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# Class 12 RD Sharma Solutions – Chapter 8 Solution of Simultaneous Linear Equations – Exercise 8.2
• Last Updated : 13 Jan, 2021
### Question 1.
2x – y + z = 0
3x + 2y – z = 0
x + 4y + 3z = 0
Solution:
Given
2x – y + z = 0
3x + 2y – z = 0
X + 4y + 3z = 0
The system can be written as
A X = 0
Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)
|A| = 2(10) + 10 + 10
|A| = 40 ≠0
Since, |A|≠0, hence x = y = z = 0 is the only solution of this homogeneous equation.
### Question 2.
2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
Solution:
Given 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
A X = 0
Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)
|A| = – 20 – 24 + 44
|A| = 0
Thus, the system has infinite solutions
Let z = k
2x – y = – 2k
5x + 3y = k
### Question 3.
3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
Solution:
Given:
3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
A X = 0
Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)
|A| = – 27 + 1 + 26
|A| = 0
Hence, the system has infinite solutions
Let z = k
3x – y = – 2k
4x + 3y = – 3k
### Question 4.
x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0
Solution:
Given:
x + y – 6z = 0
x – y + 2z = 0
– 3x + y + 2z = 0
A X = 0
Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)
|A| = – 4 – 8 + 12
|A| = 0
Hence, the system has infinite solutions
Let z = k
x + y = 6k
x – y = – 2k
### Question 5. Solve the system of homogeneous linear equations by matrix method:
x + y + z = 0
x – y – 5z = 0
x + 2y + 4z = 0
Solution:
Given:
x + y + z = 0
x – y – 5z = 0
x + 2y + 4z = 0
A X = 0
Now, |A| = 1(6) – 1(9) + 1(3)
|A| = 9 – 9
|A| = 0
Hence, the system has infinite solutions
Let z = k
x + y = –k
x – y = 5k
### Question 6. Solve the system of homogeneous linear equations by matrix method:
x + y – z = 0
x – 2y + z = 0
3x + 6y –5z = 0
Solution:
Given:
x + y – z = 0
x – 2y + z = 0
3x + 6y –5z = 0
A X = 0
Now, |A| = 1(4) – 1(–8) – 1(12)
|A| = 4 + 8 – 12
|A| = 0
Hence, the system has infinite solutions
Let z = k
x + y = –k
x – 2y = –k
### Question 7. Solve the system of homogeneous linear equations by matrix method:
3x + y – 2z = 0
x + y + z = 0
x – 2y + z =0
Solution:
Given:
3x + y – 2z = 0
x + y + z = 0
x – 2y + z =0
A X = 0
Now, |A| = 3(3) – 1(0) – 2(–3)
|A| = 9 – 0 + 6
|A| = 15 ≠0,
Hence, the given system has only trivial solutions given by x = y = z = 0.
### Question 8. Solve the system of homogeneous linear equations by matrix method:
2x + 3y – z =0
x – y – 2z = 0
3x + y + 3z = 0
Solution:
Given:
2x + 3y – z =0
x – y – 2z = 0
3x + y + 3z = 0
A X = 0
Now, |A| = 2(–3 + 2) – 3(3 + 6) – 1(4)
|A| = –2 – 27 – 4
|A| = –33 ≠0,
Hence, the given system has only trivial solutions given by x = y = z = 0.
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You have worked with lines and line segments in past courses.
Let's refresh some basic facts we know to be true about segments.
A segment (or line segment) is a part of a line that is bounded by two distinct end points. It contains every point on the line between its end points.
Postulate: A line segment is the shortest distacnce between two distinct points.
Ruler Postulate
Ruler Postulate: The points on a line can be put into a one-to-one correspondence (paired) with the real numbers. The distance between any two points is represented by the absolute value of the difference between the numbers. [Keep in mind that distances are always positive.]
The distance between D and F is | 0 - 2 | = 2. The distance between C and E is | -2 - 1 | = 3. The distance between A and C is | -3 - (-2) | = 1.
Statement: If B lies on the segment from A to C, then AB + BC = AC.
Also the converse: If AB + BC = AC, then B lies on the segment from A to C.
You may see examples where the concept of this postulate is referred to as "whole quantity", "the whole is equal to the sum of its parts" or "betweenness of points". Be sure to use the statement that your teacher wishes to follow.
In this postulate, points A, B and C are collinear, meaning they all lie on the same line.
Midpoint of a Segment
The midpoint of a segment is a unique point on the segment forming two congruent segments.
Example:
Bisector of a Segment
The bisector of a segment is a line, ray, or segment which cuts the given segment into two congruent segments. The point where the bisector crosses the segment is the midpoint of the segment.
There are an infinite number of possible bisectors of a segment passing through the midpoint.
If the bisector is also perpendicular to the segment, it is referred to as the perpendicular bisector of the segment. While there are an infinite number of possible bisectors of a segment, there is only one perpendicular bisector.
Theorem: The points on a perpendicular bisector are equidistant from the endpoints of the line segment. In the diagram, line m is the perpendicular bisector of the segment from A to B. Every point on line m is the same distance from A, as it is from B. Distances: AC = CB AD = DB AE = EB
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Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 21
Λ =1, $A^{-1}= \frac{1}{2}\begin{bmatrix} -2 & 2\\ -4 & 3 \end{bmatrix}$
Hint:
Here, we use basic concept of determinant and inverse of matrix
$A^{-1}= \frac{1}{\left | A \right |}\times Adj\left ( A \right )$
Given:
$A= \begin{bmatrix} 3 &-2 \\ 4 & -2 \end{bmatrix}$
Solution:
$A^{2}=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ 4 & -4 \end{array}\right]$
Now,
\begin{aligned} &A^{2}=\lambda A-2 I \\ &\lambda A=A^{2}+2 I \\ &\lambda A=\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]+\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \end{aligned}
\begin{aligned} &\lambda=\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \div\left[\begin{array}{cc} 3 & -2 \\ 4 & -2 \end{array}\right] \\ &\lambda=1 \\ &\text { So, } \\ &A^{2}=A-2 I \end{aligned}
Multiply $A^{-1}$ both side
\begin{aligned} &A^{-1} A \times A=A A^{-1}-2 L A^{-1} \\ &2 A^{-1}=I-A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \\ &A^{-1}=\frac{1}{2}\left[\begin{array}{ll} -2 & 2 \\ -4 & 3 \end{array}\right] \end{aligned}
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# Percentage of a percentage calculator
Percentage of a percentage Calculator enables you to multiply % by % and obtain the result. To calculate a percentage of a percentage, divide both percentages by 100 and multiply them by 100.
What will be the result of y% of x%?
## What is a Percentage
A percentage is a ratio expressed as a fraction of 100. The percentage is commonly represented by the percent sign, %, or the abbreviation “pct.” The term “per cent” is derived from the Latin word “per centum,” which means “per hundred.” The percentage is sometimes written as two words in British English: “percentage” (although percentage and percentile are reported as one word).
## How to Find Percentage of a Percentage
Sometimes you have to find the percentage of a percentage, which can be difficult. For example, converting both to decimal form and multiplying them together yields the percent of a percent.
To convert each percentage to decimal form, divide it by 100. The result is obtained by multiplying the decimals together. To convert a decimal to a percentage, multiply the result by 100.
## The formula for the Percentage of a Percentage
Use the following formula to calculate the percentage of a percentage:
Percentage of a percentage = $$(\dfrac{percent 1}{100} × \dfrac{percent 2}{100}) × 100$$
## Examples
Example 1: Find 25% of 60%.
We will start with the formula to calculate a percentage of a percentage which is:
$$P = (\dfrac{percent 1}{100} × \dfrac{percent 2}{100}) × 100$$
Now, substitute the given values in the equation.
$$P = (\dfrac{25}{100} × \dfrac{60}{100}) × 100$$
By simplifying them, we will get the following:
$$P = (0.25 × 0.6) × 100 = 0.15 × 100 = 15$$
$$P = 15\%$$
Example 2: Find 50% of 40%
$$P = (\dfrac{50}{100} × \dfrac{40}{100}) × 100$$
Simplifying,
$$P = (0.50 × 0.40 ) × 100 = 0.20 × 100 = 20$$
$$P = 20\%$$
## FAQs
How to calculate the percentage?
