showing infinity. This means that the right hand side can be increased any amount without changing the shadow price.
What is 10 by the power of 1?
Powers of 10
101=10 101=1
102=100 10-1=0.1
103=1000 10-2=0.01
104=10,000 10-3=0.001
## What does 10 to the power of 3 mean?
1000
Answer: The value of 10 raised to 3rd power i.e., 103 is 1000.
## What is the value of 2 Power 32?
Power Value
30 1,073,741,824
31 2,147,483,648
32 4,294,967,296
33 8,589,934,592
What is 10 3 to the power of 10?
Example: 10 3 = 10 × 10 × 10 = 1,000 . In words: 10 3 could be called “10 to the third power”, “10 to the power 3” or simply “10 cubed”
### What is a raised to the power calculator?
It is also known as raised to the power calculator. This calculator solves bases with both negative exponents and positive exponents. It also provides a step by step method with an accurate answer. What is an exponent?
### How do you convert negative powers of 10 to numbers?
Just remember for negative powers of 10: For negative powers of 10, move the decimal point to the left. So Negatives just go the other way. Example: What is 7.1 × 10 -3?
How many times can you multiply a number with powers of 10?
You can multiply any number by itself as many times as you want using this notation (see Exponents), but powers of 10 have a special use… Powers of 10 “Powers of 10” is a very useful way of writing down large or small numbers. Instead of having lots of zeros, you show how many powers of 10 will make that many zeros
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 11.4: Series and Their Notations
[ "article:topic", "common ratio", "Sigma Notation", "infinite series", "Annuity Problems", "authorname:openstax", "license:ccby", "showtoc:no" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
• Use summation notation.
• Use the formula for the sum of the first $$n$$ terms of an arithmetic series.
• Use the formula for the sum of the first $$n$$ terms of a geometric series.
• Use the formula for the sum of an infinite geometric series.
• Solve annuity problems.
### Using the Formula for Geometric Series
Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, $$r$$. We can write the sum of the first $$n$$ terms of a geometric series as
$$S_n=a_1+ra_1+r^2a_1+...+r^{n–1}a_1$$.
Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first $$n$$ terms of a geometric series. We will begin by multiplying both sides of the equation by $$r$$.
$$rS_n=ra_1+r^2a_1+r^3a_1+...+r^na_1$$
Next, we subtract this equation from the original equation.
$$S_n=a_1+ra_1+r^2a_1+...+r^{n–1}a_1$$
$$\underline{−rS_n=−(ra_1+r^2a_1+r^3a_1+...+r^na_1)}$$
$$(1−r)S_n=a_1−r^na_1$$
Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for $$S_n$$, divide both sides by $$(1−r)$$.
$$S_n=\dfrac{a_1(1−r^n)}{1−r}$$ $$r≠1$$
A General Note: FORMULA FOR THE SUM OF THE FIRST N TERMS OF A GEOMETRIC SERIES
A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first $$n$$ terms of a geometric sequence is represented as
$$S_n=\dfrac{a_1(1−r^n)}{1−r}$$ $$r≠1$$
How to: Given a geometric series, find the sum of the first $$n$$ terms.
1. Identify $$a_1$$, $$r$$, and $$n$$.
2. Substitute values for $$a_1$$, $$r$$, and $$n$$ into the formula $$S_n=\dfrac{a_1(1−r^n)}{1−r}$$.
3. Simplify to find $$S_n$$.
Example $$\PageIndex{4}$$: Finding the First n Terms of a Geometric Series
Use the formula to find the indicated partial sum of each geometric series.
1. $$S_{11}$$ for the series $$8 + -4 + 2 + …$$
2. $$\sum_{ 6}^{k=1}3⋅2k$$
Solution:
1. $$a_1=8$$, and we are given that $$n=11$$.
We can find $$r$$ by dividing the second term of the series by the first.
$$r=\dfrac{−4}{8}=−\dfrac{1}{2}$$
Substitute values for $$a_1$$, $$r$$, and $$n$$ into the formula and simplify.
$$S_n=\dfrac{a_1(1−r^n)}{1−r}$$
$$S_{11}=\dfrac{8(1−{(−\dfrac{1}{2})}^{11})}{1−(−\dfrac{1}{2})}≈5.336$$
2. Find $$a_1$$ by substituting $$k=1$$ into the given explicit formula.
$$a1=3⋅2^1=6$$
We can see from the given explicit formula that $$r=2$$. The upper limit of summation is $$6$$, so $$n=6$$.
Substitute values for $$a_1$$, $$r$$, and $$n$$ into the formula, and simplify.
$$S_n=\dfrac{a_1(1−r^n)}{1−r}$$
$$S_6=\dfrac{6(1−2^6)}{1−2}=378$$
Use the formula to find the indicated partial sum of each geometric series.
Exercise $$\PageIndex{4A}$$
$$S_{20}$$ for the series $$1,000 + 500 + 250 + …$$
Solution
$$≈2,000.00$$
Exercise $$\PageIndex{4B}$$
$$\sum_{k=1}^{8}3^k$$
Solution
$$9,840$$
Example $$\PageIndex{5}$$: Solving an Application Problem with a Geometric Series
At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years. Solution The problem can be represented by a geometric series with $$a_1=26,750$$; $$n=5$$; and $$r=1.016$$. Substitute values for $$a_1$$, $$r$$, and $$n$$ into the formula and simplify to find the total amount earned at the end of 5 years. $$S_n=\dfrac{a_1(1−r^n)}{1−r}$$ $$S_5=\dfrac{26,750(1−{1.016}^5)}{1−1.016}≈138,099.03$$ He will have earned a total of$138,099.03 by the end of 5 years.
Exercise $$\PageIndex{5}$$
At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years? Solution$275,513.31
### Using the Formula for the Sum of an Infinite Geometric Series
Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first $$n$$ terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is $$2+4+6+8+...$$
This series can also be written in summation notation as $$\sum_{k=1}^{\infty}2k$$, where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges.
#### Determining Whether the Sum of an Infinite Geometric Series is Defined
If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:
$$1+0.2+0.04+0.008+0.0016+...$$
The common ratio $$r = 0.2$$. As $$n$$ gets very large, the values of $$r^n$$ get very small and approach $$0$$. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with $$−1<r<1$$ approach 0; the sum of a geometric series is defined when $$−1<r<1$$.
A General Note: DETERMINING WHETHER THE SUM OF AN INFINITE GEOMETRIC SERIES IS DEFINED
The sum of an infinite series is defined if the series is geometric and $$−1<r<1$$.
How to: Given the first several terms of an infinite series, determine if the sum of the series exists.
1. Find the ratio of the second term to the first term.
2. Find the ratio of the third term to the second term.
3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
4. If a common ratio, $$r$$, was found in step 3, check to see if $$−1<r<1$$. If so, the sum is defined. If not, the sum is not defined.
Example $$\PageIndex{6}$$: Determining Whether the Sum of an Infinite Series is Defined
Determine whether the sum of each infinite series is defined.
1. $$12 + 8 + 4 + …$$
2. $$\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{3}+...$$
3. $$\sum_{k=1}^{\infty}27⋅{(\dfrac{1}{3})}^k$$
4. $$\sum_{k=1}^{\infty}5k$$
Solution:
1. The ratio of the second term to the first is $$\dfrac{2}{3}$$, which is not the same as the ratio of the third term to the second, $$\dfrac{1}{2}$$.The series is not geometric.
2. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of $$\dfrac{2}{3}$$. The sum of the infinite series is defined.
3. The given formula is exponential with a base of $$\dfrac{1}{3}$$; the series is geometric with a common ratio of $$\dfrac{1}{3}$$. The sum of the infinite series is defined.
4. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.
Determine whether the sum of the infinite series is defined.
Exercise $$\PageIndex{6A}$$
$$\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{9}{8}+...$$
Solution
The sum is defined. It is geometric.
Exercise $$\PageIndex{6B}$$
$$24+(−12)+6+(−3)+...$$
Solution
The sum of the infinite series is defined.
Exercise $$\PageIndex{6C}$$
$$\sum_{k=1}^{\infty}15⋅{(–0.3)}^k$$
Solution
The sum of the infinite series is defined.
### Finding Sums of Infinite Series
When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first $$n$$ terms of a geometric series.
$S_n=\dfrac{a_1(1−r^n)}{1−r}$
We will examine an infinite series with $$r=\dfrac{1}{2}$$. What happens to $$r^n$$ as $$n$$ increases?
$\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$
$\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}$
$\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}$
The value of $$r^n$$ decreases rapidly. What happens for greater values of $$n$$?
\begin{align} {\left(\dfrac{1}{2}\right)}^{10} &= \dfrac{1}{1,024} \\ {\left(\dfrac{1}{2}\right)}^{20} &= \dfrac{1}{1,048,576} \\ {\left(\dfrac{1}{2}\right)}^{30} &= \dfrac{1}{1,073,741,824} \end{align}
As $$n$$ gets very large, $$r^n$$ gets very small. We say that, as $$n$$ increases without bound, $$r^n$$ approaches 0. As $$r^n$$ approaches $$0$$, $$1$$, $$−r^n$$ approaches $$1$$. When this happens, the numerator approaches $$a_1$$. This give us a formula for the sum of an infinite geometric series.
FORMULA FOR THE SUM OF AN INFINITE GEOMETRIC SERIES
The formula for the sum of an infinite geometric series with $$−1<r<1$$ is
$S=\dfrac{a_1}{1−r}$
How to: Given an infinite geometric series, find its sum
1. Identify $$a_1$$ and $$r$$.
2. Confirm that $$–1<r<1$$.
3. Substitute values for $$a_1$$ and $$r$$ into the formula, $$S=\dfrac{a_1}{1−r}$$.
4. Simplify to find $$S$$.
Example $$\PageIndex{7A}$$: Finding the Sum of an Infinite Geometric Series
Find the sum, if it exists, for the following:
1. $$10+9+8+7+…$$
2. $$248.6+99.44+39.776 + …$$
3. $$\sum_{k=1}^{\infty}4,374⋅{(–\dfrac{1}{3})}^{k–1}$$
4. $$\sum_{k=1}^{\infty}\dfrac{1}{9}⋅{(\dfrac{4}{3})}^k$$
Solution:
1. There is not a constant ratio; the series is not geometric.
2. There is a constant ratio; the series is geometric. $$a_1=248.6$$ and $$r=\dfrac{99.44}{248.6}=0.4$$, so the sum exists. Substitute $$a_1=248.6$$ and $$r=0.4$$ into the formula and simplify to find the sum:
\begin{align} S&=\dfrac{a_1}{1−r} \\[5pt] &=\dfrac{248.6}{1−0.4} \\[5pt] &=414.\overline{3} \end{align}
3. The formula is exponential, so the series is geometric with $$r=–\dfrac{1}{3}$$. Find $$a_1$$ by substituting $$k=1$$ into the given explicit formula:
$$a_1=4,374⋅{(–\dfrac{1}{3})}^{1–1}=4,374$$
Substitute $$a_1=4,374$$ and $$r=−\dfrac{1}{3}$$ into the formula, and simplify to find the sum:
\begin{align} S&=\dfrac{a_1}{1−r} \\[5pt] &= \dfrac{4,374}{1−(−\dfrac{1}{3})} \\[5pt] &= 3,280.5 \end{align}
4. The formula is exponential, so the series is geometric, but $$r>1$$. The sum does not exist.
Example $$\PageIndex{7B}$$: Finding an Equivalent Fraction for a Repeating Decimal
Find an equivalent fraction for the repeating decimal $$0.\overline{3}$$
Solution
We notice the repeating decimal $$0.\overline{3}=0.333...$$ so we can rewrite the repeating decimal as a sum of terms.
$$0.\overline{3}=0.3+0.03+0.003+...$$
Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to $$0.1$$ in the second term, and the second term multiplied to $$0.1$$ in the third term.
Notice the pattern; we multiply each consecutive term by a common ratio of $$0.1$$ starting with the first term of $$0.3$$. So, substituting into our formula for an infinite geometric sum, we have
$$S_n=\dfrac{a_1}{1−r}=\dfrac{0.3}{1−0.1}=\dfrac{0.3}{0.9}=\dfrac{1}{3}$$.
Find the sum, if it exists.
Exercise $$\PageIndex{7A}$$
$$2+23+29+...$$
Solution
$$3$$
Exercise $$\PageIndex{7B}$$
$\sum_{k=1}^{\infty}0.76k+1$
Solution
The series is not geometric.
Exercise $$\PageIndex{7C}$$
$\sum_{k=1}^{\infty}{(−\dfrac{3}{8})}^k$
Solution
$−\dfrac{3}{11}$
### Solving Annuity Problems
At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with $$a_1=50$$ and $$r=100.5%=1.005$$. After the first deposit, the value of the annuity will be $$50$$. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after nn deposits using the formula for the sum of the first nn terms of a geometric series. In 6 years, there are 72 months, so $$n=72$$. We can substitute $$a_1=50$$, $$r=1.005$$, and $$n=72$$ into the formula, and simplify to find the value of the annuity after 6 years. $S_{72}=\dfrac{50(1−{1.005}^{72})}{1−1.005}≈4,320.44$ After the last deposit, the couple will have a total of $$4,320.44$$ in the account. Notice, the couple made $$72$$ payments of$50 each for a total of $$72(50) = 3,600$$. This means that because of the annuity, the couple earned $$720.44$$ interest in their college fund.
How to: Given an initial deposit and an interest rate, find the value of an annuity.
1. Determine $$a_1$$, the value of the initial deposit.
2. Determine $$n$$, the number of deposits.
3. Determine $$r$$.
• Divide the annual interest rate by the number of times per year that interest is compounded.
• Add 1 to this amount to find $$r$$.
4. Substitute values for $$a_1$$, $$r$$, and $$n$$ into the formula for the sum of the first nn terms of a geometric series, $$S_n=\dfrac{a_1(1–r^n)}{1–r}$$.
5. Simplify to find $$S_n$$, the value of the annuity after $$n$$ deposits.
Example $$\PageIndex{8}$$: Solving an Annuity Problem
A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit? Solution: The value of the initial deposit is$100, so $$a_1=100$$. A total of 120 monthly deposits are made in the 10 years, so $$n=120$$. To find $$r$$, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.
$r=1+\dfrac{0.09}{12}=1.0075$
Substitute $$a_1=100$$, $$r=1.0075$$, and $$n=120$$ into the formula for the sum of the first $$n$$ terms of a geometric series, and simplify to find the value of the annuity.
$S_{120}=\dfrac{100(1−{1.0075}^{120})}{1−1.0075}≈19,351.43$
So the account has $19,351.43 after the last deposit is made. Exercise $$\PageIndex{8}$$ At the beginning of each month,$200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?
Solution:
\$92,408.18
Media
Access these online resources for additional instruction and practice with series.
sum of the first $$n$$ terms of an arithmetic series $$S_n=\dfrac{n(a_1+a_n)}{2}$$ sum of the first $$n$$ terms of a geometric series $$S_n=\dfrac{a_1(1−r^n)}{1−r}$$⋅ $$r≠1$$ sum of an infinite geometric series with $$–1 ### Key Concepts • The sum of the terms in a sequence is called a series. • A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See Example. • The sum of the terms in an arithmetic sequence is called an arithmetic series. • The sum of the first \(n$$ terms of an arithmetic series can be found using a formula. See Example and Example.
• The sum of the terms in a geometric sequence is called a geometric series.
• The sum of the first $$n$$ terms of a geometric series can be found using a formula. See Example and Example.
• The sum of an infinite series exists if the series is geometric with $$–1<r<1$$.
• If the sum of an infinite series exists, it can be found using a formula. See ExampleExampleand Example.
• An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See Example.
### Contributors
• Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (formerly of Santa Ana College). This content produced by OpenStax and is licensed under a Creative Commons Attribution License 4.0 license.
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# Question Video: Finding the Vector Form of the Equation of the Plane given Its Normal Vector Equation Mathematics
Find the vector form of the equation of the plane that has normal vector 𝐧 = 𝐢 + 𝐣 + 𝐤 and contains the point (2, 6, 6).
02:35
### Video Transcript
Find the vector form of the equation of the plane that has normal vector 𝐧 equals 𝐢 hat plus 𝐣 hat plus 𝐤 hat and contains the point two, six, six.
Okay, so in this example, we have a plane. We’ll say this is it. And we’re told that relative to some set of coordinate axes, there’s a vector normal or perpendicular to the plane. And if we write that vector in vector form, we see that it has components of one, one, and one in the 𝑥-, 𝑦-, and 𝑧-directions, respectively. Along with this, we’re told that the plane contains a point, we’ll call it 𝑃 zero, with coordinates two, six, six. And knowing all this, we want to solve for the vector form of the plane’s equation.
To write the plane’s equation that way, we’ll want to define a vector that lies in the plane so that if we take the dot product of that vector and the normal vector 𝐧, we get zero. Here’s how we can go about doing that. First, let’s define a vector from our origin to the point 𝑃 zero. We’ll call the vector 𝐫 zero. And since it comes from the origin, it must have components two, six, six. And next, let’s do this. Let’s pick a point at random in our plane, we’ll call it 𝑃, and we’ll say this point has coordinates 𝑥, 𝑦, 𝑧.
We don’t specify what these values are, but nonetheless this point will help us because now we can draw a vector from our origin to this point 𝑃, call that vector 𝐫, which we see has components 𝑥, 𝑦, 𝑧. And then if we subtract the vector 𝐫 zero from 𝐫, and this was the whole purpose of defining 𝐫 in the first place, then we get a vector shown here, which lies in the plane, which means that this vector is indeed perpendicular to the normal vector 𝐧. And that means if we take the dot product of 𝐧 and 𝐫 minus 𝐫 zero, the result we’ll get is zero.
Another way to write this is 𝐧 dot 𝐫 is equal to 𝐧 dot 𝐫 zero. And we can now substitute in the known values for the normal vector 𝐧 and the vector 𝐫 zero. We’ll do that down here, where we see that the vector one, one, one dotted with 𝐫 is equal to one, one, one dotted with two, six, six. We see the vector 𝐫 is made up of the components 𝑥, 𝑦, 𝑧. But since we don’t know those more specifically, we’ll leave this left-hand side as it is. On the right-hand side, though, we can compute this dot product by multiplying the respective components together, giving us two plus six plus six.
And since these numbers add up to 14, we can write a simplified form of our equation like this. This is the vector form of the equation of our plane.
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# Question Video: Evaluating Combinations Using Their Properties Mathematics • 12th Grade
Evaluate ยนโตโท๐ถโโ
โ + โตยน๐ถโ.
03:16
### Video Transcript
Evaluate 157 choose 157 plus 51 choose one.
Now, to help us solve this problem, we actually have a general form. And that is ๐ choose ๐ is equal to ๐ factorial over ๐ factorial ๐ minus ๐ factorial. And we can actually use this to help us find the value of our terms. Well, when we look to evaluate 157 choose 157 plus 51 choose one, what we need to do is actually deal with each of these terms separately. So weโll just start with 157 choose 157. And this is actually in the form ๐ choose ๐ because actually our ๐ value is actually the same as our ๐ value.
And thereโs actually a special relationship that tells us that if we have ๐ choose ๐, then this is equal to one. So therefore, we can actually say that our 157 choose 157 is also gonna be equal to one. But what Iโm gonna do is actually prove this by showing how itโll work if we use the general form. So if we have 157 choose 157, then this is gonna be equal to 157 factorial, because thatโs our ๐, over ๐ factorial, which is 157 factorial, then multiplied by ๐ minus ๐, so 157 minus 157, factorial. And this is gonna be equal to 157 factorial over 157 factorial multiplied by zero factorial. Well one of the rules we have is that zero factorial is equal to one. So therefore, weโre gonna have this all equal to 157 factorial over 157 factorial which is equal to one. And so that proves what we said at the beginning which was ๐ choose ๐ โ or in this case, 157 choose 157 โ is equal to one.
Okay, brilliant. Now, letโs move on and evaluate our next term. So if we look at our second term which is 51 choose one, then yet again, this actually has a special form because if we have ๐ choose one, then this is always equal to ๐. But, weโre gonna use the general formula just to show this again. So we get 51 choose one is equal to 51 factorial over one factorial multiplied by 51 minus one factorial which is gonna be equal to 51 factorial over one factorial multiplied by 50 factorial. Well, we also know that one factorial is equal to one. So therefore, weโre gonna have 51 factorial over 50 factorial.
Well, if we think about what this actually means, itโll be 51 times 50 times 49 times 48 all divided by 50 times 49 times 48, et cetera. So therefore, if we actually divide through by our 50 factorial, weโre just left with 51. So therefore, we can say that 51 choose one gives us an answer of 51. And great, this actually fits what we thought had happened at the beginning because we said that ๐ choose one is equal to ๐. Well, ๐ was 51 in our case. So 51 choose one is equal to 51. Okay, so now what we need to do is actually evaluate 157 choose 157 plus 51 choose one. So itโs gonna be equal to one plus 51. So therefore, we can say that if we evaluate 157 choose 157 plus 51 choose one we get 52.
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# Find two numbers whose sum is 27 and product is 182.
Last updated date: 04th Aug 2024
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Hint: We will form two equations with two unknowns using the given information. We will obtain a quadratic equation in one variable by substitution. Then we will solve the quadratic equation using the quadratic formula. The solution of the quadratic formula will be one of the two required numbers. Using the solution of the quadratic equation, we will be able to find the other number.
Let the two numbers be $x$ and $y$. According to the given information, we have the following two equations,
$x+y=27$....(i)
$x\times y=182$....(ii)
From equation (i), we can write $y=27-x$. Substituting this value in equation (ii), we get
\begin{align} & x\times \left( 27-x \right)=182 \\ & \therefore 27x-{{x}^{2}}=182 \\ \end{align}
Rearranging the above equation, we get the following quadratic equation,
${{x}^{2}}-27x+182=0$
The general quadratic equation is $a{{x}^{2}}+bx+c=0$ and the quadratic formula for solving this equation is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We will compare the above quadratic equation with the general quadratic equation, so we have $a=1,\text{ }b=-27,\text{ }c=182$. We will substitute these values in the quadratic formula to find the values of $x$ in the following manner,
$x=\dfrac{-\left( -27 \right)\pm \sqrt{{{27}^{2}}-4\times 1\times 182}}{2\times 1}$
Simplifying the above equation, we get
\begin{align} & x=\dfrac{27\pm \sqrt{729-728}}{2} \\ &\Rightarrow x =\dfrac{27\pm 1}{2} \end{align}
Therefore, we get $x=\dfrac{28}{2}=14$ or $x=\dfrac{26}{2}=13$.
Now, substituting these values in equation (i), we get
for $x=14$, we have $y=27-14=13$ and for $x=13$, we have $y=27-13=14$
Therefore, the two numbers whose sum is 27 and product is 182 are 13 and 14.
Note: There is another method to solve this question which involves solving a quadratic equation with a different method. If we are familiar with the method of solving a quadratic equation by factorization, then we can reach the answer in fewer steps. The substitutions need to be done explicitly to avoid confusion.
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### Directed Line Segment, Location, Coordinate Plane, Partitions, Point M, Ratio
The key terms of this Maths chapter include Directed Line Segment, Ratio, Fountain, Coordinate Plane, Location, Endpoints, Partitions, Point M, Coordinates, Nearest Tenth, X- and Y- Coordinates of Point, Point P, Number Line, Midpoint – Mathematics Quiz
Explain why partitioning a directed line segment into a ratio of 1:2 is not the same as finding half the length of the directed line segment.
A ratio of 1:2 means that there are 3 parts in total. One part will be before the desired point, and 2 parts will be after the desired point. This is the same as finding the point that is 1/3 the length. Half the length of the segment would mean there would only be two pieces, each of equal size.
Fernando and Maria are at opposite sides of a park. They plan to meet later at a fountain that is located somewhere between them. Maria will arrive at the fountain first, before Fernando even leaves his starting location. The ratio of Maria’s starting location to the fountain to Fernando’s starting place is 5:3. What is the location of the fountain? Round to the nearest tenth, if necessary.
(-1. 0.5)
In the coordinate plane, the directed line segment from K to N has endpoints at K(-6, -2) and N(8, 3). Point L partitions the directed line segment from K to N in a ratio of 1:2. Point M partitions the directed line segment from L to N in a ratio of 3:1. What are the coordinates of point M? Round to the nearest tenth, if necessary.
##### (5.7, 2.2)
What are the x- and y- coordinates of point E, which partitions the directed line segment from A to B into a ratio of 1:2?
(0, 1)
What is the location of point G, which partitions the directed line segment from F to D into an 8:5 ratio?
1
If point P is 9/11 of the distance from M to N, then point P partitions the directed line segment from M to N into a ratio.
9:2
What are the x- and y- coordinates of point P on the directed line segment from A to B such that P is the length of the line segment from A to B?
##### (-4, -5)
What is the location of point F, which partitions the directed line segment from D to E into a 5:6 ratio?
1/11
A student wants to find point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4. Point A is at -6 and point B is at 2. The student’s work is shown.
Analyze the student’s work. Is the answer correct? Explain.
No, the student should have added 3 + 4 to get the total number of sections, and used the fraction 3/7 instead of 3/4.
What are the coordinates of point P on the directed line segment from R to Q such that P is the length of the line segment from R to Q? Round to the nearest tenth, if necessary.
(-3.5, 2.3)
Ari is swimming a 25-meter race. After swimming 6 meters, she catches up to Amanda in a ratio of 7:3 from the 6-meter mark. At what meter mark does Ari catch up to Amanda? Round to the nearest tenth, if necessary.
Ari catches up to Amanda at ___ meters.
##### 19.3
What are the x- and y-coordinates of point C, which partitions the directed line segment from A to B into the ratio 3:10? Round to the nearest tenth, if necessary.
x = ____
y = ____
-2.6
5.2
On a number line, the directed line segment from Q to S has endpoints Q at -8 and S at 12. Point R partitions the directed line segment from Q to S in a 4:1 ratio.
Which expression correctly uses the formula to find the location of point R?
B: (4/4 + 1)(12 – (-8)) + (-8)
Genevieve is cutting a 60-inch piece of ribbon into a ratio of 2:3. Since 2 inches are frayed at one end of the ribbon, she will need to start 2 inches in. This is indicated as 2 on the number line.
Where will her cut be located? Round to the nearest tenth.
She will make the cut ____ inches from the end of the ribbon.
25.2
The midpoint of a line segment partitions the line segment into a ratio of
1:1
What is the y-coordinate of the point that divides the directed line segment from J to K into a ratio of 5:1?
0
What are the x- and y- coordinates of point P on the directed line segment from K to J such that P is the length of the line segment from K to J?
(40, 96)
What are the x- and y-coordinates of point C, which partitions the directed line segment from A to B into the ratio 5:8? Round to the nearest tenth, if necessary.
(1.2, -4.7)
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# Factor Theorem
In algebraic math, the factor theorem is a theorem that establishes a relationship between factors and zeros of a polynomial. The Factor theorem is a unique case consideration of the polynomial remainder theorem. Thus the factor theorem states that a polynomial has a factor if and only if: The polynomial x – M is a factor of the polynomial f(x) if and only if f (M) = 0. Factor theorem is frequently linked with the remainder theorem, therefore do not confuse both. However, to unlock the functionality of the actor theorem, you need to explore through the remainder theorem.
What is a Factor?
You now already know about the remainder theorem. The other most crucial thing we must understand through our learning for the factor theorem is what really a “factor” is. Knowing exactly what a “factor” is not only crucial to better understand the factor theorem, in fact, to all of mathematics concepts. It is a term you will hear time and again as you head forward with your studies.
With respect to division, a factor is an expression that, when a further expression is divided by this factor, the remainder is equal to zero (0). In its simplest form, take into account the following: 5 is a factor of 20 because, when we divide 20 by 5, we obtain the whole number 4 and no remainder. Likewise, 3 is not a factor of 20 because, when we 20 divided by 3, we have 6.67, which is not a whole number.
Use of Factor Theorem to Find the Factors of a Polynomial
In practical terms, the Factor Theorem is applied to factor the polynomials “completely”. In other words, any time you do the division by a number (being a prospective root of the polynomial) and obtain a remainder as zero (0) in the synthetic division, this indicates that the number is surely a root, and hence “x minus (-) the number” is a factor.
Now Before getting to know the Factor Theorem in depth and what it means, it is imperative that you completely understand the Remainder Theorem and what factors are first.
Remainder Theorem
As mentioned above, the remainder theorem and factor theorem are intricately related concepts in algebra. The reality is the former can’t exist without the latter and vice-e-versa. That being said, let’s see what the Remainder Theorem is.
By the principle of Remainder Theorem:
• In case you divide a polynomial f(x) by (x – M), the remainder of that division is equal to f(c).
Usefulness of Remainder Theorem
The remainder theorem is particularly useful because it significantly decreases the amount of work and calculation that we would do to solve such types of mathematical problems/equations.
In absence of this theorem, we would have to face the complexity of using long division and/or synthetic division in order to have a solution for the remainder, which is both troublesome and time consuming. Moreover, an evaluation on the theories behind the remainder theorem, in addition to the visual proof of the theorem, is also quite useful.
Fun Facts
• As per the Chaldean Numerology and the Pythagorean Numerology, the numerical value of factor theorem in is: 3
• Factor theorem assures that a factor (x – M) for each root is r.
• The factor theorem does not state there is only one such factor for each root. For the fact, it is quite easy to create polynomials with arbitrary repetitions of the same root & the same factor
• Factor theorem is useful as it postulates that factoring a polynomial corresponds to finding roots.
Solved Examples
Problem 1:
Example: For a curve that crosses the x-axis at 3 points, of which one is at 2. What is the factor of 2x3−x2−7x+2?
Solution1:
The polynomial for the equation is degree 3, and could be all easy to solve. So let us arrange it first:
We can easily plot:
F (2) = 2 (2)3− (2)2− 7 (2) +2
This brings us:-
= 16−4−14+2
=12 (-14) +2
= -2 + 2
= 0
Thus! F (2) =0, so we have found a factor and a root
Therefore, (x-2) should be a factor of 2x3−x2−7x+2
Problem 2:
Find out whether x + 1 is a factor of the below given polynomial.
i). 3 x 4 + x 3 – x2 + 3x + 2
Solution 2:
Using the factor theorem,
Let f(x) = 3×4 + x3 – x2 + 3x + 2
That brings to us:-
f (–1) = 3 (–1) 4 + (–1) 3 – (–1)2 +3 (–1) + 2
= 3(1) + (–1) – 1 – 3 + 2 = 0
Hence, we conclude that (x + 1) is a factor of f (x)
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Hence, zero of polynomial q(x) is 7/2. (iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and other is a quadratic polynomial. (ii) Every polynomial is a Binomial. (D) Not defined = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) (b) Given, p(x) = 2x+5 (ii) Given, polynomial isp(y) = (y+2)(y-2) =(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x) ∴ x – 4 = 0 ⇒ x = 4 Thinking Process Question 19: Solution: Question 26: (ii) The given polynomial is 4 - y². Because the sum of any two polynomials of same degree has not always same degree. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (iv) Given polynomial h(y) = 2 y Zero of the zero polynomial is Since, remainder ≠ 0, then p(x) is not a multiple of g(x). Solution: then (5)2 = a2 + b2+ c2 + 2(10) Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39. In this class, Vivek Patriya will discuss NCERT Exemplar Polynomials Class 9 (Part - 2) .The class will be helpful for the aspirants of CBSE 9th. (iv) We have, (2x – 5) (2x² – 3x + 1) Solution: Question 23: Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3. Write the coefficient of x2 in each of the following (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. Given, polynomial is p(x) = (x – 2)2 – (x + 2)2 (iii) -4/5 is a zero of 4 – 5y (b) 5 Hence, the remainder is 50. (ii) Polynomial 3x3 is a cublic polynomial, because maximum exponent of x is 3. We know that (iii) Degree of polynomial x3 – 9x + 3x5 is 5, because the maximum exponent of x is 5. Solution: (c) 2 ⇒ y(y + 3) – 2(y + 3) = 0 (a) 4 = x(x-2)-1 (x-2)= (x-1)(x-2) Using suitable identity, evaluate the following: We have, p(x) = x4 – 2x3 + 3x2 – ox + 3a – 7 = (3x)2 + (2y)2 + (-4z)2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x) NCERT 9th Class Exemplar Problems 2021 in Pdf format are Available this web page to Download, NCERT Exemplar Problems from Class 9 for the Academic year 2021 and Exemplar also Available to Maths Subject with the Answers, NCERT Developed the Exemplar Books 2021 for 9th Class Students Education Purpose. Question 4. So, x = -1 is zero of x3 + x2 + x+1 (d) 5/2 p(2) = 4(2)2 + 2 – 2 = 16 ≠ 0 ∴ p(-1) = (-1)³ – 2(-1)² – 4(-1) -1 = x2 (x4 + x2 + 1) – 1(x4 + x2 + 1) (ii) x3-6x2+11x-6 a = 3/2. Question 19. Question 3. At x = -3, p(-3)= 3(—3)3 – 4(-3)2 + 7(-3)- 5 Show that, (a) x3 + x2 – x +1 (b) x3 + x2 + x+1 = 10-4-3= 10-7= 3 (iii) 9992 Solution: = x6 + x4 + x2 – x4 – x2 – 1 = x6 – 1, Question 34. Since p(x) is divided by x + 1, then remainder is p(-1). [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] Hence, we cannot exactly determine the degree of variable. Simplify (2x- 5y)3 – (2x+ 5y)3. = -2[r(r + 7) -6(r + 7)] Question 17. (-1)3 + (-1)2 + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true. Now, this is divided by x + 2, then remainder is p(-2). Question 6: Put 4 – 5y = 0 ⇒ y = 4/5 Thinking Process With the help of it, candidates can prepare well for the examination. Question 8: Solution: (a) x2 + y2 + 2 xy Question 4: p(x) = x – 4 If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p ( ½). Solution: Question 15: NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. => 2x-1 = 0 and x+4 = 0 When we divide p2(z) by z-3 then we get the remainder p2(3). Hence, zero of polynomial is 4. iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = (1000)2 + (1)2 – 2(1000)(1) Hence, 0 of x2-3x+2 are land 2. NCERT Class 9 Maths Unit 2 is for Polynomials. Factorise: Solution: Hence, zero of polynomial is => x = ½ and x = -4 Factorise the following (vi) Not polynomial ⇒ a2 + b2 + c2 = 25 – 20 (ii) y3 – 5y We know that, Solution: (b) 2x ⇒ -8 – 8m + 16 = 0 (ii) the coefficient of x3 (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. (c) 1 Question 12: Find p(0), p( 1) and p(-2) for the following polynomials a3+b3 + c3 = 3abc, Exercise 2.2: Short Answer Type Questions. p(x) = 10x – 4x2 – 3 Question 2: Let p(x) = 8x4 + 4x3 – 16x2 + 10x + m p1(3) = p2(3) Solution: Question 2: = (- 5x + 4y + 2z)2 (i) Polynomial (a)-6 NCERT Exemplar Problems Class 9 Maths Solutions are being updated for new academic session 2020-2021. HOTS, exemplar, and hard questions in polynomials. Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively. Solution: For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0 = (x – 2) (2x2 + x – 15) Write the coefficient of x² in each of the following = (100)2 + (1 + 2)100 + (1)(2) Question 10: BeTrained.in has solved each questions of NCERT Exemplar very thoroughly to help the students in solving any question from the book with a team of well experianced subject matter experts. = 3(3x2 – 3x – x + 1) (v) True For zero of polynomial, put p(x) = 0 Solution: = 8x³ + 2xy2 + 18xz2 + 4x2y + 6xyz – 12x2z – 4x2y – y3 – 9yz2 – 2xy2 – 3y2z + 6xyz + 12x2z + 3y2z + 27z3 + 6xyz + 9yz2 – 18xz2 (i) Given, polynomial is p(-1)=0 Hence, one of the zeroes of the polynomial p(x) is ½. NCERT Class 9 New Books for Maths Chapter 2 Polynomials are given below. = (a + b + c – a)[(a + b + c)2 + a2 + (a + b + c)a] – [(b + c) (b2 + c2– be)] Solution: ⇒ x = 0 L.H.S. Let p(x) = x3 -2mx2 +16 (i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3 = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 (ii) 6x2 + 7x – 3 lf a+b+c=9 and ab + bc + ca = 26, find a2 + b2 + c2. ⇒ a2 + b2 + c2 = 5 … (i) = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27. [using identity, (a + b)2 = a2 + b2 + 2 ab)] NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) (Hindi Medium) Ex 2.1. (d) Now, (25x2 -1) + (1 + 5x)2 ⇒ (2x)(-4) = 0 Solution: = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) (i) Since, x + y + 4 = 0, then Question 13: (i), we get (ii) 2√2a3 +8b3 -27c3 +18√2abc NCERT Books for Class 9 Maths Chapter 2 Polynomials can be of extreme use for students to understand the concepts in a simple way.Class 9th Maths NCERT Books PDF Provided will help you during your preparation for both school … Solution: (b) 9 (b) 1 Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. (b) 3x2 + 5 [polynomial and also a binomial] Thinking Process (a) -3 (b) 4 (c) 2 (d)-2 Solution: (i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1 = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3, Question 1: (vii) y³ – y (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3 [∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2] Question 5: The highest power of … Let p1(z) = az3 +4z2 + 3z-4 and p2(z) = z3-4z + o (iii) The example of trinomial of degree 2 is x2 – 4x + 3. (a) Let p (x) = 5x – 4x2 + 3 …(i) = (x – 1) [3x(x + 1) – 1(x + 1)] Solution: we get p(0) = 10(0) – 4(0)2 – 3 = 0 – 0 – 3 = -3 (b) Now, 4x2 + 8x + 3= 4x2 + 6x + 2x + 3 [by splitting middle term] (d) The degree of zero polynomial is not defined, because in zero polynomial, the coefficient of any variable is zero i.e., Ox2 or Ox5,etc. (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 (ii) The coefficient of x3 in given polynomial is 1/5. Question 8. ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 Here are all questions are solved with a full explanation and available for free to download. = [(a + b + c)3 – a3] – (b3+ c3) Give an example of a polynomial, which is = -2 (5y)3 – 30xy(2x – 5y + 2x + 5y) Hence, the value of m is 1. Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number. For zeroes of polynomial, put p(x) = 0 = 4x³ – 6x² + 2x – 10x² + 15x – 5 Because one of the exponents of the variable x is -1, which is not a whole number. NCERT Class 9 Maths Solutions develop logical thinking skills so that students cable to solve all the sums once the concept is clear. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). Question 9: (v) 3 ⇒ t = 0 and t = 2 (iii) Given, polynomial is q(x) = 2x -7 For zero of polynomial, put q(x) = 2x-7 = 0 ⇒ (-2)3 – 2m(-2)2 + 16 = 0 Which one of the following is a polynomial? ∴ Coefficient of x² in 3x² – 7x + 4 is 3. Solution: (iii) (-x + 2y-3z)2 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x). e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial. ∴ p(-3) = -143 Solution: Question 28: Question 6. We have prepared chapter wise solutions for all characters are given below. Question 6: Find the value of If the polynomials az3 +4z2 + 3z-4 and z3-4z + o leave the same remainder when divided by z – 3, find the value of a. NCERT Exemplar Solutions in Maths Classes VIII, IX and X: Get NCERT Exemplar Problem Solutions in Mathematics for classes 8 th, 9 th and 10 th for CBSE and other Students. Then it is a polynomial can have any number of zeroes Maths book covers basics fundamentals... Practice Polynomials questions and become a master of concepts Class 6, 7, 8 9... 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# Question Video: Simplifying Numerical Expressions Using the Properties of Square Roots Mathematics • 10th Grade
Simplify 12√5 + 3√5.
01:17
### Video Transcript
Simplify 12 square root five plus three square root five.
In order to add square roots, the number inside of the root must be exactly the same as the other one inside of the square root. And we have that. They’re both square root five. So we will keep that square root five and we will add the numbers in front of them.
So we will have 12 plus three and then square root five. And 12 plus three is equal to 15. This means that our answer will be 15 square root five.
Sometimes, it’s useful to think when adding and subtracting square roots is to treat them like variables. Imagine this to be 12𝑥 plus three 𝑥. So the square root fives represent like variables. They’re both 𝑥s. They’re both square root fives. So 12𝑥 plus three 𝑥 is 15𝑥. We keep the 𝑥 just like we keep the square root five as long as they’re the same.
So if one of the square root fives haven’t had been different, we wouldn’t have been able to combine them. Just like if this had been a three 𝑦, we wouldn’t be able to combine 12𝑥 and three 𝑦. We would simply have to leave them apart.
So once again, our final answer will be 15 square root five.
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# Difference between revisions of "2005 AIME I Problems/Problem 4"
## Problem
The director of a marching band asks the band members to line up in rows of four, but one is left over. Then she tries to line them up in rows of six, but three are left over. Finally, she tries to line them up in rows of seven, but four are left over.
How many members are there?
## Solution
### Solution 1
If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is $\boxed{294}$.
### Solution 2
Define the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$.
### Solution 3
The number of members is $m^2+5=n(n+7)$ for some $n$ and $m$. Multiply both sides by $4$ and complete the square to get $4m^2+69=(2n+7)^2$. Thus, we have $69=((2n+7)+2m)((2n+7)-2m)$. Since we want to maximize $n$, set the first factor equal to $69$ and the second equal to $1$. Solving gives $n=14$, so the answer is $14\cdot21=294$.
2005 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# Similar Figures
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent , and the ratios of the lengths of their corresponding sides are equal.
This common ratio is called the scale factor .
The symbol $\sim$ is used to indicate similarity.
Example 1:
In the figure below, pentagon $ABCDE\sim$ pentagon $VWXYZ$ .
(Note that the order in which you write the vertices matters; for instance, pentagon $ABCDE$ is not similar to pentagon $VZYXW$ .)
Example 2:
The two cylinders are similar. Find the scale factor and the radius of the second cylinder.