To calculate a percentage of a number, divide the number by the total value and multiply the result by 100.
$$(\dfrac{value}{total value}) 100\%$$ is the formula for calculating the percentage.
How to convert the decimal number 3.25 to a percentage?
To convert 3.25 to a percentage, multiply 3.25 by 100. We will get $$3.25 × 100 = 325\%$$
What Are Some Real-Life Percentage Examples?
The following are some real-world percentage examples:
• Nutrient percentage on a food package.
• The air comprises oxygen, carbon dioxide, nitrogen, and other gases, all measured in %.
• A percentage of your test results.
### Can't find your query?
Fill out the form below with your query and we will get back to you in 24 hours.
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# 5.1.1E: Circles (Exercises)
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Section 5.1 Exercise
1. Find the distance between the points (5,3) and (-1,-5).
2. Find the distance between the points (3,3) and (-3,-2).
3. Write an equation of the circle centered at (8 , -10) with radius 8.
4. Write an equation of the circle centered at (-9, 9) with radius 16.
5. Write an equation of the circle centered at (7, -2) that passes through (-10, 0).
6. Write an equation of the circle centered at (3, -7) that passes through (15, 13).
7. Write an equation for a circle where the points (2, 6) and (8, 10) lie along a diameter.
8. Write an equation for a circle where the points (-3, 3) and (5, 7) lie along a diameter.
9. Sketch a graph of $$\left(x-2\right)^{2} + \left(y+3\right)^{2} = 9$$.
10. Sketch a graph of $$\left(x+1\right)^{2} + \left(y-2\right)^{2} = 16$$.
11. Find the $$y$$ intercept(s) of the circle with center (2, 3) with radius 3.
12. Find the $$x$$ intercept(s) of the circle with center (2, 3) with radius 4.
13. At what point in the first quadrant does the line with equation $$y = 2x + 5$$ intersect a circle with radius 3 and center (0, 5)?
14. At what point in the first quadrant does the line with equation $$y = x + 2$$ intersect the circle with radius 6 and center (0, 2)?
15. At what point in the second quadrant does the line with equation $$y = 2x + 5$$ intersect a circle with radius 3 and center (-2, 0)?
16. At what point in the first quadrant does the line with equation $$y = x + 2$$ intersect the circle with radius 6 and center (-1,0)?
17. A small radio transmitter broadcasts in a 53 mile radius. If you drive along a straight line from a city 70 miles north of the transmitter to a second city 74 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?
18. A small radio transmitter broadcasts in a 44 mile radius. If you drive along a straight line from a city 56 miles south of the transmitter to a second city 53 miles west of the transmitter, during how much of the drive will you pick up a signal from the transmitter?
19. A tunnel connecting two portions of a space station has a circular cross-section of radius 15 feet. Two walkway decks are constructed in the tunnel. Deck A is along a horizontal diameter and another parallel Deck B is 2 feet below Deck A. Because the space station is in a weightless environment, you can walk vertically upright along Deck A, or vertically upside down along Deck B. You have been assigned to paint “safety stripes” on each deck level, so that a 6 foot person can safely walk upright along either deck. Determine the width of the “safe walk zone” on each deck. [UW]
20. A crawling tractor sprinkler is located as pictured here, 100 feet south of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves north along the hose at the rate of 1/2 inch/second. The hose is perpendicular to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk. [UW]
a. Impose a coordinate system. Describe the initial coordinates of the sprinkler and find equations of the lines forming and find equations of the lines forming the north and south boundaries of the sidewalk.
b. When will the water first strike the sidewalk?
c. When will the water from the sprinkler fall completely north of the sidewalk?
d. Find the total amount of time water from the sprinkler falls on the sidewalk.
e. Sketch a picture of the situation after 33 minutes. Draw an accurate picture of the watered portion of the sidewalk.
f. Find the area of grass watered after one hour.
21. Erik’s disabled sailboat is floating anchored 3 miles East and 2 miles north of Kingsford. A ferry leaves Kingsford heading toward Eaglerock at 12 mph. Eaglerock is 6 miles due east of Kingsford. After 20 minutes the ferry turns, heading due south. Bander is 8 miles south and 1 mile west of Eaglerock. Impose coordinates with Bander as the origin. [UW]
a. Find equations for the lines along which the ferry is moving and draw in these lines.
b. The sailboat has a radar scope that will detect any object within 3 miles of the sailboat. Looking down from above, as in the picture, the radar region looks like a circular disk. The boundary is the “edge” or circle around this disk, the interior is everything inside of the circle, and the exterior is everything outside of the circle. Give the mathematical description (an equation or inequality) of the boundary, interior and exterior of the radar zone. Sketch an accurate picture of the radar zone by determining where the line connecting Kingsford and Eaglerock would cross the radar zone.
c. When does the ferry enter the radar zone?
d. Where and when does the ferry exit the radar zone?
e. How long does the ferry spend inside the radar zone?
22. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular wheat field of radius 200 ft. The combine cuts a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north and 60 feet west of the western-most edge of the field. [UW]
a. When does Nora’s combine first start cutting the wheat?
b. When does Nora’s combine first start cutting a swath 20 feet wide?
c. Find the total amount of time wheat is being cut during this pass across the field.
d. Estimate the area of the swath cut during this pass across the field.
23. The vertical cross-section of a drainage ditch is pictured to the right. Here, R indicates in each case the radius of a circle with R = 10 feet, where all of the indicated circle centers lie along a horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is flowing into the ditch so that the level above the bottom is rising at a rate of 2 inches per minute. [UW]
a. When will the ditch be completely full?
b. Find a piecewise defined function that models the vertical cross-section of the ditch.
c. What is the width of the filled portion of the ditch after 1 hour and 18 minutes?
d. When will the filled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide?
1. 10
3. $$(x - 8)^2 + (y + 10)^2 = 8^2$$
5. $$(x - 7)^2 + (y + 2)^2 = 293$$
7. $$(x - 5)^2. + (y - 8)^2 = 13$$
9.
11. $$(0, 3 + \sqrt{5})$$ and $$(0, 3 - \sqrt{5})$$
13. (1.3416407865, 7.683281573)
15. (-1.07335, 2.8533)
17. 29.87 miles
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Equations - Solving for x
I have this problem:
$$9x^3 - 18x^2 - 4x + 8 = 0$$
However, I'm not sure how to find the values of $x$. I moved the 8 over and factor out an $x$, but the trinomial it created can't be factored. Could someone enlighten me?
-
Generally speaking, you want the right hand side to be zero. By factoring a polynomial, you're looking to write it as $(x-a)(x-b)(x-c) = 0.$ Notice that there will be a term there $-abc$ that is not a power of $x$. So, don't move the 8 over -- it won't help. Cubics are hard to factor. You might not be able to find $a,b,c$ such that they are real numbers. There is a general formula, but it is quite complicated. I'm not sure this problem can be easily solved with simple operations. – Emily Aug 21 '12 at 13:53
it would easy if you had +8 instead of having -8. Otherwise, Ross's answer is the final one. – Babak S. Aug 21 '12 at 13:54
You need the rational root theorem - there is one obvious (integer) root, and taking out this factor gives you a quadratic which is easy to factor. – Mark Bennet Aug 21 '12 at 13:56
Factor the function.
$9x^3-18x^2-4x+8=9x^2(x-2)-4(x-2)=(x-2)(9x^2-4)=(x-2)(3x-2)(3x+2)$
$(x-2)(3x-2)(3x+2)=0$
$x=2$ or $2/3$ or $-2/3$
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The rational root theorem is your friend. It says all rational roots have numerators that are factors of the constant term and denominators that are factors of the leading term. Here the numerators can be $\pm 1, \pm 2, \pm 4, \pm 8$ and the denominators can be $1,3,9$. Not too many to try. When you find one, you can divide out that root to get a quadratic. It doesn't always work (as shown with the example with -8 for a constant term) but does sometimes, often in homework.
-
I didn't actually. It was + 8 instead of - 8. – Andre Oseguera Aug 21 '12 at 13:54
@AndreOseguera: You wrote $-8$ in your primary question. Please take more attention writing the question here. :) – Babak S. Aug 21 '12 at 14:18
Yeah, sorry, I wrote it thinking I moved it over to the other side already and got all backwards. Won't happen again. :) – Andre Oseguera Aug 21 '12 at 14:20
By trial and error method I'm solving this
Divide 9x3−18x2−4x+8 by x-2 reminder is 0 and quotient is 9x2-4x
x-2 = 0 and 9x2-4x = 0
Factorizing 9x2-4x = 0, we get (3x-2)(3x+2) = 0
ie. x = 2, x = 2/3, x = -2/3 are the solutions
(Sorry I'm new to this group and don't know how to write Square of a variable, But answer must be this)
-
For formatting, you could look at meta.math.stackexchange.com/questions/1773/… Basically you put $\LaTeX$ between dollar signs. I'm not sure what this answer contributes beyond what jasoncube wrote. – Ross Millikan Aug 21 '12 at 14:36
Actually I didn't see the answer jason posted. I saw the question and just started solving it. Sorry if it was a duplicate. But thankyou for the link... – SKT Aug 21 '12 at 14:43
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Friday, September 17, 2021
# Vertical addition (m3.nbt.2) math practice
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This test question, released by the Partnership for the Assessment of Readiness for College and Career (PARCC), is discussed here in the interest of helping third-grade students, their parents, and friends understand more clearly the knowledge and skills that third graders should have in math by the end of the school year (2018 #2).
For this problem, you have to type a number into a box that represents the sum of 528 and 288. The correct answer is 816.
• If your answer is in the 700s, you forgot to “carry” the number “1” from the tens place.
• If your answer has a “0” in the tens place, you forgot to carry the “1” from the ones place.
You should always start addition problems like this from the ones place, which for whole numbers is the number on the right. The ones place of the top number is 8, and the ones place of the bottom number is also 8. Adding these gives 8 + 8, which equals 16. As you know from place value, 16 is the same as 6 plus one 10. What that means in the addition problem is that underneath the ones place in the problem write down the “6” from your answer, which is “6,” and put the one 10 in the the tens place by writing a “1” there.
Next, add the three numbers in the tens column: the 2 from the top number, the 8 from the bottom number, and the 1 that you carried over from the ones column. 2 tens + 8 tens + 1 ten = 11 tens. 11 tens is the same as 1 ten plus 1 hundred, so you now write down the 1 ten in the tens place of your answer and carry the one hundred to the hundreds place.
Finally, add the three numbers in the hundreds column: 5 hundreds + 2 hundreds + 1 hundred equals 8 hundreds. When you write the “8” in the hundreds place of the problem, you see that the resulting answer is 816.
If You Don't Understand This Problem
When we write the number “528,” what does that mean. There’s a “5” in the hundreds place, so that’s 500. There’s a “2” in the tens place, so that’s 20. And there’s an “8” in the ones place, just that’s just 8. We can write this about the number:
$528 = 500 + 20 + 8$
Likewise, with the second number, it can be broken down by place value as well.
$288 = 200 + 80 + 8$
We can then add the two numbers piece by piece. 8 + 8 in the ones place is 16, or 6 plus one 10. Then we add the tens place: 80 + 20 + the 10 from the 16 in the ones column, and that makes 110, which is actually 11 tens, or 1 ten plus 1 hundred. For now, “carry” the 1 hundred to the hundreds place and add it up there.
That makes 500 + 200 + 100 in the hundreds place, which is 800. Our final answer is:
$800 + 10 + 6 = 816$
You can learn more with this type of question by practicing different “strategies” for performing the three-digit addition. Some of the different strategies include:
• the traditional way, right to left
• an approach using just place value
• explaining that the answer is correct by subtracting one of the numbers from the sum
It would be best if students could explain what they’re doing (thinking out loud) as they solve the problems using each of these strategies.
Internet Resources
## Information for Teachers
This problem tests students’ understanding of the Common Core mathematics standard 3.NBT.A.2, which says that by the end of third grade, you should have the ability to “fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.”
Note that “fluency” means much more than being able to add and subtract quickly. The National Council of Teachers of Mathematics defines procedural fluency as “the ability to apply procedures accurately, efficiently, and flexibly; to transfer procedures to different problems and contexts; to build or modify procedures from other procedures; and to recognize when one strategy or procedure is more appropriate to apply than another.”
In that sense, using only one strategy to add three-digit numbers falls short of a demonstration of fluency, and teachers and parents should encourage students to explain different approaches to finding the arithmetic answer to these problems.
For the first six months following the publication of this article, we encourage and welcome comments about extending this lesson or information about how you have taught this lesson in your classrooms. #LearnTogether #Voxitatis
Paul Katulahttps://news.schoolsdo.org
Paul Katula is the executive editor of the Voxitatis Research Foundation, which publishes this blog. For more information, see the About page.
### Study finds increasing suicide rates among Black girls
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Why are Black adolescent girls committing suicide at increasingly higher rates in the US? Could it be an effect of racism and prejudice?
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0
# When a number is divided by 243 the quotient is 1 what is the number show step by step?
Updated: 9/25/2023
Wiki User
8y ago
The number is 243 because 243/243 = 1 and any number divided by itself is always equal to 1
Wiki User
8y ago
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Q: When a number is divided by 243 the quotient is 1 what is the number show step by step?
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Related questions
### What causes repeating sixes when you put two third on the calculator?
If you divided 2 by 3, then the quotient is 0 with remainder 2. Next, you divide 20 by 3: you get a quotient of 6 with remainder 2. At the next step you need to divided 20 by 3: quotient 6, reminder 2. Back into the loop.
5.67
250
### How can you find the factors for a number?
Step 1: Divide the number by all the numbers that are less than or equal to itself (<=) Step 2: If the reminder is 0 and the quotient is a whole number then the dividing number is a factor of our number Step 3: Proceed until the number is reached to pick up all the factors
4
### What is 12.5 divided by 6 step by step?
12.5 divided by 6 step by step = 2.0833333333333335
### What is 6 divided 10.734 step by step?
6 divided 10.734 step by step = 0.5589714924538849
0.568
### What is 0.07 divided b 0.21 step by step?
0.07 divided b 0.21 step by step = 0.33333333333333337
### How do you write 24 in binary?
We divide 24 by 2 and then the quotient obtained is also divided by 2, we continue the process until the quotient obtained is greater than 1. 24 = 2x12 + 0 12 = 2x6 + 0 6 = 2x3 + 0 3 = 2x1 + 1 The remainder obtained in last step is given highest priority and the remainder obtained in first step is given the lowest priority. The quotient obtained in the last step is written first followed by the remainders according to their priorities: So binary equivalent is 11000. (24)10 = (11000)2
### What is 30 over 9 as a mixed number?
3 and 3/9. There are three basic steps to convert an improper fraction to a mixed number: Divide the numerator by the denominator (you can use long division if you want to!) to find out what the quotient and the remainder are. If the fraction is made up of whole numbers, you will always get an integer quotient and an integer remainder. Note down what your quotient, remainder, and original denominator is. Now rewrite these three numbers in Step 2 in a mixed number format where: the quotient is the whole number next to the proper fraction the remainder is the numerator of the proper fraction the original denominator is the new denominator of the proper fraction Let’s apply these steps to our problem. What is 30 divided by 9? If you do some thinking or long division, you should get: 30 ÷ 9 → Quotient of 3 and a remainder of 3. Now that we have all the numbers we need, let’s piece together our answer: 30 ÷ 9 = 3 3/9
0.0042
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# How do you find all the zeros of f(x) = x^4 - 3.3x^3 + 2.3x^2 + 0.6x?