The height of the cylinder on the right is $\frac{1}{3}$ the height of the cylinder on the left. So, the scale factor is $\frac{1}{3}$ .
To get the radius of the smaller cylinder, divide $1.8$ by $3$ .
$1.8÷3=0.6$
So, the radius of the smaller cylinder is $0.6$ cm.
Note that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations , reflections , translations , and dilations .
Example 3:
In the figure above, the hexagon ${A}_{1}{B}_{1}{C}_{1}{D}_{1}{E}_{1}{F}_{1}$ is flipped horizontally to get ${A}_{2}{B}_{2}{C}_{2}{D}_{2}{E}_{2}{F}_{2}$ .
Then hexagon ${A}_{2}{B}_{2}{C}_{2}{D}_{2}{E}_{2}{F}_{2}$ is translated to get ${A}_{3}{B}_{3}{C}_{3}{D}_{3}{E}_{3}{F}_{3}$ .
Hexagon ${A}_{3}{B}_{3}{C}_{3}{D}_{3}{E}_{3}{F}_{3}$ is dilated by a scale factor of $\frac{1}{2}$ to get ${A}_{4}{B}_{4}{C}_{4}{D}_{4}{E}_{4}F$ .
Note that
$\begin{array}{l}{A}_{1}{B}_{1}{C}_{1}{D}_{1}{E}_{1}{F}_{1}\sim {A}_{2}{B}_{2}{C}_{2}{D}_{2}{E}_{2}{F}_{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sim {A}_{3}{B}_{3}{C}_{3}{D}_{3}{E}_{3}{F}_{3}\sim {A}_{4}{B}_{4}{C}_{4}{D}_{4}{E}_{4}{F}_{4}\end{array}$ .
That is, all four hexagons are similar. (In fact, the first three are congruent.)
Example 4:
Consider pentagon $PQRST$ on a coordinate plane.
A rotation by $180°$ about the origin takes the pentagon to $P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}$ .
Now, a dilation about the origin by a scale factor $2$ takes the pentagon $P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}$ to $P\text{'}\text{'}Q\text{'}\text{'}R\text{'}\text{'}S\text{'}\text{'}T\text{'}\text{'}$ .
Note that $PQRST\sim P\text{'}Q\text{'}R\text{'}S\text{'}T\text{'}\sim P\text{'}\text{'}Q\text{'}\text{'}R\text{'}\text{'}S\text{'}\text{'}T\text{'}\text{'}$ . That is, all three pentagons are similar. (And the first two are congruent.)
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# 30-60-90 Triangle: 4 Facts to Remember for Your Next Test
By: Mitch Ryan |
Anyone who's ever faced mind-numbing trigonometry problems and practice questions will be familiar with the Pythagorean Theorem and its square root principles in the 30-60-90 triangle. This special triangle has several short-cut rules to help new mathematicians find interior angles and linear lengths quickly.
All triangles have interior angle sums equal to 180 degrees, but special triangles (like right triangles and equilateral triangles) are the easiest to solve in geometry practice problems and other homework assignments because there is a consistent relationship between angles with the same ratio each time.
Let's look at the hypothetical triangle PQR to illustrate some of the key characteristics of a 30-60-90 triangle.
Contents
## 1. A 30-60-90 Triangle Is Always a Right Triangle
PQR is a special right triangle. All the sides and angles of this right-angled triangle (90 degrees) are uneven, unlike an equilateral triangle with equal angles or an isosceles triangle with two even legs.
However, by bisecting an isosceles triangle at the angle where the two identical sides meet, you can create two identical right triangles.
## 2. The Hypotenuse Is Twice the Length of the Smallest Side
If the problem provides one of two side lengths (the longest side, aka the hypotenuse, or the shortest side) you can easily find the other side length.
If problem states the length of the hypotenuse, you can divide that value by 2 to find the length of the shortest leg. If the problem provides the short leg dimension, you can multiply that measurement by 2 to find the hypotenuse.
## 3. The Smallest Angle Is 30 Degrees
The smallest angle of a 30-60-90 triangle is always 30 degrees and is adjacent to the longest side (hypotenuse). The other long leg leading from 30 degrees to the right angle equals the √3 times the length of the shortest side.
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# How do you solve 8x^2 - 6x + 1 = 0 using the quadratic formula?
Jan 8, 2017
First, we "play with multipliers of $8$ (1x8, 2x4, 4x2 8x1) and 1 to factor and then solve each term for $0$:
$8 {x}^{2} - 6 x + 1 = 0$
$\left(4 x - 1\right) \left(2 x - 1\right) = 0$
Now we solve each term of the factored quadratic for $0$:
Solution 1)
$4 x - 1 = 0$
$4 x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$
$4 x - 0 = 1$
$4 x = 1$
$\frac{4 x}{\textcolor{red}{4}} = \frac{1}{\textcolor{red}{4}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\cancel{\textcolor{red}{4}}} = \frac{1}{4}$
$x = \frac{1}{4}$
Solution 2)
$2 x - 1 = 0$
$2 x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$
$2 x - 0 = 1$
$2 x = 1$
$\frac{2 x}{\textcolor{red}{2}} = \frac{1}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = \frac{1}{2}$
$x = \frac{1}{2}$
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# How do I solve 0.001 = "10,000,000"/y?
Apr 11, 2017
$y$ = 10,000,000,000
#### Explanation:
Since y is at the bottom of the fraction, it can switch place with the 0.001.
$y$ = $\frac{10 , 000 , 000}{0.001}$
$y$ = 10,000,000,000
Apr 11, 2017
Convert the numbers to powers of 10, and use the exponent laws: ${10}^{a} / {10}^{b} = {10}^{a - b} .$
$y = {10}^{10} = \text{10,000,000,000} .$
#### Explanation:
Let's rewrite this question using powers of 10:
$0.001 = \frac{\text{10,000,000}}{y}$
${10}^{\text{-3}} = \frac{{10}^{7}}{y}$
When we solve this for $y$, we get
$y = \frac{{10}^{7}}{10} ^ \text{-3}$
The exponent rules tell us that ${10}^{a} / {10}^{b}$ is equal to ${10}^{a - b}$. Using this, we get
y=(10^7)/10^"-3"=10^(7-("-3"))
$\textcolor{w h i t e}{y = \frac{{10}^{7}}{10} ^ \text{-3}} = {10}^{7 + 3}$
color(white)(y=(10^7)/10^"-3")=10^10" "="10,000,000,000"
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# Vector Calculus
```16
Vector Calculus
16.1
Ve tor Fields
This chapter is concerned with applying calculus in the context of vector fields. A
two-dimensional vector field is a function f that maps each point (x, y) in R2 to a twodimensional vector hu, vi, and similarly a three-dimensional vector field maps (x, y, z) to
hu, v, wi. Since a vector has no position, we typically indicate a vector field in graphical
form by placing the vector f (x, y) with its tail at (x, y). Figure 16.1.1 shows a represenp
p
tation of the vector field f (x, y) = h−x/ x2 + y 2 + 4, y/ x2 + y 2 + 4i. For such a graph
to be readable, the vectors must be fairly short, which is accomplished by using a different
scale for the vectors than for the axes. Such graphs are thus useful for understanding the
sizes of the vectors relative to each other but not their absolute size.
Vector fields have many important applications, as they can be used to represent many
physical quantities: the vector at a point may represent the strength of some force (gravity,
electricity, magnetism) or a velocity (wind speed or the velocity of some other fluid).
We have already seen a particularly important kind of vector field—the gradient. Given
a function f (x, y), recall that the gradient is hfx (x, y), fy (x, y)i, a vector that depends on
(is a function of) x and y. We usually picture the gradient vector with its tail at (x, y),
pointing in the direction of maximum increase. Vector fields that are gradients have some
particularly nice properties, as we will see. An important example is
F=
−x
−y
−z
, 2
, 2
2
2
2
3/2
2
2
3/2
(x + y + z )
(x + y + z )
(x + y 2 + z 2 )3/2
,
419
420
Chapter 16 Vector Calculus
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
−2.0
−1.6
−1.2
−0.8
0.0
−0.4
0.4
0.8
1.2
1.6
2.0
−0.2
x
−0.4
−0.6
−0.8
−1.0
y
−1.2
−1.4
−1.6
−1.8
−2.0
Figure 16.1.1
A vector field.
which points from the point (x, y, z) toward the origin and has length
p
x2 + y 2 + z 2
1
,
= p
2
2
2
3/2
(x + y + z )
( x2 + y 2 + z 2 ) 2
which is the reciprocal of the square of the distance from (x, y, z) to the origin—in other
words, F is an “inverse square law”. The vector F is a gradient:
1
F = ∇p
,
x2 + y 2 + z 2
which turns out to be extremely useful.
Exercises 16.1.
Sketch the vector fields; check your work with Sage’s plot_vector_field function.
1. hx, yi
2. h−x, −yi
3. hx, −yi
4. hsin x, cos yi
5. hy, 1/xi
6. hx + 1, x + 3i
7. Verify equation 16.1.1.
(16.1.1)
16.2
16.2
Line Integrals
421
Line Integrals
We have so far integrated “over” intervals, areas, and volumes with single, double, and
triple integrals. We now investigate integration over or “along” a curve—“line integrals”
are really “curve integrals”.
As with other integrals, a geometric example may be easiest to understand. Consider
the function f = x + y and the parabola y = x2 in the x-y plane, for 0 ≤ x ≤ 2. Imagine
that we extend the parabola up to the surface f , to form a curved wall or curtain, as in
figure 16.2.1. What is the area of the surface thus formed? We already know one way to
compute surface area, but here we take a different approach that is more useful for the
problems to come.
6
5
2.0
1.5
4
1.0
z3
x
0.5
0.0 0
2
1
2
1
y
3
0
Figure 16.2.1
4
Approximating the area under a curve. (AP)
As usual, we start by thinking about how to approximate the area. We pick some
points along the part of the parabola we’re interested in, and connect adjacent points by
straight lines; when the points are close together, the length of each line segment will be
close to the length along the parabola. Using each line segment as the base of a rectangle,
we choose the height to be the height of the surface f above the line segment. If we add
up the areas of these rectangles, we get an approximation to the desired area, and in the
limit this sum turns into an integral.
Typically the curve is in vector form, or can easily be put in vector form; in this
example we have v(t) = ht, t2 i. Then as we have seen in section 13.3 on arc length,
the length of one of the straight line segments in the approximation is approximately
422
Chapter 16 Vector Calculus
ds = |v′ | dt =
Z
2
0
√
1 + 4t2 dt, so the integral is
Z 2
p
p
√
1
167 √
1
−
ln(4 + 17).
f (t, t ) 1 + 4t2 dt =
(t + t2 ) 1 + 4t2 dt =
17 −
48
12 64
0
2
This integral of a function along a curve C is often written in abbreviated form as
Z
f (x, y) ds.
C
EXAMPLE 16.2.1 Compute
Z
yex ds where C is the line segment from (1, 2) to (4, 7).
C
We write the line segment as a vector function: v = h1, 2i + th3, 5i, 0 ≤ t ≤ 1, or in
parametric form x = 1 + 3t, y = 2 + 5t. Then
Z
x
ye ds =
C
Z
1
(2 + 5t)e1+3t
0
p
16 √ 4 1 √
32 + 52 dt =
34e −
34 e.
9
9
All of these ideas extend to three dimensions in the obvious way.
EXAMPLE 16.2.2
Compute
Z
C
x2 z ds where C is the line segment from (0, 6, −1) to
(4, 1, 5).
We write the line segment as a vector function: v = h0, 6, −1i + th4, −5, 6i, 0 ≤ t ≤ 1,
or in parametric form x = 4t, y = 6 − 5t, z = −1 + 6t. Then
Z
C
2
x z ds =
Z
1
√ Z
(4t) (−1 + 6t) 16 + 25 + 36 dt = 16 77
2
0
√
0
1
−t2 + 6t3 dt =
56 √
77.
3
Now we turn to a perhaps more interesting example. Recall that in the simplest case,
the work done by a force on an object is equal to the magnitude of the force times the
distance the object moves; this assumes that the force is constant and in the direction of
motion. We have already dealt with examples in which the force is not constant; now we
are prepared to examine what happens when the force is not parallel to the direction of
motion.
16.2
Line Integrals
423
We have already examined the idea of components of force, in example 12.3.4: the
component of a force F in the direction of a vector v is
F·v
v,
|v|2
the projection of F onto v. The length of this vector, that is, the magnitude of the force
in the direction of v, is
F·v
,
|v|
the scalar projection of F onto v. If an object moves subject to this (constant) force, in
the direction of v, over a distance equal to the length of v, the work done is
F·v
|v| = F · v.
|v|
Thus, work in the vector setting is still “force times distance”, except that “times” means
“dot product”.
If the force varies from point to point, it is represented by a vector field F; the displacement vector v may also change, as an object may follow a curving path in two or
three dimensions. Suppose that the path of an object is given by a vector function r(t); at
any point along the path, the (small) tangent vector r′ ∆t gives an approximation to its
motion over a short time ∆t, so the work done during that time is approximately F · r′ ∆t;
the total work over some time period is then
Z
t1
t0
F · r′ dt.
It is useful to rewrite this in various ways at different times. We start with
Z
t1
t0
′
F · r dt =
Z
C
F · dr,
abbreviating r′ dt by dr. Or we can write
Z
t1
t0
′
F · r dt =
Z
t1
t0
r′
F · ′ |r′ | dt =
|r |
Z
t1
t0
′
F · T |r | dt =
Z
C
F · T ds,
using the unit tangent vector T, abbreviating |r′ | dt as ds, and indicating the path of the
object by C. In other words, work is computed using a particular line integral of the form
424
Chapter 16 Vector Calculus
we have considered. Alternately, we sometimes write
Z
Z dy
dz
dx
dt
+g
+h
F · r dt =
hf, g, hi · hx , y , z i dt =
f
dt
dt
dt
C
C
C
Z
Z
Z
Z
=
f dx + g dy + h dz =
f dx +
g dy +
h dz,
Z
′
′
′
′
C
C
C
C
and similarly for two dimensions, leaving out references to z.
EXAMPLE 16.2.3 Suppose an object moves from (−1, 1) to (2, 4) along the path
r(t) = ht, t2 i, subject to the force F = hx sin y, yi. Find the work done.
We can write the force in terms of t as ht sin(t2 ), t2 i, and compute r′ (t) = h1, 2ti, and
then the work is
Z
2
−1
2
2
ht sin(t ), t i · h1, 2ti dt =
Z
2
t sin(t2 ) + 2t3 dt =
−1
15 cos(1) − cos(4)
+
.
2
2
Alternately, we might write
Z
x sin y dx +
C
Z
y dy =
C
Z
2
2
x sin(x ) dx +
−1
Z
4
1
y dy = −
cos(4) cos(1) 16 1
+
+
−
2
2
2
2
Exercises 16.2.
1. Compute
Z
xy2 ds along the line segment from (1, 2, 0) to (2, 1, 3). ⇒
Z
sin x ds along the line segment from (−1, 2, 1) to (1, 2, 5). ⇒
Z
z cos(xy) ds along the line segment from (1, 0, 1) to (2, 2, 3). ⇒
Z
sin x dx + cos y dy along the top half of the unit circle, from (1, 0) to (−1, 0). ⇒
C
2. Compute
C
3. Compute
C
4. Compute
5. Compute
6. Compute
7. Compute
8. Compute
9. Compute
ZC
ZC
ZC
ZC
ZC
C
xey dx + x2 y dy along the line segment y = 3, 0 ≤ x ≤ 2. ⇒
xey dx + x2 y dy along the line segment x = 4, 0 ≤ y ≤ 4. ⇒
xey dx + x2 y dy along the curve x = 3t, y = t2 , 0 ≤ t ≤ 1. ⇒
xey dx + x2 y dy along the curve het , et i, −1 ≤ t ≤ 1. ⇒
hcos x, sin yi · dr along the curve ht, ti, 0 ≤ t ≤ 1. ⇒
16.3
10. Compute
11.
12.
13.
14.
Z
C
The Fundamental Theorem of Line Integrals
425
h1/xy, 1/(x + y)i · dr along the path from (1, 1) to (3, 1) to (3, 6) using straight
line segments. ⇒
Z
Compute
h1/xy, 1/(x + y)i · dr along the curve h2t, 5ti, 1 ≤ t ≤ 4. ⇒
C
Z
Compute
h1/xy, 1/(x + y)i · dr along the curve ht, t2 i, 1 ≤ t ≤ 4. ⇒
ZC
Compute
yz dx + xz dy + xy dz along the curve ht, t2 , t3 i, 0 ≤ t ≤ 1. ⇒
C
Z
Compute
yz dx + xz dy + xy dz along the curve hcos t, sin t, tan ti, 0 ≤ t ≤ π. ⇒
C
15. An object moves from (1, 1) to (4, 8) along the path r(t) = ht2 , t3 i, subject to the force
F = hx2 , sin yi. Find the work done. ⇒
16. An object moves along the line segment from (1, 1) to (2, 5), subject to the force F =
hx/(x2 + y2 ), y/(x2 + y2 )i. Find the work done. ⇒
17. An object moves along the parabola r(t) = ht, t2 i, 0 ≤ t ≤ 1, subject to the force F =
h1/(y + 1), −1/(x + 1)i. Find the work done. ⇒
18. An object moves along the line segment from (0, 0, 0) to (3, 6, 10), subject to the force F =
hx2 , y2 , z 2 i. Find the work done. ⇒
√
√
19. An object moves along the curve r(t) = h t, 1/ t, ti 1 ≤ t ≤ 4, subject to the force F =
hy, z, xi. Find the work done. ⇒
20. An object moves from (1, 1, 1) to (2, 4, 8) along the path r(t) = ht, t2 , t3 i, subject to the force
F = hsin x, sin y, sin zi. Find the work done. ⇒
21. An object moves from (1, 0, 0) to (−1, 0, π) along the path r(t) = hcos t, sin t, ti, subject to
the force F = hy2 , y2 , xzi. Find the work done. ⇒
22. Give an example of a non-trivial force field F and non-trivial path r(t) for which the total
work done moving along the path is zero.
16.3
The Fundamental Theorem of Line Integrals
One way to write the Fundamental Theorem of Calculus (7.2.1) is:
Z b
f ′ (x) dx = f (b) − f (a).
a
That is, to compute the integral of a derivative f ′ we need only compute the values of f
at the endpoints. Something similar is true for line integrals of a certain form.
THEOREM 16.3.1 Fundamental Theorem of Line Integrals
Suppose a curve
C is given by the vector function r(t), with a = r(a) and b = r(b). Then
Z
∇f · dr = f (b) − f (a),
C
provided that r is sufficiently nice.
426
Chapter 16 Vector Calculus
Proof. We write r = hx(t), y(t), z(t)i, so that r′ = hx′ (t), y ′ (t), z ′ (t)i. Also, we know
that ∇f = hfx , fy , fz i. Then
Z
Z b
Z b
′
′
′
∇f · dr =
hfx , fy , fz i · hx (t), y (t), z (t)i dt =
fx x′ + fy y ′ + fz z ′ dt.
C
a
a
By the chain rule (see section 14.4) fx x′ + fy y ′ + fz z ′ = df /dt, where f in this context
means f (x(t), y(t), z(t)), a function of t. In other words, all we have is
Z b
f ′ (t) dt = f (b) − f (a).
a
In this context, f (a) = f (x(a), y(a), z(a)). Since a = r(a) = hx(a), y(a), z(a)i, we can write
f (a) = f (a)—this is a bit of a cheat, since we are simultaneously using f to mean f (t) and
f (x, y, z), and since f (x(a), y(a), z(a)) is not technically the same as f (hx(a), y(a), z(a)i),
but the concepts are clear and the different uses are compatible. Doing the same for b, we
get
Z
Z b
∇f · dr =
f ′ (t) dt = f (b) − f (a) = f (b) − f (a).
C
a
This theorem, like the Fundamental Theorem of Calculus, says roughly that if we
integrate a “derivative-like function” (f ′ or ∇f ) the result depends only on the values of
the original function (f ) at the endpoints.
If a vector field F is the gradient of a function, F = ∇f , we say that F is a conservaR
tive vector field. If F is a conservative force field, then the integral for work, C F · dr,
is in the form required by the Fundamental Theorem of Line Integrals. This means that
in a conservative force field, the amount of work required to move an object from point a
to point b depends only on those points, not on the path taken between them.
EXAMPLE 16.3.2 An object moves in the force field
−x
−y
−z
F=
,
,
,
(x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2
along the curve r = h1 + t, t3 , t cos(πt)i as t ranges from 0 to 1. Find the work done by the
force on the object.
The straightforward way to do this involves substituting the components of r into F,
forming the dot product F · r′ , and then trying to compute the integral, but this integral is
p
extraordinarily messy, perhaps impossible to compute. But since F = ∇(1/ x2 + y 2 + z 2 )
we need only substitute:
Z
1
F · dr = p
x2 + y 2 + z 2
C
(2,1,−1)
(1,0,0)
1
= √ − 1.
6
16.3
The Fundamental Theorem of Line Integrals
427
Another immediate consequence of the Fundamental Theorem involves closed paths.
A path C is closed if it forms a loop, so that traveling over the C curve brings you back to
the starting point. If C is a closed path, we can integrate around it starting at any point
a; since the starting and ending points are the same,
Z
C
∇f · dr = f (a) − f (a) = 0.
For example, in a gravitational field (an inverse square law field) the amount of work
required to move an object around a closed path is zero. Of course, it’s only the net
amount of work that is zero. It may well take a great deal of work to get from point a to
point b, but then the return trip will “produce” work. For example, it takes work to pump
water from a lower to a higher elevation, but if you then let gravity pull the water back
down, you can recover work by running a water wheel or generator. (In the real world you
won’t recover all the work because of various losses along the way.)
To make use of the Fundamental Theorem of Line Integrals, we need to be able to
spot conservative vector fields F and to compute f so that F = ∇f . Suppose that F =
hP, Qi = ∇f . Then P = fx and Q = fy , and provided that f is sufficiently nice, we know
from Clairaut’s Theorem (14.6.2) that Py = fxy = fyx = Qx . If we compute Py and Qx
and find that they are not equal, then F is not conservative. If Py = Qx , then, again
provided that F is sufficiently nice, we can be assured that F is conservative. Ultimately,
what’s important is that we be able to find f ; as this amounts to finding anti-derivatives,
we may not always succeed.
EXAMPLE 16.3.3
First, note that
Find an f so that h3 + 2xy, x2 − 3y 2 i = ∇f .
∂
(3 + 2xy) = 2x
∂y
and
∂ 2
(x − 3y 2 ) = 2x,
∂x
so the desired f does exist. This means that fx = 3 + 2xy, so that f = 3x + x2 y + g(y);
the first two terms are needed to get 3 + 2xy, and the g(y) could be any function of y, as it
would disappear upon taking a derivative with respect to x. Likewise, since fy = x2 − 3y 2 ,
f = x2 y − y 3 + h(x). The question now becomes, is it possible to find g(y) and h(x) so
that
3x + x2 y + g(y) = x2 y − y 3 + h(x),
and of course the answer is yes: g(y) = −y 3 , h(x) = 3x. Thus, f = 3x + x2 y − y 3 .
We can test a vector field F = hP, Q, Ri in a similar way. Suppose that hP, Q, Ri =
hfx , fy , fz i. If we temporarily hold z constant, then f (x, y, z) is a function of x and y,
428
Chapter 16 Vector Calculus
and by Clairaut’s Theorem Py = fxy = fyx = Qx . Likewise, holding y constant implies
Pz = fxz = fzx = Rx , and with x constant we get Qz = fyz = fzy = Ry . Conversely, if we
find that Py = Qx , Pz = Rx , and Qz = Ry then F is conservative.
Exercises 16.3.
1. Find an f so that ∇f = h2x + y2 , 2y + x2 i, or explain why there is no such f . ⇒
2. Find an f so that ∇f = hx3 , −y4 i, or explain why there is no such f . ⇒
3. Find an f so that ∇f = hxey , yex i, or explain why there is no such f . ⇒
4. Find an f so that ∇f = hy cos x, y sin xi, or explain why there is no such f . ⇒
5. Find an f so that ∇f = hy cos x, sin xi, or explain why there is no such f . ⇒
6. Find an f so that ∇f = hx2 y3 , xy4 i, or explain why there is no such f . ⇒
7. Find an f so that ∇f = hyz, xz, xyi, or explain why there is no such f . ⇒
Z
8. Evaluate
(10x4 − 2xy3 ) dx − 3x2 y2 dy where C is the part of the curve x5 − 5x2 y2 − 7x2 = 0
C
from (3, −2) to (3, 2). ⇒
9. Let F = hyz, xz, xyi. Find the work done by this force field on an object that moves from
(1, 0, 2) to (1, 2, 3). ⇒
10. Let F = hey , xey + sin z, y cos zi. Find the work done by this force field on an object that
moves from (0, 0, 0) to (1, −1, 3). ⇒
11. Let
F=
−y
−z
−x
,
,
(x2 + y2 + z 2 )3/2 (x2 + y2 + z 2 )3/2 (x2 + y2 + z 2 )3/2
.
Find the work done by this force field on an object that moves from (1, 1, 1) to (4, 5, 6). ⇒
16.4
Green's Theorem
We now come to the first of three important theorems that extend the Fundamental Theorem of Calculus to higher dimensions. (The Fundamental Theorem of Line Integrals has
already done this in one way, but in that case we were still dealing with an essentially
one-dimensional integral.) They all share with the Fundamental Theorem the following
rather vague description: To compute a certain sort of integral over a region, we may do
a computation on the boundary of the region that involves one fewer integrations.
Note that this does indeed describe the Fundamental Theorem of Calculus and the
Fundamental Theorem of Line Integrals: to compute a single integral over an interval, we
do a computation on the boundary (the endpoints) that involves one fewer integrations,
namely, no integrations at all.
16.4
429
Green’s Theorem
THEOREM 16.4.1 Green’s Theorem If the vector field F = hP, Qi and the region
D are sufficiently nice, and if C is the boundary of D (C is a closed curve), then
Z
ZZ
∂Q ∂P
−
dA =
P dx + Q dy,
∂x
∂y
C
D
provided the integration on the right is done counter-clockwise around C.
Z
To indicate that an integral
is being done over a closed curve in the counterC I
clockwise direction, we usually write
. We also use the notation ∂D to mean the
C
I
Z
boundary of D oriented in the counterclockwise direction. With this notation,
=
.
C
∂D
We already know one case, not particularly interesting, in which this theorem is true:
I
If F is conservative, we know that the integral
F · dr = 0, because any integral of a
C
conservative vector field around a closed curve is zero. We also know in this case that
∂P/∂y = ∂Q/∂x, so the double integral in the theorem is simply the integral of the zero
function, namely, 0. So in the case that F is conservative, the theorem says simply that
0 = 0.
We illustrate the theorem by computing both sides of
Z
ZZ
4
x dx + xy dy =
y − 0 dA,
EXAMPLE 16.4.2
∂D
D
where D is the triangular region with corners (0, 0), (1, 0), (0, 1).
Starting with the double integral:
ZZ
Z 1 Z 1−x
Z 1
(1 − x)2
(1 − x)3
y − 0 dA =
y dy dx =
dx = −
2
6
0
0
0
D
1
=
0
1
.
6
There is no single formula to describe the boundary of D, so to compute the left side
directly we need to compute three separate integrals corresponding to the three sides of
the triangle, and each of these integrals we break into two integrals, the “dx” part and the
“dy” part. The three sides are described by y = 0, y = 1 − x, and x = 0. The integrals
are then
Z
Z 1
Z 0
Z 0
Z 1
Z 0
Z 0
4
4
4
x dx + xy dy =
x dx +
0 dy +
x dx +
(1 − y)y dy +
0 dx +
0 dy
∂D
0
0
1
0
0
1
1 1
1
1
= +0− + +0+0= .
5
5 6
6
Alternately, we could describe the three sides in vector form as ht, 0i, h1 − t, ti, and
h0, 1 − ti. Note that in each case, as t ranges from 0 to 1, we follow the corresponding side
430
Chapter 16 Vector Calculus
in the correct direction. Now
Z
Z
Z 1
4
4
t + t · 0 dt +
x dx + xy dy =
0
∂D
=
Z
1
4
t dt +
0
Z
1
4
0
−(1 − t) + (1 − t)t dt +
1
0
−(1 − t)4 + (1 − t)t dt =
Z
1
0 + 0 dt
0
1
.
6
In this case, none of the integrations are difficult, but the second approach is somewhat tedious because of the necessity to set up three different integrals. In different
circumstances, either of the integrals, the single or the double, might be easier to compute.
Sometimes it is worthwhile to turn a single integral into the corresponding double integral,
sometimes exactly the opposite approach is best.
Here is a clever use of Green’s Theorem: We know that areas can be computed using
double integrals, namely,
ZZ
1 dA
D
computes the area of region D. If we can find P and Q so that ∂Q/∂x − ∂P/∂y = 1, then
the area is also
Z
P dx + Q dy.
∂D
It is quite easy to do this: P = 0, Q = x works, as do P = −y, Q = 0 and P = −y/2, Q =
x/2.
EXAMPLE 16.4.3 An ellipse centered at the origin, with its two principal axes aligned
with the x and y axes, is given by
y2
x2
+
= 1.
a2
b2
We find the area of the interior of the ellipse via Green’s theorem. To do this we need a
vector equation for the boundary; one such equation is ha cos t, b sin ti, as t ranges from 0
to 2π. We can easily verify this by substitution:
x2
y2
a2 cos2 t b2 sin2 t
+
=
+
= cos2 t + sin2 t = 1.
a2
b2
a2
b2
Let’s consider the three possibilities for P and Q above: Using 0 and x gives
I
C
0 dx + x dy =
Z
0
2π
a cos(t)b cos(t) dt =
Z
2π
ab cos2 (t) dt.
0
16.4
Green’s Theorem
431
Using −y and 0 gives
I
C
−y dx + 0 dy =
Z
2π
0
−b sin(t)(−a sin(t)) dt =
Z
2π
ab sin2 (t) dt.
0
Finally, using −y/2 and x/2 gives
I
x
y
− dx + dy =
2
2
C
Z
2π
2π
=
Z
b sin(t)
a cos(t)
(−a sin(t)) dt +
(b cos(t)) dt
2
2
Z 2π
ab sin2 t ab cos2 t
ab
+
dt =
dt = πab.
2
2
2
0
−
0
0
The first two integrals are not particularly difficult, but the third is very easy, though the
choice of P and Q seems more complicated.
(0, b)
.....................................
......................
............
...........
.........
.........
........
.
.
.
.
.
.
......
...
.
.
.
.
......
.
...
.
.
.....
.
.
..
.
.....
.
.
.
.
.....
.
.
.
.
....
..
.
.
.
...
..
.
...
.
.
.
...
.
..
...
.
..
...
...
..
...
.
...
...
.
...
..
.
.
...
..
...
...
...
...
....
..
.
.....
.
.
.....
.....
.....
.....
......
.....
......
......
.
.
.
.
.......
.
.....
........
........
..........
..............
..........
.........................................................
•
•
Figure 16.4.1
A “standard” ellipse,
(a, 0)
x2
a2
+
y2
b2
= 1.
Proof of Green’s Theorem.
We cannot here prove Green’s Theorem in general, but
we can do a special case. We seek to prove that
I
P dx + Q dy =
C
ZZ
∂Q ∂P
−
dA.
∂x
∂y
D
It is sufficient to show that
I
C
P dx =
ZZ
D
∂P
dA
−
∂y
and
I
C
Q dy =
ZZ
∂Q
dA,
∂x
D
which we can do if we can compute the double integral in both possible ways, that is, using
dA = dy dx and dA = dx dy.
432
Chapter 16 Vector Calculus
ZZ
∂P
dA =
∂y
D
Z
a
b
Z
g2 (x)
g1 (x)
∂P
dy dx =
∂y
Z
b
a
P (x, g2 (x)) − P (x, g1 (x)) dx.
Here we have simply used the ordinary Fundamental Theorem of Calculus, since for the
inner integral we are integrating a derivative with respect to y: an antiderivative of ∂P/∂y
with respect to y is simply P (x, y), and then we substitute g1 and g2 for y and subtract.
H
Now we need to manipulate C P dx. The boundary of region D consists of 4 parts,
given by the equations y = g1 (x), x = b, y = g2 (x), and x = a. On the portions x = b
and x = a, dx = 0 dt, so the corresponding integrals are zero. For the other two portions,
we use the parametric forms x = t, y = g1 (t), a ≤ t ≤ b, and x = t, y = g2 (t), letting t
range from b to a, since we are integrating counter-clockwise around the boundary. The
resulting integrals give us
I
P dx =
C
Z
b
P (t, g1 (t)) dt +
a
Z
a
P (t, g2 (t)) dt =
b
Z
b
Z
b
P (t, g1 (t)) dt −
a
=
a
Z
b
P (t, g2 (t)) dt
a
P (t, g1 (t)) − P (t, g2 (t)) dt
which is the result of the double integral times −1, as desired.
The equation involving Q is essentially the same, and left as an exercise.
Exercises 16.4.
1. Compute
Z
2y dx + 3x dy, where D is described by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. ⇒
Z
xy dx + xy dy, where D is described by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. ⇒
Z
e2x+3y dx + exy dy, where D is described by −2 ≤ x ≤ 2, −1 ≤ y ≤ 1. ⇒
∂D
2. Compute
∂D
3. Compute
4. Compute
5. Compute
6. Compute
7. Compute
Z∂D
Z∂D
Z∂D
√
√
x y dx + x + y dy, where D is described by 1 ≤ x ≤ 2, 2x ≤ y ≤ 4. ⇒
(x/y) dx + (2 + 3x) dy, where D is described by 1 ≤ x ≤ 2, 1 ≤ y ≤ x2 . ⇒
Z
sin y dx + sin x dy, where D is described by 0 ≤ x ≤ π/2, x ≤ y ≤ π/2. ⇒
Z
x ln y dx, where D is described by 1 ≤ x ≤ 2, ex ≤ y ≤ ex . ⇒
∂D
9. Compute
x2 y dx + xy2 dy, where D is described by 0 ≤ x ≤ 1, 0 ≤ y ≤ x. ⇒
Z∂D
∂D
8. Compute
y cos x dx + y sin x dy, where D is described by 0 ≤ x ≤ π/2, 1 ≤ y ≤ 2. ⇒
∂D
2
16.5
10. Compute
11. Compute
12. Compute
Z
Z∂D
Z∂D
∂D
13. Evaluate
I
C
433
p
1 + x2 dy, where D is described by −1 ≤ x ≤ 1, x2 ≤ y ≤ 1. ⇒
x2 y dx − xy2 dy, where D is described by x2 + y2 ≤ 1. ⇒
y3 dx + 2x3 dy, where D is described by x2 + y2 ≤ 4. ⇒
(y − sin(x)) dx + cos(x) dy, where C is the boundary of the triangle with vertices
(0, 0), (1, 0), and (1, 2) oriented counter-clockwise. ⇒
14. Finish our proof of Green’s Theorem by showing that
16.5
Divergence and Curl
I
Q dy =
C
ZZ
∂Q
dA.
∂x
D
Divergen e and Curl
Divergence and curl are two measurements of vector fields that are very useful in a variety of
applications. Both are most easily understood by thinking of the vector field as representing
a flow of a liquid or gas; that is, each vector in the vector field should be interpreted as a
velocity vector. Roughly speaking, divergence measures the tendency of the fluid to collect
or disperse at a point, and curl measures the tendency of the fluid to swirl around the point.
Divergence is a scalar, that is, a single number, while curl is itself a vector. The magnitude
of the curl measures how much the fluid is swirling, the direction indicates the axis around
which it tends to swirl. These ideas are somewhat subtle in practice, and are beyond
the scope of this course. You can find additional information on the web, for example at
http://mathinsight.org/curl_idea and http://mathinsight.org/divergence_idea
and in many books including Div, Grad, Curl, and All That: An Informal Text on Vector
Calculus, by H. M. Schey.
Recall that if f is a function, the gradient of f is given by
∂f ∂f ∂f
∇f =
.
,
,
∂x ∂y ∂z
A useful mnemonic for this (and for the divergence and curl, as it turns out) is to let
∂ ∂ ∂
,
,
,
∇=
∂x ∂y ∂z
that is, we pretend that ∇ is a vector with rather odd looking entries. Recalling that
hu, v, wia = hua, va, wai, we can then think of the gradient as
∂f ∂f ∂f
∂ ∂ ∂
f=
,
,
,
,
,
∇f =
∂x ∂y ∂z
∂x ∂y ∂z
that is, we simply multiply the f into the vector.
434
Chapter 16 Vector Calculus
The divergence and curl can now be defined in terms of this same odd vector ∇ by
using the cross product and dot product. The divergence of a vector field F = hf, g, hi is
∂f
∂g ∂h
∂ ∂ ∂
· hf, g, hi =
,
,
+
+
.
∇·F=
∂x ∂y ∂z
∂x ∂y
∂z
The curl of F is
∇×F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
f
g
=
h
∂h ∂g
∂f
∂h ∂g ∂f
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x ∂y
.
Here are two simple but useful facts about divergence and curl.
THEOREM 16.5.1
∇ · (∇ × F) = 0.
In words, this says that the divergence of the curl is zero.
THEOREM 16.5.2
∇ × (∇f ) = 0.
That is, the curl of a gradient is the zero vector. Recalling that gradients are conservative vector fields, this says that the curl of a conservative vector field is the zero vector.
Under suitable conditions, it is also true that if the curl of F is 0 then F is conservative.
(Note that this is exactly the same test that we discussed on page 427.)
EXAMPLE 16.5.3 Let F = hez , 1, xez i. Then ∇ × F = h0, ez − ez , 0i = 0. Thus, F is
conservative, and we can exhibit this directly by finding the corresponding f .
Since fx = ez , f = xez + g(y, z). Since fy = 1, it must be that gy = 1, so g(y, z) =
y + h(z). Thus f = xez + y + h(z) and
xez = fz = xez + 0 + h′ (z),
so h′ (z) = 0, i.e., h(z) = C, and f = xez + y + C.
We can rewrite Green’s Theorem using these new ideas; these rewritten versions in
turn are closer to some later theorems we will see.
Suppose we write a two dimensional vector field in the form F = hP, Q, 0i, where P
and Q are functions of x and y. Then
∇×F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
P
Q
0
= h0, 0, Qx − Py i,
and so (∇ × F) · k = h0, 0, Qx − Py i · h0, 0, 1i = Qx − Py . So Green’s Theorem says
Z
Z
ZZ
ZZ
F · dr =
P dx + Q dy =
Qx − Py dA =
(∇ × F) · k dA. (16.5.1)
∂D
∂D
D
D
16.5
Divergence and Curl
435
Roughly speaking, the right-most integral adds up the curl (tendency to swirl) at each
point in the region; the left-most integral adds up the tangential components of the vector
field around the entire boundary. Green’s Theorem says these are equal, or roughly, that
the sum of the “microscopic” swirls over the region is the same as the “macroscopic” swirl
around the boundary.
Next, suppose that the boundary ∂D has a vector form r(t), so that r′ (t) is tangent to
the boundary, and T = r′ (t)/|r′ (t)| is the usual unit tangent vector. Writing r = hx(t), y(t)i
we get
hx′ , y ′ i
T= ′
|r (t)|
and then
hy ′ , −x′ i
N=
|r′ (t)|
is a unit vector perpendicular to T, that is, a unit normal to the boundary. Now
Z
hy ′ , −x′ i ′
F · N ds =
hP, Qi ·
|r (t)|dt =
P y ′ dt − Qx′ dt
|r′ (t)|
∂D
∂D
∂D
Z
Z
=
P dy − Q dx =
−Q dx + P dy.
Z
Z
∂D
∂D
So far, we’ve just rewritten the original integral using alternate notation. The last integral
looks just like the right side of Green’s Theorem (16.4.1) except that P and Q have traded
places and Q has acquired a negative sign. Then applying Green’s Theorem we get
Z
∂D
−Q dx + P dy =
ZZ
Px + Qy dA =
D
ZZ
D
Summarizing the long string of equalities,
Z
ZZ
F · N ds =
∇ · F dA.
∂D
∇ · F dA.
(16.5.2)
D
Roughly speaking, the first integral adds up the flow across the boundary of the region,
from inside to out, and the second sums the divergence (tendency to spread) at each point
in the interior. The theorem roughly says that the sum of the “microscopic” spreads is the
same as the total spread across the boundary and out of the region.
436
Chapter 16 Vector Calculus
Exercises 16.5.
1. Let F = hxy, −xyi and let D be given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Compute
Z
F · N ds. ⇒
Z
F · dr and
Z
F · dr and
Z
F · dr and
∂D
∂D
2
2
2. Let F = hax , by i and let D be given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Compute
Z
F · N ds. ⇒
∂D
∂D
2
2
3. Let F = hay , bx i and let D be given by 0 ≤ x ≤ 1, 0 ≤ y ≤ x. Compute
Z
F · N ds. ⇒
∂D
∂D
4. Z
Let F = hsin x cos
Z y, cos x sin yi and let D be given by 0 ≤ x ≤ π/2, 0 ≤ y ≤ x. Compute
F · dr and
F · N ds. ⇒
∂D
∂D
Z
Z
2
2
5. Let F = hy, −xi and let D be given by x + y ≤ 1. Compute
F · dr and
F · N ds.
∂D
∂D
⇒
Z
Z
6. Let F = hx, yi and let D be given by x2 + y2 ≤ 1. Compute
∂D
F · dr and
∂D
F · N ds. ⇒
7. Prove theorem 16.5.1.
8. Prove theorem 16.5.2.
9. If ∇ · F = 0, F is said to be incompressible. Show that any vector field of the form
F(x, y, z) = hf (y, z), g(x, z), h(x, y)i is incompressible. Give a non-trivial example.