May 5, 2016
Zeros of $f \left(x\right)$ are $\left\{0 , 2 , \frac{3}{2} , - \frac{1}{5}\right\}$
#### Explanation:
$f \left(x\right) = {x}^{4} - 3.3 {x}^{3} + 2.3 {x}^{2} + 0.6 x$
= $\frac{x}{10} \left(10 {x}^{3} - 33 {x}^{2} + 23 x + 6\right)$
Hence one of the zeros is $0$. Another zero could be a factor of $6$
As putting $x = 2$ in $\left(10 {x}^{3} - 33 {x}^{2} + 23 x + 6\right)$, we get
$10 \cdot {2}^{3} - 33 \cdot {2}^{2} + 23 \cdot 2 + 6 = 80 - 132 + 46 + 6 = 0$
Hence, $2$ is another zero and dividing $10 {x}^{3} - 33 {x}^{2} + 23 x + 6$ by $\left(x - 2\right)$, we get
$10 {x}^{3} - 33 {x}^{2} + 23 x + 6 = \left(x - 2\right) \left(10 {x}^{2} - 13 x - 3\right)$
or $\left(x - 2\right) \left(10 {x}^{2} - 15 x + 2 x - 3\right)$
or $\left(x - 2\right) \left(5 x \left(2 x - 3\right) + 1 \cdot \left(2 x - 3\right)\right)$
= $\left(x - 2\right) \left(5 x + 1\right) \left(2 x - 3\right)$
Hence other zeros are given by $2 x - 3 = 0$ and $5 x + 1 = 0$ i.e. are $\frac{3}{2}$ and $- \frac{1}{5}$
Hence zeros of $f \left(x\right)$ are $\left\{0 , 2 , \frac{3}{2} , - \frac{1}{5}\right\}$
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# Difference between revisions of "2008 AMC 12B Problems/Problem 23"
## Problem
The sum of the base-$10$ logarithms of the divisors of $10^n$ is $792$. What is $n$?
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$
## Solutions
### Solution 1
Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$. Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$. For any factor $2^a \times 5^b$, there will be another factor $2^{n-a} \times 5^{n-b}$. Note this is not true if $10^n$ is a perfect square. When these are added, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$. $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.
There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$. We then plug in answer choices and arrive at the answer $\boxed {11}$
### Solution 2
We are given $$\log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792$$ The property $\log(ab) = \log(a)+\log(b)$ now gives $$\log_{10}(d_1 d_2\cdot\ldots d_k) = 792$$ The product of the divisors is (from elementary number theory) $a^{d(n)/2}$ where $d(n)$ is the number of divisors. Note that $10^n = 2^n\cdot 5^n$, so $d(n) = (n + 1)^2$. Substituting these values with $a = 10^n$ in our equation above, we get $n(n + 1)^2 = 1584$, from whence we immediately obtain $\framebox[1.2\width]{(A)}$ as the correct answer.
### Solution 3
For every divisor $d$ of $10^n$, $d \le \sqrt{10^n}$, we have $\log d + \log \frac{10^n}{d} = \log 10^n = n$. There are $\left \lfloor \frac{(n+1)^2}{2} \right \rfloor$ divisors of $10^n = 2^n \times 5^n$ that are $\le \sqrt{10^n}$. After casework on the parity of $n$, we find that the answer is given by $n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}$.
### Solution 4
The sum is $$\sum_{p=0}^{n}\sum_{q=0}^{n} \log(2^p5^q) = \sum_{p=0}^{n}\sum_{q=0}^{n}(p\log(2)+q\log(5))$$ $$= \sum_{p=0}^{n} ((n+1)p\log(2) + \frac{n(n+1)}{2}\log(5))$$ $$= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5)$$ $$= \frac{n(n+1)^2}{2}$$ Trying for answer choices we get $n=11$
## Alternative thinking
After arriving at the equation $n(n+1)^2 = 1584$, notice that all of the answer choices are in the form $10+k$, where $k$ is $1-5$. We notice that the ones digit of $n(n+1)^2$ is $4$, and it is dependent on the ones digit of the answer choices. Trying $1-5$ for $n$, we see that only $1$ yields a ones digit of $4$, so our answer is $11$.
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# Lesson Notes By Weeks and Term - Primary 3
2-D Shapes
TERM: 2nd Term
WEEK: 5
CLASS: Primary 3
AGE: 8 years
DURATION: 5 periods of 40 minutes each
DATE:
SUBJECT: Mathematics
TOPIC: 2-D shapes
SPECIFIC OBJECTIVES: At the end of the lesson, the pupils should be able to
1. Describe, sort and compare circles.
2. Describe, sort and compare triangles.
3. Describe, sort and compare squares.
INSTRUCTIONAL TECHNIQUES: Explanation, question and answer, demonstration, practical, assessments
INSTRUCTIONAL MATERIALS: String and sticks. scrap paper, advertisement flyers (collect and bring from home), videos from source-
PERIOD 1: Assessments
PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1ORALASSESSMENTS The teacher asks questions on time and calendars Pupils respond and participate STEP 2DISCUSSION The teacher discusses all the methods used by some learners in the oral assessments(some of the questions are solved on the board by the learners) and addresses any misconceptions that may have risen Pupils pay attention and participate STEP 3WRITTEN ASSESSMENTS 1 Write half past 7 in digital time.2 Write quarter to 9 in digital time.3 Circle the clock that shows quarter past two.4. show half past 3 in the afternoon on an analogue clock.5 Show quarter to 10 on an analogue and a digital clock.6 How much time passed between 2 o’clock and half past four in the afternoon? Pupils attempt their class work STEP 4SUMMARY The teacher marks the written assessments, corrects were necessary and commends the pupils
PERIOD 2: Circles
PRESENTATION
TEACHER’S ACTIVITY
PUPIL’S ACTIVITY
STEP 1
MENTAL MATHS
The teacher begins the lesson with some mental calculations
Calculate
1 5 × 10 =
2 2 × 10 =
3 7 × 10 =
4 1 × 10 =
5 4 × 10 =
6 3 × 10 =
7 10 × 10 =
8 0 × 10 =
9 6 × 10 =
10 8 × 10 =
Pupils respond and participate
STEP 2
CONCEPT
DEVELOPMENT
The teacher
• Describes a circle by saying, I am thinking about a shape. It is round and has no straight sides. What shape am I thinking about? (a circle)
• Ensures that you model the correct use of vocabulary.
CLASS ACTIVITY
The teacher
Takes the learners outside to a place where the learners can draw in the sand.
Lets the learners get into groups of 5.
- What shape did we discuss in class? ( a circle)
- What are the features of a circle? (It is round and has no straight sides.)
Instructs the class:
In your groups find a stick.
- Each of you draw a different size circle in the sand.
Asks: Who drew the biggest circle? Who drew the smallest circle?
Gives each group a piece of string.
Using the string make a circle.
- Which was easier to use to make a circle? A stick or string? Discuss this with your learners (no right answer here).
Returns to the classroom.
ACTIVITY II
The teacher ask the learners to
Draws a circle. Draw three more, but all of them should look different.
-Discuss with the class what is different about each one. (The differences could be in the
size of the shapes. Circles do not look different if their ‘orientation’ is changed.)
Pupils pay attention and participate
STEP 3
CLASS-WORK
1 Draw 3 different sized circles in the table below.
Small circle Bigger circle Biggest circle
2. Draw circles in different positions in the table below.
Circle at the top Circle in the middle Circle at the bottom
3 Use 6 circles to create a picture.
Pupils attempt their class work
STEP 4
HOME-WORK
Find and draw 3 objects that are circular in your home.
Pupils attempt their class work
STEP 5
SUMMARY
The tacher summarizes by reminding the pupils how to describe, compare and draw circles.
She marks their class works, makes corrections where necessary and commends them positively
PERIOD 3: Triangles
PERIOD 4: Squares
PRESENTATION TEACHER’S ACTIVITY PUPIL’S ACTIVITY STEP 1MENTAL MATHS The teacher begins the lesson with some mental calculationsCalculate1 1 × 1 = 2 1 × 2 = 3 2 × 2 = 4 2 × 3 = 5 3 × 4 = 6 3 × 5 =7 3 x 6 = 8 4 × 5 =9 5 × 1 =10 10 × 2 = Pupils respond and participate STEP 2CONCEPTDEVELOPMENT The teacher• Describes a square by saying, I am thinking about a shape.It has 4 sides and 4 corners. All the sides are the same length and all the corners are the same size. What shape am I thinking about? (a square)• Ensures that you model the correct use of vocabulary. CLASS ACTIVITYThe teacher• Takes the learners outside to a place where the learners can draw in the sand.• Lets the learners get into groups of 5.What shape did we discuss in class? ( a square)- What are the features of a square? (It has 4 equal sides and 4 corners)• Instructs the class:In your groups find a stick.- Each of you draw a different size square in the sand.• Asks: Who drew the biggest square? Who drew the smallest square?• Gives each group a piece of string.- Using the string make a square. Encourage the learners to use the correct vocabulary.- Which was easier to use to make a square? A stick or string? Discuss this with your learners (no right answers here).• Returns to the classroom. Pupils pay attention and participate STEP 3CLASS-WORK 1 Draw a square. Draw three more, but all should look different.(Answers will vary. The differences could be in the size and/or orientation ofthe shapes.)2 Draw a picture that is made up of 8 different sized squares.3 Draw three squares:a A square with 4 cm sides.b A square with 7 cm sides.c A square with 10 cm sides. Pupils attempt their class work STEP 4HOME-WORK Find and draw 3 objects that are square in your home. Pupils attempt their class work STEP 5SUMMARY The teacher summarizes by reminding the pupils how to describe and compare squares. She marks their class works, makes corrections where necessary and commends them positively
PERIOD 5: Weekly Test/consolidations
TEACHER’S ACTIVITY: The teacher revises all the concepts treated from period 1-4 and gives the pupils follow through exercises, quiz and tests . She marks the exercises, makes corrections and commends the pupils positively.
PUPIL’S ACTIVITY: The pupils work on the worksheets and exercises given by the teacher individually
CONSOLIDATION
1 Complete the table:
2. Draw a car using triangles, circles and squares.
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Least Common Multiples
1 / 6
# Least Common Multiples - PowerPoint PPT Presentation
Least Common Multiples. Multiples. Multiples are the product of a number and any whole number. LCM- least common multiple- the least multiple common to all numbers. Find by: List the multiples Ex 1: 6 and 9: 6: 6,12,18,24,30, 36,42,48 9: 9,18,27,36, 45,54,63,72,81 LCM: 18.
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### Least Common Multiples
Multiples
• Multiples are the product of a number and any whole number.
• LCM- least common multiple- the least multiple common to all numbers.
• Find by:
• List the multiples
• Ex 1: 6 and 9:
• 6: 6,12,18,24,30, 36,42,48
• 9: 9,18,27,36, 45,54,63,72,81
• LCM: 18
Ex. 2: Find by method 2:
• Find the prime factorization of two numbers. Write each prime factor the greatest number of times it occurs for any of the digits.
• 4,14: 4= 2x2; 14= 2x7: So 2 occurs twice, 7 occurs once:
• So 2x2x7=28, 28 is the least common multiple.
Uses of LCM
• Finding LCM of algebraic terms:
• Ex:3
• 35x^2 and 175xy
• 35x^2= 5*7*x*x 175 = 7*5*5*x*y
• LCM=5*5*7*x*x*y
LCD: Least Common Denominator
• The LCD will be the LCM of the Denominator:
• Ex. 4:
• 1/12, -3/16, 5/18
• Find LCM of 12, 16,18:
• 12=2*3*2
• 16=2*2*2*2
• 18=2*3*3
• Multiply 2*2*2*2*3*3=144
You can also used the LCD for ALGEBRAIC fractions
• Ex. 5: Find LCD for the following
• 6c/5b; 2c/15b, 7c/25b
• 5b=5*b
• 15b=3*5*b
• 25b=5*5*b
• 3*5*5*b=75b
• LCD is 75b
• DO 1-10 in notebook now.
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Algebra Tutorials! Thursday 25th of July
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
# Linear Equations and Inequalities in One Variable
Example:
Correct the error on the right side of each of the problems.
1. 5(x + 3) = 5x + 3 Incorrect Distributive property 5(x + 3) = 5x + 5 · 3 Correct Distributive property = 5x + 15 Multiplication
2. 7(4 + y) = 4 + 7y Incorrect Distributive property 7(4 + y) = 7 · 4 + 7y Correct Distributive property = 28 + 7y Multiplication
3. 2(7a + 4) – 5 = 2(7a) + 2(4) – 2 · 5 Incorrect Distributive property 2(7a + 4) – 5 = 2(7a) + 2(4) – 5 Correct Distributive property = 14 a + 8 – 5 Multiplication = 14a + 3 Addition
## Replacement Values
The value of an expression is found by replacing a variable with a given number.
Example:
Find the value of the following expressions by replacing a variable with the given number.
Expression Value of the Variable Value of the Expression 2x+ 3 x = 5 2 (5) + 3 = 13 t2 - 5t t = - 2 (-2)2 – 5(-2) = 4 + 10 = 14 x2 – 5xy + y2 x = 3 y = - 4 (3)2 – 5(3)(- 4) + (- 4)2 = 9 + 60 + 16 = 85
Sequences: Replace n by the given value:
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28
Q:
A sum is being lent at 20 % per annum compound interest.what is the ratio of increase in amount of 4th year to 5th year?
A) 4:5 B) 5:4 C) 5:6 D) can't be determined
Explanation:
Q:
Simple interest on a certain sum at 7% per annum for 4 years is Rs. 2415. What will be the compound interest on the same principal at 4% per annum in two years?
A) Rs. 704 B) Rs. 854 C) Rs. 893 D) Rs. 914
Explanation:
We know that,
From given data, P = Rs. 8625
Now, C.I =
2 152
Q:
Find the compound interest on Rs. 6,500 for 4 years if the rate of interest is 10% p.a. for the first 2 years and 20% per annum for the next 2 years?
A) Rs. 3845 B) Rs. 4826 C) Rs. 5142 D) Rs. 4415
Explanation:
We know the formula for calculating
The compound interest where P = amount, r = rate of interest, n = time
Here P = 5000, r1 = 10, r2 = 20
Then
C = Rs. 4826.
5 111
Q:
What is the difference between the compound interests on Rs. 5000 for 11⁄2 years at 4% per annum compounded yearly and half-yearly?
A) Rs. 1.80 B) Rs. 2.04 C) Rs. 3.18 D) Rs. 4.15
Explanation:
Compound Interest for 1 12 years when interest is compounded yearly = Rs.(5304 - 5000)
Amount after 112 years when interest is compounded half-yearly
Compound Interest for 1 12 years when interest is compounded half-yearly = Rs.(5306.04 - 5000)
Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000)= 5306.04 - 5304 = Rs. 2.04
4 192
Q:
The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 56. Then the sum is :
A) Rs. 3680 B) Rs. 2650 C) Rs. 1400 D) Rs. 1170
Explanation:
We know thatThe Difference between Compound Interest and Simple Interest for n years at R rate of interest is given by
Here n = 2 years, R = 20%, C.I - S.I = 56
8 195
Q:
Find the compound interest on Rs. 2680 at 8% per annum for 2 years ?
A) Rs. 664.21 B) Rs. 548.68 C) Rs. 445.95 D) Rs. 692.57
Explanation:
We know Compound Interest = C.I. = P1+r100t - 1
Here P = 2680, r = 8 and t = 2
C.I. = 26801 + 81002-1= 268027252-12= 26802725+12725-1= 2680 5225×225
= (2680 x 52 x 2)/625
= 445.95
Compound Interest = Rs. 445.95
9 237
Q:
The compound interest on Rs. 8000 for 3 year at 10% p.a. is
A) 2648 B) 2145 C) 2587 D) 2784
Explanation:
8000 × 33.1% = 2648
11 417
Q:
Compound interest on a certain sum of money at 20% per annum for 2 years is Rs.5995. What is the SI on the same money at 8% per annum for 6 years ?
A) Rs. 5989 B) Rs. 6540 C) Rs. 7844 D) Rs. 6789
Explanation:
Given C.I = 5984, R = 20% , T = 2yrs
5984 = $\inline \fn_jvn \small C.I=P\left [ \left ( 1&plus;\left ( \frac{20}{100}\right )^{2} \right )-1 \right ]$
=> P = (5995x25)/11
P = Rs. 13625
Now S.I = PTR/100
SI = (13625 x 8 x 6)/100 = Rs. 6540
19 1354
Q:
At a certain rate of interest the compound interest of 3 years and simple interest of 5 years for certain sum of money is respectively Rs. 1513.2 and Rs. 2400. Find the common rate of interest per annum ?