16.6
Ve tor Fun tions for Surfa es
We have dealt extensively with vector equations for curves, r(t) = hx(t), y(t), z(t)i. A
similar technique can be used to represent surfaces in a way that is more general than the
equations for surfaces we have used so far. Recall that when we use r(t) to represent a
curve, we imagine the vector r(t) with its tail at the origin, and then we follow the head
of the arrow as t changes. The vector “draws” the curve through space as t varies.
Suppose we instead have a vector function of two variables,
r(u, v) = hx(u, v), y(u, v), z(u, v)i.
As both u and v vary, we again imagine the vector r(u, v) with its tail at the origin, and
its head sweeps out a surface in space. A useful analogy is the technology of CRT video
screens, in which an electron gun fires electrons in the direction of the screen. The gun’s
direction sweeps horizontally and vertically to “paint” the screen with the desired image.
In practice, the gun moves horizontally through an entire line, then moves vertically to the
next line and repeats the operation. In the same way, it can be useful to imagine fixing a
16.6
Vector Functions for Surfaces
437
value of v and letting r(u, v) sweep out a curve as u changes. Then v can change a bit, and
r(u, v) sweeps out a new curve very close to the first. Put enough of these curves together
and they form a surface.
EXAMPLE 16.6.1 Consider the function r(u, v) = hv cos u, v sin u, vi. For a fixed value
of v, as u varies from 0 to 2π, this traces a circle of radius v at height v above the x-y
plane. Put lots and lots of these together,and they form a cone, as in figure 16.6.1.
3
3
2
z
2
z
1
−2
0
−2
0
0
y
2
1
−2
0
−2
0
x
2
Figure 16.6.1
0
y
2
x
2
Tracing a surface.
EXAMPLE 16.6.2 Let r = hv cos u, v sin u, ui. If v is constant, the resulting curve is a
helix (as in figure 13.1.1). If u is constant, the resulting curve is a straight line at height
u in the direction u radians from the positive x axis. Note in figure 16.6.2 how the helixes
and the lines both paint the same surface in a different way.
This technique allows us to represent many more surfaces than previously.
EXAMPLE 16.6.3
The curve given by
r = h(2 + cos(3u/2)) cos u, (2 + cos(3u/2)) sin u, sin(3u/2)i
is called a trefoil knot. Recall that from the vector equation of the curve we can compute
the unit tangent T, the unit normal N, and the binormal vector B = T × N; you may
want to review section 13.3. The binormal is perpendicular to both T and N; one way to
interpret this is that N and B define a plane perpendicular to T, that is, perpendicular
to the curve; since N and B are perpendicular to each other, they can function just as i
438
Chapter 16 Vector Calculus
Figure 16.6.2
Tracing a surface. (AP)
and j do for the x-y plane. Of course, N and B are functions of u, changing as we move
along the curve r(u). So, for example, c(u, v) = N cos v + B sin v is a vector equation for
a unit circle in a plane perpendicular to the curve described by r, except that the usual
interpretation of c would put its center at the origin. We can fix that simply by adding c
to the original r: let f = r(u) + c(u, v). For a fixed u this draws a circle around the point
r(u); as u varies we get a sequence of such circles around the curve r, that is, a tube of
radius 1 with r at its center. We can easily change the radius; for example r(u) + ac(u, v)
gives the tube radius a; we can make the radius vary as we move along the curve with
r(u) + g(u)c(u, v), where g(u) is a function of u. As shown in figure 16.6.3, it is hard to see
that the plain knot is knotted; the tube makes the structure apparent. Of course, there is
nothing special about the trefoil knot in this example; we can put a tube around (almost)
any curve in the same way.
Figure 16.6.3
Tubes around a trefoil knot, with radius 1/2 and 3 cos(u)/4. (AP)
16.6
Vector Functions for Surfaces
439
We have previously examined surfaces given in the form f (x, y). It is sometimes
useful to represent such surfaces in the more general vector form, which is quite easy:
r(u, v) = hu, v, f (u, v)i. The names of the variables are not important of course; instead
of disguising x and y, we could simply write r(x, y) = hx, y, f (x, y)i.
We have also previously dealt with surfaces that are not functions of x and y; many
of these are easy to represent in vector form. One common type of surface that cannot be
represented as z = f (x, y) is a surface given by an equation involving only x and y. For
example, x + y = 1 and y = x2 are “vertical” surfaces. For every point (x, y) in the plane
that satisfies the equation, the point (x, y, z) is on the surface, for every value of z. Thus,
a corresponding vector form for the surface is something like hf (u), g(u), vi; for example,
x + y = 1 becomes hu, 1 − u, vi and y = x2 becomes hu, u2 , vi.
Yet another sort of example is the sphere, say x2 + y 2 + z 2 = 1. This cannot be written
in the form z = f (x, y), but it is easy to write in vector form; indeed this particular
surface is much like the cone, since it has circular cross-sections, or we can think of it as
a tube around a portion of the z-axis, with a radius that varies depending on where along
√
√
the axis we are. One vector expression for the sphere is h 1 − v 2 cos u, 1 − v 2 sin u, vi—
this emphasizes the tube structure, as it is naturally viewed as drawing a circle of radius
√
1 − v 2 around the z-axis at height v. We could also take a cue from spherical coordinates,
and write hsin u cos v, sin u sin v, cos ui, where in effect u and v are φ and θ in disguise.
It is quite simple in Sage to plot any surface for which you have a vector representation.
Using different vector functions sometimes gives different looking plots, because Sage in
effect draws the surface by holding one variable constant and then the other. For example,
you might have noticed in figure 16.6.2 that the curves in the two right-hand graphs are
superimposed on the left-hand graph; the graph of the surface is just the combination of
the two sets of curves, with the spaces filled in with color.
Here’s a simple but striking example: the plane x + y + z = 1 can be represented
quite naturally as hu, v, 1 − u − vi. But we could also think of painting the same plane by
choosing a particular point on the plane, say (1, 0, 0), and then drawing circles or ellipses
(or any of a number of other curves) as if that point were the origin in the plane. For
example, h1 − v cos u − v sin u, v sin u, v cos ui is one such vector function. Note that while
it may not be obvious where this came from, it is quite easy to see that the sum of the
x, y, and z components of the vector is always 1. Computer renderings of the plane using
these two functions are shown in figure 16.6.4.
Suppose we know that a plane contains a particular point (x0 , y0 , z0 ) and that two
vectors u = hu0 , u1 , u2 i and v = hv0 , v1 , v2 i are parallel to the plane but not to each other.
We know how to get an equation for the plane in the form ax + by + cz = d, by first
computing u × v. It’s even easier to get a vector equation:
r(u, v) = hx0 , y0 , z0 i + uu + vv.
440
Chapter 16 Vector Calculus
3
−1.0
2
−0.5
0.0
1.0
0.5
−1
1
0.5
1.0
0.0
0
0
0
1
−0.5
1−1
−1.0
2
0
−1
1
Figure 16.6.4
Two representations of the same plane. (AP)
The first vector gets to the point (x0 , y0 , z0 ) and then by varying u and v, uu + vv gets to
every point in the plane.
Returning to x + y + z = 1, the points (1, 0, 0), (0, 1, 0), and (0, 0, 1) are all on the
plane. By subtracting coordinates we see that h−1, 0, 1i and h−1, 1, 0i are parallel to the
plane, so a third vector form for this plane is
h1, 0, 0i + uh−1, 0, 1i + vh−1, 1, 0i = h1 − u − v, v, ui.
This is clearly quite similar to the first form we found.
We have already seen (section 15.4) how to find the area of a surface when it is defined
in the form f (x, y). Finding the area when the surface is given as a vector function is very
similar. Looking at the plots of surfaces we have just seen, it is evident that the two sets
of curves that fill out the surface divide it into a grid, and that the spaces in the grid are
approximately parallelograms. As before this is the key: we can write down the area of a
typical little parallelogram and add them all up with an integral.
Suppose we want to approximate the area of the surface r(u, v) near r(u0 , v0 ). The
functions r(u, v0 ) and r(u0 , v) define two curves that intersect at r(u0 , v0 ). The derivatives of r give us vectors tangent to these two curves: ru (u0 , v0 ) and rv (u0 , v0 ), and then
ru (u0 , v0 ) du and rv (u0 , v0 ) dv are two small tangent vectors, whose lengths can be used
as the lengths of the sides of an approximating parallelogram. Finally, the area of this
parallelogram is |ru × rv | du dv and so the total surface area is
Z bZ d
|ru × rv | du dv.
a
c
16.6
Vector Functions for Surfaces
441
EXAMPLE 16.6.4 We find the area of the surface hv cos u, v sin u, ui for 0 ≤ u ≤ π
and 0 ≤ v ≤ 1; this is a portion of the helical surface in figure 16.6.2. We compute
ru = h−v sin u, v cos u, 1i and rv = hcos u, sin u, 0i. The cross product of these two vectors
√
is hsin u, − cos u, vi with length 1 + v 2 , and the surface area is
Z
0
π
Z
1
0
p
√
√
π
2
2 + 1)
π
ln(
+
.
1 + v 2 dv du =
2
2
Exercises 16.6.
1. Describe or sketch the surface with the given vector function.
a. r(u, v) = hu + v, 3 − v, 1 + 4u + 5vi
b. r(u, v) = h2 sin u, 3 cos u, vi
c. r(s, t) = hs, t, t2 − s2 i
d. r(s, t) = hs sin 2t, s2 , s cos 2ti
2. Find a vector function r(u, v) for the surface.
a. The plane that passes through the point (1, 2, −3) and is parallel to the vectors h1, 1, −1i
and h1, −1, 1i.
b. The lower half of the ellipsoid 2x2 + 4y2 + z 2 = 1.
c. The part of the sphere of radius 4 centered at the origin that lies between the planes
z = −2 and z = 2.
3. Find the area of the portion of x + 2y + 4z = 10 in the first octant. ⇒
4. Find the area of the portion of 2x + 4y + z = 0 inside x2 + y2 = 1. ⇒
5. Find the area of z = x2 + y2 that lies below z = 1. ⇒
p
6. Find the area of z = x2 + y2 that lies below z = 2. ⇒
7. Find the area of the portion of x2 + y2 + z 2 = a2 that lies in the first octant. ⇒
8. Find the area of the portion of x2 + y2 + z 2 = a2 that lies above x2 + y2 ≤ b2 , b ≤ a. ⇒
9. Find the area of z = x2 − y2 that lies inside x2 + y2 = a2 . ⇒
10. Find the area of z = xy that lies inside x2 + y2 = a2 . ⇒
11. Find the area of x2 + y2 + z 2 = a2 that lies above the interior of the circle given in polar
coordinates by r = a cos θ. ⇒
p
12. Find the area of the cone z = k x2 + y2 that lies above the interior of the circle given in
polar coordinates by r = a cos θ. ⇒
13. Find the area of the plane z = ax + by + c that lies over a region D with area A. ⇒
p
14. Find the area of the cone z = k x2 + y2 that lies over a region D with area A. ⇒
15. Find the area of the cylinder x2 + z 2 = a2 that lies inside the cylinder x2 + y2 = a2 . ⇒
16. The surface f (x, y) can be represented with the vector function hx, y, f (x, y)i. Set up the
surface area integral using this vector function and compare to the integral of section 15.4.
442
16.7
Chapter 16 Vector Calculus
Surfa e Integrals
In the integral for surface area,
Z
b
a
Z
d
|ru × rv | du dv,
c
the integrand |ru ×rv | du dv is the area of a tiny parallelogram, that is, a very small surface
area, so it is reasonable to abbreviate it dS; then a shortened version of the integral is
ZZ
1 · dS.
D
We have already seen that if D is a region in the plane, the area of D may be computed
with
ZZ
1 · dA,
D
so this is really quite familiar, but the dS hides a little more detail than does dA.
Just as we can integrate functions f (x, y) over regions in the plane, using
ZZ
f (x, y) dA,
D
so we can compute integrals over surfaces in space, using
ZZ
f (x, y, z) dS.
D
In practice this means that we have a vector function r(u, v) = hx(u, v), y(u, v), z(u, v)i for
the surface, and the integral we compute is
Z
a
b
Z
c
d
f (x(u, v), y(u, v), z(u, v))|ru × rv | du dv.
That is, we express everything in terms of u and v, and then we can do an ordinary double
integral.
EXAMPLE 16.7.1 Suppose a thin object occupies the upper hemisphere of x2 + y 2 +
z 2 = 1 and has density σ(x, y, z) = z. Find the mass and center of mass of the object.
(Note that the object is just a thin shell; it does not occupy the interior of the hemisphere.)
16.7
Surface Integrals
443
We write the hemisphere as r(φ, θ) = hcos θ sin φ, sin θ sin φ, cos φi, 0 ≤ φ ≤ π/2 and
0 ≤ θ ≤ 2π. So rθ = h− sin θ sin φ, cos θ sin φ, 0i and rφ = hcos θ cos φ, sin θ cos φ, − sin φi.
Then
rθ × rφ = h− cos θ sin2 φ, − sin θ sin2 φ, − cos φ sin φi
and
|rθ × rφ | = | sin φ| = sin φ,
since we are interested only in 0 ≤ φ ≤ π/2. Finally, the density is z = cos φ and the
integral for mass is
Z 2π Z π/2
cos φ sin φ dφ dθ = π.
0
0
By symmetry, the center of mass is clearly on the z-axis, so we only need to find the
z-coordinate of the center of mass. The moment around the x-y plane is
Z
2π
0
Z
π/2
z cos φ sin φ dφ dθ =
0
Z
2π
0
Z
π/2
cos2 φ sin φ dφ dθ =
0
2π
,
3
so the center of mass is at (0, 0, 2/3).
Now suppose that F is a vector field; imagine that it represents the velocity of some
fluid at each point in space. We would like to measure how much fluid is passing through
a surface D, the flux across D. As usual, we imagine computing the flux across a very
small section of the surface, with area dS, and then adding up all such small fluxes over D
with an integral. Suppose that vector N is a unit normal to the surface at a point; F · N
is the scalar projection of F onto the direction of N, so it measures how fast the fluid is
moving across the surface. In one unit of time the fluid moving across the surface will fill a
volume of F · N dS, which is therefore the rate at which the fluid is moving across a small
patch of the surface. Thus, the total flux across D is
ZZ
D
F · N dS =
ZZ
F · dS,
D
defining dS = N dS. As usual, certain conditions must be met for this to work out; chief
among them is the nature of the surface. As we integrate over the surface, we must choose
the normal vectors N in such a way that they point “the same way” through the surface.
For example, if the surface is roughly horizontal in orientation, we might want to measure
the flux in the “upwards” direction, or if the surface is closed, like a sphere, we might want
to measure the flux “outwards” across the surface. In the first case we would choose N to
have positive z component, in the second we would make sure that N points away from the
444
Chapter 16 Vector Calculus
origin. Unfortunately, there are surfaces that are not orientable: they have only one side,
so that it is not possible to choose the normal vectors to point in the “same way” through
the surface. The most famous such surface is the Möbius strip shown in figure 16.7.1. It
is quite easy to make such a strip with a piece of paper and some tape. If you have never
done this, it is quite instructive; in particular, you should draw a line down the center of
assigned to the points of the Möbius strip, there will be normal vectors very close to each
other pointing in opposite directions.
Figure 16.7.1
A Möbius strip. (AP)
Assuming that the quantities involved are well behaved, however, the flux of the vector
field across the surface r(u, v) is
ZZ
D
F · N dS =
ZZ
D
ru × rv
F·
|ru × rv | dA =
|ru × rv |
ZZ
F · (ru × rv ) dA.
D
In practice, we may have to use rv × ru or even something a bit more complicated to make
sure that the normal vector points in the desired direction.
p
EXAMPLE 16.7.2 Compute the flux of F = hx, y, z 4 i across the cone z = x2 + y 2 ,
0 ≤ z ≤ 1, in the downward direction.
We write the cone as a vector function: r = hv cos u, v sin u, vi, 0 ≤ u ≤ 2π and
0 ≤ v ≤ 1. Then ru = h−v sin u, v cos u, 0i and rv = hcos u, sin u, 1i and ru × rv =
16.7
Surface Integrals
445
hv cos u, v sin u, −vi. The third coordinate −v is negative, which is exactly what we desire,
that is, the normal vector points down through the surface. Then
Z
2π
0
Z
1
0
2π
hx, y, z i · hv cos u, v sin u, −vi dv du =
Z
=
Z
=
Z
4
Z
1
2π
Z
1
2π
Z
1
0
0
0
0
0
xv cos u + yv sin u − z 4 v dv du
0
v 2 cos2 u + v 2 sin2 u − v 5 dv du
v 2 − v 5 dv du =
π
.
3
Exercises 16.7.
1. Find the center of mass of an object that occupies the upper hemisphere of x2 + y2 + z 2 = 1
and has density x2 + y2 . ⇒
2. Find the center of
p mass of an object that occupies the surface z = xy, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
and has density 1 + x2 + y2 . ⇒
p
3. Find the center of mass of an object that occupies the surface z = x2 + y2 , 1 ≤ z ≤ 4 and
has density x2 z. ⇒
4. Find the centroid of the surface of a right circular cone of height h and base radius r, not
including the base. ⇒
ZZ
5. Evaluate
h2, −3, 4i · N dS, where D is given by z = x2 + y2 , −1 ≤ x ≤ 1, −1 ≤ y ≤ 1,
D
oriented up. ⇒
ZZ
6. Evaluate
hx, y, 3i · N dS, where D is given by z = 3x − 5y, 1 ≤ x ≤ 2, 0 ≤ y ≤ 2, oriented
up. ⇒
7. Evaluate
⇒
8. Evaluate
D
ZZ
hx, y, −2i · N dS, where D is given by z = 1 − x2 − y2 , x2 + y2 ≤ 1, oriented up.
ZZ
hxy, yz, zxi · N dS, where D is given by z = x + y2 + 2, 0 ≤ x ≤ 1, x ≤ y ≤ 1,
D
D
oriented up. ⇒
ZZ
9. Evaluate
hex , ey , zi · N dS, where D is given by z = xy, 0 ≤ x ≤ 1, −x ≤ y ≤ x, oriented
up. ⇒
10. Evaluate
up. ⇒
D
ZZ
hxz, yz, zi · N dS, where D is given by z = a2 − x2 − y2 , x2 + y2 ≤ b2 , oriented
D
11. A fluid has density 870 kg/m3 and flows with velocity v = hz, y2 , x2 i, where distances are
in meters and the components of v are in meters per second. Find the rate of flow outward
through the portion of the cylinder x2 + y2 = 4, 0 ≤ z ≤ 1 for which y ≥ 0. ⇒
446
Chapter 16 Vector Calculus
12. Gauss’s Law says that the net charge, Q, enclosed by a closed surface, S, is
ZZ
Q = ǫ0
E · N dS
where E is an electric field and ǫ0 (the permittivity of free space) is a known constant; N is
oriented outward. Use Gauss’s Law to find the charge contained in the cube with vertices
(±1, ±1, ±1) if the electric field is E = hx, y, zi. ⇒
16.8
Stokes's Theorem
Recall that one version of Green’s Theorem (see equation 16.5.1) is
Z
ZZ
F · dr =
(∇ × F) · k dA.
∂D
D
Here D is a region in the x-y plane and k is a unit normal to D at every point. If D is
instead an orientable surface in space, there is an obvious way to alter this equation, and
it turns out still to be true:
THEOREM 16.8.1 Stokes’s Theorem
Provided that the quantities involved are
sufficiently nice, and in particular if D is orientable,
Z
ZZ
F · dr =
(∇ × F) · N dS,
∂D
D
if ∂D is oriented counter-clockwise relative to N.
Note how little has changed: k becomes N, a unit normal to the surface, and dA
becomes dS, since this is now a general surface integral. The phrase “counter-clockwise
relative to N” means roughly that if we take the direction of N to be “up”, then we go
around the boundary counter-clockwise when viewed from “above”. In many cases, this
description is inadequate. A slightly more complicated but general description is this:
imagine standing on the side of the surface considered positive; walk to the boundary and
turn left. You are now following the boundary in the correct direction.
p
EXAMPLE 16.8.2 Let F = hexy cos z, x2 z, xyi and the surface D be x = 1 − y 2 − z 2 ,
oriented in the positive x direction. It quickly becomes apparent that the surface integral
in Stokes’s Theorem is intractable, so we try the line integral. The boundary of D is the
unit circle in the y-z plane, r = h0, cos u, sin ui, 0 ≤ u ≤ 2π. The integral is
Z 2π
Z 2π
xy
2
0 du = 0,
he cos z, x z, xyi · h0, − sin u, cos ui du =
0
because x = 0.
0
16.8
Stokes’s Theorem
447
EXAMPLE 16.8.3 Consider the cylinder r = hcos u, sin u, vi, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2,
oriented outward, and F = hy, zx, xyi. We compute
ZZ
∇ × F · N dS =
D
Z
∂D
F · dr
in two ways.
First, the double integral is
Z
2π
0
Z
2
0
h0, − sin u, v − 1i · hcos u, sin u, 0i dv du =
Z
2π
Z
2
0
0
− sin2 u dv du = −2π.
The boundary consists of two parts, the bottom circle hcos t, sin t, 0i, with t ranging
from 0 to 2π, and hcos t, sin t, 2i, with t ranging from 2π to 0. We compute the corresponding integrals and add the results:
Z
2π
0
2
− sin t dt +
Z
0
2π
− sin2 t + 2 cos2 t = −π − π = −2π,
as before.
An interesting consequence of Stokes’s Theorem is that if D and E are two orientable
surfaces with the same boundary, then
ZZ
(∇ × F) · N dS =
D
Z
∂D
F · dr =
Z
∂E
F · dr =
ZZ
(∇ × F) · N dS.
E
Sometimes both of the integrals
ZZ
(∇ × F) · N dS
and
D
Z
∂D
F · dr
are difficult, but you may be able to find a second surface E so that
ZZ
(∇ × F) · N dS
E
has the same value but is easier to compute.
EXAMPLE 16.8.4 In example 16.8.2 the line integral was easy to compute. But we
might also notice that another surface E with the same boundary is the flat disk y 2 +z 2 ≤ 1,
448
Chapter 16 Vector Calculus
given by r = h0, v cos u, v sin ui. The normal is rv × ru = hv, 0, 0i. We compute the curl:
∇ × F = hx − x2 , −exy sin z − y, 2xz − xexy cos zi.
Since x = 0 everywhere on the surface,
(∇ × F) · N = h0, − sin z − y, 0i · hv, 0, 0i = 0,
so the surface integral is
ZZ
0 dS = 0,
E
as before. In this case, of course, it is still somewhat easier to compute the line integral,
avoiding ∇ × F entirely.
EXAMPLE 16.8.5 Let F = h−y 2 , x, z 2 i, and let the curve C be the intersection of the
cylinder x2 + y 2 = 1 withZ the plane y + z = 2, oriented counter-clockwise when viewed
from above. We compute
F · dr in two ways.
C
First we do it directly: a vector function for C is r = hcos u, sin u, 2 − sin ui, so
r = h− sin u, cos u, − cos ui, and the integral is then
′
Z
0
2π
2
2
y sin u + x cos u − z cos u du =
Z
0
2π
sin3 u + cos2 u − (2 − sin u)2 cos u du = π.
To use Stokes’s Theorem, we pick a surface with C as the boundary; the simplest
such surface is that portion of the plane y + z = 2 inside the cylinder. This has vector
equation r = hv cos u, v sin u, 2 − v sin ui. We compute ru = h−v sin u, v cos u, −v cos ui,
rv = hcos u, sin u, − sin ui, and ru × rv = h0, −v, −vi. To match the orientation of C we
need to use the normal h0, v, vi. The curl of F is h0, 0, 1 + 2yi = h0, 0, 1 + 2v sin ui, and the
surface integral from Stokes’s Theorem is
Z
0
2π
Z
1
(1 + 2v sin u)v dv du = π.
0
In this case the surface integral was more work to set up, but the resulting integral is
somewhat easier.
16.8
Stokes’s Theorem
449
Proof of Stokes’s Theorem.
We can prove here a special case of Stokes’s Theorem,
which perhaps not too surprisingly uses Green’s Theorem.
Suppose the surface D of interest can be expressed in the form z = g(x, y), and let
F = hP, Q, Ri. Using the vector function r = hx, y, g(x, y)i for the surface we get the
surface integral
ZZ
ZZ
∇ × F · dS =
hRy − Qz , Pz − Rx , Qx − Py i · h−gx , −gy , 1i dA
D
E
=
ZZ
−Ry gx + Qz gx − Pz gy + Rx gy + Qx − Py dA.
E
Here E is the region in the x-y plane directly below the surface D.
For the line integral, we need a vector function for ∂D. If hx(t), y(t)i is a vector
function for ∂E then we may use r(t) = hx(t), y(t), g(x(t), y(t))i to represent ∂D. Then
Z
Z b
Z b
dx
dx
dy
dz
dy
∂z dx ∂z dy
P
F · dr =
P
dt.
+Q
+ R dt =
+Q
+R
+
dt
dt
dt
dt
dt
∂x dt
∂y dt
a
∂D
a
using the chain rule for dz/dt. Now we continue to manipulate this:
Z b
dy
∂z dx ∂z dy
dx
dt
+ Q +R
+
P
dt
dt
∂x dt
∂y dt
a
Z b ∂z dy
∂z dx
dt
+ Q+R
=
P +R
∂x dt
∂y dt
a
Z ∂z
∂z
=
P +R
dx + Q + R
dy,
∂x
∂y
∂E
which now looks just like the line integral of Green’s Theorem, except that the functions
P and Q of Green’s Theorem have been replaced by the more complicated P + R(∂z/∂x)
and Q + R(∂z/∂y). We can apply Green’s Theorem to get
ZZ
Z ∂z
∂z
∂
∂z
∂
∂z
dx + Q + R
dy =
Q+R
−
P +R
dA.
P +R
∂x
∂y
∂x
∂y
∂y
∂x
∂E
E
Now we can use the chain rule again to evaluate the derivatives inside this integral, and it
becomes
ZZ
Qx + Qz gx + Rx gy + Rz gx gy + Rgyx − (Py + Pz gy + Ry gx + Rz gy gx + Rgxy ) dA
E
=
ZZ
Qx + Qz gx + Rx gy − Py − Pz gy − Ry gx dA,
E
which is the same as the expression we obtained for the surface integral.
450
Chapter 16 Vector Calculus
Exercises 16.8.
1. Let F = hz, x, yi. The plane z = 2x + 2y − 1 and the paraboloid z = x2 + y2 intersect in a
closed curve. Stokes’s Theorem implies that
ZZ
I
ZZ
(∇ × F) · N dS =
F · dr =
(∇ × F) · N dS,
C
D1
D2
where the line integral is computed over the intersection C of the plane and the paraboloid,
and the two surface integrals are computed over the portions of the two surfaces that have
boundary C (provided, of course, that the orientations all match). Compute all three integrals. ⇒
2. Let D be the portion of z Z=Z 1 − x2 − y2 above the x-y plane, oriented up, and let F =
hxy2 , −x2 y, xyzi. Compute
(∇ × F) · N dS. ⇒
D
3. Let D be Z
the portion of z = 2x + 5y inside x2 + y2 = 1, oriented up, and let F = hy, z, −xi.
Compute
F · dr. ⇒
I∂D
4. Compute
x2 z dx + 3x dy − y3 dz, where C is the unit circle x2 + y2 = 1 oriented counterC
clockwise. ⇒
5. Let D be the portion of z = px + qy + r over a region in the x-y plane that
Z has area A,
oriented up, and let F = hax + by + cz, ax + by + cz, ax + by + czi. Compute
F · dr. ⇒
∂D
6. Let D be any surface and letZF = hP (x), Q(y), R(z)i (P depends only on x, Q only on y,
and R only on z). Show that
F · dr = 0.
∂D
Z
f ∇g + g∇f · dr = 0, where r describes a closed curve C to which Stokes’s
7. Show that
C
Theorem applies. (See theorems 12.4.1 and 16.5.2.)
16.9
The Divergen e Theorem
The third version of Green’s Theorem (equation 16.5.2) we saw was:
Z
ZZ
F · N ds =
∇ · F dA.
∂D
D
With minor changes this turns into another equation, the Divergence Theorem:
THEOREM 16.9.1 Divergence Theorem
Under suitable conditions, if E is a
region of three dimensional space and D is its boundary surface, oriented outward, then
ZZ
ZZZ
F · N dS =
∇ · F dV.
D
E
16.9
451
The Divergence Theorem
Proof. Again this theorem is too difficult to prove here, but a special case is easier. In
the proof of a special case of Green’s Theorem, we needed to know that we could describe
the region of integration in both possible orders, so that we could set up one double integral
using dx dy and another using dy dx. Similarly here, we need to be able to describe the
three-dimensional region E in different ways.
We start by rewriting the triple integral:
ZZZ
ZZZ
ZZZ
ZZZ
ZZZ
∇ · F dV =
(Px + Qy + Rz ) dV =
Px dV +
Qy dV +
Rz dV.
E
E
E
E
E
The double integral may be rewritten:
ZZ
ZZ
ZZ
ZZ
ZZ
F · N dS =
(P i + Qj + Rk) · N dS =
P i · N dS +
Qj · N dS +
Rk · N dS.
D
D
D
D
D
To prove that these give the same value it is sufficient to prove that
ZZ
ZZZ
P i · N dS =
Px dV,
D
E
ZZ
Qj · N dS =
ZZZ
Qy dV, and
Rk · N dS =
ZZZ
Rz dV.
D
ZZ
(16.9.1)
E
D
E
Not surprisingly, these are all pretty much the same; we’ll do the first one.
We set the triple integral up with dx innermost:
ZZZ
E
Px dV =
ZZ Z
B
g2 (y,z)
Px dx dA =
g1 (y,z)
ZZ
P (g2 (y, z), y, z) − P (g1 (y, z), y, z) dA,
B
where B is the region in the y-z plane over which we integrate. The boundary surface of
E consists of a “top” x = g2 (y, z), a “bottom” x = g1 (y, z), and a “wrap-around side”
that is vertical to the y-z plane. To integrate over the entire boundary surface, we can
integrate over each of these (top, bottom, side) and add the results. Over the side surface,
the vector N is perpendicular to the vector i, so
ZZ
ZZ
P i · N dS =
0 dS = 0.
side
side
Thus, we are left with just the surface integral over the top plus the surface integral
over the bottom. For the top, we use the vector function r = hg2 (y, z), y, zi which gives
452
Chapter 16 Vector Calculus
ry × rz = h1, −g2y , −g2z i; the dot product of this with i = h1, 0, 0i is 1. Then
ZZ
P i · N dS =
top
ZZ
P (g2 (y, z), y, z) dA.
B
In almost identical fashion we get
ZZ
ZZ
P i · N dS = −
P (g1 (y, z), y, z) dA,
B
bottom
where the negative sign is needed to make N point in the negative x direction. Now
ZZ
ZZ
ZZ
P i · N dS =
P (g2 (y, z), y, z) dA −
P (g1 (y, z), y, z) dA,
B
D
B
which is the same as the value of the triple integral above.
EXAMPLE 16.9.2 Let F = h2x, 3y, z 2i, and consider the three-dimensional volume
inside the cube with faces parallel to the principal planes and opposite corners at (0, 0, 0)
and (1, 1, 1). We compute the two integrals of the divergence theorem.
The triple integral is the easier of the two:
1
Z
1
Z
0
0
Z
1
2 + 3 + 2z dx dy dz = 6.
0
The surface integral must be separated into six parts, one for each face of the cube. One
face is z = 0 or r = hu, v, 0i, 0 ≤ u, v ≤ 1. Then ru = h1, 0, 0i, rv = h0, 1, 0i, and
ru × rv = h0, 0, 1i. We need this to be oriented downward (out of the cube), so we use
h0, 0, −1i and the corresponding integral is
Z
1
0
Z
1
2
0
−z du dv =
Z
1
0
Z
1
0 du dv = 0.
0
Another face is y = 1 or r = hu, 1, vi. Then ru = h1, 0, 0i, rv = h0, 0, 1i, and ru × rv =
h0, −1, 0i. We need a normal in the positive y direction, so we convert this to h0, 1, 0i, and
the corresponding integral is
Z
0
1
Z
0
1
3y du dv =
Z
1
0
Z
1
3 du dv = 3.
0
The remaining four integrals have values 0, 0, 2, and 1, and the sum of these is 6, in
agreement with the triple integral.
16.9
The Divergence Theorem
453
EXAMPLE 16.9.3 Let F = hx3 , y 3 , z 2 i, and consider the cylindrical volume x2 +y 2 ≤ 9,
0 ≤ z ≤ 2. The triple integral (using cylindrical coordinates) is
Z
2π
Z
3
Z
(3r 2 + 2z)r dz dr dθ = 279π.
0
0
0
2
For the surface we need three integrals. The top of the cylinder can be represented
by r = hv cos u, v sin u, 2i; ru × rv = h0, 0, −vi, which points down into the cylinder, so we
convert it to h0, 0, vi. Then
Z
2π
0
Z
3
3
3
3
3
hv cos u, v sin u, 4i · h0, 0, vi dv du =
0
Z
2π
0
Z
3
4v dv du = 36π.
0
The bottom is r = hv cos u, v sin u, 0i; ru × rv = h0, 0, −vi and
Z
2π
0
Z
3
3
0
3
3
3
hv cos u, v sin u, 0i · h0, 0, −vi dv du =
Z
0
2π
Z
3
0 dv du = 0.
0
The side of the cylinder is r = h3 cos u, 3 sin u, vi; ru × rv = h3 cos u, 3 sin u, 0i which does
point outward, so
Z
2π
0
Z
2
0
h27 cos3 u, 27 sin3 u, v 2 i · h3 cos u, 3 sin u, 0i dv du
=
Z
2π
0
Z
2
81 cos4 u + 81 sin4 u dv du = 243π.
0
The total surface integral is thus 36π + 0 + 243π = 279π.
Exercises 16.9.
1. Using F = h3x, y3 , −2z 2 i and the region bounded by x2 + y2 = 9, z = 0, and z = 5, compute
both integrals from the Divergence Theorem. ⇒
2
2
2
2. Let E be Z
the
Z volume described by 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c, and F = hx , y , z i.
Compute
F · N dS. ⇒
∂E
3. Let E be the volume described
Z Z by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and F =
x+y
h2xy, 3xy, ze
i. Compute
F · N dS. ⇒
∂E
4. Let E be the
Z Z volume described by 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ x + y, and F = hx, 2y, 3zi.
Compute
F · N dS. ⇒
∂E
454
Chapter 16 Vector Calculus
5. Let E be the volume described by x2 + y2 + z 2 ≤ 4, and F = hx3 , y3 , z 3 i. Compute
ZZ
F·
∂E
N dS. ⇒
p
6. Let E be the hemisphere described by 0 ≤ z ≤ 1 − x2 − y2 , and
ZZ
p
p
p
2
2
2
2
2
2
2
2
2
F · N dS. ⇒
F = h x + y + z , x + y + z , x + y + z i. Compute
∂E
2
2
2
2
7. Let
Z Z E be the volume described by x + y ≤ 1, 0 ≤ z ≤ 4, and F = hxy , yz, x zi. Compute
F · N dS. ⇒
∂E
8. Let E be the solid cone above the x-yZ Zplane and inside z = 1 −
p
F · N dS. ⇒
hx cos2 z, y sin2 z, x2 + y2 zi. Compute
p
x2 + y2 , and F =
∂E
9. Prove the other two equations in the display 16.9.1.
10. Suppose D is a closed surface, and that D and F are sufficiently nice. Show that
ZZ
(∇ × F) · N dS = 0
D
where N is the outward pointing unit normal.
11. Suppose D is a closed surface, D is sufficiently nice, and F = ha, b, ci is a constant vector
field. Show that
ZZ
F · N dS = 0
D
where N is the outward pointing unit normal.
12. We know that the volume of a region E may often be computed as
ZZZ
dx dy dz. Show that
E
1
this volume may also be computed as
3
ZZ
∂E
unit normal to ∂E.
hx, y, zi · N dS where N is the outward pointing
```
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# Colorado - Grade 1 - Math - Geometry - Partitioning Into Equal Shares - 1.G.3
### Description
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
• Standard ID - 1.G.3
• Subjects - Math Common Core
### Keywords
• Math
• Geometry
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
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# Theorems on Straight Lines and Plane
Here we will discuss about the theorems on straight lines and plane using step-by-step explanation on how to proof the theorem.
Theorem: If a straight line is perpendicular to each of two intersecting straight lines at their point of intersection, it is also perpendicular to the plane in which they lie.
Let the straight line OP be perpendicular to each of two intersecting straight lines OM and ON at their point of intersection O and XY be the plane in which OM and ON lie. We are to prove that the straight line OP is perpendicular to the plane XY.
Construction: Through O draw any straight line OC in the XY plane and take any point C on it. Now, complete the parallelogram OACB in the XY plane by drawing lines CB and CA parallel to OM and ON respectively. Join AB, which cuts OC at D. Join PA, PB and PD.
Proof: Since OACB is a parallelogram and its two diagonals AB and OC intersect at D, hence D is the mid-point of AB (Since, diagonals of a parallelogram bisect each other).
Therefore, PD is a median of the triangle APB; hence, by Apollonius theorem we get,
AP² + BP² = 2 (AD² + PD²) . . . (1)
Again, OC is a median of the triangle OAB; hence, by the same theorem we get,
OA² + OB² = 2 (AD² + OD²) . . . (2)
Subtracting (2) from (1) we get,
(AP² - OA² ) + (BP² - OB² ) = 2 (PD² - OD² ) . . . (3)
Now, OP is perpendicular to both OA and OB.
Therefore, AP² = OA² + OP²
or, AP² – OA² = OP² . . . (4)
and BP² = OB² + OP ²
or, BP ² - OB² = OP² . . . (5)
From (3), (4) and (5) we get,
OP² + OP² = 2 (PD² - OD²)
or, 2. OP ² = 2 (PD² - OD²)
or, OP ² = PD² - OD²
or, OP ² + OD² = PD²
Therefore, ∠POD (i.e., ∠POC) is a right angle.
Therefore, OP is perpendicular to OC at O. But OC is any straight line through O in the plane XY. Therefore, OP is perpendicular to the plane XY at O.
Examples:
1. O is a point in the plane of the triangle ABC; if X be a point outside the plane such that PO is perpendicular to both OA and OB and if XA = XB = XC, show that O is the circum-centre of the triangle ABC.
Since XO is perpendicular to both OA and OB at their point of intersection O, hence, XO is perpendicular to the plane of the triangle ABC. Therefore, XO is perpendicular to OC.
Now, in triangles XOA and POB we have
XA = XB (given), XO is common and ∠XOA = ∠XOB (each being a right angle)
Therefore, triangles XOA and XOB are congruent.
Therefore, OA = OB . . . (1)
Similarly, in triangles XOA and XOC we have,
XA = XC (given), XO is common and ∠XOA = ∠XOC = 1 rt. angle.
Therefore, triangles POA and POC are congruent
Therefore, OA = OC . . . (2)
From (1) and (2) we get, OA = OB = OC
Therefore, O is the circum-centre of the triangle ABC.
2. The straight line PQ is perpendicular to a plane ; in this plane the straight line QT is perpendicular to a straight line RS at T. Show that RT is perpendicular to the plane containing PT and QT.
Let PQ be perpendicular to the plane XY at Q. In XY plane, draw QT perpendicular to a straight line RQ, T being the foot of the perpendicular. Join PR, QR and PT.
It is required to prove that RT is perpendicular to the plane containing PT and QT.
Since PQ is perpendicular to the plane XY and the lines QR and QT lie in this plane, hence PQ is perpendicular to both QR and QT. Therefore, from the right-angled △ PQR we get,
PQ² + QR² = PR²
or, PQ² = PR² - QR² . . . (1)
Again, from the right-angled △ PQT we get,
QT² = PQ² + QT² = PR² – QR² + QT² [using (1)]
= PR² - (QR² - QT²)
= PR² - RT²
[Since, QT ⊥ RT Therefore QR² = QT² + RT² or, QR² – QT² = RT²] Or, TR ² = QT ² + RT²
Therefore, PT ⊥ RT i.e., RT is perpendicular to PT.
Again, RT is perpendicular to QT (given). Thus, RT is perpendicular to both PT and QT.
Therefore, RT is perpendicular to the place containing PT and QT.
3. ABC is a triangle right – angled at C.P is a point outside the plane ABC such that PA = PB = PC. If D be the mid-point of AB, prove that PD is perpendicular to CD. Show also that PD is perpendicular to the plane of the triangle ABC.
By question ACB = 1 rt and D is the mid-point of the hypotenuse AB in ABC.
Therefore, AD = BD = CD.
Now, in triangle PDA and PDB we have
PA = PB (given), AD = BD and PD is common. Therefore, the triangle is congruent.
Therefore PDA = PDB = ½ ∙ 2 rt. Angles
= 1 rt. Angle.
i.e., PD is perpendicular to DA
Again, in triangle PDA and PDC we have,
PA = PC (given), AD = DC and PD is common.
Therefore, the triangles are congruent.
Therefore, PDC = PDA = 1 rt. Angle.
i.e., PD is perpendicular to DC.
Therefore, PD is perpendicular to both DA and CD i.e., PD is perpendicular to the plane containing DA and DC i.e., it is perpendicular to the plane of the triangle ABC.
`
Geometry
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Derive the Trigonometric Identities (Complex Plane)
This example demonstrates how to derive the trigonometric identities using the geometry of the complex plane. This is a very elegant way to derive trigonometric identities, but it requires an understanding of complex numbers and Euler’s Formula.
Background
Given an angle as input, Euler’s formula returns a point on the unit circle in the complex plane. The formula relates complex input to the exponential function to the trigonometric functions sine and cosine as shown in the figure below.