A) 5% B) 6% C) 4% D) 3%
Explanation:
Given compound interest for 3 years = Rs. 1513.2
and simple interest for 5 years = Rs. 2400
Now, we know that C.I =
=> 1513.2 = ...........(A)
And S.I = PTR/100
=> 2400 = P5R/100 ..................(B)
By solving (A) & (B), we get
R = 5%.
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
Difference between revisions of "2017 AIME II Problems/Problem 6"
Problem
Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.
Solution 1
Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$. The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$. Rearranging, $s^2-(2n+85)^2=843$. By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$. It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$. The two values of $n$ that satisfy one of the equations are $168$ and $27$. Summing these together, the answer is $168+27=\boxed{195}$.
Solution 2
Clearly, the result when $n$ is plugged into the given expression is larger than $n$ itself. Let $x$ be the positive difference between that result and $n$, so that $\sqrt{n^2+85n+2017}=n+x$. Squaring both sides and canceling the $n^2$ terms gives $85n+2017=2xn+x^2$. Combining like terms, $(85-2x)n=x^2-2017$, so
$$n=\frac{x^2-2017}{85-2x}.$$
Since $n$ is positive, there are two cases, which are simple (luckily). Remembering that $x$ is a positive integer, then $x^2-2017$ and $85-2x$ are either both positive or both negative. The smallest value for which $x^2>2017$ is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that $x<45$ (from the numerator) and $85-2x<0$, which means $x>42$. This only gives two solutions, $x=43, 44$. Plugging these into the expression for $n$, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\boxed{195}$.
Solution 3 (Abuse the discriminant)
Let the integer given by the square root be represented by $x$. Then $0 = n^2 + 85n + 2017 - x^2$. For this to have rational solutions for $n$ (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)
Thus, $b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2$ for some integer $y$. Then $4x^2 - 843 = y^2$. Rearranging this equation yields that $843 = (2x+y)(2x-y)$. Noticing that there are 2 factor pairs of $843$, namely, $1*843$ and $3*281$, there are 2 systems to solve for $x$ and $y$ that create rational $n$. These yield solutions $(x,y)$ of $(211, 421)$ and $(71, 139)$.
The solution to the initial quadratic in $n$ must then be $\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}$. Noticing that for each value of $x$ that has rational solutions for $n$, the corresponding value of the square root of the discriminant is $y$, the formula can be rewritten as $n = \frac{-85 \pm y}{2}$. One solution is $\frac{421 - 85}{2} = 168$ and the other solution is $\frac{139 - 85}{2} = 27$. Thus the answer is $168 + 27 = \boxed{195}$ as both rational solutions are integers.
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# 6th Class Mathematics Line, Area and Triangle Geometrical Basic Shapes
## Geometrical Basic Shapes
Category : 6th Class
### Geometrical Basic Shapes
Point
A dot which indicates position but not dimension is called a point. A point does not have length, breadth and height. P and Q are points in the figure below.
In the picture above, P and Q are points In the picture above two points lie on the same line and therefore called collinear points. Points do not lie on the same line are called non collinear points.
Line
A set of infinite points which can be extended infinite distance in both sides is called a line.
In the picture above, M is a line
Features of the Line
(i) The length of a line is infinite.
(ii) It has no terminal, therefore, can be extended infinitive in both directions.
(iii) It is made up of infinite points.
Line Segment
A line of fix length is called line segment.
In the picture above, P and Q is a line segment and represented by $\overline{PQ}.$
Features of the Line Segment
(i) A Line segment has fix length.
(ii) It has two end points.
Ray
A ray is defined as the line that can be extended infinite in one direction.
In the picture above, end point of terminal point Ray AB is represented as AB
Angle
Angle is formed between two rays which having common end point. Symbol of angle $=\angle$
Vertex or common end point = 0 Arms of angle $\angle AOB$ = OA and OB the name of the above angle can be $\angle \text{ }AOB$or $\angle BOA$ The unit of measurement of an angle is degree ${{(}^{o}})$
Types of Angles
There are various types of angles which are the following:
Acute Angle
The angle between 0° and 90° is called an acute angle.
In the picture above $\angle AOB$or $\angle BOA$ is an acute angle. The inclined arm OA on the horizontal is less inclined than vertical line. ${{10}^{o}},{{30}^{o}},{{60}^{o}},{{80}^{o}}$ are acute angles.
Right Angle
An angle of ${{90}^{o}}$ is called right angle.
Obtuse Angle
An angle whose measure is between ${{90}^{o}}$ and ${{180}^{o}}$ is called an obtuse angle.
Straight Angle
An angle whose measure is ${{180}^{o}}$is called straight angle. Or an angle formed between two opposite rays is called straight angle.
Reflex Angle
An angle whose measure is more than ${{180}^{o}}$ and less then ${{360}^{o}}$ is called reflex angle.
Complementary angle
Two angles whose sum is ${{90}^{o}}$ is called the complimentary of each other ${{60}^{o}}+{{30}^{o}}={{90}^{o}}$
Complementary angle of ${{60}^{o}}={{90}^{o}}-{{60}^{o}}={{30}^{o}}$
Complementary angle of ${{30}^{o}}={{90}^{o}}-{{30}^{o}}={{60}^{o}}$
Complementary angle of $\text{ }\!\!\theta\!\!\text{ }=90-\text{ }\!\!\theta\!\!\text{ }$
The complementary angle of a given angle
$={{90}^{o}}-$ the given angle.
Therefore, if thesum of two angles is 90° then they are called complementary angles to each other.
Supplementary Angle
Two angles whose sum is ${{180}^{o}}$ is called supplementary angles. ${{120}^{o}}+{{60}^{o}}={{180}^{o}}$
Supplementary Angle of ${{120}^{o}}={{180}^{o}}-{{120}^{o}}={{60}^{o}}$
Supplementary Angle of ${{60}^{o}}={{180}^{o}}-{{60}^{o}}={{120}^{o}}$
The supplementary angle of the given angle $=\text{ }{{180}^{o}}-$ the given angle Supplementary angle of $\text{ }\!\!\theta\!\!\text{ }={{180}^{o}}-\text{ }\!\!\theta\!\!\text{ }$
Equal Angles
If the measurement of two angles is equal then they are called equal angles.
In the picture above, the measurement of the angles is equal. Therefore, $\angle AOB$and $\angle PQR$are equal angles.
Two angles are said to be adjacent even if they have common vertex and one common arm. In $\angle AOC$
Vertex = O
Arms = OA, OC
Arms = OC, OB
Hence vertex 0 and arm OC is common so these two angles AOC and COB are adjacent angles.
Vertically Opposite Angle
Two angles are said to be vertically opposite angles if they are formed by opposite rays having common vertex.
Vertically opposite angles are formed by two lines which intersects each other.
In the picture above, $\angle AOD$and $\angle BOC$are vertically opposite angles. They are formed by opposite rays and they have same or common end point 0.
Similarly, $\angle AOC$and $\angle BOD$are vertically opposite angles.
The measurement of an angle is 180°. Which one of the following types of angle is this?
(a) Right angle
(b) Straight angle
(c) Reflex angle
(d) All of these
(e) None of these
Explanation
A straight angle has a measurement of 180° . Which one of the following is the complementary angle of 30°?
(a) ${{30}^{o}}$
(b) ${{60}^{o}}$
(c) ${{90}^{o}}$
(d) All of these
(e) None of these
Explanation
Complementary angle of ${{30}^{o}}={{90}^{o}}-{{30}^{o}}={{60}^{o}}$
In the options given below the pair of angles are given. Find the complementary pair.
(a) ${{51}^{o}},{{25}^{o}}$
(b) ${{61}^{o}},{{29}^{o}}$
(c) ${{45}^{o}},{{55}^{o}}$
(d) All of these
(e) None of these
Explanation
If the sum of pair of angles is ${{90}^{o}}$ then the pair is called complementary pair of angles. The sum of angles $~{{61}^{o}}+{{29}^{o}}={{90}^{o}}.$ Hence the pair of angles ${{61}^{o}},{{29}^{o}}$ is a complementary pair of angles.
Choose the pair of supplementary angle from the options given below?
(a) ${{100}^{o}},{{200}^{o}}$
(b) ${{105}^{o}},{{75}^{o}}$
(c) ${{30}^{o}},{{180}^{o}}$
(d) All of these
(e) None of these
Explanation
The sum of angles of the pair ${{105}^{o}},{{75}^{o}}={{105}^{o}}+{{75}^{o}}={{180}^{o}}.$Hence, the pair of angles ${{105}^{o}},{{75}^{o}}$ is a supplementary pair of angles.
Two figures are given below. Which one of the following options is correct for the adjacent angles if the line segment OB divides the AOC into two equal parts?
(a) AOB adjacent $\angle BOC$and $\angle TOR$adjacent $\angle ROP$
(b) $\angle TOR$adjacent$\angle AOB$and $\angle ROP$adjacent $\angle AOB$
(c) $\angle POT$adjacent$\angle AOC$
(d) All of these
(e) None of these
Explanation
Two angles said to be adjacent even if they have one common arm and one common vertex.
Hence,
$\angle AOB$adjacent $\angle BOC$and $\angle TOR$adjacent $\angle ROP.$
#### Other Topics
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# 4²+7²+...+(3n+1)²=S, what is S?
## Related Questions
"You rent four movies and return them a day late. The regular rental fee is xdollars per movie. You are charged a total of (4x+8) dollars. a. Factor the expression using the cake method (GCF).(4x+8) = __________ Don't put spaces when you type your answer. b. The fee for returning a movie 1 day late is \$__________"
Step-by-step explanation:
Let's call the amount of dollars per movie.
The following expression represents charges in dollars:
If we use the cake method, it would be like the following:
| 4 8
2 | 2 4
2 | 1 2
So, the Greatest Common Factor is 2 times 2, which is 4. GCF = 4.
a. The factored expression would be:
Notice that the charge per movie is \$(x+2), because the factor 4 represents the total number of movies.
Therefore, in this case, the fee per movie is x + 2. (b).
a. 4(x + 2)
b. \$2.00
Step-by-step explanation:
The 'cake method' is another method that can be used to find greatest common factor (GCF) and least common multiple (LCM). In this method, you form an 'upside down layered cake' to factor your numbers:
Look at what two factors you can multiply to find those numbers and put the common number on the outside and the other number down below:
2 I 4 8 I
2 I 2 4 I
1 I 1 2 I
Now, multiply the numbers on the left side of the cake (2 x 2 x 1), GCF = 4.
With the GCF = 4, put this number outside the parenthesis and factor: 4(x + 2). Since 'x' represents the cost of one movie rental, then the fee for returning a movie 1 day late is \$2.00.
BRAINLIEST!! Find the mean of the data set. If necessary, round to the nearest tenth. 18, 14, 8, 22, 25, 8 A. 15.8
B. 14.8
C. 16.1
D. 17.3
Step-by-step explanation:
The reason is because you add all of then and then divide it by how many number there is 18+14+8+22+25+8= 95
then you divide by 6 which is 15.833333 rounded to the nearest tenth is 15.8
Show how you can solve the equation 3x=9 by multiplying each side by the reciprocal of 3.
The final value of x for the equation is 3
What are linear equations?
Linear equations help in representing the relationship between variables such as x, y, and z, and are expressed in exponents of one degree. In these linear equations, we use algebra, starting from the basics such as the addition and subtraction of algebraic expressions.
Given here: The equation as 3x=9
Now multiplying both sides by 1/3 we get
3x/3=9/3
x=3
Hence, The final value of x for the equation is 3
brainly.com/question/29739212
#SPJ5
The reciprocal of 3 is just . So you would have ×3x=9 and you'd get x=3
. Curt is installing a picket fence around his garden which is 45.45 ft by 14.63 ft. When he gets to the hardware store he finds that each picket fence slat is two feet wide. If Curt does not want any spaces between each slat, use estimation to determine how many slats of fencing he should buy.
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# 180 Days of Math for Fifth Grade Day 46 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 46 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 46
Directions Solve each problem.
Question 1.
97 – 54 = _____
Explanation:
Subtract the smallest number 54 from the biggest number 97,
to get the difference.
Question 2.
Explanation:
Multiply the numbers 7 and 7 to get the product.
7 x 7 = 49
Question 3.
How many groups of 4 are in 56?
______________
Explanation:
Question 4.
Is 1,528 greater than or less than 1,258?
______________
1,528 is greater than 1,258
Explanation:
First compare the biggest place value,
if the place value is same then move to the next one.
Here when we compare hundreds place value 1,528 is greater.
Question 5.
$$\frac{1}{5}$$ of 20 is ____.
Explanation:
$$\frac{1}{5}$$ of 20
=$$\frac{20}{5}$$
=4
Question 6.
20 – 3 × 6 = ___
Explanation:
20 – 3 × 6
=20 – 18
= 2
Question 7.
45 ÷ = 9
Explanation:
Let the unknown number be x
45 ÷ x = 9
x = 45 ÷ 9
x = 5
Question 8.
What is the length of this line?
______________
2.75in
Explanation:
The smallest unit of measurement in length is inches.
Measure the object “0” is at one of the ends.
Then, look for the last full inch before the opposite end of the object .
Question 9.
Name the polygon that is created by the cross-section.
Cuboid
Explanation:
A cuboid is a three-dimensional solid shape that has 6 faces, 8 vertices, and 12 edges.
Question 10.
What fraction of the people represented in the circle graph chose chocolate?
_____________________
50%
Explanation:
From the above pie chart the fraction of the people represented in the circle graph choose chocolate is 50 %
As the pie chart is divided into 4 parts.
Question 11.
If you spin the spinner, what is the probability you will land on green?
_____________________
Explanation:
There are three colors in the spinner and are of equally divided.
So, the probability of landing on green,
$$\frac{1}{3}$$ = 0.33
Question 12.
Edward spends 1$$\frac{1}{2}$$ hours at soccer practice every Monday, Wednesday, and Friday. He spends 2 hours at the game on Saturday. How much of Edward’s time is spent on soccer each week?
_____________________
Edward spends 1$$\frac{1}{2}$$ hours at soccer practice every Monday, Wednesday, and Friday.
= 1$$\frac{1}{2}$$ x 3
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# Chapter 2 Describing Data with Numerical Measurements
## Presentation on theme: "Chapter 2 Describing Data with Numerical Measurements"— Presentation transcript:
Chapter 2 Describing Data with Numerical Measurements
General Objectives: Graphs are extremely useful for the visual description of a data set. However, they are not always the best tool when you want to make inferences about a population from the information contained in a sample. For this purpose, it is better to use numerical measures to construct a mental picture of the data. ©1998 Brooks/Cole Publishing/ITP
Specific Topics 1. Measures of center: mean, median, and mode
2. Measures of variability: range, variance, and standard deviation 3. Tchebysheff’s Theorem and the Empirical Rule 4. Measures of relative standing: z-scores, percentiles, quartiles, and the interquartile range 5. Box plots ©1998 Brooks/Cole Publishing/ITP
2.1 and 2.2 Describing a Set of Data with Numerical Measures and Measures of Center
Definition: Numerical descriptive measures associated with a population of measurements are called parameters. Those computed from sample measurements are called statistics. Definition: The arithmetic mean or average of a set is equal to the sum of the measurements divided by n. Notation: Sample mean: Population mean: m ©1998 Brooks/Cole Publishing/ITP
Figure 2.2 Character Dotplot
Example 2.1 Use a dotplot to display the n = 5 measurements 2, 9, 11, 5, 6. Find the sample mean of these observations, and compare its value with what you might consider the “center” of these observations on the dotplot. Solution The dotplot in Figure 2.2 seems to be centered between 6 and 8. To find the sample mean, we can calculate The statistic = 6.6 is the balancing point or fulcrum shown on the dotplot. It does seem to mark the center of the data. Figure 2.2 Character Dotplot ©1998 Brooks/Cole Publishing/ITP
Definition: The median m of a set of n measurements is the value of x that falls in the middle position when the measurements are ordered from smallest to largest. The median is less sensitive to extreme values or outliers than the mean. The value .5(n + 1) indicates the position of the median in the ordered data set. ©1998 Brooks/Cole Publishing/ITP
Figure 2.3(a) Relative frequency distribution showing the effect of extreme values on the mean and median ©1998 Brooks/Cole Publishing/ITP
The midpoint of the modal class is taken as the mode.