The formula can be visualized as a number in the complex plane on the unit circle. When given the angle and the complex constant as input, the exponential function returns a complex number on the unit circle.
The reason this formula is useful when deriving trigonometric identities is because of the properties of the exponential function.
Derive Sum of Two Angles Identities (Complex Plane)
This example derives the sum of two angles identities using the properties of complex numbers and Euler’s formula.
Steps
1. The sum of two angles (alpha) and (beta) form a complex number on the unit circle in the complex plane. From Euler’s formula, we know this complex number can be expressed using cosine and sine.
This complex number can be visualized as a point in the complex plane, where the horizontal coordinate is equal to and the vertical coordinate is equal to .
2. Distribute the complex constant for both angles.
Apply the addition property of the exponential function to expand the right-hand side.
3. Apply Euler’s formula to both sides of the equation.
Multiply the right-hand side of the equation.
Substitute for on the right-hand side.
The geometry of the expressions on the right-hand can side can be visualized by drawing the trigonometry of the two angles on the unit circle.
These lengths could be solved-for, as shown in this example, but here they are the natural result of the multiplication of two complex numbers.
4. Equate the real part and complex parts together.
Divide both sides of the second expression by to get the two summation identities.
Derive Difference of Two Angles Identities (Complex Plane)
This example demonstrates how the difference of two angles identities can be derived using the properties of complex numbers and Euler’s formula. The two difference identities are shown below:
Steps
1. Observe that the angle (theta) forms the complex number on the unit circle given by Euler’s formula.
This complex number is visualized by the figure below:
2. Model the difference of the two angles (alpha) and (beta) as input to the exponential function.
Distribute the complex constant for both angles.
3. Apply the addition property of the exponential function to expand to right-hand side.
4. Apply Euler’s formula to both sides of the equation where the exponential function appears. Note, when applying Euler’s formula for the negative angle the following symmetry can be observed.
This results in the following expression.
Multiply the right-hand side of the equation.
Substitute for in the expression.
Simplify and group the real and complex parts of the right-hand side.
5. Equate the real parts of the complex number and the complex parts of the number.
Divide both sides of the second expression by to finish deriving the two summation identities.
Derive Double Angles Identities (Complex Plane)
This example demonstrates how to derive the double angle identities using the properties of complex numbers in the complex plane. The double-angle identities are shown below.
Steps
1. Observe that the angle (theta) forms the complex number on the unit circle given by Euler’s formula.
This complex number is visualized by the figure below:
2. Model the double angle (alpha) as the input to the exponential function.
Distribute the complex constant .
3. Apply the addition property of the exponential function to expand to right-hand side.
4. Apply Euler’s formula to both sides of the equation where the exponential function appears.
Multiply the right-hand side of the equation.
Substitute for in the expression.
5. Equate the real parts and complex parts of the two complex numbers.
This gives us the cosine double angle identity.
And the sine double angle identity.
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Challenge #1 Solution
Read on for the solution to Challenge #1… but give it a try first if you haven’t already!
So let’s find a value for that strange infinite beast from Challenge #1, which is known as an infinite continued fraction.
We’ll start by using a time-tested problem solving technique: NAME AND CONQUER. (In other words, one of the first steps toward getting a handle on a difficult or complex expression is often to simply give it a name.) Let’s call our infinite continued fraction x:
$\displaystyle x = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}$
And now for the “creative leap” I hinted at in the challenge: notice that if we look only at the expression in the denominator of the first fraction, it looks like this:
$1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}$
But wait a minute! That’s just the same thing as x itself. In other words, x seems to contain entire copies of itself as subparts! Since the denominator of the first fraction is equal to the expression as a whole, we can replace the denominator by x, like this:
$\displaystyle x = 1 + \frac{1}{x}$
You can see why it was so helpful to give our expression a name! Now we can proceed to solve for x. First, we multiply both sides of the equation by x:
$\displaystyle x^2 = x + 1$
Now, moving everything to the left side, we get a quadratic equation which we can solve using the quadratic formula:
$\begin{array}{rcl} x^2 - x - 1 & = & 0 \\ x & = & \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} \\ x & = & \frac{1 \pm \sqrt{5}}{2} \end{array}$
The positive solution, $\frac{1 + \sqrt{5}}{2} \approx 1.618$, is the one we want — this is the “value” of the infinite continued fraction. One way to think about this is that if you try various “stopping points” you will see that the values you get seem to be getting closer and closer to 1.618 (more on this later):
$\begin{array}{rcl} 1 & = & 1 \\ 1 + \frac{1}{1} & = & 2 \\ 1 + \cfrac{1}{1 + \cfrac{1}{1}} & = & 1.5 \\ 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1}}} & = & 1.666\dots \end{array}$
and so on. It turns out that this is actually a very famous — and beautiful — number with a special name: the golden ratio, or sometimes the golden mean. It is so special and famous that it has its own special symbol, phi (pronounced “fie”, not “fee”), which looks like this: $\phi$. (Phi is the twenty-first letter of the Greek alphabet.)
$\displaystyle 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}} = \frac{1 + \sqrt{5}}{2} = \phi = 1.6180339887\dots$
In subsequent posts I’ll explore some of the reasons that the golden ratio is such a beautiful and famous number — it shows up in a lot of different places, some quite unexpected! See Challenge #2 for more infinite expressions to evaluate…
Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in famous numbers, golden ratio, infinity, solutions. Bookmark the permalink.
7 Responses to Challenge #1 Solution
1. Dylan says:
What problems might arise when you try and determine the 200th term of the fraction?
2. Ashley says:
What problems arise when you try to work out the 200th term in the infinite continued fraction used in challenge 1
3. Brent says:
Dylan, Ashley: I’m not entirely sure what you’re asking, but I can try to answer.
Here’s how you can work out successive approximations to the infinite continued fraction. Start with the number 1. Then, at each step, find the reciprocal of the current term, and add 1. If you keep track of this as a fraction, there are no problems; all the arithmetic involves integers. So
1/1 + 1 = 2
1/2 + 1 = 1/2 + 2/2 = 3/2
1/(3/2) + 1 = 2/3 + 3/3 = 5/3
1/(5/3) + 1 = 3/5 + 5/5 = 8/5
and so on. Now, doing this by hand out to the 200th term would be extremely tedious and error-prone. So you would probably write a computer program. You might run into problems here if you use a language with limited-precision integers, since the 200th term expressed as a fraction involves rather large integers. In fact, the 200th term is 734544867157818093234908902110449296423351 over 453973694165307953197296969697410619233826. But if you use a language with arbitrary-precision integers (such as Ruby, Haskell, J, Mathematica, or even Java’s BigInteger class), this is no problem. Another problem would arise if you try to use a floating-point (decimal) data type to compute the 200th term instead of explicit fractions; standard floating-point data types do not have enough precision to show any difference between (say) the 199th and 200th terms.
4. Brian says:
For all you guys obviously working on your IB portfolios:
I think that all the teacher wants to know about the 200th term is that you understand the continued fraction has no decimal or fractal equivalent, at least not using real numbers (Hint!). All I said is that by the time one calculates the 200th term, it’s obvious the value won’t solidify and therefore it must be expressed in another fashion. This assumption is further evidenced by the fact the next question asks you to find an exact value for the fraction.
5. rongrong says:
i dont really get what you meant bythe continued fraction no decimal or fractal equivlen. and why the value will not solidfy and how can it e expressed in another fashion?
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# Graphing inequalities (x-y plane) review
We graph inequalities like we graph equations but with an extra step of shading one side of the line. This article goes over examples and gives you a chance to practice.
The graph of a two-variable linear inequality looks like this:
It's a line with one side shaded to indicate which x-y pairs are solutions to the inequality.
In this case, we can see that the origin left parenthesis, 0, comma, 0, right parenthesis is a solution because it is in the shaded part, but the point left parenthesis, 4, comma, 4, right parenthesis is not a solution because it is outside of the shaded part.
Want a video introduction to graphing inequalities? Check out this video.
### Example 1
We want to graph 4, x, plus, 8, y, is less than or equal to, minus, 24.
So, we put it in slope-intercept form:
\begin{aligned}4x+8y&\leq -24\\\\ 8y&\leq -4x-24\\\\ y&\leq-\dfrac{4}{8}x-3\\\\ y&\leq-\dfrac{1}{2}x-3 \end{aligned}
Notice:
• We shade below (not above) because y is less than (or equal to) the other side of the inequality.
• We draw a solid line (not dashed) because we're dealing with an "or equal to" inequality. The solid line indicates that points on the line are solutions to the inequality.
Want to see another example but in video form? Check out this video.
### Example 2
We want to graph minus, 12, x, minus, 4, y, is less than, 5.
So, we put it in slope-intercept form:
\begin{aligned}-12x-4y&< 5\\\\ -4y&< 12x+5\\\\ y&>-3x-\dfrac{5}{4} \end{aligned}
Notice:
• We shade above (not below) because y is greater than the other side of the inequality.
• We draw a dashed line (not solid) because we aren't dealing with an "or equal to" inequality. The dashed line indicates that points on the line are not solutions of the inequality.
### Example 3
We're given a graph and asked to write the inequality.
Looking at the line, we notice:
• y-intercept is start color #7854ab, minus, 2, end color #7854ab
• Slope is start fraction, delta, y, divided by, delta, x, end fraction, equals, start fraction, 4, divided by, 1, end fraction, equals, start color #e07d10, 4, end color #e07d10
The slope-intercept form of the inequality is
y, space, question mark, space, start color #e07d10, 4, end color #e07d10, x, start color #7854ab, minus, 2, end color #7854ab
where the "?" represents the unknown inequality symbol.
Notice:
• The graph is shaded above (not below), so y is greater than the other side of the inequality.
• The graph has a dashed line (not solid), so we aren't dealing with an "or equal to" inequality.
Therefore, we should use the greater than symbol.
y, is greater than, 4, x, minus, 2
Want to see another example but in video form? Check out this video.
## Practice
Problem 1
• Current
Which graph represents 8, x, minus, 5, y, is less than, 3?
## Want to join the conversation?
• Can anybody give me some tips on this subject, it's still kind of confusing. Thanks
• Hi! I know this is late and that you 100% won't see this comment, BUT I like to help and LOVE math. So, here's my tip: when looking to find the graph of an inequality, look at inequality sign first. If it has a line directly below it, it is deemed inclusive, indicating a solid line. If there is no line under the inequality sign, it is deemed non-inclusive, indicating a dashed line. Then, look at the the y term--not y-intercept. Take note of it's value. If it is a negative you are going to want to flip the direction of the sign. For instance, if you have the linear inequality -5y>8x+1, you might initially assume that the solutions to the inequality will be represented by shading the half plane that is above the y-intercept 1, but this is incorrect. In order to isolate the y variable we have to divide it by -5, along with other expression of the inequality (8x+1). Hence, we FLIP the original greater than sign (>) to a less than sign (<), which changes the entire format of the graph (or at least the solutions to the problem).
• Am i the only human here? or is it all just bots?
• There are lots of humans here. We're not bots.
• How do you graph x>= -2 , and why do you graph it vertically? and how do you know which side to shade? thank you, this is so confusing.
• x = -2 is a vertical line. It contains all points on the xy-plane where the value of x is -2.
To graph x ≥ -2, you have to know that ≥ is the greater than or equal to symbol.
The equal part means you'll need to use a solid line on the boundary itself (x = -2).
The greater than part means you'll need to shade the side of the line that has values of x that are more than -2. On an x-axis that is scaled and numbered properly, all the numbers more than -2 are clearly labeled on the right side of the vertical line.
That's how you know which side to shade!
• How does this help us in the real world?
• if you wanna be a financialist
• Can anyone answer this for me:
Choose the graph of inequality x > -2
• to find the graph of an inequality it is just like finding the graph of en equation. So first pretend it is x = -2
Now, the two extra steps are look at if it is just greater than or less than, or if it is also equal to. if it is greater than or less than the line of the graph is dashed. if it is greater than or equal to OR less than or equal to than it is a solid line like in a normal equation.
The second step is then to find where you shade in. With a linear equation it's super easy. If this were y > -2 you would shade above the line, so on the positive side. if it were y < -2 you would shade below the line, so on the negative side. So look at yours, check if you want to shade on the positive side or neagative side of the line, then determine which is which.
Can you handle it from there?
• How do you graph with two inequalities??
• Just plot both lines on the graph and make sure to use the right y-intercept and if it's not an equal to sign make the line dotted
• Why is Khan Academy so hard to learn? I don't feel that it is helping me in any way.
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One tool that can be used is Quadratic congruence solver. We can help me with math work.
## The Best Quadratic congruence solver
There is Quadratic congruence solver that can make the technique much easier. R is a useful tool for solving for radius. Think of it like a ruler. If someone is standing in front of you, you can use your hand to measure their height and then use the same measurement to determine the radius of their arm. For example, if someone is 5 feet tall and has an arm that is 6 inches long, their radius would be 5 inches. The formula for calculating radius looks like this: [ ext{radius} = ext{length} imes ext{9} ] It's really just making the length times 9. So, if they're 6 inches tall and their arm is 6 inches long, their radius would be 36 inches. Using R makes sense when you are trying to solve for any other dimension besides length - such as width or depth. If a chair is 4 feet wide and 3 feet deep, then its width would be equal to half its depth (2 x 3 = 6), so you could easily calculate its width by dividing 2 by 1.5 (6 ÷ 2). But if you were trying to figure out the chair's height instead of its width, you would need an actual ruler to measure the distance between the ground and the seat. The solution to this problem would be easier with R than without it.
One important thing to remember about solving absolute value equations is that you can only use addition and subtraction operations when solving them. You can’t use multiplication or division to solve absolute value equations because those operations change the number in the equation rather than just finding its absolute value. To solve absolute value equations, all you have to do is add or subtract one number from both sides of the equation until you get 0 on one side and then subtract that number from both sides again until you get 0 on both sides. Example: Find the absolute value of 6 + 4 = 10 Subtracting 4 from both sides gives us 2 math>egin{equation} ext{Absolute Value} end{equation} The absolute value of a number x is the distance between 0 and x, or egin{equation}label{eq:absv} ext{x}} Therefore, egin
The best world problem solver is the one that can see a problem at its earliest stages and come up with solutions before it becomes an emergency. While most of us think of problem solving as a matter of coming up with solutions to specific problems, the best problem solvers see problems as opportunities. They are alert to the warning signs that something is going wrong and are able to anticipate what might happen next. They are also skilled at connecting seemingly unconnected dots and anticipating how things will play out. The best world problem solvers are careful observers who look for patterns everywhere. They can quickly spot when a situation is out of whack and why. They are also skilled at predicting how people are likely to react in certain situations. These skills set them apart from everyone else, allowing them to take action before things get out of hand. In short, they are the best world problem solvers because they see all sides of a situation, know how to anticipate it, and how to deal with it before it becomes an emergency.
Math word problems are an important part of pre-school and elementary school math classes. These problems help students practice problem solving, logic, and math vocabulary. In addition, they provide a fun way for students to practice spelling words like “more” and “to.” With that being said, it can be difficult for parents to find free math word problems apps that work well with their kids. Fortunately, there is an app that solves this problem: Math Word Pro. This app offers a wide variety of math word problems that cover different subjects like grammar, geometry, and fractions. Each problem includes a learning objective, step-by-step instructions on how to solve the problem, and tips on how to improve your math skills. This makes Math Word Pro an ideal app for both children and adults who need help with math word problems.
Linear equations are mathematical equations that have one variable in terms of the other. For example, if you have a 2x2 table, an equation could be written as 2 + 2 = 4. This equation could be used to put together the pieces of the puzzle by adding or subtracting the corresponding numbers. If you have a 3x3 table, an equation could be written as 3 + 3 = 6. An important thing to remember about linear equations is that they are always true (assuming they make sense). As you can see in the examples above, this means that if you add or subtract variables, you will always get the same answer. The only way to get a different result is if there is a typo or some other mistake in your math.
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Ivanka Watson
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# Factors of 578: Prime Factorization, Methods, and Examples
578 is an even composite number which implies it has more than 2 factors. The factors of 577 are the numbers that follow the divisibility rule. The divisibility rules say that one number is divisible by the other if the remainder we get on their division is zero.Â
578 has both positive and negative factors that are the additive inverse of each other.
### Factors of 578
Here are the factors of number 578.
Factors of 578: Â 1, 2, 17, 34, 289 and 578
### Negative Factors of 578
The negative factors of 578Â are similar to its positive aspects, just with a negative sign.
Negative Factors of 578: – 1, -2, -17, -34, -289 and -578
### Prime Factorization of 578
The prime factorization of 578Â is the way of expressing its prime factors in the product form.
Prime Factorization: 2 x 17$^2$Â
In this article, we will learn about the factors of 578 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 578?
The factors of 578 are  1, 2, 17, 34, 289 and 578. These numbers are the factors as they do not leave any remainder when divided by 578.
The factors of 589 are classified as prime numbers and composite numbers. The prime factors of the number 578 can be determined using the prime factorization technique.
## How To Find the Factors of 578?
You can find the factors of 578Â by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 578, create a list containing the numbers that are exactly divisible by 578 with zero remainders. One important thing to note is that 1 and 578 are the 578’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 578 are determined as follows:
$\dfrac{578}{1} = 578$
$\dfrac{578}{2} = 289$
$\dfrac{578}{17} = 34$
$\dfrac{578}{34} = 17$
$\dfrac{578}{289} = 2$
$\dfrac{578}{578} = 1$
Therefore, Â 1, 2, 17, 34, 289, and 578 are the factors of 578.
### Total Number of Factors of 578
For 578, there are 6Â positive factors and 6Â negative ones. So in total, there are 12 factors of 578.Â
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 578 is given as:
Factorization of 578 is  1 x 2 x 17 x 17.
The exponent of 1 and 2 is 1 and 17 is 2.
Adding 1 to each and multiplying them together results in 12 .
Therefore, the total number of factors of 578 is 12. 6 is positive, and 6 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 578 by Prime Factorization
The number 578 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 578 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 578, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 578Â can be expressed as:
578 = 2 x 17 x 17
## Factors of 578 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 578, the factor pairs can be found as:
1 x 578 = 578
2 x 289 = 578Â
17 x 34 = 578Â
The possible factor pairs of 578 are given as (1, 578), (1, 289), and (17, 34 ).
All these numbers in pairs, when multiplied, give 578 as the product.
The negative factor pairs of 578 are given as:
-1 x -578 = 578Â
-2 x -289 = 578Â
-17 x -34 = 578Â
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -17, -34, -289 and -578 are called negative factors of 578.
The list of all the factors of 578, including positive as well as negative numbers, is given below.
Factor list of 578: 1, -1, 2, -2, 17, -17, 34, and -34
## Factors of 578 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 578 are there?
### Solution
The total number of Factors of 578 is 6.
Factors of 578 are 1, 2, 17, 34, 289, and 578.
### Example 2
Find the factors of 578 using prime factorization.
### Solution
The prime factorization of 578 is given as:
578 $\div$ 2 = 289Â
289 $\div$ 17 = 17Â
17 $\div$ 17 = 1Â
So the prime factorization of 578 can be written as:
2 x 17$^2$ = 578
Factors of 577 | Factors List | Factors of 579
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# If a^2 - 6b^2 - ab = 0, what is value of b/a ?
Aug 14, 2018
$- \frac{1}{2} , \mathmr{and} , \frac{1}{3}$.
#### Explanation:
Given that, ${a}^{2} - 6 {b}^{2} - a b = 0$.
$\therefore 6 {b}^{2} + a b - {a}^{2} = 0$.
$\therefore \underline{6 {b}^{2} + 3 a b} - \underline{2 a b - {a}^{2}} = 0$.
$\therefore 3 b \left(2 b + a\right) - a \left(2 b + a\right) = 0$.
$\therefore \left(2 b + a\right) \left(3 b - a\right) = 0$.
$\therefore 2 b + a = 0 , \mathmr{and} , 3 b - a = 0$.
$\therefore 2 b = - a , \mathmr{and} , 3 b = a$.
$\therefore \frac{b}{a} = - \frac{1}{2} , \mathmr{and} , \frac{b}{a} = \frac{1}{3}$.
Aug 14, 2018
$\frac{b}{a} = \frac{1}{3} \mathmr{and} \frac{b}{a} = - \frac{1}{2}$
#### Explanation:
Here ,
${a}^{2} - 6 {b}^{2} - a b = 0$
Dividing each term by ${a}^{2} \ne 0$
$1 - 6 {b}^{2} / {a}^{2} - \frac{b}{a} = 0$
$\implies 6 {\left(\frac{b}{a}\right)}^{2} + \frac{b}{a} - 1 = 0$
For simplicity take $\frac{b}{a} = x$
$\therefore 6 {x}^{2} + x - 1 = 0$
$\implies 6 {x}^{2} + 3 x - 2 x - 1 = 0$
$\implies 3 x \left(2 x + 1\right) - 1 \left(2 x + 1\right) = 0$
$\implies \left(3 x - 1\right) \left(2 x + 1\right) = 0$
$\implies 3 x - 1 = 0 \mathmr{and} 2 x + 1 = 0$
$\implies x = \frac{1}{3} \mathmr{and} x = - \frac{1}{2}$
Subst. back $x = \frac{b}{a}$
$\therefore \frac{b}{a} = \frac{1}{3} \mathmr{and} \frac{b}{a} = - \frac{1}{2}$
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# Casework
In combinatorics, casework is a counting method where one splits a problem into several parts, counts these cases individually, then adds together each part's total. Casework is a very general problem-solving approach, and as such has wide applicability.
## Examples
Here are some examples of problems that are solved by breaking them down into several cases. While the problems presented here can be completely solved by casework, most problems in the wild cannot. However, in a sizeable amount of math problems (not just combinatorics), casework is a crucial intermediate step.
While there are problems where casework produces the most elegant solution, in those where a shorter answer exists, casework may be considered brute force. This is especially true if that alternative solution uses complementary counting.
### Example 1
How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters.
Solution: We divide the problem into cases, based on how long the word is.
Case 1: The word is one letter long. Clearly, there are $5$ of these words.
Case 2: The word is two letters long. Constructing the set of these words, there are $5$ options for the first letter and $5$ options for the second letter, so there are $5^2 = 25$ of these words.
Case 3: The word is three letters long. By similar logic as above, we have $5$ options for the first letter, $5$ options for the second, and $5$ options for the third. Then there are $5^3 = 125$ of these letters.
Adding all our cases up, there are $5 + 25 + 125 = 155$ words that are less than four letters long and contain only the letters A, B, C, D, and E. $\square$
### Example 2
How many positive integers satisfy the equation $x^4 + y < 70$?
Solution: We use casework, based on the value of x.
Case 1: $x=1$. Then this problem becomes $y<69$, so there are $68$ possible values for $y$, and thus $68$ solutions when $x$ is one.
Case 2: $x=2$. Then this problem becomes $y<54$, so there are $53$ possible values for $y$, and thus $53$ solutions when $x$ is two.
If $x=3$, the problem becomes $y<-11$. But the question mandates that $y$ is positive, so there are no solutions when $y \geq 3$. Thus, there are $68+53=121$ positive integer solutions to the equation. $\square$
### Example 3
2004 AIME II Problem 4: How many positive integers less than 10,000 have at most two different digits?
Solution: Let $A$ and $B$ be the two digits of the number. Use casework, based on how many digits the number has.
Case 1: The number is one digit. All numbers in this category satisfy the given condition, so there are $9$ of these.
Case 2: The number is two-digit. Again, all numbers in this category have two different digits, so there are $99$ of these.
Case 3: The number is three-digit. The possible cases in this category are $AAA, AAB, ABA,$ and $BAA.$ Using a constructive approach, there are $9$ digits for what the first number can be, zero excluded to keep the number three-digit. The second digit can be $9$ digits, with zero included and the first digit's number removed. Then there are $9 + 3 \cdot 81 = 252$ of these numbers.
Case 4: The number is four-digit. The possible cases in this category are $AAAA, AAAB, AABA, ABAA, BAAA, AABB, ABAB,$ and $ABBA.$ Using the same logic as before, the first digit has $9$ options, as does the second. Then there are $9 + 7 \cdot 81 = 576$ of these numbers.
Thus, there are $9 + 90 + 252 + 495 = 927$ integers less than $10,000$ that have at most two different digits. $\square$
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# Solve the Bernoulli Differential equations. xy′ - y = xy^5
Solve the Bernoulli Differential equations. $xy\prime -y=x{y}^{5}$
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Here the given differential equation is
$x{y}^{{}^{\prime }}-y=x{y}^{5}$
which also can be written as
${y}^{{}^{\prime }}-\frac{y}{x}={y}^{5}$.
Let .The above De reduces to
$-\frac{{y}^{5}}{4}\frac{dv}{dx}-\frac{y}{x}={y}^{5}$
$⇒-\frac{1}{49}\frac{dv}{dx}-\frac{{y}^{-4}}{x}=1$
$⇒\frac{dv}{dx}+\frac{4v}{x}=-4$
$⇒{x}^{4}\frac{dv}{dx}+4v{x}^{3}=-4{x}^{4}$
$⇒\frac{d}{dx}\left(v{x}^{4}\right)=-4{x}^{4}$
$⇒v{x}^{4}=-4\int {x}^{4}dx+c$
$⇒v{x}^{4}=-\frac{4{x}^{5}}{5}+c$
$⇒v=-\frac{4x}{5}+c{x}^{-4}$
$⇒{y}^{-4}=-\frac{4x}{5}+c{x}^{-4}$
$⇒y={\left(-\frac{4x}{5}+c{x}^{-4}\right)}^{4}$
where c is a arbitrary constant.
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# Problem of the Week
## Updated at Oct 28, 2013 9:02 AM
How can we solve for the derivative of $$\frac{\sin{x}}{\cos^{2}x}$$?
Below is the solution.
$\frac{d}{dx} \frac{\sin{x}}{\cos^{2}x}$
1 Use Quotient Rule to find the derivative of $$\frac{\sin{x}}{\cos^{2}x}$$. The quotient rule states that $$(\frac{f}{g})'=\frac{f'g-fg'}{{g}^{2}}$$.$\frac{\cos^{2}x(\frac{d}{dx} \sin{x})-\sin{x}(\frac{d}{dx} \cos^{2}x)}{\cos^{4}x}$2 Use Trigonometric Differentiation: the derivative of $$\sin{x}$$ is $$\cos{x}$$.$\frac{\cos^{3}x-\sin{x}(\frac{d}{dx} \cos^{2}x)}{\cos^{4}x}$3 Use Chain Rule on $$\frac{d}{dx} \cos^{2}x$$. Let $$u=\cos{x}$$. Use Power Rule: $$\frac{d}{du} {u}^{n}=n{u}^{n-1}$$.$\frac{\cos^{3}x-\sin{x}\times 2\cos{x}(\frac{d}{dx} \cos{x})}{\cos^{4}x}$4 Use Trigonometric Differentiation: the derivative of $$\cos{x}$$ is $$-\sin{x}$$.$\frac{1}{\cos{x}}+\frac{2\sin^{2}x}{\cos^{3}x}$Done1/cos(x)+(2*sin(x)^2)/cos(x)^3
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# 180 Days of Math for Fifth Grade Day 102 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 102 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 102
Directions: Solve each problem.
Question 1.
Subtract 48 from 96.
48
Explanation:
Question 2.
18 • 6 = __________
108
Explanation:
Question 3.
44
Explanation:
Question 4.
Is 67,106 less than 76,106?
Yes.
Explanation:
67,106 comes 1st in the number sequence than 76,106.
We can subtract the numbers also.
Question 5.
Write the percentage for the shaded part on the hundreds square.
35%
Explanation:
The shaded part is of 35 squares out of 100 squares. so we can say it as 35% of shaded part.
Question 6.
72 ÷ 12 + 15 = ___________
21
Explanation:
Given, 72 ÷ 12 + 15
First let us calculate 72 ÷ 12
72 ÷ 12 = 6
6 + 15
= 21
Question 7.
16
Explanation:
Given, 38 + ______ = 54
Let us consider the blank space be ‘x’.
38 + x = 54
Subtract 38 on both sides.
38 + x – 38 = 54 – 38
x = 16
Question 8.
4 feet = __________ inches
48 inches
Explanation:
1 foot = 12 inches
4 feet = 4 × 12
= 48
Question 9.
Do intersecting lines meet at a 90° angle?
Yes.
Explanation:
Yes, Two intersecting lines when meet at 90 degrees angle are called Perpendicular lines.
Question 10.
If membership in the football club increases by 25%, how many members will it have?
Insufficient data.
Question 11.
In a game, the probability that a spinner will land on a 3 is $$\frac{2}{5}$$. How many times would you expect to land on 3 if you spin the spinner 15 times?
6
Explanation:
The probability that a spinner will land on a 3 = (2/5)
The probability to land on 3 when we spin 15 times = 15 ×(2/5)
= 6
Question 12.
Mom bought 3 pounds of bananas at the store. The bananas cost 89 cents a pound. If she paid for the bananas with a five-dollar bill, how much change did she get back?
Explanation:
The cost of bananas cost per pound = 89 cents
The cost of 3 pound bananas = 3 × 89
= 267 cents
As 1 dollar = 100 cents, The coast of 3 pounds of bananas =2.67 dollars
Mom gave 5 dollars bill, the amount she gets back = 5.00 – 2.67
= 2.33
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# How to Explain Ratios
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People often explain ratios as comparisons between two things. It is a statement that shows how one item compares to another. Ratios are written in several different ways, including as fractions. They are also written with the word βtoβ or with a colon. Ratios are used in many different ways. They are taught to children in school as a way of comparing things and they are commonly used in the business world through financial ratios.
Compare two items. Ratios are used as a means of comparison. For example, if you have four apples and six oranges, you can create a ratio of 4 to 6. It can also be written 4/6 or 4:6.
First comes first. In all ratios, the first item mentioned is the first number of the ratio. All ratios have two numbers. For example if you have a ratio of men to women that is 10/15, the 10 represents the number of men because it is listed first.
Simplify ratios. Because ratios are written as fractions, they can be simplified. To simplify the ratio from above that is 10/15, divide both numbers by 5 to get 2/3. 2/3 is an equivalent fraction to 10/15. This ratio simplified means that there are two men for every three women. This is equivalent to having 10 men for every 15 women.
Calculate possibilities. For this example, assume that the ratio remains consistent, but now you have 50 men. You can calculate how many women you should have based on knowing how many men you have. To calculate this, determine what number is multiplied by the 10 men to get 50 and then multiply that some number times the 15 women and the answer will be the number of women you have. To get from 10 to 50, you need to multiply the 10 by five. Multiply the 15 by five to get 75. The new ratio is now 50/75.
Learn how businesses use ratios. Businesses often use ratios to compare performance information from financial statements. A common ratio is the current ratio which is calculated by dividing the current assets by the current liabilities. When companies use financial ratios, they compare the answers to prior periods' answers to determine whether the ratio improved. They also compare the numbers to industry standards to determine how well their company is performing compared to others in the same business.
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Smartick is a fun way to learn math!
Aug05
# Adding and Subtracting Decimal Numbers Using Money as an Example
To put the addition and subtraction of decimal numbers into practice with money, it is essential to have previous knowledge about decimals and their representation with money. Once we have mastered this knowledge, it’s time to start solving problems.
## Steps to add and subtract decimal numbers:
1. Place the numbers in columns in such a way that the ones coincide with the ones, the tens with the tens, etc.
2. Add or subtract as if they were whole numbers.
3. Finally, place the decimal (the period) in the result, separating the whole part from the decimal.
When we use money on a daily basis, we should remember that money is made up of bills which correspond to the whole part, and cents which make up the decimal.
### Example of addition with decimal numbers
This summer I want to go on a beach vacation. To get from my city to Narragansett Beach it costs $5.16 and to visit Water Wizz Water Park it costs$7.40. How much am I going to spend if I decide to go to the beach one day and the park the next?
In order to complete this problem, we begin by sorting the coins according to their amounts on the table and following the steps listed above.
After completing the problem we get the sum – $12.56. The result will be represented by twelve$1 coins, five 10¢ coins, and six 1¢ coins:
### Example of subtraction with decimal numbers
Otto has $12.56 that he has saved throughout the month for his first pet. When he arrives at the store he decides to buy a small hamster that costs$7.50. How much money will he have left over after the purchase?
We represent this subtraction problem in the following way:
We have obtained $5.06 as a result. Therefore we will have five$1 coins and six 1¢ coins:
There are many different coins and bills. Instead of using five $1 coins, we could use one$5 bill and the result would still be the same.
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## 1. Linear equation
A linear equation has the following form :
In $(1)$, $a_i$ are called the coefficients and $x_i$ are called the variables (also called unknowns).
#### Example I
$(2)$ is a linear equation.
#### Example II
$(3)$ is not a linear equation.
## 2. Linear system
A linear system (also called system of linear equations) gathers several linear equations which are related to each other.
There are many methods for solving such systems. We are going to see the addition method only whose goal is to eliminate a variable by adding two equations.
#### Example III
In order to eliminate a variable by adding $(4)$ and $(5)$ together, we can for instance multiply them by $1$ and $-2$, respectively. By doing so, we get $(6)$ and $(7)$.
Let’s solve $(8)$ :
$-8y = 16 \iff$ $y = -2$
Let’s replace $y$, for example in $(5)$ (it could be any other) :
$x +2\cdot \underbrace{-2}_{y} = -7 \iff x -4 = -7 \iff$ $x = -3$
#### Example IV
Let’s eliminate $x$ by multiplying $(9)$ by $1$ and $(10)$ by $1$, and thus getting $(12)$ and $(13)$ (below left). Let’s one more time eliminate $x$ by multiplying $(10)$ by $-3$ and $(11)$ by $1$, and thus getting ($15$) and ($16$) (below right).
Let’s eliminate $y$ by multiplying $(14)$ by $1$ and $(17)$ by $2$. By doing so, we get $(18)$ and $(19)$.
Let’s solve $(20)$ :
$20z = 60 \iff$ $z=3$
Let’s replace $z$ for example in $(17)$ :
$y + 11\cdot \underbrace{3}_{z} = 31 \iff y +33 = 31 \iff$ $y = -2$
Let’s replace $y$ and $z$ for example in $(10)$ (it could be any other) :
$x + \underbrace{-2}_{y} -3\cdot \underbrace{3}_{z} = -12 \iff x + -11 = -12 \iff$ $x = -1$
## Recapitulation
The point is to eliminate a variable by adding two equations together. For that purpose it is necessary to find the right factors. The addition generates a new equation which in turns can be used for the elimination of another variable.
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# Similar Shape Problems
In this worksheet, students solve problems involving similar shapes where the scale factor is known or can be found.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: 2D Shapes: Triangles, Quadrilaterals and Polygons
Difficulty level:
### QUESTION 1 of 10
Similar shapes have matching corresponding angles and corresponding sides which are in the same ratio.
In these two triangles, the angles match and you will notice that the second triangle has sides which are double the length of the sides of the first triangle.
The ratio of the bases is 3:6, which reduces like a fraction to 1:2.
The ratio of the heights is 7:14, which reduces like a fraction to 1:2.
Example
These two rectangles are similar.
Find the missing length.
The ratio of the bases is 4:8, which reduces to 1:2.
The big rectangle's sides are 2 times the matching sides of the small rectangle.
The missing side is 2 × 5 = 10 cm.
Look at these similar rectangles.
What is the ratio of their lengths, reduced to its lowest terms?
5:10
1:2
3:6
10:5
Look at these similar rectangles.
What is the ratio of their lengths, reduced to its lowest terms?
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
What is the ratio of their lengths, reduced to its lowest terms?
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
• Question 1
Look at these similar rectangles.
What is the ratio of their lengths, reduced to its lowest terms?
1:2
EDDIE SAYS
3:6 and 5:10 reduce like a fraction to 1:2.
• Question 2
Look at these similar rectangles.
What is the ratio of their lengths, reduced to its lowest terms?
1:2
EDDIE SAYS
7:14 and 5:10 reduce like a fraction to 1:2.
• Question 3
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
5
EDDIE SAYS
Ratio is 1:2, so missing length is half of 10.
• Question 4
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
2
EDDIE SAYS
Ratio is 1:4, so missing length is a quarter of 8.
• Question 5
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
21
EDDIE SAYS
Ratio is 1:3, so missing length is 7 × 3 = 21.
• Question 6
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
3
EDDIE SAYS
Ratio is 1:5, so missing length is 15 ÷ 5 = 3.
• Question 7
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
10
EDDIE SAYS
Ratio is 1:2, so missing length is double 5.
• Question 8
These two shapes are similar.
What is the ratio of their lengths, reduced to its lowest terms?
1:6
EDDIE SAYS
2:12 and 7:42 reduce to 1:6 like fractions.
• Question 9
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
25
EDDIE SAYS
Ratio is 1:5, so missing length is 5 × 5 = 25.
• Question 10
These two shapes are similar.
State the missing length in cm.
(Just write the number.)
21
EDDIE SAYS
Ratio is 1:3, so missing length is 3 × 7 = 21.
---- OR ----
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# Algebraic Expressions and Identities
### 9.1 What are Expressions?
In earlier classes, we have already become familiar with what algebraic expressions (or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x2, 4xy + 7 etc.
You can form many more expressions. As you know expressions are formed from variables and constants. The expression 2y – 5 is formed from the variable y and constants 2 and 5. The expression 4xy + 7 is formed from variables x and y and constants 4 and 7.
We know that, the value of y in the expression, 2y – 5, may be anything. It can be
2, 5, –3, 0, etc.; actually countless different values. The value of an expression changes with the value chosen for the variables it contains. Thus as
y takes on different values, the value of 2y – 5 goes on changing. When y = 2, 2y – 5 = 2(2) – 5 = –1; when y = 0, 2y – 5 = 2 × 0 –5 = –5, etc. Find the value of the expression 2y – 5 for the other given values of y.
Number line and an expression:
Consider the expression x + 5. Let us say the variable x has a position X on the number line;
X may be anywhere on the number line, but it is definite that the value of x + 5 is given by a point P, 5 units to the right of X. Similarly, the value of x – 4 will be 4 units to the left of X and so on.
What about the position of 4x and 4x + 5?
The position of 4x will be point C; the distance of C from the origin will be four times the distance of X from the origin. The position D of 4x + 5 will be 5 units to the right of C.
### TRY THESE
1. Give five examples of expressions containing one variable and five examples of expressions containing two variables.
2. Show on the number line x, x – 4, 2x + 1, 3x – 2.
### 9.2 Terms, Factors and Coefficients
Take the expression 4x + 5. This expression is made up of two terms, 4x and 5. Terms are added to form expressions. Terms themselves can be formed as the product of factors. The term 4x is the product of its factors 4 and x. The term 5 is made up of just one factor, i.e., 5.
### TRY THESE
Identify the coefficient of each term in the expression
x2y2 – 10x2y + 5xy2 – 20.
The expression 7xy – 5x has two terms 7xy and –5x. The term 7xy is a product of factors 7, x and y. The numerical factor of a term is called its numerical coefficient or simply coefficient. The coefficient in the term 7xy is 7 and the coefficient in the term –5x is –5.
### 9.3 Monomials, Binomials and Polynomials
Expression that contains only one term is called a monomial. Expression that contains two terms is called a binomial. An expression containing three terms is a trinomial and so on. In general, an expression containing, one or more terms with non-zero coefficient (with variables having non negative integers as exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.
Examples of monomials: 4x2, 3xy, –7z, 5xy2, 10y, –9, 82mnp, etc.
Examples of binomials: a + b, 4l + 5m, a + 4, 5 –3xy, z2 – 4y2, etc.
Examples of trinomials: a + b + c, 2x + 3y – 5, x2yxy2 + y2, etc.
Examples of polynomials: a + b + c + d, 3xy, 7xyz – 10, 2x + 3y + 7z, etc.
### TRY THESE
1. Classify the following polynomials as monomials, binomials, trinomials.
z + 5, x + y + z, y + z + 100, abac, 17
2. Construct
(a) 3 binomials with only x as a variable;
(b) 3 binomials with x and y as variables;
(c) 3 monomials with x and y as variables;
(d) 2 polynomials with 4 or more terms.
### 9.4 Like and Unlike Terms
Look at the following expressions:
7x, 14x, –13x, 5x2, 7y, 7xy, –9y2, –9x2, –5yx
Like terms from these are:
(i) 7x, 14x, –13x are like terms.
(ii) 5x2 and –9x2 are like terms.
(iii) 7xy and –5yx are like terms.
Why are 7x and 7y not like?
Why are 7x and 7xy not like?
Why are 7x and 5x2 not like?
### TRY THESE
Write two terms which are like
(i) 7xy (ii) 4mn2 (iii) 2l
### 9.5 Addition and Subtraction of Algebraic Expressions
In the earlier classes, we have also learnt how to add and subtract algebraic expressions. For example, to add 7x2 – 4x + 5 and 9x – 10, we do
7x2 – 4x + 5
+ 9x – 10
7x2 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate row. While doing so we write like terms one below the other, and add them, as shown. Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below the other, we have
7xy + 5yz – 3zx
+ 4yz + 9zx – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy + 9yz + 3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y in the second expression and 5x in the third expression, are carried over as they are, since they have no like terms in the other expressions.
Example 2: Subtract 5x2 – 4y2 + 6y – 3 from 7x2 – 4xy + 8y2 + 5x – 3y.
Solution:
7x2 – 4xy + 8y2 + 5x – 3y
5x2 – 4y2 + 6y – 3
(–) (+) (–) (+)
2x2 – 4xy + 12y2 + 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse. Thus subtracting –3 is the same as adding +3. Similarly, subtracting 6y is the same as adding – 6y; subtracting – 4y2 is the same as adding 4y2 and so on. The signs in the third row written below each term in the second row help us in knowing which operation has to be performed.
### Exercise 9.1
1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy (ii) 1 + x + x2 (iii) 4x2y2 – 4x2y2z2 + z2
(iv) 3 – pq + qrrp (v) (vi) 0.3a – 0.6ab + 0.5b
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3, 5x – 4y + 3xy, 4z – 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
(i) abbc, bcca, caab (ii) ab + ab, bc + bc, c a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2 (iv) l2 + m2, m2 + n2, n2 + l2,2lm + 2mn + 2nl
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq2 + 5p2q
### 9.6 Multiplication of Algebraic Expressions: Introduction
(i) Look at the following patterns of dots.
(ii) Can you now think of similar other situations in which two algebraic expressions have to be multiplied?
Ameena gets up. She says, “We can think of area of a rectangle.” The area of a rectangle is l × b, where l is the length, and b is breadth. If the length of the rectangle is increased by 5 units, i.e., (l + 5) and breadth is decreased by 3 units , i.e., (b – 3) units, the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a rectangular box is given by the product of its length, breadth and height).
(iv) Sarita points out that when we buy things, we have to carry out multiplication. For example, if
price of bananas per dozen = ` p
and for the school picnic bananas needed = z dozens,
then we have to pay = ` p × z
Suppose, the price per dozen was less by ` 2 and the bananas needed were less by 4 dozens.
Then, price of bananas per dozen = ` (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ` (p – 2) × (z – 4)
### TRY THESE
Can you think of two more such situations, where we may need to multiply algebraic expressions?
[Hint: Think of speed and time;
Think of interest to be paid, the principal and the rate of simple interest; etc.]
### 9.7 Multiplying a Monomial by a Monomial
##### 9.7.1 Multiplying two monomials
In all the above examples, we had to carry out multiplication of two or more quantities. If the quantities are given by algebraic expressions, we need to find their product. This means that we should know how to obtain this product. Let us do this systematically. To begin with we shall look at the multiplication of two monomials.
We begin with
4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y = –15xy
Some more useful examples follow.
(iv) 5x × 4x2 = (5 × 4) × (x × x2)
= 20 × x3 = 20x3
Note that 5 × 4 = 20
i.e., coefficient of product = coefficient of first monomial × coefficient of second monomial;
and x × x2 = x3
i.e., algebraic factor of product
= algebraic factor of first monomial
× algebraic factor of second monomial.
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x2yz
Observe how we collect the powers of different variables in the algebraic parts of the two monomials. While doing so, we use the rules of exponents and powers.
##### 9.7.2 Multiplying three or more monomials
Observe the following examples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x2y2 × 6x3y3 = (4xy × 5x2y2) × 6x3y3 = 20x3y3 × 6x3y3 = 120x3y3 × x3y3
= 120 (x3 × x3) × (y3 × y3) = 120x6 × y6 = 120x6y6
It is clear that we first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.
We can find the product in other way also.
4xy × 5x2y2 × 6x3 y3
= (4 × 5 × 6) × (x × x2 × x3) × (y × y2 × y3)
= 120 x6y6
### TRY THESE
Find 4x × 5y × 7z
First find 4x × 5y and multiply it by 7z;
or first find 5y × 7z and multiply it by 4x.
Is the result the same? What do you observe?
Does the order in which you carry out the multiplication matter?
Example 3: Complete the table for area of a rectangle with given length and breadth.
Solution:
Example 4: Find the volume of each rectangular box with given length, breadth and height.
Solution: Volume = length × breadth × height
Hence, for (i) volume = (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz
for (ii) volume = m2n × n2p × p2m
= (m2 × m) × (n × n2) × (p × p2) = m3n3p3
for (iii) volume = 2q × 4q2 × 8q3
= 2 × 4 × 8 × q × q2 × q3 = 64q6
### Exercise 9.2
1. Find the product of the following pairs of monomials.
(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p3, – 3p
(v) 4p, 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
3. Complete the table of products.
4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2 (iv) a, 2b, 3c
5. Obtain the product of
(i) xy, yz, zx (ii) a, – a2, a3 (iii) 2, 4y, 8y2, 16y3
(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp
### 9.8 Multiplying a Monomial by a Polynomial
##### 9.8.1 Multiplying a monomial by a binomial
Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ?
Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law,
3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x
We commonly use distributive law in our calculations. For example:
7 × 106 = 7 × (100 + 6)
= 7 × 100 + 7 × 6 (Here, we used distributive law)
= 700 + 42 = 742
7 × 38 = 7 × (40 – 2)
= 7 × 40 – 7 × 2 (Here, we used distributive law)
= 280 – 14 = 266
Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x
and 5xy × (y2 + 3) = (5xy × y2) + (5xy × 3) = 5xy3 + 15xy.
What about a binomial × monomial? For example, (5y + 2) × 3x = ?
We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a
Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.
### TRY THESE
Find the product (i) 2x (3x + 5xy) (ii) a2 (2ab – 5c)
##### 9.8.2 Multiplying a monomial by a trinomial
Consider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law;
3p × (4p2 + 5p + 7) = (3p × 4p2) + (3p × 5p) + (3p × 7)
= 12p3 + 15p2 + 21p
Multiply each term of the trinomial by the monomial and add products.
Observe, by using the distributive law, we are able to carry out the multiplication term by term.
### TRY THESE
Find the product: (4p2 + 5p + 7) × 3p
Example 5: Simplify the expressions and evaluate them as directed:
(i) x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2
Solution:
(i) x (x – 3) + 2 = x2 – 3x + 2
For x = 1, x2 – 3x + 2 = (1)2 – 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0
(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y2 – 21y – 3y + 12 – 63
= 6y2 – 24y – 51
For y = –2, 6y2 – 24y – 51 = 6 (–2)2 – 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21
(i) 5m (3 – m) and 6m2 – 13m (ii) 4y (3y2 + 5y – 7) and 2 (y3 – 4y2 + 5)
Solution:
(i) First expression = 5m (3 m) = (5m × 3) – (5m × m) = 15m – 5m2
Now adding the second expression to it,15m – 5m2 + 6m2 – 13m = m2 + 2m
(ii) The first expression = 4y (3y2 + 5y – 7) = (4y × 3y2) + (4y × 5y) + (4y × (–7))
= 12y3 + 20y2 – 28y
The second expression = 2 (y3 – 4y2 + 5) = 2y3 + 2 × (– 4y2) + 2 × 5
= 2y3 – 8y2 + 10
Adding the two expressions, 12y3 + 20y2 – 28y
+ 2y3 – 8y2 + 10
14y3 + 12y2 – 28y + 10
Example 7: Subtract 3pq (pq) from 2pq (p + q).
Solution: We have 3pq (pq) = 3p2q – 3pq2 and
2pq (p + q) = 2p2q + 2pq2
Subtracting, 2p2q + 2pq2
3p2q – 3pq2
– +
p2q + 5pq2
### Exercise 9.3
1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r (ii) ab, ab (iii) a + b, 7a2b2 (iv) a2 – 9, 4a
(v) pq + qr + rp, 0
2. Complete the table.
First expression Second expression Product
3. Find the product.
Observe, every term in one binomial multiplies every term in the other binomial.
(i) (a2) × (2a22) × (4a26) (ii)
(iii) (iv) x × x2 × x3 × x4
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x = .
(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = – 1.
5. (a) Add: p ( pq), q ( qr) and r ( rp)
(b) Add: 2x (zxy) and 2y (zyx)
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (ab + c) from 4c ( a + b + c )
### 9.9 Multiplying a Polynomial by a Polynomial
##### 9.9.1 Multiplying a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this step-by-step, as we did in earlier cases, following the distributive law of multiplication,
(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
Observe, every term in one binomial multiplies every term in the other binomial
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6a2 + 9ab + 8ba + 12b2
= 6a2 + 17ab + 12b2 (Since ba = ab)
When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be present. But two of these are like terms, which are combined, and hence we get 3 terms. In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.
Example 8: Multiply
(i) (x – 4) and (2x + 3) (ii) (xy) and (3x + 5y)
Solution:
(i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)
= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2x2 + 3x – 8x – 12
= 2x2 – 5x – 12 (Adding like terms)
(ii) (xy) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y)
= (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y)
= 3x2 + 5xy – 3yx – 5y2 = 3x2 + 2xy – 5y2 (Adding like terms)
Example 9: Multiply
(i) (a + 7) and (b – 5) (ii) (a2 + 2b2) and (5a – 3b)
Solution:
(i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)
= ab – 5a + 7b – 35
[using the distributive law]
Note that there are no like terms involved in this multiplication.
(ii) (a2 + 2b2) × (5a – 3b) = a2 (5a – 3b) + 2b2 × (5a – 3b)
= 5a3 – 3a2b + 10ab2 – 6b3
##### 9.9.2 Multiplying a binomial by a trinomial
In this multiplication, we shall have to multiply each of the three terms in the trinomial by each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may reduce to 5 or less, if the term by term multiplication results in like terms. Consider
× = a × (a2 + 3a + 5) + 7 × (a2 + 3a + 5)
= a3 + 3a2 + 5a + 7a2 + 21a + 35
= a3 + (3a2 + 7a2) + (5a + 21a) + 35
= a3 + 10a2 + 26a + 35 (Why are there only 4 terms in the final result?)
Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.
Solution: We have
(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)
= 2a2 – 3ab + ac + 2ab – 3b2 + bc
= 2a2ab – 3b2 + bc + ac (Note, –3ab and 2ab
are like terms)
and (2a – 3b) c = 2ac – 3bc
Therefore,
(a + b) (2a – 3b + c) – (2a – 3b) c = 2a2ab – 3b2 + bc + ac – (2ac – 3bc)
= 2a2ab – 3b2 + bc + ac – 2ac + 3bc
= 2a2ab – 3b2 + (bc + 3bc) + (ac – 2ac)
= 2a2 – 3b2ab + 4bcac
### Exercise 9.4
1. Multiply the binomials.
(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
(vi)
2. Find the product.
(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7xy)
(iii) (a2 + b) (a + b2) (iv) (p2q2) (2p + q)
3. Simplify.
(i) (x2 – 5) (x + 5) + 25 (ii) (a2 + 5) (b3 + 3) + 5
(iii) (t + s2) (t2s)
(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(xy) (vi) (x + y)(x2xy + y2)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + bc)
### 9.10 What is an Identity?
Consider the equality (a + 1) (a +2) = a2 + 3a + 2
We shall evaluate both sides of this equality for some value of a, say a = 10.
For a = 10, LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132
RHS = a2 + 3a + 2 = 102 + 3 × 10 + 2 = 100 + 30 + 2 = 132
Thus, the values of the two sides of the equality are equal for a = 10.
Let us now take a = –5
LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2) = (– 4) × (–3) = 12
RHS = a2 + 3a + 2 = (–5)2 + 3 (–5) + 2
= 25 – 15 + 2 = 10 + 2 = 12
Thus, for a = –5, also LHS = RHS.
We shall find that for any value of a, LHS = RHS. Such an equality, true for every value of the variable in it, is called an identity. Thus,
(a + 1) (a + 2) = a2 + 3a + 2 is an identity.
An equation is true for only certain values of the variable in it. It is not true for all values of the variable. For example, consider the equation
a2 + 3a + 2 = 132
It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc.
Try it: Show that a2 + 3a + 2 = 132 is not true for a = –5 and for a = 0.
### 9.11 Standard Identities
We shall now study three identities which are very useful in our work. These identities are obtained by multiplying a binomial by another binomial.
Let us first consider the product (a + b) (a + b) or (a + b)2.
(a + b)2 = (a + b) (a + b)
= a(a + b) + b (a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2 (since ab = ba)
Thus (a + b)2 = a2 + 2ab + b2 (I)
Clearly, this is an identity, since the expression on the RHS is obtained from the LHS by actual multiplication. One may verify that for any value of a and any value of b, the values of the two sides are equal.
Next we consider (ab)2 = (ab) (ab) = a (ab) – b (ab)
We have = a2abba + b2 = a2 – 2ab + b2
or (ab)2 = a2 – 2ab + b2 (II)
Finally, consider (a + b) (ab). We have (a + b) (ab) = a (ab) + b (ab)
= a2ab + bab2 = a2b2 (since ab = ba)
or (a + b) (ab) = a2b2 (III)
The identities (I), (II) and (III) are known as standard identities.
### TRY THESE
1. Put – b in place of b in Identity (I). Do you get Identity (II)?
We shall now work out one more useful identity.
(x + a) (x + b) = x (x + b) + a (x + b)
= x2 + bx + ax + ab
or (x + a) (x + b) = x2 + (a + b) x + ab (IV)
### TRY THESE
1. Verify Identity (IV), for a = 2, b = 3, x = 5.
2. Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?
3. Consider, the special case of Identity (IV) with a = – c and b = – c. What do you get? Is it related to Identity (II)?
4. Consider the special case of Identity (IV) with b = – a. What do you get? Is it related to Identity (III)?
We can see that Identity (IV) is the general form of the other three identities also.
### 9.12 Applying Identities
We shall now see how, for many problems on multiplication of binomial expressions and also of numbers, use of the identities gives a simple alternative method of solving them.
Example 11: Using the Identity (I), find (i) (2x + 3y)2 (ii) 1032
Solution:
(i) (2x + 3y)2 = (2x)2 + 2(2x) (3y) + (3y)2 [Using the Identity (I)]
= 4x2 + 12xy + 9y2
We may work out (2x + 3y)2 directly.
(2x + 3y)2 = (2x + 3y) (2x + 3y)
= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)
= 4x2 + 6xy + 6 yx + 9y2 (as xy = yx)
= 4x2 + 12xy + 9y2
Using Identity (I) gave us an alternative method of squaring (2x + 3y). Do you notice that the Identity method required fewer steps than the above direct method? You will realise the simplicity of this method even more if you try to square more complicated binomial expressions than (2x + 3y).
(ii) (103)2 = (100 + 3)2
= 1002 + 2 × 100 × 3 + 32 (Using Identity I)
= 10000 + 600 + 9 = 10609
We may also directly multiply 103 by 103 and get the answer. Do you see that Identity (I) has given us a less tedious method than the direct method of squaring 103? Try squaring 1013. You will find in this case, the method of using identities even more attractive than the direct multiplication method.
Example 12: Using Identity (II), find (i) (4p – 3q)2 (ii) (4.9)2
Solution:
(i) (4p – 3q)2 = (4p)2 – 2 (4p) (3q) + (3q)2 [Using the Identity (II)]
= 16p2 – 24pq + 9q2
Do you agree that for squaring (4p – 3q)2 the method of identities is quicker than the direct method?
(ii) (4.9)2 = (5.0 – 0.1)2 = (5.0)2 – 2 (5.0) (0.1) + (0.1)2
= 25.00 – 1.00 + 0.01 = 24.01
Is it not that, squaring 4.9 using Identity (II) is much less tedious than squaring it by direct multiplication?
Example 13: Using Identity (III), find
(i) (ii) 9832 – 172 (iii) 194 × 206
Solution:
(i) =
=
Try doing this directly. You will realise how easy our method of using Identity (III) is.
(ii) 9832 – 172 = (983 + 17) (983 – 17)
[Here a = 983, b = 17, a2b2 = (a + b) (ab)]
Therefore, 9832 – 172 = 1000 × 966 = 966000
(iii) 194 × 206 = (200 – 6) × (200 + 6) = 2002 – 62
= 40000 – 36 = 39964
Example 14: Use the Identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following:
(i) 501 × 502 (ii) 95 × 103
Solution:
(i) 501 × 502 = (500 + 1) × (500 + 2) = 5002 + (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2 = 251502
(ii) 95 × 103 = (100 – 5) × (100 + 3) = 1002 + (–5 + 3) × 100 + (–5) × 3
= 10000 – 200 – 15 = 9785
### Exercise 9.5
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7)
(iv) (3a) (3a) (v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2 + b2) (– a2 + b2) (vii) (6x – 7) (6x + 7) (viii) (– a + c) (– a + c)
(ix) (x) (7a – 9b) (7a – 9b)
2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y) (vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
3. Find the following squares by using the identities.
(i) (b – 7)2 (ii) (xy + 3z)2 (iii) (6x2 – 5y)2
(iv) (v) (0.4p – 0.5q)2 (vi) (2xy + 5y)2
4. Simplify.
(i) (a2b2)2 (ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab2c (vii) (m2n2m)2 + 2m3n2
5. Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2 (ii) (9p – 5q)2 + 180pq = (9p + 5q)2
(iii) + 2mn =
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (ab) (a + b) + (bc) (b + c) + (ca) (c + a) = 0
6. Using identities, evaluate.
(i) 712 (ii) 992 (iii) 1022 (iv) 9982
(v) 5.22 (vi) 297 × 303 (vii) 78 × 82 (viii) 8.92
(ix) 10.5 × 9.5
7. Using a2b2 = (a + b) (ab), find
(i) 512 – 492 (ii) (1.02)2 – (0.98)2 (iii) 1532 – 1472
(iv) 12.12 – 7.92
8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8
### What have we Discussed?
1. Expressions are formed from variables and constants.
2. Terms are added to form expressions. Terms themselves are formed as product of factors.
3. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials respectively. In general, any expression containing one or more terms with non-zero coefficients (and with variables having non- negative integers as exponents) is called a polynomial.
4. Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same.
5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms.
6. There are number of situations in which we need to multiply algebraic expressions: for example, in finding area of a rectangle, the sides of which are given as expressions.
7. A monomial multiplied by a monomial always gives a monomial.
8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial.
9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial). Note that in such multiplication, we may get terms in the product which are like and have to be combined.
10. An identity is an equality, which is true for all values of the variables in the equality.
On the other hand, an equation is true only for certain values of its variables. An equation is not an identity.
11. The following are the standard identities:
(a + b)2 = a2 + 2ab + b2 (I)
(ab)2 = a2 – 2ab + b2 (II)
(a + b) (ab) = a2b2 (III)
12. Another useful identity is (x + a) (x + b) = x2 + (a + b) x + ab (IV)
13. The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.
|
##### Integration by Parts
Integration by parts method is used when we want to integrate the product of two functions. When finding the derivative of the product of two functions we use the product rule, and since the integral is the reverse of derivative then integration by parts is the reverse of the product rule.
Let’s explain this step by step
If we have two differentiable functions $\displaystyle u(x)$ and $\displaystyle v(x)$ and we want to find the derivative of their product we use the product rule:
$\displaystyle (u\cdot v{)}’=u\cdot {v}’+v\cdot {u}’$ or written like $\displaystyle d(uv)=udv+vdu$
By rearranging this rule we can write:
$\displaystyle udv=d(uv)-vdu$
By integrating both sides we get:
$\displaystyle \int{{udv=\int{{d(uv)-\int{{vdu}}}}}}$
Simplifying, we get the integration by part formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
This formula helps us to turn a complicated integral into a simple one. The only thing we should be careful and choose wisely is ‘u’ and ‘dv’.
When should you choose ‘u’?
There are some common expressions when you should always choose that as ‘u’.
Inverse trigonometric Functions:
$\displaystyle \arcsin x,\arccos x,\arctan x………$
Logarithmic Functions:
$\displaystyle \log x,\ln x,{{\log }_{2}}2x,….{{\log }_{a}}nx$
Algebraic Functions:
$\displaystyle x,{{x}^{2}},{{x}^{3}},5{{x}^{4}},….,a{{x}^{n}}$
Trigonometric Functions:
$\displaystyle \sin x,\cos x,\tan x……..\sin nx,\cos nx$
Exponential Functions:
$\displaystyle {{e}^{x}},{{e}^{{3x}}}{{,3}^{{-2x}}},……,{{e}^{{nx}}},{{a}^{{nx}}}$
This list is written from the top to the least priority.
Example 1: Find the integral $\displaystyle \int{x}{{e}^{{2x}}}dx$?
In this kind of expression we should always remember that the power of x has a priority over the exponential expression.
So let’s note $\displaystyle u=x$ than $\displaystyle du=dx$
What’s left would be dv, so $\displaystyle dv={{e}^{{2x}}}dx$
Integrating it we obtain v:
$\displaystyle \int{{dv}}=\int{{{{e}^{{2x}}}}}dx$
$\displaystyle v=\frac{1}{2}{{e}^{{2x}}}$
Now we substitute them into the integration by parts formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{x}{{e}^{{2x}}}dx=x\cdot \frac{1}{2}{{e}^{{2x}}}-\int{{\frac{1}{2}{{e}^{{2x}}}dx}}$
Then try to find the integral:
$\displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}x{{e}^{{2x}}}-\frac{1}{2}\int{{{{e}^{{2x}}}dx}}$
$\displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}x{{e}^{{2x}}}-\frac{1}{4}{{e}^{{2x}}}+C$
$\displaystyle \int{x}{{e}^{{2x}}}dx=\frac{1}{2}{{e}^{{2x}}}(x-\frac{1}{2})+C$
Note! When choosing u we should choose the one that makes du/dx on a simpler form than u.
Example 2: Find the integral $\displaystyle \int{{3x\sin 2xdx}}$?
So let’s note $\displaystyle u=3x$ than $\displaystyle du=3dx$
What’s left would be dv, so $\displaystyle dv=\sin 2xdx$
Integrating it we obtain v:
$\displaystyle \int{{dv=\int{{\sin 2xdx}}}}$
$\displaystyle v=-\frac{1}{2}\cos 2x$
Now we substitute them into the integration by parts formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{{3x\sin 2xdx}}=3x\cdot -\frac{1}{2}\cos 2x-\int{{-\frac{1}{2}\cos 2x\cdot 3dx}}$
Then try to find the integral:
$\displaystyle \int{{3x\sin 2xdx}}=-\frac{{3x}}{2}\cos 2x+\frac{3}{2}\int{{\cos 2xdx}}$
$\displaystyle \int{{3x\sin 2xdx}}=-\frac{{3x}}{2}\cos 2x+\frac{3}{4}\sin 2x$
Example 3: Find the integral $\displaystyle \int{{x\ln 2xdx}}$?
From the priorities above it’s obvious whose gone be the u and dv.
So let’s note $\displaystyle u=\ln 2x$ than$\displaystyle du=\frac{2}{x}dx$
What’s left would be dv, so $\displaystyle dv=xdx$
Integrating it we obtain v:
$\displaystyle \int{{dv=\int{{xdx}}}}$
$\displaystyle v=\frac{{{{x}^{2}}}}{2}$
Now we substitute them into the integration by parts formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{{x\ln 2xdx=\ln 2x}}\cdot \frac{{{{x}^{2}}}}{2}-\int{{\frac{{{{x}^{2}}}}{2}\cdot \frac{2}{x}dx}}$
Then try to find the integral:
$\displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}\ln 2x}}{2}}}-\int{{xdx}}$
$\displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}\ln 2x}}{2}}}-\frac{{{{x}^{2}}}}{2}+C$
$\displaystyle \int{{x\ln 2xdx=\frac{{{{x}^{2}}}}{2}}}(\ln x-1)+C$
Example 4: Find the integral $\displaystyle \int{{2\text{arctgxdx}}}$
So let’s note $\displaystyle u=\arctan x$ than $\displaystyle du=\frac{1}{{1+{{x}^{2}}}}dx$
What’s left would be dv, so $\displaystyle dv=2dx$
Integrating it we obtain v:
$\displaystyle \int{{dv=\int{{2dx}}}}$
$\displaystyle v=2x$
Now we substitute them into the integration by parts formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{{2\text{arctgxdx}=}}2x\arctan x-\int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx$
Then try to find the integral:
We use the u-substitution method to find the integral $\displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx$
We substitute $\displaystyle u=1+{{x}^{2}}$ than $\displaystyle du=2xdx$
Finding the integral by substituting:
$\displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx=\int{{\frac{{du}}{u}}}$
$\displaystyle \int{{\frac{{du}}{u}}}=\ln u+C$
Now we substitute back the $\displaystyle u=1+{{x}^{2}}$
$\displaystyle \int{{\frac{{2x}}{{1+{{x}^{2}}}}}}dx=\ln (1+{{x}^{2}})+C$
Example 5: Find the integral$\displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx$?
Based on the priorities of the table:
So let’s note $\displaystyle u={{x}^{2}}$ than$\displaystyle du=2xdx$
What’s left would be dv, so $\displaystyle dv={{2}^{x}}dx$
Integrating it we obtain v:
$\displaystyle \int{{dv=\int{{{{2}^{x}}dx}}}}$
$\displaystyle v=\frac{{{{2}^{x}}}}{{\ln 2}}$
Now we substitute them into the integration by parts formula:
$\displaystyle \int{{udv=uv-\int{{vdu}}}}$
$\displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx={{x}^{2}}\cdot \frac{{{{2}^{x}}}}{{\ln 2}}-\int{{\frac{{{{2}^{x}}}}{{\ln 2}}\cdot }}2xdx$
Then try to find the integral:
$\displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{2}{{\ln 2}}\int{{x\cdot {{2}^{x}}}}dx$
For the integral we are going to use again the integration by parts:
$\displaystyle \int{{x\cdot {{2}^{x}}}}dx=x\cdot \frac{{{{2}^{x}}}}{{\ln 2}}-\int{{\frac{{{{2}^{x}}}}{{\ln 2}}dx}}$
$\displaystyle \int{{x\cdot {{2}^{x}}}}dx=\frac{{{{2}^{x}}x}}{{\ln 2}}-\frac{{{{2}^{x}}}}{{{{{\ln }}^{2}}2}}$
Now we obtain the integral:
$\displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{2}{{\ln 2}}(\frac{{{{2}^{x}}x}}{{\ln 2}}-\frac{{{{2}^{x}}}}{{{{{\ln }}^{2}}2}})+C$
$\displaystyle \int{{{{2}^{x}}\cdot {{x}^{2}}}}dx=\frac{{{{2}^{x}}{{x}^{2}}}}{{\ln 2}}-\frac{{{{2}^{x}}^{{+1}}x}}{{{{{\ln }}^{2}}2}}+\frac{{{{2}^{x}}^{{+1}}}}{{{{{\ln }}^{3}}2}}+C$
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# 4.11 Chapter 4 Example Solutions
## 4.2 Example Solutions
### Example 1: Factoring the Greatest Common Factor
1. First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of $x^3$, $x^2$, and $x$ is $x$. (Note that the GCF of a set of expressions in the form $x^n$ will always be the exponent of lowest degree.) And the GCF of $y^3$, $y^2$, and $y$ is $y$. Combine these to find the GCF of the polynomial, $3xy$.
Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that $3xy\left(2x^2y^2\right)=6x^3y^3$, $3xy\left(15xy\right)=45x^2y^2$, and $3xy\left(7\right)=21xy$.
Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
$\left(3xy\right)\left(2x^2y^2+15xy+7\right)$
2. $\left(b^2-a\right)\left(x+6\right)$
## 4.3 Example Solutions
### Example 1: Factoring a Trinomial with Leading Coefficient 1
1. We have a trinomial with leading coefficient $1$, $b=2$, and $c=−15$. We need to find two numbers with a product of $−15$ and a sum of $2$. In Table 1, we list factors until we find a pair with the desired sum.
Table 1
Factors of -15 Sum of Factors
$1, -15$ $-14$
$-1, 15$ $14$
$3, -5$ $-2$
$-3, 5$ $2$
Now that we have identified $p$ and $q$ as $−3$ and $5$, write the factored form as $\left(x-3\right)\left(x+5\right)$.
2. $\left(x−6\right)\left(x−1\right)$
## 4.4 Example Solutions
### Example 1: Factoring a Trinomial by Grouping
1. We have a trinomial with $a=5$, $b=7$, and $c=−6$. First, determine $ac=−30$. We need to find two numbers with a product of $−30$ and a sum of $7$. In Table 2, we list factors until we find a pair with the desired sum.
Table 2
Factors of -30 Sum of Factors
1, -30 -29
-1, 30 29
2, -15 -13
-2, 15 13
3, -10 -7
-3, 10 7
So $p=−3$ and $q=10$.
2. a. $\left(2x+3\right)\left(x+3\right)$
b. $\left(3x−1\right)\left(2x+1\right)$
## 4.5 Example Solutions
### Example 1: Factoring a Perfect Square Trinomial
1. Notice that $25x^2$ and $4$ are perfect squares because $25x^2 = (5x)^2$ and $4 = 2^2$. Then check to see if the middle term is twice the product of $5x$ and $2$. The middle term is, indeed, twice the product: $2(5x)(2) = 20x$. Therefore, the trinomial is a perfect square trinomial and can be written as $(5x+2)^2$.
2. $\left(7x-1\right)^2$
## 4.6 Example Solutions
### Example 1: Factoring a Difference of Squares
1. Notice that $9x^2$ and $25$ are perfect squares because $9x^2 = (3x)^2$ and $25 = 5^2$. The polynomial represents a difference of squares and can be written as $(3x + 5)(3x - 5)$.
2. $\left(9y-10\right)\left(9y-10\right)$
## 4.7 Example Solutions
### Example 1: Factoring a Sum of Cubes
1. Notice that $x^3$ and $512$ are cubes because $8^3 = 512$. Rewrite the sum of cubes as $(x + 8)(x^2 - 8x +64)$.
2. $\left(6a+b\right)\left(36a^2-6ab+b^2\right)$
### Example 2: Factoring a Difference of Cubes
1. Notice that $8x^3$ and $125$ are cubes because $8x^3 = (2x)^3$ and $125 = 5^3$. Write the difference of cubes as $(2x - 5)(4x^2 + 10x + 25)$.
2. $\left(10x-1\right)\left(100x^2-10x+1\right)$
## 4.8 Example Solutions
### Example 1: Factoring Expressions with Fractional or Negative Exponents
1. Factor out the term with the lowest value of the exponent. In this case, that would be ${(x\;+\;2)}^\frac{-1}3$.
${(x\;+\;2)}^\frac{-1}3(3x\;+\;4(x\;+2))\;\;\;\;\;\;\;\;\;\;\;\;\;Factor\;out\;the\;GCF.\\{(x\:+\;2)}^\frac{-1}3(3x\;+4x\;+8)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Simplify.\\{(x\;+\;2)}^\frac{-1}3(7x\;+8)$
2. $\left(5a-1\right)^\frac{-1}{4}\left(17a-2\right)$
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# Solve the following inequality:${{\log }_{2x}}({{x}^{2}}-5x+6)<1$
Last updated date: 06th Aug 2024
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Hint: We should take care about logarithm inequalities by taking consideration of the logarithm. That is, if the base is greater than 1, the inequality sign doesn’t change. However, if base if less than 1, the inequality sign is reversed.
Thus, we should solve two cases concerning logarithm bases separately. (namely, 2x>1 and 2x<1)
Case 1: 2x > 1
\begin{align} & {{\log }_{2x}}({{x}^{2}}-5x+6)<1 \\ & {{\log }_{2x}}({{x}^{2}}-5x+6)<{{\log }_{2x}}(2x) \\ \end{align}
Here, we don’t have to change the inequality sign since, 2x > 1,
\begin{align} & {{x}^{2}}-5x+6<2x \\ & {{x}^{2}}-7x+6<0 \\ & (x-1)(x-6)<0\text{ -- (1)} \\ \end{align}
From (1), we have 1 < x < 6 -- (A)
Also, since, 2x > 1
We have, x > $\dfrac{1}{2}\text{ -- (B)}$
Thus, we need to satisfy (A) and (B) together, since for Case 1 to be true, both these conditions should be true.
Thus, doing so, we would have,
1 < x < 6 (Since, this inequality satisfies both (A) and (B) both together).
Case 2: 2x < 1
\begin{align} & {{\log }_{2x}}({{x}^{2}}-5x+6)<1 \\ & {{\log }_{2x}}({{x}^{2}}-5x+6)<{{\log }_{2x}}(2x) \\ \end{align}
Here, we have to change the inequality sign since, 2x < 1,
\begin{align} & {{x}^{2}}-5x+6>2x \\ & {{x}^{2}}-7x+6>0 \\ & (x-1)(x-6)>0\text{ -- (2)} \\ \end{align}
From (2), we have, x < 1 and x > 6 -- (C)
Further, we also have, 2x < 1
So, x < $\dfrac{1}{2}\text{ -- (D)}$
Thus, we need to satisfy (C) and (D) together, since for Case 2 to be true, both these conditions should be true.
Thus, doing so, we would have,
x < $\dfrac{1}{2}$(Since, this inequality satisfies both (C) and (D) both together)
Combining the inequalities from Case 1 and Case 2, we get,
1 < x < 6 -- (I) or
x < $\dfrac{1}{2}$ -- (II)
Thus, (I) and (II) satisfy the above inequality in the problem
Note: A common mistake made is not considering case 2, where 2x < 1. Thus, it is important to remember that when solving logarithm inequalities, when the logarithm base is less than 1, the inequality sign changes. Another important thing to keep in mind is when solving a particular case (say 2x > 1), one should take the intersection of both the inequality and 2x > 1. (that is, both the inequalities should satisfy together).
To confirm the obtained inequalities, one can put the values obtained from the final answer into the question. Example, we got 1
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# Math Insight
### Introduction to probability distributions
Math 2241, Spring 2022
Name:
ID #:
Due date: April 6, 2022, 11:59 p.m.
Table/group #:
Group members:
Total points: 1
1. When you flip a fair coin, the two possible outcomes, heads ($H$) or tails ($T$), have equal probability.
$P(H) =$
$P(T) =$
1. If an experiment consisted flipping one coin and recording the number of heads observed, we could let the random variable $N$ be the observed number of heads. In this case, the two possible outcomes are the event $N=0$ of observing no heads and the vent $N=1$ of observing one head. To capture the possible values of $N$ and their probabilities, we can use a probability distribution function, or probability mass function, which we'll denote by $f_N(n)$. The function is defined by $$f_N(n) = P(N = n).$$ Note how $N$ and $n$ represent different objects. $N$ is the random variable and $n$ is a number (in this case $0$ or $1$) that could be the value of $N$. The definition of $f_N$ is shorthand for \begin{align*} f_N(0) &= P(N=0)\\ f_N(1) &= P(N=1)\\ f_N(2) &= P(N=2)\\ &\ldots \end{align*} Given that we are flipping just one coin, we can easily fill in the values.
$f_N(0) =$
$f_N(1) =$
$f_N(2) =$
$f_N(3) =$
In fact, since we can get at most one head, if $n>1$, we know that $f_N(n) =$
.
We can graph the probability distribution $f_N(n)$ using either a bar graph or a line graph.
Feedback from applet
Point heights:
Click the “toggle” button to switch between a bar graph and a line graph. The bar graph is more intuitive, as it is similar to a histogram of actual experiments (see below). When we want to graph a large number of events or plot multiple graphs together, the simpler line graph works better.
2. A similar experiment is to flip a coin twice and let $N$ be the observed number of heads. We define the probability distribution function (or probability mass function) $f_N$ in the same way: $$f_N(n) = P(N = n).$$ The values are only slightly trickier to fill in.
$f_N(0) =$
$f_N(1) =$
$f_N(2) =$
$f_N(3) =$
In fact, since we can get at most two heads, if $n>2$, we know that $f_N(n) =$
.
Graph the probability distribution $f_N(n)$.
Feedback from applet
Point heights:
3. A another experiment is to flip a coin four times and let $N$ be the observed number of heads. We define the probability distribution function (or probability mass function) $f_N$ in the same way: $$f_N(n) = P(N = n).$$ This time, it's a bit trickier to determine the probabilities.
$f_N(0) =$
$f_N(1) =$
$f_N(2) =$
$f_N(3) =$
$f_N(4) =$
Since we can get at most four heads, if $n>4$, we know that $f_N(n) =$
.
Graph the probability distribution $f_N(n)$.
Feedback from applet
Point heights:
4. If you flip a coin ten times and let $N$ be the number of heads, the probability distribution function $f_N(n)$ looks like the following bar graph.
What is the probability of getting getting seven heads out of the ten coin flips? $f_N(7) \approx$
What is the probability of getting just one head? $f_N(1) \approx$
This probability distribution of the number of heads is common probability distribution, so it is given a special name: the binomial distribution.
Obtaining the binomial distribution in R (Show)
2. A probability distribution of a random variable tell us the probability of getting any particular value. We can also repeatedly generate a random number and determine the frequency we obtain each value.
1. We can use R to randomly (actually pseudorandomly) flip coins for us so that we simulate coin flip experiments. You can enter rbinom(1,1,0.5) to random obtain a 0 or 1 with probability 0.5. If we let 1 correspond to $H$ and 0 correspond to $T$, we can view this as the result of a coin flip. Enter the command rbinom(1,1,0.5) over and over again to see that it randomly gives a 0 or a 1.
We can use the same function to flip a bunch of coins at once. To do our 10 coin-flip experiment, enter the command rbinom(10,1,0.5) and see that you will get a vector of 10 numbers, each which are 0 or 1. To count the number of heads, you can just add up all those 1's. In R, type sum(rbinom(10,1,0.5)) to perform 10 coin flips and add up the number of H's. You should get a number between 0 and 10. However, as shown above a 0 or 10, or even a 1 or 9, would be unlikely, so you'll usually get a number between 2 and 8.
Repeat this experiment 7 times by repeatedly running the command. Enter your results, separated by commas.
In fact, the binomial distribution, which is the distribution that rbinom is sampling from, already includes the process of adding up the number of heads. The second argument is the number of coin flips. Instead of typing the command sum(rbinom(10,1,0.5)), you can type the simpler command rbinom(1,10,0.5) to get the same result.
Now we can repeat the experiment of flipping 10 coins and counting the heads multiple times with a single command. To do this experiment 100 times, enter the command rbinom(100,10,0.5).
2. To use R to perform the four coin-flip experiment 10 times, what R command should you type?
Assign the result these experiments to the vector results by entering
results=_＿
Then, you can display a histogram of the results by typing the command:
hist(results, breaks=-0.5:4.5)
(The argument -0.5:4.5 is shorthand for the vector c(-0.5,0.5,1.5,2.5,3.5,4.5). It tells R that the breakpoints between intervals for the histogram are the half-integers. In this way, each bar will be centered at the integer -- a better visualization of the fact that each bar represents integer values.)
Use the below applet to draw the histogram you obtained.
Feedback from applet
Non-negative bar heights:
Sum of bar heights:
If we let $N$ be the number of heads observed in each experiment, the $x$-axis of the above histogram represents $N$. The height of each bar represents the number of times you observed the given number of heads, i.e., the frequency that $N$ was the given value. To better compare with the probability distributions of the previous question, we can divide the bar height by 10 (the number of experiments) so that the bar heights represent relative frequency. You can produce such a histogram in R by setting the probability flag to true,
hist(results, breaks=-0.5:4.5, probability=TRUE)
though it isn't too hard to divide by 10 in your head. (With this command, R labels the $y$-axes as “Density,” but for our case, “Relative frequency” is a better description of the bar heights.)
Redraw the histogram, this time with relative frequencies.
Feedback from applet
Non-negative bar heights:
Represents probability distribution:
Sum of bar heights:
(The bar heights should be exactly 1/10th of the heights from the previous histogram, though we don't check for that here.)
Is your relative frequency histogram close to the probability distribution you calculated for the four coin-flip experiment in the previous problem?
Rerun the set of experiments by typing the two commands multiple times.
results=_＿ hist(results, breaks=-0.5:4.5, probability=TRUE)
Do the relative frequency histograms change much?
3. Increase the number of repetitions of the four coin-flip experiment. To run a set of 100 experiments, you can enter the command results=rbinom(100, 4, 0.5). Then, you can use the same histogram commands to plot histograms in terms of frequency or relative frequency.
The R function dbinom gives the probability distribution for the binomial random variable, the one we determined in the previous question. As a comparison, you can plot a bar graph of the probability distribution with these commands.
n=0:4 barplot(dbinom(n,4,0.5), names.arg=n)
The comparison will be easier to see if you can plot the probability distribution directly on top of the histogram from the experiments. Since it is hard to visualize two bar plots on top of each other, we'll change the plot of the probability distribution to a line graph with points. Assuming you have already defined n=0:4, the following set of commands will run a set of 100 experiments, plot a histogram, and then plot the probability distribution on top of it with red points connected by lines.
results=rbinom(100, 4, 0.5)
hist(results, breaks=-0.5:4.5, probability=TRUE)
lines(n,dbinom(n,4,0.5),col='red', lwd=5, type='b')
Run these commands multiple times to see how the results change as you repeat the set of 100 four-coin-flip experiments. Keep in mind that the red points and lines do not change. They may appear to move up or down, but that illusion is simply due to the scale of the $y$-axis changing based on the histogram.
What do you observe about the relationship between the height of the red points (representing the probability distribution) and the height of the bars (representing the relative frequency of each value of $N$ from the experiments)?
Increase the number of experiments in each set to 1000, 10,000, or even 100,000. What happens when you create the relative frequency histograms from a large set of experiments?
This result illustrates how the relative frequency histograms approach the probability distribution as you increase the number of samples (or number of experiments in each set). You could think of the probability distribution as being the limit of the relative frequencies as the number of samples approaches infinity.
3. Let's experiment with coins is to count how many heads you can obtain in a row. Continue flipping a coin until you obtain your first tail and let $N$ be the number of heads before that tail. What is the probability distribution $f_N(n)$ of the random variable $N$?
1. First of all, what valid options for $N$? The smallest value that $N$ could be is
. Is there a largest possible value of $N$?
The probability that the first $13,462,305$ flips result in a head before you see your first tail is
. Therefore, $N$ could be any non-negative integer. It would just be extremely unlikely to be a large number.
2. What is the probability that the first coin flip is a tail, $T$?
Therefore, the probability that $N$ is zero is $f_N(0) = P(N=0)=$
.
3. What is the probability that the first coin flip is a head, $H$?
In terms of the random variable $N$, this probability is
$=＿$.
4. So far, we've introduced two events, which we labeled in terms of the random variable $N$.
• $N=0$: the event that the first coin flip was $T$, so that we obtained zero heads.
• $N \ge 1$: the event that the first coin flip was $H$, so that we know we obtained at least one head.
Since we keep flipping coins only if we obtain an $H$, we continue with more events only conditioned on the event $N \ge 1$. In that case, we consider two more events.
• : the event that the second coin flip was $T$, so we obtained exactly
• : the event that the second coin flip was $H$, so we know we obtained at least
(Online, enter $\ge$ as >= or as the symbol ≥.)
The probabilities of these two events, conditioned on the event $N \ge 1$, are simple since they involve just one more coin flip.
• $P(N=1\,|\,N \ge 1) =$
• $P(N \ge 2 \,|\,N \ge 1) =$
From the definition of conditional probability, recall that $P(A,B) = P(A\,|\,B)P(B)$. Substituting the event $N \ge 1$ for $B$ and either event $N=1$ or $N \ge 2$ for $A$, we can calculate that
• $P(N=1, N \ge 1) = P(N=1\,|\,N \ge 1) P(N \ge 1) =$
$\times$
$=$
• $P(N \ge 2, N \ge 1) = P(N \ge 2\,|\,N \ge 1) P(N \ge 1) =$
$\times$
$=$
But now, the notation is a bit silly. If we've obtained exactly one head ($N=1$), then obviously we've obtained at least one head ($N \ge 1$). Similarly, if we've obtained at least two heads ($N \ge 2$), then obviously we've obtained at least one head ($N \ge 1$). The events $N=1$ and $N \ge 2$ cannot occur without the event $N \ge 1$. Hence, $P(N=1)=P(N=1, N \ge 1)$ and $P(N \ge 2) = P(N \ge 2, N \ge 1)$. We summarize by giving the probabilities of obtaining exactly one head and of obtaining at least two heads.
• $f_N(1) = P(N=1) =$
• $P(N \ge 2) =$
5. We can repeat this procedure to include one more coin flip and calculate the probability that $N=2$ (if third coin flip is $T$) and that $N \ge 3$ (if third coin flip is $H$).
Calculate the probabilities of these events conditioned on the fact that we already flipped heads twice in a row.
• $P(N=2 \,|\, N \ge 2)=$
• $P(N \ge 3 \,|\, N \ge 2)=$
Multiply by $P(N \ge 2)$ to determine the probabilities of obtaining exactly two heads or obtaining at least three heads.
• $f_N(2) = P(N=2)=$
• $P(N \ge 3)=$
6. Notice to get from $P(N \ge 1)$ to $P(N \ge 2)$ we multiply by $\frac{1}{2}$, the probability of getting one more $H$. Similarly, we multiply by $\frac{1}{2}$ to get from $P(N \ge 2)$ to $P(N \ge 3)$. In general, the event $N \ge n$ corresponds to flipping $H$ $n$ times in a row, so the probability $P(N \ge n)$ is the product of $n$ $\frac{1}{2}$'s. $P(N \ge n)$ is the exponential function
$P(N \ge n) =$
.
(Online enter $a^b$ as a^b.)
To know that we obtained exactly $n$ heads, we need to flip a coin one more time and this time obtain a $T$. We need to multiply by one more $\frac{1}{2}$ (the probability of $T$) to determine the probability distribution function $f_N(n)$.
$f_N(n) = P(N=n) =$
(Online, you'll need to use parentheses after the ^ to put a n+1 in the exponent.)
7. Use the below applet to sketch the probability distribution $f_N(n)$ of the number $N$ of consecutive heads. The distribution continues for all non-negative integers $n$, but we just plot $0 \le n \le 5$.
Feedback from applet
Point heights:
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# What Is The Square Root Of A Complex Number? (2 Methods To Solve)
We know how to find the square root of a real number, but what about the square root of a complex number or an imaginary number? As it turns out, we can do it, but there is a little more to it.
So, what is the square root of a complex number? The square root of a complex number Z is a complex number S that satisfies Z = S2. Note that -S (the negative of S) is also a square root of Z. We can use polar form to find the square root of a complex number. For an imaginary number bi, the square roots are √(b/2) + i√(b/2) and -√(b/2) – i√(b/2).
Of course, we can also use algebra to solve for the square root of a complex number. However, the formulas can get a little tricky.
In this article, we’ll talk about how to find the square root of a complex number and the square root of an imaginary number, using a formula and polar form.
Let’s get going.
## What Is The Square Root Of A Complex Number?
The square root of a complex number Z is another complex number S that satisfies the equation
• Z = S2
In other words, S is the square root of Z.
We are already familiar with the square roots of real numbers:
• If R > 0, then the square root of R is √R.
• If R = 0, then the square root of R is 0.
• If R < 0, then the square root of R is i√R.
Remember that a real number is really a complex number a + bi where b = 0.
The trouble in finding square roots emerges when we try to take the square root of a complex number whose imaginary part is nonzero (that is, a number a + bi where b is not zero).
Let’s start with how to find the square root of i, the imaginary unit.
### What Is The Square Root Of i?
The square root of i is the complex number √(1/2) + i√(1/2). There is a second square root of i, which is the negative of this first root: -√(1/2) – i√(1/2).
Here is one way to find the square root of i with algebra.
The square root of i is a complex number, so we’ll call it a + bi. By the definition of a square root:
• (a + bi)2 = i [definition of the square root of i]
• (a + bi)(a + bi) = i
• a2 + abi + abi + b2i2 = i [used FOIL to expand the product of two binomials]
• a2 + 2abi + b2i2 = i [combine like terms]
• a2 + 2abi – b2 = i [used the definition of i as the square root of -1, or i2 = -1]
• (a2 – b2) + 2abi = 0 + 1i [separate real and imaginary parts on each side]
Now we will write two equations: one for the real parts on each side of the equation, and one for the imaginary parts on each side of the equation:
• a2 – b2 = 0 [set the real parts equal]
• 2abi = 1i [set the imaginary parts equal]
The first equation (real parts) has a difference of squares on the left side, which factors easily:
• (a2 – b2) = 0
• (a + b)(a – b) = 0 [factor the left side as a difference of squares]
This leaves us with a + b = 0 or a – b = 0. Thus, either a = -b or a = b.
We will try both of these cases in the second equation:
Case 1: a = -b
• 2abi = 1i
• 2ab = 1
• 2(-b)b = 1
• -2b2 = 1
• b2 = -1/2
• b = +i√(1/2) or – i√(1/2)
This does not work, since b must be a real number. Now trying the second case:
Case 2: a = b
• 2abi = 1i
• 2ab = 1
• 2(b)b = 1
• 2b2 = 1
• b2 = 1/2
• b = +√(1/2) or -√(1/2)
Since a = b, we have two possible solutions for the square root of i:
• √(1/2) + i√(1/2)
• -√(1/2) – i√(1/2)
So the imaginary unit, i, has two distinct square roots. Note that any complex number (except zero) has two distinct square roots, which are negatives of each other.
Now, we can use the square root of i to find the square root of any imaginary number.
### What Is The Square Root Of An Imaginary Number?
The square root of an imaginary number bi is the complex number √(b/2) + i√(b/2). There is a second square root of I, which is the negative of this first root: -√(b/2) – i√(b/2).
To verify this, we can simply square this complex number by using FOIL, combining like terms, and simplifying by using i2 = -1:
• (√(b/2) + i√(b/2))2
• =(√(b/2) + i√(b/2))( √(b/2) + i√(b/2))
• =b/2 + ib/2 + ib/2 + i2b/2 [use FOIL on the product of binomials]
• =b/2 + 2bi/2 + i2b/2 [combine like terms]
• =b/2 + 2bi/2 – b/2 [used the definition of i as the square root of -1, or i2 = -1]
• =2bi/2 [combine like terms]
• =bi [cancel 2]
When we square the negative root, -√(b/2) – i√(b/2), we will get the same result, since the negatives will becomes positive when squared.
So, the two square roots of the imaginary number bi are:
• √(b/2) + i√(b/2)
• -√(b/2) – i√(b/2)
### How To Find The Square Root Of A Complex Number (2 Methods)
What if we have a complex number a + bi, where both a and b are not zero? In some cases, finding the square root is a little more difficult when using algebra – in that case, we might need to use a different method.
#### Method 1: Find The Square Root Of A Complex Number Using Algebra (FOIL & Solve A System Of Equations)
To find the square root of a complex number c + di, we can use a similar method to the one we used earlier to find the square root of i.
The square root of a + bi is a complex number, so we’ll call it c + di. By the definition of a square root:
• (c + di)2 = a + bi [definition of the square root of i]
• (c + di)(c + di) = a + bi
• c2 + cdi + cdi + d2i2 = a + bi [used FOIL to expand the product of two binomials]
• c2 + 2cdi + b2i2 = a + bi [combine like terms]
• c2 + 2cdi – b2 = a + bi [used the definition of i as the square root of -1, or i2 = -1]
• (c2 – d2) + 2cdi = a + bi [separate real and imaginary parts on each side]
Now we will write two equations: one for the real parts on each side of the equation, and one for the imaginary parts on each side of the equation:
• c2 – d2 = a [set the real parts equal]
• 2cdi = bi [set the imaginary parts equal]
Solving the second equation for c, we get c = b/2d. Using this in the first equation, we get:
• c2 – d2 = a
• (b/2d)2 – d2 = a [c = b/2d]
• b2/4d2 – d2 = a
• b2 – 4d4 = 4ad2
• 4d4 + 4ad2 – b2 = 0 [put all terms on one side]
• 4x2 + 4ax – b2 = 0 [make the change of variables x = d2]
Now we need to solve this quadratic equation in x, with coefficients A = 4, B = 4a, and C = -b2. Using the quadratic formula gives us:
x = -1/2(a + √(a2 + b2)) and x = -1/2(a – √(a2 + b2))
Since x = d2, we would need to take the square root of each of these expressions to find d.
Since d is a real number, we need to take the square root of the positive number, which is x = -1/2(a – √(a2 + b2)).
So, d = √(-1/2(a – √(a2 + b2))) and d = -√(-1/2(a – √(a2 + b2))).
We can also find c by using the equation c = b/2d.
##### Example: Find The Square Root of 4 + 3i.
In this case, we have the complex number 4 + 3i with a = 4 and b = 3.
Then:
• d = √(-1/2(a – √(a2 + b2)))
• d = √(-1/2(4 – √(42 + 32)))
• d = √(-1/2(4 – √(16 + 9)))
• d = √(-1/2(4 – √(25)))
• d = √(-1/2(-1))
• d = √(1/2)
We can also find c with:
• c = b/2d
• c = 3 / 2(√(1/2))
• c = 3/√2
So, the square root of 4 + 3i is c + di = 3/√2 + i√(1/2).
We can verify this by squaring 3/√2 + i√(1/2) to get 3 + 4i.
The negative of this complex number, -3/√2 – i√(1/2), is also a square root of 3 + 4i.
#### Method 2: Find The Square Root Of A Complex Number Using Polar Form (Modulus & Argument)
To find the square root of a complex number, it is much easier to convert to polar form first and then take the square root.
To convert a complex number a + bi to polar form, we need to calculate both the modulus and the argument.
The modulus R (a real number) is given by the equation
• R = √(a2 + b2)
The argument A (an angle) is given by the equation
• A = tan-1(b/a)
The polar form of the complex number a + bi is
• ReiA
where R is the modulus √(a2 + b2) and A is the argument tan-1(b/a). In other words,
• a + bi = ReiA
Note: remember that a and A are not necessarily the same number!
Once we have polar form, we can find the square root. We use half of A as our new angle and take the square root of the modulus R as our new modulus.
So, the square root of a + bi = ReiA is:
• (√R)*eiA/2
The negative of this number, -(√R)*eiA/2, is also a square root of a + bi.
Let’s look at an example of how this works.
##### Example: Find The Square Root of 4 + 3i Using Polar Form
For the complex number 4 + 3i, we have a = 4 and b = 3.
The modulus R is:
• R = √(a2 + b2)
• R = √(42 + 32)
• R = √(16 + 9)
• R = √(25)
• R = 5
The argument A is:
• A = tan-1(b/a)
• A = tan-1(3/4)
Then the polar form of the complex number 4 + 3i is:
• ReiA
• =5e0.6435i
To take the square root, we take the square root of the modulus and half of the argument to get:
• (√5)* e0.32175i
• 2.2121e0.1024i
Converting this back to a complex number, we get approximately 2.1213 + 0.7071i, which corresponds to our solution from earlier: 3/√2 + i√(1/2).
## Conclusion
Now you know how to find the square root of an imaginary number and the square root of a complex number. You also know how to convert from a complex number to polar form.
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# Find asymptotically equivalent function for $\ln \binom{n^2}{2n}$
How can I find an asymptotically equivalent function for $$\ln \binom{n^2}{2n}$$?
I think I need to use the definition of the binomial coefficient and Stirling's approximation, but I'm not sure what to do after this.
• Where did you get to after using Stirling and the binomial coefficient? That expression may actually be fully simplified even though it may look complicated. Oct 2 '19 at 9:03
In the same spirit an in answer and comment, we can even get a very good approximation.
Writing $$\binom{n^2}{2 n}=\frac{(n^2)!}{(2n)! \, (n^2-2n)!}$$ $$\log \left(\binom{n^2}{2 n}\right)=\log\left((n^2)!\right)-\log\left((2n)!\right)-\log\left((n^2-2n)!\right)$$ Using Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply to each term and continue with Taylor series for large values of $$n$$. This should give $$\log \left(\binom{n^2}{2 n}\right)=2 n \left(\log (n)+\log \left(\frac{e}{2}\right)\right)-\log \left(2 e^2 \sqrt{\pi }\right)-\log{(\sqrt n)}-\frac 3 {8n}+O\left(\frac{1}{n^2}\right)$$
As shown below, this is even quite good for very small values of $$n$$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 3 & 4.49297 & 4.43082 \\ 4 & 9.49277 & 9.46265 \\ 5 & 15.0177 & 14.9999 \\ 6 & 20.9595 & 20.9478 \\ 7 & 27.2466 & 27.2383 \\ 8 & 33.8286 & 33.8224 \\ 9 & 40.6676 & 40.6628 \\ 10 & 47.7345 & 47.7306 \end{array} \right)$$
By Stirling approximation $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$ we have that
$$\binom{n^2}{2n}=\frac{n^2!}{(2n)!(n^2-2n)!}\sim \frac{\sqrt{2 \pi n^2}\left(\frac{n^2}{e}\right)^{n^2}}{\sqrt{2 \pi 2n}\left(\frac{2n}{e}\right)^{2n}\sqrt{2 \pi (n^2-2n)}\left(\frac{n^2-2n}{e}\right)^{n^2-2n}}$$
$$\sim\frac{\sqrt{2 \pi }n^{2n^2+1}e^{-n^2}}{2\sqrt \pi n^\frac12 2^{2n}n^{2n}e^{-2n}\cdot \sqrt{2 \pi}nn^{2n^2-4n}e^{-n^2+2n}}=$$
$$=\frac1{2\sqrt \pi} \cdot {n^{2n-\frac12}} \cdot 2^{-2n}$$
and therefore
$$\ln \binom{n^2}{2n}\sim n\ln n$$
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# 9.5: Classifying Polygons
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Sculpture
Courtesy of Martin Fuchs
Marc, Isaac and Isabelle continue to work on their design for the skatepark. Isabelle loves art and thinks that adding some sculpture to the entrance of the skatepark could be cool way to integrate art into the design. Marc and Isaac agree and the three decide to visit a sculpture garden to get ideas. Once they decide on what they want to create, they hope that Mr. Craven, the art teacher, will help them create it with some other classmates.
Upon visiting the sculpture garden, the three notice immediately that there are many different shapes in each sculpture.
Their favorite sculpture is pictured above. Isabelle liked the three dimensional aspect of the sculpture, but did not like that it was all made of triangles.
“Let’s design one with all kinds of polygons,” Marc suggests as they head home.
“That’s a great idea! Which ones should we use?” Isabelle asks.
“What is a polygon anyway?” Isaac interrupts.
Marc and Isabelle look at him. Isaac has not been paying attention in math class.
Before Marc and Isabelle fill in Isaac, what do you know about polygons? Can you define them? Which ones should the trio use in their sculpture? Pay attention in this lesson and you will learn all about polygons.
What You Will Learn
By the end of this lesson, you will be able to demonstrate the following skills.
• Classify polygons.
• Distinguish between regular and irregular polygons.
• Relate sides of polygons to the sum of the interior angles.
• Relate sides of polygons to the number of diagonals form a vertex.
Teaching Time
I. Classify Polygons
This lesson begins talking about polygons in specific detail. In the last two lessons, we worked with triangles and with quadrilaterals. Triangles and quadrilaterals are also polygons; we just haven’t been describing them in this way yet. This lesson will help you to understand how to identify polygons as well as learn some valuable information about them. Polygons are everywhere in the world around us and you will be working with polygons in many ways for a long time.
What is a polygon?
A polygon is a simple closed figure formed by three or more segments. A triangle is a polygon and a quadrilateral is a polygon too. Here are three pictures of polygons.
Polygons
You can see that all three of these figures are simple closed figures that are created by three or more line segments.
Not a Polygon
These figures are not polygons. A polygon does not have a curve in it. The first two figures have curves in them. The third figure is not a closed figure. The last figure has sides that overlap. A polygon does not have sides that overlap.
There are several different types of polygons. Some of them you may have heard of before.
What are some different types of polygons?
1. Triangle – has three sides
2. Quadrilateral – has four sides
3. Pentagon – has five sides
4. Hexagon – has six sides
5. Heptagon – has seven sides
6. Octagon – has eight sides
7. Nonagon – has nine sides
8. Decagon – has ten sides
These polygons can be seen in real life all the time. Look at the following pictures and determine which polygon is pictured.
1.
2.
3.
Take a few minutes to check your work
II. Distinguish Between Regular and Irregular Polygons
Now that you have been introduced to the different types of polygons, it is time to learn about classifying polygons. All polygons can be classified as regular or irregular polygons. You have to understand the difference between a regular or irregular polygon to classify each shape. Let’s learn how we can tell the difference between them.
What is a regular polygon?
A regular polygon is a polygon where all of the side lengths are equal. In other words, the polygon is an equilateral polygon where all the side lengths are congruent.
Let’s look at an example.
Example
This triangle is a regular triangle. All three side lengths are congruent. Here is an example of an irregular polygon.
Example
By counting the sides, you can see that this is a five sided figure. It is a pentagon. However, the sides are not congruent. Therefore, it is an irregular pentagon.
Irregular polygons have side lengths that are not congruent.
Now it’s time for you to practice. Name each figure as a regular or irregular polygon.
1.
2.
Take a few minutes to check your work with a friend.
III. Relate Sides of Polygons to the Number of Diagonals from a Vertex
We can divide polygons into triangles using diagonals. This becomes very helpful when we try to figure out the sum of the interior angles of a polygon other than a triangle or a quadrilateral.
Look at the second piece of information in this box. The sum of the interior angles of a quadrilateral is \begin{align*}360^\circ\end{align*}. Why is this important? You can divide a quadrilateral into two triangles using diagonals. Each triangle is \begin{align*}180^\circ\end{align*}, so the sum of the interior angles of a quadrilateral is \begin{align*}360^\circ\end{align*}.
Let’s look at an example of this.
Example
Here is one diagonal in the quadrilateral. We can only draw one because otherwise the lines would cross. A diagonal is a line segment in a polygon that joins two nonconsecutive vertices. A consecutive vertex is one that is next to another one, so a nonconsecutive vertex is a vertex that is not next to another one.
How do we use this with other polygons?
We can divide up other polygons using diagonals and figure out the sum of the interior angles.
Example
Here is a hexagon that has been divided into triangles by the diagonals. You can see here that there are four triangles formed. If sum of the interior angles of each triangle is equal to \begin{align*}180^\circ\end{align*}, and we have four triangles, then the sum of the interior angles of a hexagon is:
\begin{align*}4(180) = 720^\circ\end{align*}
We can follow this same procedure with any other polygon.
What if we don’t have the picture of the polygon? Is there another way to figure out the number of triangles without drawing in all of the diagonals? The next section will show you how using a formula with the number of sides in a polygon can help you in figuring out the sum of the interior angles.
IV. Relate Sides of Polygons to Sum of Interior Angles
To better understand how this works, let’s look at a table that shows us the number of triangles related to the number of sides in a polygon.
Do you see any patterns?
The biggest pattern to notice is that the number of triangles is 2 less than the number of sides. Why is this important? Well, if you know that the sum of the interior angles of one triangle is equal to 180 degrees and if you know that there are three triangles in a polygon, then you can multiply the number of triangles by 180 and that will give you the sum of the interior angles.
Here is the formula.
\begin{align*}x =\end{align*} number of sides
\begin{align*}(x - 2)180 =\end{align*} sum of the interior angles
You can take the number of sides and use that as \begin{align*}x\end{align*}. Then solve for the sum of the interior angles.
Let’s try this out.
Example
What is the sum of the interior angles of a decagon?
A decagon has ten sides. That is our \begin{align*}x\end{align*} measurement. Now let’s use the formula.
\begin{align*}(x - 2)180 & = (10 - 2)180 \\ 8(180) & = 1440^\circ\end{align*}
Our answer is that there are \begin{align*}1440^\circ\end{align*} in a decagon.
Try a few of these on your own.
1. The sum of the interior angles of a pentagon
2. The sum of the interior angles of a heptagon
Check your work with a neighbor. Did you use the formula to solve for the sum of the interior angles?
## Real Life Example Completed
The Sculpture
Courtesy of Martin Fuchs
Here is the original problem once again. Read the problem and underline any important information.
Marc, Isaac and Isabelle continue to work on their design for the skatepark. Isabelle loves art, and thinks that adding some sculpture to the entrance of the skatepark could be cool way to integrate art into the design. Marc and Isaac agree and the three decide to visit a sculpture garden to get ideas. Once they decide on what they want to create, they hope that Mr. Craven, the art teacher, will help them create it with the help of some other classmates.
Upon visiting the sculpture garden, the three notice immediately that there are many different shapes in each sculpture.
Their favorite sculpture is pictured above. Isabelle liked the three dimensional aspect of the sculpture, but did not like that it was all made of triangles.
“Let’s design one with all kinds of polygons,” Marc suggests as they head home.
“That’s a great idea! Which ones should we use?” Isabelle asks.
“What is a polygon anyway?” Isaac interrupts.
Marc and Isabelle look at him. Isaac has not been paying attention in math class.
Marc, Isabelle and Isaac want to design a sculpture of polygons. A polygon is a closed figure made up of at least three line segments.
Once they fill Isaac in on how to define a polygon, the three students begin to list out different types of polygons.
Triangle
Square
Rectangle
Pentagon
Hexagon
Heptagon
Octagon
Nonagon
Decagon
After a lot of negotiation, here is a rough sketch of their sculpture design. Can you identify each polygon?
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Polygon
A simple closed figure formed by three or more line segments.
Pentagon
five sided polygon
Hexagon
six sided polygon
Heptagon
seven sided polygon
Octagon
eight sided polygon
Nonagon
nine sided polygon
Decagon
ten sided polygon
Regular Polygon
polygon with all sides congruent
Irregular Polygon
a polygon where all of the side lengths are not congruent
Congruent
exactly the same or equal
Diagonal
a line segment in a polygon that connects nonconsecutive vertices
Nonconsecutive
not next to each other
## Technology Integration
Other Videos:
1. http://www.mathplayground.com/mv_polygon_angle_sum.html – A very clear video on finding the sum of the interior angles of a polygon using diagonals and triangles
## Time to Practice
Directions: Determine whether or not each image is a polygon. If yes, write polygon, if no, write not a polygon.
1.
2.
3.
4.
5.
6.
7.
Directions: For numbers 8 – 14, go back and use each figure in 1 – 7. Explain why it is or why it is not a polygon.
8.
9.
10.
11.
12.
13.
14.
Directions: Look at each image and name the type of polygon pictured.
15.
16.
17.
18.
19.
20.
Directions: Name the number of diagonals in each polygon.
21.
22.
23.
Directions: Use the formula to name the sum of the interior angles of each polygon.
24. Hexagon
25. Nonagon
26. Decagon
27. Pentagon
### Notes/Highlights Having trouble? Report an issue.
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# Factoring out quotient of an expression.
I am following the MIT Open Courseware on Single Variable Calculus and in the first lesson when taking the derivative of a simple function I found myself confused because of my knowledge gap in factoring expressions. I'm confused on this particular step.
$$\frac{\delta F}{\delta x} = \frac {\frac{1}{x_{0}+ \delta x} - \frac{1}{x_0}}{ \delta x} = \frac{1}{\delta x}{ (\frac{x_{0} - (x_{0} + \delta x)}{(x_{0} +\delta x){x_{0}}})}$$
How does the factoring out of the quotient happen? Could you explain it step by step?
$\delta_x$ is in the denominator. Therefore it can be factored out. And the common denominator of $x_0$ and $x_0+\delta_x$is the product of both: $x_0\cdot (x_0+\delta_x )$. Therefore $\frac{1}{x_0+\delta_x}$ has to be expanded by $x_0$ and $\frac{1}{x_0}$ has to be expanded by $x_0+\delta_x$:
$\frac{1}{x_0+\delta_x}\cdot \frac{x_0}{x_0}-\frac{1}{x_0} \cdot \frac{x_0+\delta_x}{x_0+\delta_x}=\frac{x_0-(x_0+\delta_x)}{x_0\cdot (x_0+\delta_x )}$
The numerator becomes $-\delta x$. That is what cancels with the $\frac1{\delta x}$ outside the brackets.
Then you are left with $\frac{-1}{(x_0+\delta x)x_0}$, which is continuous at $\delta x=0$, so you can just let $\delta x=0$, and get $\frac{-1}{x_0^2}$
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# WHICH X SATISFIES AN EQUATION
## About "Which x satisfies an equation"
Which x satisfies an equation :
A solution to an equation is a value of a variable that makes the equation true.
How to check which x satisfies an equation :
• Apply the first value instead of x in the given equation.
• If the applied value satisfies the given equation,the we can decide that particular value is a solution.
• Otherwise we have to apply the next values.
Let us see some example problems based on the above concept.
Note :
Every linear equations will have only one value for the variable that we found in the given equation.
Every quadratic equations will have two values for the variable that we found in the given equation.
Like wise according to the degree of the polynomial, we will have number of solutions.
Example 1 :
What value of z is a solution to this equation?
z + 11 = 17
Options :
(a) z = 5 (b) z = 6
Solution :
Let us apply the first value of z that is 5 instead of z in the given equation.
5 + 11 = 17
16 ≠ 17
Since the above value do not satisfies the given equation, we can decide that z = 5 is not the solution.
Now apply the second value of z that is 6 instead of z in the given equation.
6 + 11 = 17
17 = 17
Since the above value satisfies the given equation, we can decide that z = 5 is the solution.
Hence 5 is the solution of the given equation.
Example 2 :
What value of p is a solution to this equation?
3P + 4 = 25
Options :
(a) P = 2 (b) P = 7
Solution :
Let us apply the first value of P that is 2 instead of P in the given equation.
3P + 4 = 25
3(2) + 4 = 25
6 + 4 = 25
10 ≠ 25
Since the above value do not satisfies the given equation, we can decide that P = 2 is not the solution.
Now apply the second value of P that is 7 instead of z in the given equation.
3P + 4 = 25
3(7) + 4 = 25
21 + 4 = 25
25 = 25
Since the above value satisfies the given equation, we can decide that p = 7 is the solution.
Hence 7 is the solution of the given equation.
Example 3 :
What value of x is a solution to this equation?
5x + 3 = 17 - 2x
Options :
(a) x = 2 (b) x = -1
Solution :
Let us apply the first value of x that is 2 instead of x in the given equation.
5(2) + 3 = 17 - 2(2)
10 + 3 = 17 - 4
13 = 13
Since the value satisfies the given equation, we donot have to check the second value. Because every linear equation will have only one solution.
Hence 2 in the solution of the given equation.
Example 4 :
What value of x is a solution to this equation?
12 - x = 8
Options :
(a) x = -4 (b) x = 4
Solution :
Let us apply the first value of x that is -4 instead of x in the given equation.
12 - (-4) = 8
12 + 4 = 8
16 ≠ 8
Now apply the second value of x that is 4 instead of x in the given equation.
12 - 4 = 8
8 = 8
Hence the solution of the given equation is 4.
Example 5 :
What value of q is a solution to this equation?
q/6 = 7
Options :
(a) q = 42 (b) q = 1
Solution :
Let us apply the first value of q that is 42 instead of q in the given equation.
42/6 = 7
7 = 7
Hence the solution of the given equation is 42.
After having gone through the stuff given above, we hope that the students would have understood "Which x satisfies an equation".
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# What are equilateral, scalene, isosceles and right-angled triangles?
Children learn to classify triangles as equilateral, isosceles, scalene or right-angled in KS2 geometry. Our guide for parents explains everything you need to know about triangles in primary school, from working out the area to calculating the internal angles.
## What is a triangle?
A triangle is a polygon with three sides and three angles. It is a 2D shape.
## When do children learn about triangles in primary school?
Children begin learning about 2D shapes in Year 1, where they learn to recognise and name circles, triangles, squares and rectangles.
In Year 2, children will learn how to describe the properties of 2D shapes; for example, they will be able to say of a triangle: ‘It has three sides’.
They will also be asked to recognise line symmetry in shapes; for example, this triangle has a vertical line of symmetry:
In Year 3, children start learning about right angles. For example: they may be given the following shapes and asked to say which have right angles:
Answer: only the first one does.
In Year 4, children will be asked to compare shapes, based on their properties and sizes. For example: they may be asked to put ticks on each of the following shapes that have parallel lines (they will have learnt about parallel lines in Year 3):
Answer: the second and third shapes have parallel lines, triangles cannot have parallel lines.
Children will also learn about acute and obtuse angles, and may be asked to recognise them in shapes, for example they may be asked how many acute and obtuse angles this triangle has:
Answer: the top angle is obtuse (bigger than a right angle) and the bottom two angles are acute (smaller than right angles).
This learning will feed into learning how to classify triangles under the following names:
## What is an equilateral triangle?
An equilateral triangle has three equal sides and three equal angles.
## What is a scalene triangle?
A scalene triangle's three sides are all unequal.
## What is an isosceles triangle?
An isosceles triangle has two equal sides and two equal angles.
## What is a right-angled triangle?
A right-angled triangle has an angle that measures 90º.
In Year 5, children continue their learning of acute and obtuse angles within shapes.
In Year 6, children are taught how to calculate the area of a triangle. There is a basic formula for this, which is:
base x height
__________
2
This means that you multiply the measurement of the base by the height, and then divide this answer by 2.
For example, this dark green triangle has a base of 6cm and a height of 4cm. We multiply these to make 24cm and then divide this by 2 to make the area which is 12cm². (If we didn't divide by 2 we'd be calculating the area of a rectangle, represented below by the total green area.)
Children in Year 6 also move onto finding unknown angles in triangles.
They are taught that the internal (inside) angles of a triangle always total 180º. They may be given a diagram like this (not drawn to scale):
The child would need to work out that the two angles shown equal 70º. Therefore, for the three angles to total 180º, the third angle must be 110º.
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# NCERT Solutions for Chapter 14 Statistics Class 9 Maths
Chapter Name NCERT Solutions for Chapter 14 Statistics Class Class 9 Topics Covered Statistics and its LimitationsData, its types and its CharacteristicsFrequency of dataMean, Median and Mode Related Study Materials NCERT Solutions for Class 9 MathsNCERT Solutions for Class 9Revision Notes for Chapter 14 Statistics Class 9 MathsImportant Questions for Chapter 14 Statistics Class 9 MathsMCQ for for Chapter 14 Statistics Class 9 Maths
## Short Revision for Ch 14 Statistics Class 9 Maths
1. Data are collected for a definite purpose.
2. Primary data is obtained by the investigator himself or herself.
3. Secondary data is obtained from a source which already had the information stored.
4. Methods of graphical representation of data are :
(i) Bar graph
(ii) Histogram
(iii) Frequency polygon.
5. Mean, median and mode are the three measures of central tendency.
6. Mean is computed by adding all the values of the observations and dividing it by the total number of observations.
7. Mean is denoted by x .
8. Mean of n observations,
9. Mean is also calculated by,
10. Median is the value of the middle - most observation(s).
11. If the number n of observations is odd, then median = value of [(n+1)/2]th observation.
12. Mode is the most frequently occurring observation.
13. Mode = 3 median - 2 mean.
### Exercise 14.1
1. Give five examples of data that you can collect from your day-to-day life.
Solution
1. Number of students in each section of our school.
2. Height of each student of our class.
3. Temperature each day during the month.
4. Number of languages a student of our class knows.
5. Number of metro routes in Delhi and NCR.
2. Classify the data in Q.1 above as primary or secondary data.
Solution
1. Primary data.
2. Primary data.
3. Secondary data.
4. Primary data.
5. Secondary data.
### Exercise 14.2
1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
As the frequency of blood group O is highest, i.e., 12 and that of AB is shortest, hence O is the most common and AB is the rarest blood group.
2. The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
(i) 11 engineers each have distance 5 - 10 km and 10 - 15 km from residence to the place of work.
(ii) One engineer each has distance between 20 - 25 km and 25 - 30 km from residence to the place of work.
3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
(i) The given data is about the month of September (rainy season).
(ii) Range = 99.2 - 84.9 = 14.3
4The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table?
5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of Sulphur dioxide more than 0.11 parts per million?
Solution
(i) Grouped Frequency Distribution Table:
6. Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
7. The value of π up to 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week?
9. A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
### Exercise 14.3
1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %):
S.No. Causes Female fatality rate (%) 1. Reproductive health conditions 31.8 2. Neuropsychiatric conditions 25.4 3. Injuries 12.4 4. Cardiovascular conditions 4.3 5. Respiratory conditions 4.1 6. Other causes 22.0
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution
(i) Bar graph representing the causes of female fatality rate (in %)
(ii) Reproductive health conditions is the major cause of illness and death of women.
(iii) Two factors are uneducated and poor background.
2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.
S.No. Section Number of girls per thousand boys 1. Scheduled Caste (SC) 940 2. Scheduled Tribe (ST) 970 3. Non SC/ST 920 4. Backward districts 950 5. Non-backward districts 920 6. Rural 930 7. Urban 910
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution
(i) Bar graph representing the number of girls (to the nearest ten) per thousand boys in different section of society.
(ii) The highest percentage of girls per thousand boys and the lowest percentage of boys per thousand boys in different sections of Indian society are from ST and urban sections respecively.
3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:
Political party A B C D E F Seats won 75 55 37 29 10 37
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
4. The length of 40 leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:
S.No. Length (in mm) Number of leaves 1. 118 – 126 3 2. 127 – 135 5 3. 136 – 144 9 4. 145 – 153 12 5. 154 – 162 5 6. 163 – 171 4 7. 172 – 180 2
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
5. The following table gives the life times of 400 neon lamps:
Life Time (in hours) Number of lamps 300 – 400 14 400 – 500 56 500 – 600 60 600 – 700 86 700 – 800 74 800 – 900 62 900 – 1000 48
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
6. The following table gives the distribution of students of two sections according to the marks obtained by them :
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below :
Represent the data of both the teams on the same graph by frequency polygons.
[Hint. First make the class intervals continuous.]
Solution
Let us represent the data into continuous class - intervals.
8. A random survey of the number of children of various age groups playing in a park was found as follows:
Draw a histogram to represent the data above.
9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
### Exercise 14.4
1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x+2, 72, 78, 84, 95
4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution
Arranging the data in ascending order, we have
14, 14, 14, 14, 17, 18, 18, 18 22,23, 25 28
We notice 14 occurs maximum number of times.
Hence, the mode is 14.
5. Find the mean salary of 60 workers of a factory from the following table:
6. Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Solution
(i) Mean height of the students of a class.
(ii) Median weight of a pen, a book, a rubber band, a match box and a chair.
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# Partial Derivatives
A Partial Derivative is a derivative where we hold some variables constant. Like in this example:
### Example: a function for a surface that depends on two variables x and y
When we find the slope in the x direction (while keeping y fixed) we have found a partial derivative.
Or we can find the slope in the y direction (while keeping x fixed).
Let's first think about a function of one variable (x):
f(x) = x2
We can find its derivative using the Power Rule:
f’(x) = 2x
But what about a function of two variables (x and y):
f(x, y) = x2 + y3
We can find its partial derivative with respect to x when we treat y as a constant (imagine y is a number like 7 or something):
f’x = 2x + 0 = 2x
Explanation:
• the derivative of x2 (with respect to x) is 2x
• we treat y as a constant, so y3 is also a constant (imagine y=7, then 73=343 is also a constant), and the derivative of a constant is 0
To find the partial derivative with respect to y, we treat x as a constant:
f’y = 0 + 3y2 = 3y2
Explanation:
• we now treat x as a constant, so x2 is also a constant, and the derivative of a constant is 0
• the derivative of y3 (with respect to y) is 3y2
That is all there is to it. Just remember to treat all other variables as if they are constants.
### Holding A Variable Constant
So what does "holding a variable constant" look like?
### Example: the volume of a cylinder is V = π r2 h
We can write that in "multi variable" form as
f(r, h) = π r2 h
For the partial derivative with respect to r we hold h constant, and r changes:
f’r = π (2r) h = 2πrh
(The derivative of r2 with respect to r is 2r, and π and h are constants)
It says "as only the radius changes (by the tiniest amount), the volume changes by 2πrh"
It is like we add a skin with a circle's circumference (2πr) and a height of h.
For the partial derivative with respect to h we hold r constant:
f’h = π r2 (1)= πr2
(π and r2 are constants, and the derivative of h with respect to h is 1)
It says "as only the height changes (by the tiniest amount), the volume changes by πr2"
It is like we add the thinnest disk on top with a circle's area of πr2.
Let's see another example.
### Example: The surface area of a square prism.
The surface includes the top and bottom with areas of x2 each, and 4 sides of area xy each:
f(x, y) = 2x2 + 4xy
fx = 4x + 4y
fy = 0 + 4x = 4x
### Three or More Variables
We can have 3 or more variables. Just find the partial derivative of each variable in turn while treating all other variables as constants.
### Example: The volume of a cube with a square prism cut out from it.
f(x, y, z) = z3 − x2y
f’x = 0 − 2xy = −2xy
f’y = 0 − x2 = −x2
f’z = 3z2 − 0 = 3z2
When there are many x's and y's it can get confusing, so a mental trick is to change the "constant" variables into letters like "c" or "k" that look like constants.
### Example: f(x, y) = y3sin(x) + x2tan(y)
It has x's and y's all over the place! So let us try the letter change trick.
With respect to x we can change "y" to "k":
f(x, y) = k3sin(x) + x2tan(k)
f’x = k3cos(x) + 2x tan(k)
But remember to turn it back again!
f’x = y3cos(x) + 2x tan(y)
Likewise with respect to y we turn the "x" into a "k":
f(x, y) = y3sin(k) + k2tan(y)
f’y = 3y2sin(k) + k2sec2(y)
f’y = 3y2sin(x) + x2sec2(y)
But only do this if you have trouble remembering, as it is a little extra work.
Notation: we have used f’x to mean "the partial derivative with respect to x", but another very common notation is to use a funny backwards d (∂) like this:
∂f∂x = 2x
Which is the same as:
f’x = 2x
is called "del" or "dee" or "curly dee"
So ∂f ∂x can be said "del f del x"
### Example: find the partial derivatives of f(x, y, z) = x4 − 3xyz using "curly dee" notation
f(x, y, z) = x4 − 3xyz
∂f∂x = 4x3 − 3yz
∂f∂y = −3xz
∂f∂z = −3xy
You might prefer that notation, it certainly looks cool.
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# What are the zeroes of the polynomial $x\left( x-2 \right)\left( x+3 \right)$?
Last updated date: 04th Aug 2024
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Hint: In this question, we are given a polynomial $x\left( x-2 \right)\left( x+3 \right)$ whose zeroes we have to find, which means we have to find roots of this polynomial. For this, we equate the polynomial equal to zero, since the polynomial is already factorized so we just put all factors to zero and find different values of x. These values will satisfy the polynomial when equal to zero.
Complete step-by-step solution
Before jumping into finding the solution, let us first understand the meaning of zeroes of polynomials. For any polynomial, there are some values of the variable which makes polynomial zero. These values are called zeroes of the polynomial. These are also called roots of the polynomial. The degree of the polynomial determines the number of roots of the polynomial.
We are given polynomial as $x\left( x-2 \right)\left( x+3 \right)$
Let us first simplify it to find its degree and hence, the number of zeroes it has. Simplifying we get:
$\Rightarrow x\left( {{x}^{2}}-2x+3x-6 \right)={{x}^{3}}+{{x}^{2}}-6x$
Hence, the given polynomial has three roots.
Let us take the original polynomial given which is $x\left( x-2 \right)\left( x+3 \right)$
Since this polynomial is already factorized, so, it will be easy to find zeros by just putting all factors to zero.
Putting all factors to zero one by one we get:
$\Rightarrow x=0$.
$\Rightarrow x-2=0$ simplified as $x = 2$
$\Rightarrow x+3=0$ simplified as $x = -3.$
Hence, we have obtained three different zeroes which are 0, 2, and -3. Putting any of them in a polynomial will give us the value of the polynomial as zero.
Note: Students should try to factorize the polynomial first for finding the zeros of polynomials. Polynomials for which factorization is not possible, we can solve them by various methods depending upon the degree of example. We can use the splitting middle term or discriminant method for a polynomial of degree 2, the hit and trial method can be used for polynomials of degree 3. Students should take care that in the equation $x\left( x-2 \right)\left( x+3 \right)=0$ they should not just take x on the other side and eliminate it. Doing this will give only two roots.
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# How do you find a unit vector that is perpendicular to both the vector u = 0,2,1 and v = 1, -1, 1?
Sep 15, 2016
$\pm \frac{1}{\sqrt{14}} \left(3 , 1 , - 2\right)$, for opposite directions.
#### Explanation:
If vector $u = \left({u}_{1} , {u}_{2} , {u}_{3}\right) \mathmr{and} v = \left({v}_{1} , {v}_{2} , {v}_{3}\right)$, then
vectors $\pm u X v = \pm \left({u}_{2} {v}_{3} - {u}_{3} {v}_{2} , {u}_{3} {v}_{1} - {u}_{1} {v}_{3} , {u}_{1} {v}_{2} - {u}_{2} {v}_{1}\right)$
are perpendicular to both $u \mathmr{and} v$, in the opposite directions.
Here, $u \left(0 , 2 , 1\right) \mathmr{and} v = \left(1 , - 1 , 1\right)$. So,
$\pm u X v$
=+-((2)(1)-(1)(-1), (1)(1)-(0)((1), (0)(-1)-((2)(1))
$= \pm \left(3 , 1 , - 2\right)$.
For unit vectors,
divide by the modulus $| \left(3 , 1 , - 2\right) | = \sqrt{{3}^{2} + {\left(1\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{14}$,
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# MA.7.DP.1.4
Use proportional reasoning to construct, display and interpret data in circle graphs.
### Clarifications
Clarification 1: Data is limited to no more than 6 categories.
General Information
Subject Area: Mathematics (B.E.S.T.)
Strand: Data Analysis and Probability
Status: State Board Approved
## Benchmark Instructional Guide
### Terms from the K-12 Glossary
• Circle Graph
• Data
• Proportional Relationships
### Vertical Alignment
Previous Benchmarks
Next Benchmarks
### Purpose and Instructional Strategies
In grade 6, students worked with solving problems using ratio relationships. In grade 7, students apply their knowledge of ratios to solve problems involving proportions, including using proportional reasoning to construct, display and interpret categorical data in circle graphs. In high school, students will select an appropriate method to represent data, depending on whether it is numerical or categorical data and on whether it is univariate or bivariate.
• Circle graphs can be used to show how categories represent part of a whole, or compositions. Totals are represented as percentages totaling 100%, which illustrates the percentage breakdown of items and visually represents a comparison. Circle graphs are not effective, however, when there are too many categories.
• Students should be able to identify strengths and limitations in showcasing data within a circle graph.
• Instruction begins with data sets out of 100 to allow for easier calculations of percentages.
• Students should brainstorm how they might split up their circle into the needed percentages (MTR.5.1).
• For example, students can slice a circle into 4 equal parts to show students the 4 right angles at the center which total 360°. Then emphasize using proportional relationships to determine the central angle sizes needed based on the percentage size of each “slice” of the circle.
• Students should collect their own data with which to create a circle graph (MTR.7.1).
• For example, have students count colored candy/snacks or survey other students in the room about their favorite color, favorite sport or favorite genre of music/movies.
• Use protractors or online software to assist in creating circle graphs accurately.
### Common Misconceptions or Errors
• Students may incorrectly use the percent of a category for the central angle degrees instead of finding the degrees by using a proportion.
• Students may incorrectly round or make other errors in calculations that will lead to the circle graph sections not totaling 100%.
### Strategies to Support Tiered Instruction
• Teacher models several examples to work through with students, showing how to set up the proportion to find the central angle degrees, referencing patterns for students to discuss.
• For example, if students need to determine the angle measure that corresponds to 21%, the proportion below can be used.
$\frac{\text{21}}{\text{100}}$ = $\frac{\text{x}}{\text{360}}$
• Teacher models and works through several problems while discussing aloud how to properly round when having to total to 100%, reinforcing to students to work through the problems carefully as to not make computation errors.
• Teacher models using computer-based software to create circle graphs to verify how to properly round.
• Teacher provides students with fill in the blank examples working from percent of a category using proportions to find the central angle degrees.
• Teacher provides several completed examples of problems where rounding was needed for students to reference while working through multiple problems together.
• For students incorrectly using a protractor, provide students with a circle and allow them to measure sections then find the percent.
• Teacher models using fraction circle manipulatives to support converting fractions to percentages.
A group of friends has been given \$800 to host a party. They must decide how much money will be spent on food, drinks, paper products, music and decorations.
• Part A. As a group, develop two options for the friends to choose from regarding how to spend their money. Decide how much to spend in each area and create a circle graph for each option to represent your choices.
• Part B. Mikel presented the circle graph below with his recommendations on how to spend the money. How much did he choose to spend on food and drinks? How much did he choose to spend on music?
### Instructional Items
Instructional Item 1
Circle Point High School surveyed its students to determine which mode of transportation they use to get to and from school. Create and label a circle graph based on the results given below.
*The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.
## Related Courses
This benchmark is part of these courses.
1205040: M/J Grade 7 Mathematics (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1205050: M/J Accelerated Mathematics Grade 7 (Specifically in versions: 2014 - 2015, 2015 - 2020, 2020 - 2022, 2022 and beyond (current))
1204000: M/J Foundational Skills in Mathematics 6-8 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
7812020: Access M/J Grade 7 Mathematics (Specifically in versions: 2014 - 2015, 2015 - 2018, 2018 - 2019, 2019 - 2022, 2022 and beyond (current))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MA.7.DP.1.AP.4: Use proportional reasoning to interpret data in a pie chart.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Lesson Plans
The Watergate Effect part 3:
Students will create a circle graph to display categorical data of the public presidential approval rates after the Supreme Court Case United States v. Nixon. Students will graph results independently and compare them to the circle graphs created during the Watergate Effect Part 1 Lesson (Resource ID#: 208926) and the Watergate Effect Part 2 Lesson (Resource ID#: 210122) to discuss the trend of the data over the entirety of the Supreme Court case.
Type: Lesson Plan
The Watergate Effect part 2:
Students will create a circle graph to display categorical data of the public presidential approval rates during the Supreme Court Case United States v. Nixon. Students will graph results in pairs/groups and compare them to the circle graph created during the Watergate Effect Part 1 Lesson (Resource ID#: 208926).
Type: Lesson Plan
Create a Circle Graph to Represent Percentages:
Students will compare each region's percentage of seats in the U.S. House of Representatives to other regions and the whole. Students will calculate central angle degrees and create a circle graph to represent the percentages. The civics standard will be the real-world example used to apply the concept of displaying data to the Legislative Branch of government in this integrated lesson plan.
Type: Lesson Plan
The Watergate Effect Part 1:
Students will create a circle graph to display categorical data of the public presidential approval rates of Richard Nixon before the Supreme Court Case United States v. Nixon. Students will calculate percentages and central angle degrees to graph results in pairs/groups and analyze the results in this integrated lesson plan.
Type: Lesson Plan
Use Circle Graphs to Analyze International Organizations:
Students will analyze international organizations using proportional reasoning to construct circle graphs while examining the purpose of international organizations and the United States’ participation in this integrated lesson plan.
Type: Lesson Plan
Graphing Data:
This is lesson 2 in a mini unit of 3 lessons. Students will analyze data collected from students, teachers, and principals to decide whether cell phone usage should be allowed in the classroom. They will be receiving data from fictional surveys of teachers and principals. Students will use the given data to choose and create an appropriate graphical representation.
Type: Lesson Plan
Introduction to Voting and Graphing Data:
The students will vote on whether cell phones should be allowed in the classroom or not. They will use this data to select the appropriate way of graphing the results. The teacher will give sample data from other teachers and principals for students to review. The correlation will be relating students to local voting in this integrated lesson plan.
Type: Lesson Plan
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark.
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# How do you differentiate f(x)=tan((1/cos(7x))^2) using the chain rule?
Nov 10, 2015
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 {\sec}^{2} \left({\sec}^{2} \left(7 x\right)\right) \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$
#### Explanation:
Okay, so, we have
$y = \tan \left({\sec}^{2} \left(7 x\right)\right)$
Call ${\sec}^{2} \left(7 x\right) = u$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \tan \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dx}} {\sec}^{2} \left(7 x\right)$
Call $\sec \left(7 x\right) = v$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dv}} {v}^{2} \frac{\mathrm{dv}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dx}} \left(\sec \left(7 x\right)\right)$
Call $7 x = w$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \frac{d}{\mathrm{dw}} \sec \left(w\right) \frac{\mathrm{dw}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(w\right) \sec \left(w\right) \frac{d}{\mathrm{dx}} \left(7 x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(w\right) \sec \left(w\right)$
Substituting $w$ back to $x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 v \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(7 x\right) \sec \left(7 x\right)$
Substituting $v$ back to $x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 \cdot {\sec}^{2} \left(u\right) \cdot \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$
Substituing $u$ back to $x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 14 {\sec}^{2} \left({\sec}^{2} \left(7 x\right)\right) \tan \left(7 x\right) {\sec}^{2} \left(7 x\right)$
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### Course: Integrated math 2>Unit 11
Lesson 8: Inscribed shapes problem solving
Example showing supplementary opposite angles in inscribed quadrilateral.
## Want to join the conversation?
• I don't quite get how khan got 90 degrees for the small arc.
• We know that the measure of an arc is DOUBLE the measure of the Inscribed Angle. In this case, the angle WIL is inscribed by the blue arc. The measure given for this angle is 45 degrees.
So, the measure of the blue arc is 45*2=90 degrees.
I hope you got that now.
• Okay, this is basically turning my mind to mush. How are the angles always supplementary? Where did he get 90 degrees from? I'm confused, please help me.
• Angles for inscribed quadrilaterals are always supplementary. It's one of those weird facts of math that they don't normally explain, like why you divide before adding, etc. If you do some research, you can likely find some proof for why inscribed quadrilaterals are always supplementary, but it's not something normally covered in math courses. Sal got the 90 degrees by multiplying 45 by 2. This is because of one of the theorems that states that inscribed angles are 1/2 of the degree measure of the arc they intercept. Hope this helped.
• I still not understanding the math its really stressing to learn about the subject but am getting there
• You can just do 180-45 because the opposite angles of a quadrilateral have a sum of 180 degrees.
• is the opposite of angle WDL a valid inscribed angle
• WIL, ILD, LDW and DWI are all inscribed angles
An inscribed angle is the angle formed from the intersection of two chords, and a chord is a line segment that has each end point on the side of the circle somewhere.
So there are 4 chords, WI, IL, LD and DW and each place they intersect forms an inscribed angle.
I assume by opposite you mean WIL, but all angles there are inscribed angles.
Let me know if this did not help.
• at in video he talks about the sums being supplementary. are they Always supplementary? how can you tell if they are or not?
• From what Sal said in the video after that point, he implies that they are always that way and he is going to prove it in a later video.
• Wait, the angle is not inscribed? What?
• At what point exactly in the video you're getting confused? Let us know so we can help!
(1 vote)
• I thought triangles could only add up to 180 not 289
• It is not triangle there it's an quadrilateral
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# Mastering the Conversion: Understanding Microcoulombs to Coulombs
In the realm of electrical engineering, understanding the conversion from microcoulombs to coulombs is paramount. The concept of charge is fundamental to this discipline, and comprehending how to convert between these units is essential for accurate calculations and practical applications. In this comprehensive guide, we delve into the intricacies of this conversion, providing clarity and insights for engineers, students, and enthusiasts alike.
## Unraveling the Basics: What are Microcoulombs and Coulombs?
Before delving into the conversion process, it’s crucial to grasp the fundamentals of microcoulombs and coulombs. Microcoulombs and coulombs are units of electric charge, with microcoulombs representing a smaller quantity compared to coulombs.
### Microcoulombs:
Microcoulombs, denoted by the symbol µC, are a metric unit of electric charge equal to one millionth of a coulomb (1 µC = 10^-6 C). They are commonly used to measure small amounts of charge in electrical circuits and electronic devices.
### Coulombs:
Coulombs, represented by the symbol C, are the standard unit of electric charge in the International System of Units (SI). One coulomb is equivalent to the charge transported by a constant current of one ampere in one second.
## Understanding the Conversion Process
Converting microcoulombs to coulombs involves a straightforward mathematical calculation. Since microcoulombs are a fraction of a coulomb, the conversion requires multiplying the quantity of microcoulombs by a conversion factor to obtain the equivalent value in coulombs.
### Conversion Formula:
To convert microcoulombs (µC) to coulombs (C), use the following formula:
�=µ�106C=106µC
Where:
• C is the charge in coulombs.
• µC is the charge in microcoulombs.
## Practical Examples:
Let’s illustrate the conversion process with a few practical examples:
### Example 1:
Problem: Convert 500 µC to coulombs.
Solution: �=500µ�106=0.0005�C=106500µC=0.0005C
### Example 2:
Problem: Convert 2.5 × 10^6 µC to coulombs.
Solution: �=2.5×106µ�106=2.5�C=1062.5×106µC=2.5C
## Applications in Electrical Engineering
Understanding the conversion from microcoulombs to coulombs is crucial in various applications within the field of electrical engineering:
• Circuit Analysis: When analyzing electrical circuits, engineers often encounter charge values expressed in microcoulombs. Converting these values to coulombs enables accurate calculations of current, voltage, and other parameters.
• Capacitance Calculations: Capacitors store charge in the form of coulombs. By converting microcoulombs to coulombs, engineers can determine the capacitance of a capacitor and its performance in a circuit.
• Electrostatics: In studies related to electrostatics and static electricity, understanding charge quantities in coulombs is essential for analyzing phenomena such as electric fields, potential differences, and electric flux.
## Conclusion
Mastering the conversion from microcoulombs to coulombs is indispensable for anyone working in the field of electrical engineering. By understanding this fundamental concept, engineers can accurately analyze circuits, calculate capacitance, and explore various aspects of electrostatics with precision and confidence.
By following the simple conversion formula and applying it to practical examples, engineers can streamline their calculations and ensure the accuracy of their results, paving the way for innovation and advancement in the field of electrical engineering.
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# Solve for: f(x)=2sqrt(x)
## Expression: $f\left( x \right)=2\sqrt{ x }$
Substitute $y$ for $f\left( x \right)$ into the equation
$y=2\sqrt{ x }$
Take the natural logarithm of both sides of the equation
$\ln\left({y}\right)=\ln\left({2\sqrt{ x }}\right)$
Simplify the expression
$\ln\left({y}\right)=\ln\left({2}\right)+\frac{ 1 }{ 2 } \times \ln\left({x}\right)$
Differentiate both sides of the equation implicitly with respect to $x$
$\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({y}\right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2}\right)+\frac{ 1 }{ 2 } \times \ln\left({x}\right) \right)$
Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({y}\right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }$ to find the derivative
$\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2}\right)+\frac{ 1 }{ 2 } \times \ln\left({x}\right) \right)$
Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f+g \right)=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( g \right)$
$\frac{ \mathrm{d} }{ \mathrm{d}y} \left( \ln\left({y}\right) \right) \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2}\right) \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 1 }{ 2 } \times \ln\left({x}\right) \right)$
Use $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({x}\right) \right)=\frac{ 1 }{ x }$ to find derivative
$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \ln\left({2}\right) \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 1 }{ 2 } \times \ln\left({x}\right) \right)$
Find the derivative
$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=0+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 1 }{ 2 } \times \ln\left({x}\right) \right)$
Find the derivative
$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=0+\frac{ 1 }{ 2 } \times \frac{ 1 }{ x }$
Simplify the expression
$\frac{ 1 }{ y } \times \frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ 1 }{ 2x }$
Multiply both sides of the equation by $y$
$\frac{ \mathrm{d}y }{ \mathrm{d}y }=y \times \frac{ 1 }{ 2x }$
Substitute the initial equation $y=2\sqrt{ x }$ to express the derivative only in terms of $x$
$\frac{ \mathrm{d}y }{ \mathrm{d}y }=2\sqrt{ x } \times \frac{ 1 }{ 2x }$
Simplify the expression
$\frac{ \mathrm{d}y }{ \mathrm{d}y }=\frac{ \sqrt{ x } }{ x }$
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## Problem 298
Problem 298
Study the following diagram:
i. Characteristics of addition that are different from multiplication
ii. Characteristics that are common to both operations
iii. Characteristics of multiplication that are different from addition
a. Complete the diagram to illustrate how the characteristics of multiplication and addition of whole numbers compare.
b. Prepare a similar diagram for subtraction and division.
Solution
a.
i. The identity symbol for addition is the zero and identity symbol for multiplication is one.
ii. Multiplication and addition are associative, commutative, closure and have an identity property
iii. Multiplication has a distributive and zero property, addition does not.
b.
## Problem 297
Problem 297
Use the distributive property of multiplication over addition to find each product:
a. (20 + 5) * 3
b. 4 * (5 + 6)
C. (x + 10) * (3x + 2)
Solution
a. (20 + 5) * 3 = (20 * 3) + (5 * 3) = 60 + 15 = 75
b. 4*(5 + 6) = (4 * 5) + (4 * 6) = 20 + 24 = 44
c. (x + 10) * (3x + 2) = (x)(3x) + (x)(2) + (10)(3x) + (10)(2) = 3x2 + 2x + 30x + 20 = 3x2 + 32x + 20
## Problem 296
Problem 296
Properties of addition may be investigated in relation to sets other then the set of whole numbers. Use the set of multiples of 3, {0 ,3,6,9,….} to answer the following questions, and then justify answers:
a. Is this set closed under addition?
b. Does this set have an additive identity element?
c. Is addition with this set commutative?
d. Is addition with this set associative?
Solution
a. yes
b. yes
c. yes
d. yes
## Problem 295
Problem 295
Use the definition of subtraction to determine the subtraction equation that are related to each of the following addition equations:
a. 8 + 7 = n
b. 14 + x = 25
c. r + s = t
Solution
a. 8 = n-7
b. x = 25-14
c. r = t-s
## Problem 294
Problem 294
Use sets to verify the following answers:
a. 4 + 5 = 9
b. 0 + 6 = 6
c. 10 + 2 = 12
Solution
a. {11,12,13,14}+{15,16,17,18,19} = {11,12,13,14,15,16,17,18,19}
b. {}+{a,b,c,d,e,f} = {a,b,c,d,e,f}
c. {1,2,3,4,5,6,7,8,9,10}+{11,12} ={1,2,3,4,5,6,7,8,9,10,11,12}
## Problem 293
Problem 293
Decide whether the following statements are true or false and justify your answer:
a. If n(A) < n(B), then A B.
b. If n(A) n(B) , then A B.
c. If A B, then n(A)< n(B).
d. If A B, then n(A) n(B).
Solution
a. False, elements are not same in both sets.
b. False , not equal or equivalent
c. True , equal or equivalent
d. True , equal or equivalent
## Problem 292
Problem 292
a. If two sets are equal, are they necessarily equivalent?
Why or why not?
b. If two sets are equivalent, are they necessarily equal?
Why or why not?
c. Draw a diagram of and write a statement describing the relationship between the ideas of two sets being equivalent and the idea of two sets being equal.
Solution
a. Yes, equal means both have same element and all equal sets are always equivalent also.
b. NO, equivalent shows number of elements are same may be they are same but not necessarily.
c. Equal sets have same elements and same number of elements
Equivalent sets have same number of elements but not same elements.
Equal A={1,2,3} , B = {1,2,3}
Equivalent A= {1,2,3}, B= {A,B,C}
## Problem 291
Problem 291
How are the ideas of subsets and proper subsets used in counting to identify relationships between whole numbers?
Solution
It tell us which is less and which is greater than between the whole numbers. If a proper set is equivalent to a other set it means that other set is less than the first one.
## Problem 290
Problem 290
Use sets to verify that 8 > 6.
Solution
{a ,b ,c ,d ,e ,f ,g ,h} > {a ,b ,c ,d ,e ,f}
Now we can see first set has 2 additional elements, so 8 > 6
## Problem 289
Problem 289
Use the following sets to complete the given statements in as many ways as possible:
A = {a, b, c, d}
B = {e , f}
C = {b ,a , c, d}
D = {1 , 2 , 3, 4}
a. Sets _____and____are not equivalent and not equal.
b. Sets_____and____are not equivalent but are equal.
c. Sets_____and ____are equivalent but not equal.
d. Sets_____and____are equivalent and equal.
Solution
a. A and B , B and C , B and D
b. No any combination.
c. A and D, C and D
d. A and C
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# 2(4d +4)=d +1I'm trying help my daughter with her math... this is the last one im stuck on. Please help.
samhouston | Certified Educator
Begin by using the Distributive Property on the left side.
2(4d + 4) = d + 1
8d + 8 = d + 1
Next subtract d from both sides.
7d + 8 = 1
Next, subtract 8 from both sides.
7d = -7
Finally, divide both sides by 7.
d = -1
taangerine | Student
Distribute the 2 into the 4d and 4.
8d+8=d+1
Combine like terms
7d=-7
Divide 7 to isolate x.
d=-1
givingiswinning | Student
2(4d +4)=d +1
8d + 8 = d+1
8d - d = 1 - 8
7d = -7
d = -1
atyourservice | Student
2(4d +4)=d +1
First use the distributive property
8d + 8 = d+1
combine like terms by moving the terms that look alike to the same side
8d - d = 1 - 8
7d = -7
divide by 7 because we are trying to solve b
`(7d) / 7 = -7/7`
d = -1
delrae86 | Student
Distribute first so 8d +8 = d + 1
next subract d from both sides so now 7d + 8 = 1
next subract 8 from both sides so now 7d = -7
last divide both sides by 7 and get d = -1.
ehyman-kdhs | Student
• First, multiply the 2 into the equation on the left hand side
2(4d + 4) = d + 1
becomes
2*4*d + 2*4 = d + 1
8d + 8 = d + 1
• next, subtract the 1 from (d + 1) both sides
8d + 8 = d + 1
- 8 - 8
8d = d - 7
• then, subtract the d on the right hand side from both sides
8d = d - 7
-d -d
you are left with 7d = -7
• divide both sides by 7 to be left with just d
7d/7 = d
-7/7 = -1
• and you are left with d = -1
leeroyjkns | Student
Thank you! I ended up doing it the hard way. I never thought I would have to do algebre again :)
8d+8=d+1
8d+8-d=1
7d+8=1
7d=-8+1
7d=-7
7d/7=-7/7
d=-7/7
d=-1
Thanks again, love this site!
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# Laboratory: Numeric Values
Summary: We explore some of the kinds of numbers and procedures that many implementations of Scheme, including Racket, support.
## Exercises
### Exercise 1: Checking Answers
a. Determine what type DrRacket gives for the square root of two, computed by (sqrt 2). Is it exact or inexact? Real? Rational? An integer?
b. How do we know that the answer it gives us is correct? (What does “correct” mean when the answer is irrational?) We could check by squaring the value, as in
> (* (sqrt 2) (sqrt 2))
Better yet, we could subtract that result from 2, as in
> (- 2 (* (sqrt 2) (sqrt 2)))
c. What do the results of these experiments suggest? Why do you think you got the answer you got?
d. Do you expect to have the same problem as in the previous exercise if you compute the square root of 4 rather than the square root of 2? Why or why not?
### Exercise 2: Bounds
Suppose that we’ve defined val as a number, lower as 0, and upper as 100. Consider the following definition.
> (define bounded-val (min (max val lower) upper))
a. Suppose val is 25. What value will this definition associate with bounded-val?
b. Suppose val is 211. What value will this definition associate with bounded-val?
c. Suppose val is -25. What value will this definition associate with bounded-val?
d. Explain, in your own words, what the definition computes when lower is 0 and upper is 100.
e. Suppose we redefined lower to -10 and upper to 10 and then redid each of a-c. What results would we get?
f. Explain, in your own words, what this definition computes in terms of lower and upper.
### Exercise 3: Remainder
As the reading suggests, the remainder procedure computes the amount “left over” after you divide one number by another. The reading suggests that remainder provides an interesting alternative to using max and min to limit the values of functions.
a. What value do you expect each of the following to produce? Write down answers! Do not just type the code into DrRacket.
> (remainder 8 3)
> (remainder 3 8)
> (remainder 8 8)
> (remainder 9 8)
> (remainder 16 8)
> (remainder 827 8)
> (remainder 0 8)
> (remainder -8 8)
> (remainder -7 8)
> (remainder -9 8)
> (remainder -1 8)
b. Check your answers experimentally, one at a time. If you find that any of your answers don’t match what Scheme does, try to figure out why (asking your professor or a tutor if you need help), and then rethink your remaining answers before checking them experimentally.
### Exercise 4: From Reals to Integers
As the reading on numbers suggests, Scheme provides four functions that convert real numbers to nearby integers: floor, ceiling, round, and truncate. The reading also claims that there are differences between all four.
To the best of your ability, figure out what each does, and what distinguishes it from the other three. In your tests, you should try both positive and negative numbers, numbers close to integers and numbers far from integers. (Numbers whose fractional part is 0.5 are about as far from an integer as any real number can be.)
Once you have figured out answers, check the notes on this problem.
### Exercise 5: Exploring Rationals
DrRacket’s implementation of Scheme permits you to treat any real number as a rational number, which means we can get the numerator and denominator of any real number. Let’s explore what numerator and denominator that implementation uses for a variety of values.
a. Determine the numerator and denominator of the rational representation of the square root of 2.
b. Determine the numerator and denominator of the rational representation of 1.5.
c. Determine the numerator and denominator of the rational representation of 1.2.
d. Determine the numerator and denominator of 6/5.
If you are puzzled by some of the later answers, you may want to read the notes on this problem. Note that we will not expect you to regularly figure out these strange numerators and denominators.
### Exercise 6: Large Numbers
Many programming languages have limits on the size of the numbers they represent. In some cases, if the number is large enough, they approximate it. In other cases, if the number is large enough, the calculations you do with the error are erroneous. (You’ll learn why in a subsequent course.) And in still others, the language treats large enough values as the special value “infinity”.
See what happens if you try to have DrRacket compute with some very large exact integers. You may find the expt function helpful. Then see what happens if you try convert those integers to inexact values. Here are two examples to start with, but you should try more.
> (define x (expt 2 100))
> (define ex (exact->inexact x))
See what happens if you try to have DrRacket compute with some very small exact rational numbers (say, 1 divided by one of those large numbers). Then see what happens if you convert those rational numbers to inexact values.
### Exercise 7: Complex Numbers
Wev’ve seen that Scheme provides integers, rationals, reals, exact, and inexact numbers, many think that these are more kinds of numbers than you would ever need. But, believe it or not, it provides even more.
What value do you think Racket will give for (sqrt -4)?
Although you may have been told at one time that “negative numbers don’t have square roots”, that’s only true if we restrict ourselves to real roots. As you’ve just discovered, Racket supports “complex numbers”, which have not only a real component, but also an imaginary component represented by i, the square root of negative 1. Complex numbers have many uses, including representing points on the plane. We may revisit them later this semester. For now, we’re just introducing them to help you understand the depth of Scheme (and to warn you that you will sometimes get answers when you expect errors.
## For Those with Extra Time
### Extra 1: Rounding, Revisited
You may recall that we have a number of mechanisms for rounding real numbers to integers. But what if we want to round not to an integer, but to only two digits after the decimal point? Scheme does not include a built-in operation for doing that kind of rounding. Nonetheless, it is fairly straightforward.
Suppose we have a value, val. Write instructions that give val rounded to the nearest hundredth. For example,
> (define val 22.71256)
> (your-instructions val)
22.71
> (define val 10.7561)
> (your-instructions val)
10.76
### Extra 2: Rounding, Re-Revisited
In a problem above, you wrote instructions for rounding a real number to two digits after the decimal place. While such rounding is useful, it is even more useful to be able to round to an arbitrary number of digits after the decimal point.
Suppose precision is a non-negative integer and val is a real value. Write instructions for rounding val to use only precision digits after the decimal point.
> (your-instructions ... val ... precision ...)
As you write your instructions, you may find the expt function useful. (expt b p) computes bp.
## Notes
### Notes on Exercise 4: From Reals to Integers
Here are the ways we tend to think of the four functions:
(floor r) finds the largest integer less than or equal to r. Some would phrase this as “floor rounds down”.
(ceiling r) finds the smallest integer greater than or equal to r. Some would phrase this as “ceiling rounds up”.
(truncate r) removes the fractional portion of r, the portion after the decimal point.
(round r) rounds r to the nearest integer. It rounds up if the decimal portion is greater than 0.5 and it rounds down if the decimal portion is less than 0.5. If the decimal portion equals 0.5, it rounds toward the even number.
> (round 1.5)
2
> (round 2.5)
2
> (round 7.5)
8
> (round 8.5)
8
> (round -1.5)
-2
> (round -2.5)
-2
It’s pretty clear that floor and ceiling differ: If r has a fractional component, then (floor r) is one less than (ceiling r).
It’s also pretty clear that round differs from all of them, since it can round in two different directions.
We can also tell that truncate is different from ceiling, at least for positive numbers, because ceiling always rounds up, and removing the fractional portion of a positive number causes us to round down.
So, how do truncate and floor differ? As the previous paragraph implies, they differ for negative numbers. When you remove the fractional component of a negative number, you effectively round up. (After all, -2 is bigger than -2.2.) However, floor always rounds down.
Why does Scheme include so many ways to convert reals to integers? Because experience suggests that if you leave any of them out, some programmer will need that precise conversion.
### Notes on Exercise 5: Exploring Rationals
The underlying Scheme implementation seems to represent the fractional part of many numbers as the ratio of some number and 4503599627370496, which happens to be 252. (Most computers like powers of 2.) By using a large denominator, it helps ensure that representations are as accurate as possible.
If you are energetic, you might scour the Web to find out why they use an exponent of 52.
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# How to solve for x as an exponent
It’s important to keep them in mind when trying to figure out How to solve for x as an exponent. We can solve math word problems.
## How can we solve for x as an exponent
Do you need help with your math homework? Are you struggling to understand concepts How to solve for x as an exponent? There are many ways to solve quadratic functions, but one of the most popular methods is known as the quadratic formula. This formula is based on the fact that any quadratic equation can be rewritten in the form of ax^2 + bx + c = 0. The quadratic formula then states that the roots of the equation are given by: x = (-b +/- sqrt(b^2 - 4ac)) / (2a). In other words, the roots of a quadratic equation are always symmetrical around the axis of symmetry, which is given by x = -b/(2a). To use the quadratic formula, simply plug in the values of a, b, and c into the formula and solve for x. Keep in mind that there may be more than one root, so be sure to check all possible values of x. If you're struggling to remember the quadratic formula, simply Google it or look it up in a math textbook. With a little practice, you'll be solvingquadratics like a pro!
Absolute value is a concept in mathematics that refers to the distance of a number from zero on a number line. The absolute value of a number can be thought of as its magnitude, or how far it is from zero. For example, the absolute value of 5 is 5, because it is five units away from zero on the number line. The absolute value of -5 is also 5, because it is also five units away from zero, but in the opposite direction. Absolute value can be represented using the symbol "| |", as in "|5| = 5". There are a number of ways to solve problems involving absolute value. One common method is to split the problem into two cases, one for when the number is positive and one for when the number is negative. For example, consider the problem "find the absolute value of -3". This can be split into two cases: when -3 is positive, and when -3 is negative. In the first case, we have "|-3| = 3" (because 3 is three units away from zero on the number line). In the second case, we have "|-3| = -3" (because -3 is three units away from zero in the opposite direction). Thus, the solution to this problem is "|-3| = 3 or |-3| = -3". Another way to solve problems involving absolute value is to use what is known as the "distance formula". This formula allows us to calculate the distance between any two points on a number line. For our purposes, we can think of the two points as being 0 and the number whose absolute value we are trying to find. Using this formula, we can say that "the absolute value of a number x is equal to the distance between 0 and x on a number line". For example, if we want to find the absolute value of 4, we would take 4 units away from 0 on a number line (4 - 0 = 4), which tells us that "the absolute value of 4 is equal to 4". Similarly, if we want to find the absolute value of -5, we would take 5 units away from 0 in the opposite direction (-5 - 0 = -5), which tells us that "the absolute value of -5 is equal to 5". Thus, using the distance formula provides another way to solve problems involving absolute value.
This can be simplified to x=log32/log8. By using the Powers Rule, you can quickly and easily solve for exponents. However, it is important to note that this rule only works if the base of the exponent is 10. If the base is not 10, you will need to use a different method to solve for the exponent. Nevertheless, the Powers Rule is a useful tool that can save you time and effort when solving for exponents.
Homework help answers can be found in many places. The internet is a great resource for finding Homework help answers. Many websites offer Homework help answers for free. Homework help answers can also be found in books, encyclopedias, and magazines. The library is a good place to start looking for Homework help answers. Homework help answers can also be found by talking to a teacher or tutor. Homework help answers should be used to supplement learning, not replace it.
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# The circumcentre of the triangle with vertices (0, 0), (3, 0) and (0, 4) is?(a) (1, 1)(b) $\left( 2,\dfrac{3}{2} \right)$ (c) $\left( \dfrac{3}{2},2 \right)$ (d) None of these
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Hint: First, before proceeding for this, we must draw the triangle in the graph so that we get to know what this triangle looks like. Then, we should be aware of the fact that in a right angled triangle, the value of the circumcentre of the triangle is the midpoint of the hypotenuse. Then, by using midpoint formula for the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ as $x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$ , we get the desired answer.
In this question, we are supposed to find the circumcentre of the triangle with vertices (0, 0), (3, 0) and (0, 4).
So, before proceeding for this, we must draw the triangle in the graph so that we get to know what this triangle looks like.
So, we can clearly see from the figure that the graph of the triangle with vertices (0, 0), (3, 0) and (0, 4) is a right angled triangle.
Now, we should be aware of the fact that in a right angled triangle, the value of the circumcentre of the triangle is the midpoint of the hypotenuse.
Now, we need to use the midpoint formula to calculate the circumcentre of the given triangle.
So, midpoint formula for the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are given by:
$x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2},y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$
So, now the points $\left( {{x}_{1}},{{y}_{1}} \right)$ from the figure is (3, 0) and points $\left( {{x}_{2}},{{y}_{2}} \right)$ is (0, 4).
Then, by using the values of the above points, we get the coordinates of the midpoint of hypotenuse of the triangle as:
$x=\dfrac{3+0}{2},y=\dfrac{0+4}{2}$
Then, solve the above expression to get the value of coordinates as:
\begin{align} & x=\dfrac{3}{2},y=\dfrac{4}{2} \\ & \Rightarrow x=\dfrac{3}{2},y=2 \\ \end{align}
So, the values of the midpoints of the hypotenuse is $\left( \dfrac{3}{2},2 \right)$ which is also the value of circumcentre of the right angled triangle with vertices as (0, 0), (3, 0) and (0, 4).
So, the correct answer is “Option C”.
Note: Now, to solve these types of questions we need to be careful with the coordinates as the first coordinate is always x coordinate and second coordinate is always y-coordinate. If we confuse between them and select y as first coordinate then we get our answer as option (b) which is a wrong answer. So, we must be careful about the coordinate system.
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1 is what percent of 85?
(1 is 1.17647 percent of 85)
1 is 1.17647 percent of 85. Explanation: What does 1 percent or 1% mean?
Percent (%) is an abbreviation for the Latin “per centum”, which means per hundred or for every hundred. So, 1% means 1 out of every 100.
Methods to calculate "1 is what percent of 85" with step by step explanation:
Method 1: Diagonal multiplication to calculate 1 is what percent of 85.
1. Step 1: For 85, our answer will be 1
2. Step 2: For 100, our answer will be x
3. Step 3: 85*x = 1*100 (In Step 1 and 2 see colored text; Diagonal multiplications will always be equal)
4. Step 4: x = 1*100/85 = 100/85 = 1.17647
Method 2: Same side division to calculate 1 is what percent of 85
1. Step 1: For 85, our answer will be 1
2. Step 2: For 100, our answer will be x
3. Step 3: 1/x = 85/100 (In Step 1 and 2, see colored text; Same side divisions will always be equal)
4. Step 4: x/1 = 100/85
5. Step 5: x = 1*100/85 = 100/85 = 1.17647
Percentage examples
Percentages express a proportionate part of a total. When a total is not given then it is assumed to be 100. E.g. 1% (read as 1 percent) can also be expressed as 1/100 or 1:100.
Example: If 1% (1 percent) of your savings are invested in stocks, then 1 out of every 100 dollars are invested in stocks. If your savings are \$10,000, then a total of 1*100 (i.e. \$100) are invested in stocks.
Differences between percentages and percentage points
Let s take an imaginary example: In 2020, 90 percent of the population could use Inernet, and in 1990 only 10 percent could use the Internet. One can thus say that from 1990 to 2020, the Inernet usage increased by 80 percentage points although Internet usage grew at much higher pace (it increased by 900 percent from 10 to 90). In short percentages indicate ratios, not differences.
Percentage-point differences are one way to express a probability of something happening. Consider a drug that cures a given disease in 80 percent of all cases, while without the drug, the disease heals on its own in only 65 percent of cases. The drug reduces absolute risk by 15 percentage points.
Scholarship programs to learn math
Here are some of the top scholarships available to students who wish to learn math.
Examples to calculate "What is the percent decrease from X to Y?"
When you purchase through links on our website, we may earn an affiliate commission from platforms like Amazon.
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# Question Video: Finding the Solution Set of Quadratic Equations Involving Proportion Mathematics • 7th Grade
Given that π¦/π§ = π§/π = π/π = 4/11, determine the solution set of the equation π¦π₯Β² β 2π§π₯ + π = 0.
03:57
### Video Transcript
Given that π¦ over π§ equals π§ over π equals π over π, which equals four elevenths, determine the solution set of the equation π¦π₯ squared minus two π§π₯ plus π equals zero.
Now, whilst it might not look like it, we actually have a quadratic equation here. π¦, negative two π§, and π are real constants. And so if we can find the values of these constants using the information about the relationship between π¦, π§, π, and π, weβll be able to solve the quadratic equation by either factoring or using the quadratic formula.
So here is the relation weβve been given. Weβre told that π¦ and π§ are in the same proportion to one another as π§ and π such that π¦ over π§ equals four elevenths as does π§ over π. Now, in fact, we also know that the same relationship holds for π over π. But weβre not given any information about π in the rest of the question. So weβre going to ignore this.
Instead, weβll use these two equations to find values for π¦, negative two π§, and π. In fact, consider the first equation. Letβs make π¦ the subject by multiplying both sides by π§. So π¦ equals four elevenths π§. Weβre going to make π the subject in our second equation. To do so, we begin by multiplying by π and then dividing both sides by four elevenths. When we divide by a fraction, of course, we multiply by the reciprocal of that fraction. So π§ divided by four over 11 is 11 over four π§.
Now, we see that we have expressions for π¦ and π in terms of π§. And this is useful because if we substitute them into our quadratic equation, we will get an equation purely in terms of π₯ and π§. So letβs see what that looks like. π¦π₯ squared becomes four elevenths π§π₯ squared. And we replace π with 11 over four π§. So our quadratic equation is four elevenths π§π₯ squared minus two π§π₯ plus 11 over four π§ equals zero.
Now, we know that π§ itself cannot be equal to zero since when we divide some number π¦ by π§, we get a constant four elevenths. This means we can divide this entire equation by π§. And so we have a quadratic equation four elevenths π₯ squared minus two π₯ plus 11 over four equals zero. To get rid of the fractions here, we can multiply through by the lowest common multiple of 11 and 44, and thatβs 44. Four elevenths times 44 becomes four times four, which is 16. Negative two times 44 is negative 88. And 11 over four times 44 is the same as 11 times 11, which is 121. And this of course is all equal to zero.
We can then factor the expression on the left-hand side. And we do so by noting that both 16π₯ squared and 121 are square numbers. So we might deduce that each expression contains four π₯ and 11. In fact, if we have four π₯ minus 11 times four π₯ minus 11, we get 16π₯ squared and 121 as required. But four π₯ times negative 11 plus negative 11 times four π₯ gives us negative 88π₯. So our quadratic equation can now be written as four π₯ minus 11 squared equals zero.
Letβs solve for π₯. Since we have zero on the right-hand side, we can simply take the square root. So four π₯ minus 11 equals zero. Adding 11 to both sides, we get four π₯ equals 11. And finally, dividing through by four, and we get π₯ is equal to 11 over four.
And so there is simply one solution to the equation. Using set notation, we say itβs the set containing the single element 11 over four.
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# 13.7: Extreme Values and Saddle Points
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Learning Objectives
• Use partial derivatives to locate critical points for a function of two variables.
• Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
• Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.
One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.
## Critical Points
For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.
Definition: Critical Points
Let $$z=f(x,y)$$ be a function of two variables that is differentiable on an open set containing the point $$(x_0,y_0)$$. The point $$(x_0,y_0)$$ is called a critical point of a function of two variables $$f$$ if one of the two following conditions holds:
1. $$f_x(x_0,y_0)=f_y(x_0,y_0)=0$$
2. Either $$f_x(x_0,y_0) \; \text{or} \; f_y(x_0,y_0)$$ does not exist.
Example $$\PageIndex{1}$$: Finding Critical Points
Find the critical points of each of the following functions:
1. $$f(x,y)=\sqrt{4y^2−9x^2+24y+36x+36}$$
2. $$g(x,y)=x^2+2xy−4y^2+4x−6y+4$$
Solution:
a. First, we calculate $$f_x(x,y) \; \text{and} \; f_y(x,y):$$
\begin{align*} f_x(x,y) =\dfrac{1}{2}(−18x+36)(4y^2−9x^2+24y+36x+36)^{−1/2} \\ =\dfrac{−9x+18}{\sqrt{4y^2−9x^2+24y+36x+36}} \end{align*}
\begin{align*} f_y(x,y) =\dfrac{1}{2}(8y+24)(4y^2−9x^2+24y+36x+36)^{−1/2} \\ =\dfrac{4y+12}{\sqrt{4y^2−9x^2+24y+36x+36}} \end{align*}.
Next, we set each of these expressions equal to zero:
\begin{align*} \dfrac{−9x+18}{\sqrt{4y^2−9x^2+24y+36x+36}} =0 \\ \dfrac{4y+12}{\sqrt{4y^2−9x^2+24y+36x+36}} =0. \end{align*}
Then, multiply each equation by its common denominator:
\begin{align*} −9x+18 =0 \\ 4y+12 =0. \end{align*}
Therefore, $$x=2$$ and $$y=−3,$$ so $$(2,−3)$$ is a critical point of $$f$$.
We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:
$4y^2−9x^2+24y+36x+36=0. \label{critical1}$
Equation \ref{critical1} represents a hyperbola. We should also note that the domain of $$f$$ consists of points satisfying the inequality
$4y^2−9x^2+24y+36x+36≥0.$
Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:
\begin{align*} 4y^2−9x^2+24y+36x+36 =0 \\ 4y^2−9x^2+24y+36x =−36 \\ 4y^2+24y−9x^2+36x =−36 \\ 4(y^2+6y)−9(x^2−4x) =−36 \\ 4(y^2+6y+9)−9(x^2−4x+4) =−36−36+36 \\ 4(y+3)^2−9(x−2)^2 =−36.\end{align*}
Dividing both sides by $$−36$$ puts the equation in standard form:
\begin{align*} \dfrac{4(y+3)^2}{−36}−\dfrac{9(x−2)^2}{−36} =1 \\ \dfrac{(x−2)^2}{4}−\dfrac{(y+3)^2}{9} =1. \end{align*}
Notice that point $$(2,−3)$$ is the center of the hyperbola.
Thus, the critical points of the function $$f$$ are $$(2, -3)$$ and all points on the hyperbola, $$\dfrac{(x−2)^2}{4}−\dfrac{(y+3)^2}{9}=1$$.
b. First, we calculate $$g_x(x,y)$$ and $$g_y(x,y)$$:
\begin{align*} g_x(x,y) =2x+2y+4 \\ g_y(x,y) =2x−8y−6. \end{align*}
Next, we set each of these expressions equal to zero, which gives a system of equations in $$x$$ and $$y$$:
\begin{align*} 2x+2y+4 =0 \\ 2x−8y−6 =0. \end{align*}
Subtracting the second equation from the first gives $$10y+10=0$$, so $$y=−1$$. Substituting this into the first equation gives $$2x+2(−1)+4=0$$, so $$x=−1$$.
Therefore $$(−1,−1)$$ is a critical point of $$g$$. There are no points in $$\mathbb{R}^2$$ that make either partial derivative not exist.
Figure $$\PageIndex{1}$$ shows the behavior of the surface at the critical point.
Exercise $$\PageIndex{1}$$
Find the critical point of the function $$f(x,y)=x^3+2xy−2x−4y.$$
Hint
Calculate $$f_x(x,y)$$ and $$f_y(x,y)$$, then set them equal to zero.
The only critical point of $$f$$ is $$(2,−5)$$.
The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.
Definition: Global and Local Extrema
Let $$z=f(x,y)$$ be a function of two variables that is defined and continuous on an open set containing the point $$(x_0,y_0).$$ Then $$f$$ has a local maximum at $$(x_0,y_0$$ if
$f(x_0,y_0)≥f(x,y)$
for all points $$(x,y)$$ within some disk centered at $$(x_0,y_0)$$. The number $$f(x_0,y_0)$$ is called a local maximum value. If the preceding inequality holds for every point $$(x,y)$$ in the domain of $$f$$, then $$f$$ has a global maximum (also called an absolute maximum) at $$(x_0,y_0).$$
The function $$f$$ has a local minimum at $$(x_0,y_0)$$ if
$f(x_0,y_0)≤f(x,y)$
for all points $$(x,y)$$ within some disk centered at $$(x_0,y_0)$$. The number $$f(x_0,y_0)$$ is called a local minimum value. If the preceding inequality holds for every point $$(x,y)$$ in the domain of $$f$$, then $$f$$ has a global minimum (also called an absolute minimum) at $$(x_0,y_0)$$.
If $$f(x_0,y_0)$$ is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).
In Calculus 1, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.
Fermat’s Theorem for Functions of Two Variables
Let $$z=f(x,y)$$ be a function of two variables that is defined and continuous on an open set containing the point $$(x_0,y_0)$$. Suppose $$f_x$$ and $$f_y$$ each exists at $$(x_0,y_0)$$. If f has a local extremum at $$(x_0,y_0)$$, then $$(x_0,y_0)$$ is a critical point of $$f$$.
## Second Derivative Test
Consider the function $$f(x)=x^3.$$ This function has a critical point at $$x=0$$, since $$f'(0)=3(0)^2=0$$. However, $$f$$ does not have an extreme value at $$x=0$$. Therefore, the existence of a critical value at $$x=x_0$$ does not guarantee a local extremum at $$x=x_0$$. The same is true for a function of two or more variables. One way this can happen is at a saddle point. An example of a saddle point appears in the following figure.
Figure $$\PageIndex{3}$$: Graph of the function $$z=x^2−y^2$$. This graph has a saddle point at the origin.
In this graph, the origin is a saddle point. This is because the first partial derivatives of f$$(x,y)=x^2−y^2$$ are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to $$y=0$$ is $$z=x^2$$ (a parabola opening upward), but the vertical trace corresponding to $$x=0$$ is $$z=−y^2$$ (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.
Given the function $$z=f(x,y),$$ the point $$\big(x_0,y_0,f(x_0,y_0)\big)$$ is a saddle point if both $$f_0(x_0,y_0)=0$$ and $$f_y(x_0,y_0)=0$$, but $$f$$ does not have a local extremum at $$(x_0,y_0).$$
The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant $$D$$ that replaces $$f''(x_0)$$ in the second derivative test for a function of one variable.
Second Derivative Test
Let $$z=f(x,y)$$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $$(x_0,y_0)$$. Suppose $$f_x(x_0,y_0)=0$$ and $$f_y(x_0,y_0)=0.$$ Define the quantity
$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−\big(f_{xy}(x_0,y_0)\big)^2.$
Then:
1. If $$D>0$$ and $$f_{xx}(x_0,y_0)>0$$, then f has a local minimum at $$(x_0,y_0)$$.
2. If $$D>0$$ and $$f_{xx}(x_0,y_0)<0$$, then f has a local maximum at $$(x_0,y_0)$$.
3. If $$D<0$$, then $$f$$ has a saddle point at $$(x_0,y_0)$$.
4. If $$D=0$$, then the test is inconclusive.
See Figure $$\PageIndex{4}$$.
To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.
Problem-Solving Strategy: Using the Second Derivative Test for Functions of Two Variables
Let $$z=f(x,y)$$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $$(x_0,y_0).$$ To apply the second derivative test to find local extrema, use the following steps:
1. Determine the critical points $$(x_0,y_0)$$ of the function $$f$$ where $$f_x(x_0,y_0)=f_y(x_0,y_0)=0.$$ Discard any points where at least one of the partial derivatives does not exist.
2. Calculate the discriminant $$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−\big(f_{xy}(x_0,y_0)\big)^2$$ for each critical point of $$f$$.
3. Apply the four cases of the test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.
Example $$\PageIndex{2}$$: Using the Second Derivative Test
Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:
1. $$f(x,y)=4x^2+9y^2+8x−36y+24$$
2. $$g(x,y)=\dfrac{1}{3}x^3+y^2+2xy−6x−3y+4$$
Solution:
a. Step 1 of the problem-solving strategy involves finding the critical points of $$f$$. To do this, we first calculate $$f_x(x,y)$$ and $$f_y(x,y)$$, then set each of them equal to zero:
\begin{align*} f_x(x,y) =8x+8 \\ f_y(x,y) =18y−36. \end{align*}
Setting them equal to zero yields the system of equations
\begin{align*} 8x+8 =0 \\ 18y−36 =0. \end{align*}
The solution to this system is $$x=−1$$ and $$y=2$$. Therefore $$(−1,2)$$ is a critical point of $$f$$.
Step 2 of the problem-solving strategy involves calculating $$D.$$ To do this, we first calculate the second partial derivatives of $$f:$$
\begin{align*} f_{xx}(x,y) =8 \\ f_{xy}(x,y) =0 \\ f_{yy}(x,y) =18. \end{align*}
Therefore, $$D=f_{xx}(−1,2)f_{yy}(−1,2)−\big(f_{xy}(−1,2)\big)^2=(8)(18)−(0)^2=144.$$
Step 3 states to apply the four cases of the test to classify the function's behavior at this critical point.
Since $$D>0$$ and $$f_{xx}(−1,2)>0,$$ this corresponds to case 1. Therefore, $$f$$ has a local minimum at $$(−1,2)$$ as shown in the following figure.
Figure $$\PageIndex{5}$$: The function $$f(x,y)$$ has a local minimum at $$(−1,2,−16).$$ Note the scale on the $$y$$-axis in this plot is in thousands.
b. For step 1, we first calculate $$g_x(x,y)$$ and $$g_y(x,y)$$, then set each of them equal to zero:
\begin{align*} g_x(x,y) =x^2+2y−6 \\ g_y(x,y) =2y+2x−3. \end{align*}
Setting them equal to zero yields the system of equations
\begin{align*} x^2+2y−6 =0 \\ 2y+2x−3 =0. \end{align*}
To solve this system, first solve the second equation for $$y$$. This gives $$y=\dfrac{3−2x}{2}$$. Substituting this into the first equation gives
\begin{align*} x^2+3−2x−6 =0 \\ x^2−2x−3 =0 \\ (x−3)(x+1) =0. \end{align*}
Therefore, $$x=−1$$ or $$x=3$$. Substituting these values into the equation $$y=\dfrac{3−2x}{2}$$ yields the critical points $$\left(−1,\frac{5}{2}\right)$$ and $$\left(3,−\frac{3}{2}\right)$$.
Step 2 involves calculating the second partial derivatives of $$g$$:
\begin{align*} g_{xx}(x,y) =2x \\ g_{xy}(x,y) =2\\ g_{yy}(x,y) =2. \end{align*}
Then, we find a general formula for $$D$$:
\begin{align*} D(x_0, y_0) =g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)−\big(g_{xy}(x_0,y_0)\big)^2 \\ =(2x_0)(2)−2^2\\ =4x_0−4.\end{align*}
Next, we substitute each critical point into this formula:
\begin{align*} D\left(−1,\tfrac{5}{2}\right) =(2(−1))(2)−(2)^2=−4−4=−8 \\ D\left(3,−\tfrac{3}{2}\right) =(2(3))(2)−(2)^2=12−4=8. \end{align*}
In step 3, we note that, applying Note to point $$\left(−1,\frac{5}{2}\right)$$ leads to case $$3$$, which means that $$\left(−1,\frac{5}{2}\right)$$ is a saddle point. Applying the theorem to point $$\left(3,−\frac{3}{2}\right)$$ leads to case $$1$$, which means that $$\left(3,−\frac{3}{2}\right)$$ corresponds to a local minimum as shown in the following figure.
Exercise $$\PageIndex{2}$$
Use the second derivative to find the local extrema of the function
$f(x,y)=x^3+2xy−6x−4y^2. \nonumber$
Hint
Follow the problem-solving strategy for applying the second derivative test.
$$\left(\frac{4}{3},\frac{1}{3}\right)$$ is a saddle point, $$\left(−\frac{3}{2},−\frac{3}{8}\right)$$ is a local maximum.
## Absolute Maxima and Minima
When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.
Extreme Value Theorem
A continuous function $$f(x,y)$$ on a closed and bounded set $$D$$ in the plane attains an absolute maximum value at some point of $$D$$ and an absolute minimum value at some point of $$D$$.
Now that we know any continuous function $$f$$ defined on a closed, bounded set attains its extreme values, we need to know how to find them.
Finding Extreme Values of a Function of Two Variables
Assume $$z=f(x,y)$$ is a differentiable function of two variables defined on a closed, bounded set $$D$$. Then $$f$$ will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:
1. The values of $$f$$ at the critical points of $$f$$ in $$D$$.
2. The values of $$f$$ on the boundary of $$D$$.
The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of $$D$$, then it is located at an interior point of $$D$$. But an interior point $$(x_0,y_0)$$ of $$D$$ that’s an absolute extremum is also a local extremum; hence, $$(x_0,y_0)$$ is a critical point of $$f$$ by Fermat’s theorem. Therefore the only possible values for the global extrema of $$f$$ on $$D$$ are the extreme values of $$f$$ on the interior or boundary of $$D$$.
Problem-Solving Strategy: Finding Absolute Maximum and Minimum Values
Let $$z=f(x,y)$$ be a continuous function of two variables defined on a closed, bounded set $$D$$, and assume $$f$$ is differentiable on $$D$$. To find the absolute maximum and minimum values of $$f$$ on $$D$$, do the following:
1. Determine the critical points of $$f$$ in $$D$$.
2. Calculate $$f$$ at each of these critical points.
3. Determine the maximum and minimum values of $$f$$ on the boundary of its domain.
4. The maximum and minimum values of $$f$$ will occur at one of the values obtained in steps $$2$$ and $$3$$.
Finding the maximum and minimum values of $$f$$ on the boundary of $$D$$ can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example. The same approach can be used for other shapes such as circles and ellipses.
If the boundary of the set $$D$$ is a more complicated curve defined by a function $$g(x,y)=c$$ for some constant $$c$$, and the first-order partial derivatives of $$g$$ exist, then the method of Lagrange multipliers can prove useful for determining the extrema of $$f$$ on the boundary which is introduced in Lagrange Multipliers.
Example $$\PageIndex{3}$$: Finding Absolute Extrema
Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:
1. $$f(x,y)=x^2−2xy+4y^2−4x−2y+24$$ on the domain defined by $$0≤x≤4$$ and $$0≤y≤2$$
2. $$g(x,y)=x^2+y^2+4x−6y$$ on the domain defined by $$x^2+y^2≤16$$
Solution:
a. Using the problem-solving strategy, step $$1$$ involves finding the critical points of $$f$$ on its domain. Therefore, we first calculate $$f_x(x,y)$$ and $$f_y(x,y)$$, then set them each equal to zero:
\begin{align*} f_x(x,y) =2x−2y−4 \\f_y(x,y) =−2x+8y−2. \end{align*}
Setting them equal to zero yields the system of equations
\begin{align*} 2x−2y−4 =0\\−2x+8y−2 =0. \end{align*}
The solution to this system is $$x=3$$ and $$y=1$$. Therefore $$(3,1)$$ is a critical point of $$f$$. Calculating $$f(3,1)$$ gives $$f(3,1)=17.$$
The next step involves finding the extrema of $$f$$ on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:
$$L_1$$ is the line segment connecting $$(0,0)$$ and $$(4,0)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=0$$ for $$0≤t≤4$$. Define $$g(t)=f\big(x(t),y(t)\big)$$. This gives $$g(t)=t^2−4t+24$$. Differentiating $$g$$ leads to $$g′(t)=2t−4.$$ Therefore, $$g$$ has a critical value at $$t=2$$, which corresponds to the point $$(2,0)$$. Calculating $$f(2,0)$$ gives the $$z$$-value $$20$$.
$$L_2$$ is the line segment connecting $$(4,0)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=4,y(t)=t$$ for $$0≤t≤2.$$ Again, define $$g(t)=f\big(x(t),y(t)\big).$$ This gives $$g(t)=4t^2−10t+24.$$ Then, $$g′(t)=8t−10$$. g has a critical value at $$t=\frac{5}{4}$$, which corresponds to the point $$\left(0,\frac{5}{4}\right).$$ Calculating $$f\left(0,\frac{5}{4}\right)$$ gives the $$z$$-value $$27.75$$.
$$L_3$$ is the line segment connecting $$(0,2)$$ and $$(4,2)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=2$$ for $$0≤t≤4.$$ Again, define $$g(t)=f\big(x(t),y(t)\big).$$ This gives $$g(t)=t^2−8t+36.$$ The critical value corresponds to the point $$(4,2).$$ So, calculating $$f(4,2)$$ gives the $$z$$-value $$20$$.
$$L_4$$ is the line segment connecting $$(0,0)$$ and $$(0,2)$$, and it can be parameterized by the equations $$x(t)=0,y(t)=t$$ for $$0≤t≤2.$$ This time, $$g(t)=4t^2−2t+24$$ and the critical value $$t=\frac{1}{4}$$ correspond to the point $$\left(0,\frac{1}{4}\right)$$. Calculating $$f\left(0,\frac{1}{4}\right)$$ gives the $$z$$-value $$23.75.$$
We also need to find the values of $$f(x,y)$$ at the corners of its domain. These corners are located at $$(0,0),(4,0),(4,2)$$ and $$(0,2)$$:
\begin{align*} f(0,0) =(0)^2−2(0)(0)+4(0)^2−4(0)−2(0)+24=24 \\f(4,0) =(4)^2−2(4)(0)+4(0)^2−4(4)−2(0)+24=24 \\ f(4,2) =(4)^2−2(4)(2)+4(2)^2−4(4)−2(2)+24=20\\ f(0,2) =(0)^2−2(0)(2)+4(2)^2−4(0)−2(2)+24=36. \end{align*}
The absolute maximum value is $$36$$, which occurs at $$(0,2)$$, and the global minimum value is $$20$$, which occurs at both $$(4,2)$$ and $$(2,0)$$ as shown in the following figure.
b. Using the problem-solving strategy, step $$1$$ involves finding the critical points of $$g$$ on its domain. Therefore, we first calculate $$g_x(x,y)$$ and $$g_y(x,y)$$, then set them each equal to zero:
\begin{align*} g_x(x,y) =2x+4 \\ g_y(x,y) =2y−6. \end{align*}
Setting them equal to zero yields the system of equations
\begin{align*} 2x+4 =0 \\ 2y−6 =0. \end{align*}
The solution to this system is $$x=−2$$ and $$y=3$$. Therefore, $$(−2,3)$$ is a critical point of $$g$$. Calculating $$g(−2,3),$$ we get
$g(−2,3)=(−2)^2+3^2+4(−2)−6(3)=4+9−8−18=−13.$
The next step involves finding the extrema of g on the boundary of its domain. The boundary of its domain consists of a circle of radius $$4$$ centered at the origin as shown in the following graph.
The boundary of the domain of g can be parameterized using the functions $$x(t)=4\cos t,\, y(t)=4\sin t$$ for $$0≤t≤2π$$. Define $$h(t)=g\big(x(t),y(t)\big):$$
\begin{align*} h(t) =g\big(x(t),y(t)\big) \\ =(4\cos t)^2+(4\sin t)^2+4(4\cos t)−6(4\sin t) \\ =16\cos^2t+16\sin^2t+16\cos t−24\sin t\\ =16+16\cos t−24\sin t. \end{align*}
Setting $$h′(t)=0$$ leads to
\begin{align*} −16\sin t−24\cos t =0 \\ −16\sin t =24\cos t\\\dfrac{−16\sin t}{−16\cos t} =\dfrac{24\cos t}{−16\cos t} \\\tan t =−\dfrac{4}{3}. \end{align*}
\begin{align*} −16\sin t−24\cos t =0 \\ −16\sin t =24\cos t\\\dfrac{−16\sin t}{−16\cos t} =\dfrac{24\cos t}{−16\cos t} \\\tan t =−\dfrac{3}{2}. \end{align*}
This equation has two solutions over the interval $$0≤t≤2π$$. One is $$t=π−\arctan (\frac{3}{2})$$ and the other is $$t=2π−\arctan (\frac{3}{2})$$. For the first angle,
\begin{align*} \sin t =\sin(π−\arctan(\tfrac{3}{2}))=\sin (\arctan (\tfrac{3}{2}))=\dfrac{3\sqrt{13}}{13} \\ \cos t =\cos (π−\arctan (\tfrac{3}{2}))=−\cos (\arctan (\tfrac{3}{2}))=−\dfrac{2\sqrt{13}}{13}. \end{align*}
Therefore, $$x(t)=4\cos t=−\frac{8\sqrt{13}}{13}$$ and $$y(t)=4\sin t=\frac{12\sqrt{13}}{13}$$, so $$\left(−\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)$$ is a critical point on the boundary and
\begin{align*} g\left(−\tfrac{8\sqrt{13}}{13},\tfrac{12\sqrt{13}}{13}\right) =\left(−\tfrac{8\sqrt{13}}{13}\right)^2+\left(\tfrac{12\sqrt{13}}{13}\right)^2+4\left(−\tfrac{8\sqrt{13}}{13}\right)−6\left(\tfrac{12}{\sqrt{13}}{13}\right) \\ =\frac{144}{13}+\frac{64}{13}−\frac{32\sqrt{13}}{13}−\frac{72\sqrt{13}}{13} \\ =\frac{208−104\sqrt{13}}{13}≈−12.844. \end{align*}
For the second angle,
\begin{align*} \sin t =\sin (2π−\arctan (\tfrac{3}{2}))=−\sin (\arctan (\tfrac{3}{2}))=−\dfrac{3\sqrt{13}}{13} \\ \cos t =\cos (2π−\arctan (\tfrac{3}{2}))=\cos (\arctan (\tfrac{3}{2}))=\dfrac{2\sqrt{13}}{13}. \end{align*}
Therefore, $$x(t)=4\cos t=\frac{8\sqrt{13}}{13}$$ and $$y(t)=4\sin t=−\frac{12\sqrt{13}}{13}$$, so $$\left(\frac{8\sqrt{13}}{13},−\frac{12\sqrt{13}}{13}\right)$$ is a critical point on the boundary and
\begin{align*} g\left(\tfrac{8\sqrt{13}}{13},−\tfrac{12\sqrt{13}}{13}\right) =\left(\tfrac{8\sqrt{13}}{13}\right)^2+\left(−\tfrac{12\sqrt{13}}{13}\right)^2+4\left(\tfrac{8\sqrt{13}}{13}\right)−6\left(−\tfrac{12\sqrt{13}}{13}\right) \\ =\dfrac{144}{13}+\dfrac{64}{13}+\dfrac{32\sqrt{13}}{13}+\dfrac{72\sqrt{13}}{13}\\ =\dfrac{208+104\sqrt{13}}{13}≈44.844. \end{align*}
The absolute minimum of g is $$−13,$$ which is attained at the point $$(−2,3)$$, which is an interior point of D. The absolute maximum of g is approximately equal to 44.844, which is attained at the boundary point $$\left(\frac{8\sqrt{13}}{13},−\frac{12\sqrt{13}}{13}\right)$$. These are the absolute extrema of g on D as shown in the following figure.
Exercise $$\PageIndex{3}$$:
Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function
$f(x,y)=4x^2−2xy+6y^2−8x+2y+3 \nonumber$
on the domain defined by $$0≤x≤2$$ and $$−1≤y≤3.$$
Hint
Calculate $$f_x(x,y)$$ and $$f_y(x,y)$$, and set them equal to zero. Then, calculate $$f$$ for each critical point and find the extrema of $$f$$ on the boundary of $$D$$.
The absolute minimum occurs at $$(1,0): f(1,0)=−1.$$
The absolute maximum occurs at $$(0,3): f(0,3)=63.$$
Example $$\PageIndex{4}$$: Profitable Golf Balls
Pro-$$T$$ company has developed a profit model that depends on the number $$x$$ of golf balls sold per month (measured in thousands), and the number of hours per month of advertising $$y$$, according to the function
$z=f(x,y)=48x+96y−x^2−2xy−9y^2,$
where $$z$$ is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is $$50,000$$, and the maximum number of hours of advertising that can be purchased is $$25$$. Find the values of $$x$$ and $$y$$ that maximize profit, and find the maximum profit.
Solution
Using the problem-solving strategy, step $$1$$ involves finding the critical points of $$f$$ on its domain. Therefore, we first calculate $$f_x(x,y)$$ and $$f_y(x,y),$$ then set them each equal to zero:
\begin{align*} f_x(x,y) =48−2x−2y \\ f_y(x,y) =96−2x−18y. \end{align*}
Setting them equal to zero yields the system of equations
\begin{align*} 48−2x−2y =0 \\ 96−2x−18y =0. \end{align*}
The solution to this system is $$x=21$$ and $$y=3$$. Therefore $$(21,3)$$ is a critical point of $$f$$. Calculating $$f(21,3)$$ gives $$f(21,3)=48(21)+96(3)−21^2−2(21)(3)−9(3)^2=648.$$
The domain of this function is $$0≤x≤50$$ and $$0≤y≤25$$ as shown in the following graph.
$$L_1$$ is the line segment connecting $$(0,0)$$ and $$(50,0),$$ and it can be parameterized by the equations $$x(t)=t,y(t)=0$$ for $$0≤t≤50.$$ We then define $$g(t)=f(x(t),y(t)):$$
\begin{align*} g(t) =f(x(t),y(t)) \\ =f(t,0)\\ =48t+96(0)−y^2−2(t)(0)−9(0)^2\\ =48t−t^2. \end{align*}
Setting $$g′(t)=0$$ yields the critical point $$t=24,$$ which corresponds to the point $$(24,0)$$ in the domain of $$f$$. Calculating $$f(24,0)$$ gives $$576.$$
$$L_2$$ is the line segment connecting and $$(50,25)$$, and it can be parameterized by the equations $$x(t)=50,y(t)=t$$ for $$0≤t≤25$$. Once again, we define $$g(t)=f\big(x(t),y(t)\big):$$
\begin{align*} g(t) =f\big(x(t),y(t)\big)\\ =f(50,t)\\ =48(50)+96t−50^2−2(50)t−9t^2 \\ =−9t^2−4t−100. \end{align*}
This function has a critical point at $$t =−\frac{2}{9}$$, which corresponds to the point $$(50,−29)$$. This point is not in the domain of $$f$$.
$$L_3$$ is the line segment connecting $$(0,25)$$ and $$(50,25)$$, and it can be parameterized by the equations $$x(t)=t,y(t)=25$$ for $$0≤t≤50$$. We define $$g(t)=f\big(x(t),y(t)\big)$$:
\begin{align*} g(t) =f\big(x(t),y(t)\big)\\ =f(t,25) \\ =48t+96(25)−t^2−2t(25)−9(25^2) \\ =−t^2−2t−3225. \end{align*}
This function has a critical point at $$t=−1$$, which corresponds to the point $$(−1,25),$$ which is not in the domain.
$$L_4$$ is the line segment connecting $$(0,0)$$ to $$(0,25)$$, and it can be parameterized by the equations $$x(t)=0,y(t)=t$$ for $$0≤t≤25$$. We define $$g(t)=f\big(x(t),y(t)\big)$$:
\begin{align*} g(t) =f\big(x(t),y(t)\big) \\ =f(0,t) \\ =48(0)+96t−(0)^2−2(0)t−9t^2 \\ =96t−t^2. \end{align*}
This function has a critical point at $$t=\frac{16}{3}$$, which corresponds to the point $$\left(0,\frac{16}{3}\right)$$, which is on the boundary of the domain. Calculating $$f\left(0,\frac{16}{3}\right)$$ gives $$256$$.
We also need to find the values of $$f(x,y)$$ at the corners of its domain. These corners are located at $$(0,0),(50,0),(50,25)$$ and $$(0,25)$$:
\begin{align*} f(0,0) =48(0)+96(0)−(0)^2−2(0)(0)−9(0)^2=0\\f(50,0) =48(50)+96(0)−(50)^2−2(50)(0)−9(0)^2=−100 \\f(50,25) =48(50)+96(25)−(50)^2−2(50)(25)−9(25)^2=−5825 \\ f(0,25) =48(0)+96(25)−(0)^2−2(0)(25)−9(25)^2=−3225. \end{align*}
The maximum critical value is $$648$$, which occurs at $$(21,3)$$. Therefore, a maximum profit of $$648,000$$ is realized when $$21,000$$ golf balls are sold and $$3$$ hours of advertising are purchased per month as shown in the following figure.
## Key Concepts
• A critical point of the function $$f(x,y)$$ is any point $$(x_0,y_0)$$ where either $$f_x(x_0,y_0)=f_y(x_0,y_0)=0$$, or at least one of $$f_x(x_0,y_0)$$ and $$f_y(x_0,y_0)$$ do not exist.
• A saddle point is a point $$(x_0,y_0)$$ where $$f_x(x_0,y_0)=f_y(x_0,y_0)=0$$, but $$(x_0,y_0)$$ is neither a maximum nor a minimum at that point.
• To find extrema of functions of two variables, first find the critical points, then calculate the discriminant and apply the second derivative test.
## Key Equations
• Discriminant
$$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2$$
## Glossary
critical point of a function of two variables
the point $$(x_0,y_0)$$ is called a critical point of $$f(x,y)$$ if one of the two following conditions holds:
1. $$f_x(x_0,y_0)=f_y(x_0,y_0)=0$$
2. At least one of $$f_x(x_0,y_0)$$ and $$f_y(x_0,y_0)$$ do not exist
discriminant
the discriminant of the function $$f(x,y)$$ is given by the formula $$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−(f_{xy}(x_0,y_0))^2$$
given the function $$z=f(x,y),$$ the point $$(x_0,y_0,f(x_0,y_0))$$ is a saddle point if both $$f_x(x_0,y_0)=0$$ and $$f_y(x_0,y_0)=0$$, but $$f$$ does not have a local extremum at $$(x_0,y_0)$$
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# Order of magnitude Physics
In physics, we come across quantities that vary over a wide range. For example, we deal with both massive planetary objects like planets and galaxies, as well as very microscopic particles like the nucleus of an atom.
No matter how massive or small a physical quantity is, we need magnitude to describe it.
For a physical quantity, magnitude means its numerical value and the units in which it is expressed.
For example, if we say Rohan travelled a distance of 100 Km, then 100 is the numerical value of the physical quantity distance, and Km is the unit of measurement for distance.
In this case, 100 km is a short distance compared to much greater distances between the sun and the earth.
These huge distances are difficult to remember. Instead of memorizing the proper distance, we need to have an approximate idea of the size of the quantity in question.
## What is Order of magnitude?
The order of magnitude gives us a rough estimate of the size of a physical quantity, regardless of how large or small its magnitude is. It describes roughly how big a number to the power of 10 is. So, to express such a broad range of numbers, we employ the power of the ten approaches.
If two numbers differ by one order of magnitude, one is roughly ten times greater. They differ by a factor of around 100 if they differ by two orders of magnitude.
### Definition
The order of magnitude of a physical quantity is that power of 10 which is closest to its magnitude.
### Calculate the order of magnitude.
Let us look at the order of magnitude formula that can be used to find the order of magnitude of a number.
#### Formula
The order of magnitude of a number $N$ can be expressed as relation $$N=n\times 10^x$$
If $0.5< n\leq5$, then $x$ will be the order of magnitude of $N$.
This condition requires us to convert our number $N$ to a decimal (in a format of $n\times 10^x$) so that our number $n$ in decimals falls between $0.5$ and $5$.
After converting our number to the power of 10 i.e., $x$ gives the order of magnitude of that number.
So, from this condition, we have two simple rules for finding the order of magnitude for calculating the order of magnitude.
## How to find order of magnitude
Rule 1:- If we have the first digit of our number between 1 and 4 then decimal comes after the first digit.
Example 1:- Find the order of magnitude of number 2134
We can write this number in the form of $N=n10^x$ which is
$2134=2.134\times 10^3$
here, $x=+3$
so, 3 is the order of magnitude of the number 2134.
Example 2:- Find the order of magnitude of number 0.003548.
We can write this number in the form of $N=n10^x$ which is
$0.003548=3.584\times 10^{-3}$
here, $x=-3$
so, $-3$ is the order of magnitude of number 0.003548.
Rule 2:- If we have the first digit of our number between 5 and 9 then the decimal comes before the first digit.
Example 3:- Find the order of magnitude of number 6678.
We can write this number in the form of $N=n10^x$ that is
$6678=.6678\times 10^4$
here, $x=+4$
so, $+4$ is the order of magnitude of the number 6678.
Example 4:- Find the order of magnitude of number 0.005348.
We can write this number in the form of $N=n10^x$ that is
$0.005348=0.5348\times 10^{-2}$
here, $x=-2$
so, $-2$ is the order of magnitude of the number 0.005348.
See also What is the Unit of Moment of Inertia in physics?
## Solved questions
Question 1
What is the typical order of magnitude of the base current for a BJT?
(a) $10 ^{-8}$
(b) $10 ^{-9}$
(c) $10^{-6}$
(d) $10^{-3}$
Solution:-
The correct answer is (c) $10^{-6}$
Because the base current is much smaller than the collector and emitter currents, it is normally measured in microamperes.
Question 2
What is the order of magnitude of the seconds present in a day?
Solution:-
We know that
1 day = 24 hours
1 hour = 60 min
1 min = 60 sec
Number of seconds in a day are
$1\,\, day = 24 \times 60 \times 60 = 86400\,\, sec$
In terms of $N=n\times 10^x$
1 day $=0.864\times 10^5 sec$
Hence, the order of magnitude is 5.
Question 3
Find the order of magnitude of a light year.
Solution
One light year is defined as the distance covered by light in one year.
We know that speed of light in vacuum is
$c=3\times 10^8 m/s$
Thus,
1 light year $=3\times 10^8 m/s \times 1\, year$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 3\times 10^8 m/s \times (365\times 24 \times 3600\, sec)=9.5 \times 10^{15}m$
According to rule 2 of order of magnitude $9.5>5$.
In terms of $N=n\times 10^x$
1 light year $=.95 \times 10^{16}$
Thus, order of magnitude of one light year is $16$.
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# this is what u do on your first problem....first u take out the greatest common factor that would go into all of them...
4n^2-20n+24...now what number will go into all of them...well 4 will...4 will go into 4,20,and 24...so.
it would be like this..4(n^2-5n+6)...then theirs one more step...4(n-2)(n-3)
4x^2-22x+10 would be...2(2x^2-11x+5)...then u factor again...2(2x-1)(x-5)
1/2x^2-1/2would be 1/2(x^2-1)...then u factor again...1/2(x+1)(x-1)
Factor:
4n²-20n+24
4x²-22x+10
1/2x²-1/2
On each, take out the common factor
2(2n²-10n+12)
2(2x²-11x+ 5)
1/2 (x²-1)
Now you should be able to factor the inside.
Thanks, but that's where I get stuck. I thought like for the first one that you have to multipy two numbers together to get 12 and they have to add to -10. I can't think of 2 numbers, or am I just doing it wrong?
9 months ago
## No worries! Factoring can be a bit tricky, but with practice, it becomes easier.
Let's take a closer look at your first problem:
2(2n² - 10n + 12)
To factor the trinomial 2n² - 10n + 12, we need to find two numbers that multiply to give the constant term (12) and add up to the coefficient of the linear term (-10).
In this case, we are looking for two numbers that multiply to give 12 and add up to -10. One way to determine these numbers is by listing all possible factor pairs of 12:
1 and 12
2 and 6
3 and 4
Now, let's see if any of these pairs add up to -10.
1 + 12 = 13
2 + 6 = 8
3 + 4 = 7
None of these pairs adds up to -10. Therefore, we cannot factor the trinomial using integer factors.
However, there is another technique we can use called factoring by grouping. Here's how it works:
1. Split the middle term, -10n, into two terms such that their coefficients multiply to give the product of the coefficient of the square term (2) and the constant term (12). In this case, that product is 24.
2. Rewrite the trinomial, grouping the terms properly:
2n² - 10n + 12 = 2n² - 4n - 6n + 12
3. Now, factor by grouping:
2n² - 4n - 6n + 12 = 2n(n - 2) - 6(n - 2)
4. Notice that we now have a common binomial term, (n - 2). We can factor it out:
2n(n - 2) - 6(n - 2) = (n - 2)(2n - 6)
So, the factored form of the trinomial 2(2n² - 10n + 12) is (n - 2)(2n - 6).
I hope this helps! Let me know if there's anything else you need assistance with.
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# Common Core: 6th Grade Math : Make Tables of Equivalent Ratios, Find Missing Values, and Plot Values on a Coordinate Plane: CCSS.Math.Content.6.RP.A.3a
## Example Questions
### Example Question #31 : Grade 6
At a local microchip factory, there are managers for every workers. How many managers are needed for workers?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem, we will create a table of proportions using the following ratio.
If we solve for the table, then we can find the number of managers needed for .
The factory will need .
### Example Question #32 : Grade 6
At a local microchip factory, there are managers for every workers. How many managers are needed for workers?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem, we will create a table of proportions using the following ratio.
If we solve for the table, then we can find the number of managers needed for .
The factory will need .
### Example Question #31 : Grade 6
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #31 : Grade 6
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #11 : Make Tables Of Equivalent Ratios, Find Missing Values, And Plot Values On A Coordinate Plane: Ccss.Math.Content.6.Rp.A.3a
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #12 : Make Tables Of Equivalent Ratios, Find Missing Values, And Plot Values On A Coordinate Plane: Ccss.Math.Content.6.Rp.A.3a
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #721 : Ssat Middle Level Quantitative (Math)
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #51 : Numbers And Operations
At a car production company, manufacturers place tires and transmission on every car in the production line. A manager orders tires, how many transmissions should he order?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, the manager should order .
### Example Question #42 : Grade 6
Traffic from the suburbs and farms into a city typically follows an observable pattern. On any given morning there are cars for every trucks. On one particular busy morning there are trucks. How many cars are sitting in traffic?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, there are .
### Example Question #721 : Ssat Middle Level Quantitative (Math)
Traffic from the suburbs and farms into a city typically follows an observable pattern. On any given morning there are cars for every trucks. On one particular busy morning there are trucks. How many cars are sitting in traffic?
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem we must make a table of ratios. In the question we are given the base ratio:
We can use this ratio to make a table.
According to the table, there are .
|
# Leetcode – Two Sum IV – Input is a BST Solution in Python
## Introduction
Binary Search Trees (BSTs) are a widely-used data structure in computer science. They allow efficient storage, retrieval, and deletion of data in a sorted manner. In this article, we are going to delve into solving the “Two Sum IV – Input is a BST” problem from Leetcode. This problem combines the concept of the Two Sum problem with the properties of a Binary Search Tree. We will explore the problem in detail, discuss various approaches for solving it, and provide Python code for each solution.
## Problem Statement
Given the root of a Binary Search Tree and a target number k, return true if there exist two elements in the BST such that their sum is equal to the given target.
### Example:
Input: root = [5,3,6,2,4,null,7], k = 9
Output: true
Explanation: The BST is:
5
/ \
3 6
/ \ \
2 4 7
5 + 4 = 9, so return true.
### Note:
• The number of nodes in the tree is in the range [1, 10^4].
• -10^4 <= Node.val <= 10^4
• root is guaranteed to be a valid binary search tree.
• -10^5 <= k <= 10^5
## Solution
### 1. In-Order Traversal and Two Pointers
One way to approach this problem is to use an in-order traversal to retrieve the elements of the BST in ascending order. Once we have the elements in sorted order, we can use the two-pointer technique to check for two elements whose sum equals the target k.
1. Perform an in-order traversal to get the elements in ascending order.
2. Use two pointers, one at the beginning and one at the end of the array.
3. Move the pointers inward until they meet, checking if the sum of the elements they point to is equal to k.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def find_target(self, root, k):
def in_order_traversal(node):
if not node:
return []
return in_order_traversal(node.left) + [node.val] + in_order_traversal(node.right)
elements = in_order_traversal(root)
left, right = 0, len(elements) - 1
while left < right:
current_sum = elements[left] + elements[right]
if current_sum == k:
return True
elif current_sum < k:
left += 1
else:
right -= 1
return False
#### Time Complexity:
• O(n), where n is the number of nodes in the BST.
#### Space Complexity:
• O(n), due to storing the elements of the BST in an array.
### 2. Using a Hash Set
We can solve this problem in a single pass through the BST using a hash set. During the traversal, for each node, we check if the difference between the target value k and the current node’s value is in the hash set. If it is, we have found two elements that add up to k.
1. Perform any tree traversal (in-order, pre-order, post-order).
2. For each node, check if k - node.val is in the hash set.
3. If not, add node.val to the hash set.
class Solution:
def find_target(self, root, k):
def dfs(node, seen):
if not node:
return False
complement = k - node.val
if complement in seen:
return True
return dfs(node.left, seen) or dfs(node.right, seen)
return dfs(root, set())
#### Time Complexity:
• O(n), where n is the number of nodes in the BST.
#### Space Complexity:
• O(n), due to storing the elements in a hash set.
## Conclusion
In this article, we explored the “Two Sum IV – Input is a BST” problem on Leetcode and discussed two different approaches in Python: In-Order Traversal with Two Pointers and Hash Set. While both solutions have the same time and space complexity, the Hash Set approach is arguably more concise and requires only a single pass through the BST.
|
# A woman wants to construct a box with a base length that is twice the base width. The material to build the top and bottom is \$9/meter squared, and the material to build the sides is \$6/meter squared. If the woman wants the box to have a volume of 70 meters cubed, determine the dimensions of the box that will minimize the cost of production. What is the minimum cost? \$924.56, \$865.98, \$687.45, or \$727.97?
The minimum cost is approximately \$727.97.
We are asked to find the minimum cost of a box subject to the following constraints: the volume is 70 cubic meters, the length is twice the width for the top and bottom, the cost of the material for the top and bottom is \$9 per square meter, and the cost for the side material is \$6 per square meter.
Let w be the width of the bottom, so 2w is the length, and h is the height of the box. The surface area of the box is `SA=2(w)(2w)+2(h)(2w)+2(h)(w)`
or `SA=4w^2+6hw`
Then, the cost of the materials will be `C=4w^2(9)+6(6hw)=36w^2+36hw`
We would like to have this in terms of a single variable. We use the fact that the volume is 70 to write C in terms of w:
`70=w(2w)(h) => h=35/w^2`
So, `C=36w^2+1260/w`
We can use calculus to find the minimum. The domain of the function C(w) is positive reals, so C is defined on its domain. The minimum, assuming there is one, will occur when the first derivative is zero.
`C'(w)=72w-1260/w^2` If C'=0, then
`72w=1260/w^2 => w^3=17.5 => w~~2.5962`
Then, `l=2w~~5.1925, h=35/w^2~~5.1925`
So, the minimum cost is `C=36w^2+1260/w~~727.9739`
or approximately \$727.97
(See attachment for the graph of C.)
A quick check shows
C(1)=1296
C(2)=774
C(3)=744
C(4)=891
C(5)=1152
so, the answer is reasonable. Also `2(2.5962)^2(5.1925)~~69.9975`, so it fits the requirements.
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