Definition: The mode is the category that occurs most frequently, or the most frequently occurring value of x. When measurements on a continuous variable have been grouped as a frequency or relative frequency histogram, the class with the highest frequency is called the modal class. The midpoint of the modal class is taken as the mode. ©1998 Brooks/Cole Publishing/ITP
Figure 2.4(a) Relative frequency histograms for the milk data
Figure 2.4(b) Relative frequency histograms for the GPA data
2.3 Measures of Variability
Variability or dispersion is a very important characteristic of data. See Figure 2.5 for examples of variability or dispersion of data. Definition: The range, R, of a set of n measurements is defined as the difference between the largest and the smallest measurements. See Figure 2.6 for an example of two relative frequency distributions that have the same range but very different shape and variability. Figure 2.7 shows the deviations of points from the mean. Table 2.1 shows a computation of for the data in Figure 2.7. ©1998 Brooks/Cole Publishing/ITP
Figure 2.5(a) Variability or dispersion of data
Figure 2.5(b) Variability or dispersion of data
Figure 2.6(a) Distributions with equal range and unequal variability
Figure 2.6(b) Distributions with equal range and unequal variability
Figure 2.7 Showing the deviations of points from the mean
The population variance is denoted by s 2 and is given by the formula
Definition: The variance of a population of N measurements is defined to be the average of the squares of the deviations of the measurements about their mean m. The population variance is denoted by s 2 and is given by the formula This measure will be relatively large for highly variable data and relatively small for less variable data. Definition: The variance of a sample of n measurements is defined to be the sum of the squared deviations of the measurements about their mean divided by (n -1). ©1998 Brooks/Cole Publishing/ITP
The sample variance is denoted by s 2 and is given by the formula:
Definition: The standard deviation of a set of measurements is equal to the positive square root of the variance. Notation: n: number of measurements in the sample s 2: sample variance : sample standard deviation ©1998 Brooks/Cole Publishing/ITP
The shortcut method for calculating s 2 :
where = sum of the squares of the individual measurements and = square of the sum of the individual measurements. ©1998 Brooks/Cole Publishing/ITP
Points to remember about variance and standard deviation:
. Points to remember about variance and standard deviation: - The value of s is always greater than or equal to zero. - The larger the value of s 2 or s, the greater the variability of the data set. - If s 2 or s is equal to zero, all measurements must have the same value. - The standard deviation s is computed in order to have a measure of variability measured in the same units as the observations. ©1998 Brooks/Cole Publishing/ITP
- The interval (m ± 1s) contains approximately 68% of the measurements
Empirical Rule: Given a distribution of measurements that is approximately mound-shaped: - The interval (m ± 1s) contains approximately 68% of the measurements - The interval (m ± 2s) contains approximately 95% of the measurements. - The interval (m ± 3s) contains almost all of the measurements. The Empirical Rule applies to data with a normal distribution and many other types of data. Use the Empirical Rule when the data distribution is roughly mound-shaped. ©1998 Brooks/Cole Publishing/ITP
2.6 Measures of Relative Standing
Definition: The sample z score is a measure of relative standing defined by A z-score measures the distance between an observation and the mean, measured in units of standard deviation. An outlier is an unusually large or small observation. z-scores between -2 and +2 are highly likely. Z-scores exceeding 3 in absolute value are very unlikely. ©1998 Brooks/Cole Publishing/ITP
Definition: A set of n measurements on the variable x has been arranged in order of magnitude.The pth percentile is the value of x that exceeds p% of the measurements and is less than the remaining (100 - p)%. Example 2.13 Suppose you have been notified that your score of 610 on the Verbal Graduate Record Examination placed you at the 60th percentile in the distribution of scores. Where does your score of 610 stand in relation to the scores of others who took the examination? Solution Scoring at the 60th percentile means that 60% of all examina-tion scores were lower than yours and 40% were higher. ©1998 Brooks/Cole Publishing/ITP
The median is the same as the 50th percentile.
The 25th and 75th percentiles are called the lower and upper quartiles. Figure 2.12 ©1998 Brooks/Cole Publishing/ITP
The second quartile is the median.
Definition: A set of n measurements on the variable x has been arranged in order of magnitude. The lower quartile (first quartile), Q1, is the value of x that exceeds one-fourth of the measurements and is less than the remaining 3/4. The second quartile is the median. The upper quartile (third quartile), Q 3, is the value of x that exceeds three-fourths of the measurements and is less than one-fourth. ©1998 Brooks/Cole Publishing/ITP
The upper quartile, Q 3, is the value of x in the position .75(n + 1).
When the measurements are arranged in order of magnitude, the lower quartile, Q1, is the value of x in the position .25(n +1). The upper quartile, Q 3, is the value of x in the position .75(n + 1). When these positions are not integers, the quartiles are found by interpolation, using the values in the two adjacent positions. Definition: The interquartile range (IQR) for a set of measurements is the difference between the upper and lower quartiles; that is, IQR = Q 3 - Q 1. ©1998 Brooks/Cole Publishing/ITP
2.7 The Box Plot From a box plot, you can quickly detect any skewness in the shape of the distribution and see whether there are any outliers in the data set. To construct a box plot: 1. Calculate the median, the upper and lower quartiles, and the IQR. 2. Draw a horizontal line representing the scale of measurement. 3. Form a box above the line with the ends at Q 1 and Q 3 . 4. Draw a vertical line through the box at the location of m, the median. ©1998 Brooks/Cole Publishing/ITP
Calculate lower and upper fences as follows:
Inner fences: Q (IQR) and Q (IQR) - Measurements outside the lower and upper fences are called suspect outliers. - Whiskers extend to the largest and smallest measurements inside the fences. To finish the box plot: - Locate the largest and smallest values using the scale along the horizontal axis, and connect them to the box with horizontal lines called whiskers. - Any suspect outliers are marked with an asterisk (*). ©1998 Brooks/Cole Publishing/ITP
Figure 2. 15 shows the various values associated with the box plot
Figure 2.15 shows the various values associated with the box plot. Example 2.15 exhibits the calculations for and the plotting of a box plot. Figure 2.15 Skewed distributions usually have a long whisker in the direction of the skewness, and the median line is drawn away from the direction of the skewness. ©1998 Brooks/Cole Publishing/ITP
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# NCERT Solutions and Notes for Class 6 Maths Chapter 9: Data Handling (Free PDF)
In this chapter, students will understand everything related to Data Handling. The chapter is very important is it lays the foundation for understanding Data Handling for complex calculations. Students will learn concepts like recording and organizing data. After completing this chapter, students will be able to make a pictograph for the given data set and interpret its results. Chapter 9 of Class 6 Maths will be useful for students in all their higher study lessons. Read through for CBSE NCERT Class 6 Maths Data Handling: Chapter 9 Notes and Important Exercise Questions.
## CBSE NCERT Class 6 Maths Data Handling: Chapter 9 Notes – PDF Download
Below we have given topic-wise notes for the CBSE NCERT Class 6 Maths Data Handling: Chapter 9. We have also provided a downloadable free PDF at the end of these notes so you can download and take a printout to study later when you need quick revision before going to the exam hall.
### Topic 1: What is Data?
Data is a collection of numbers gathered to give some information.
### Topic 2: Organisation of Data
The organization of data involves separating and arranging different data based on some given/desired parameters.
### Topic 3: Pictograph
A pictograph represents data through pictures of objects. It helps answer the questions on the data at a glance.
### Topic 4: Bar Graph
Bars of uniform width can be drawn horizontally or vertically with equal spacing between them and then the length of each bar represents the given number. Such a method of representing data is called a bar diagram or a bar graph.
## Important Questions in NCERT Class 6 Maths Chapter 9: PDF Download
Below we have provided some important exercise questions and their solutions from Class 6 Maths Data Handling: Chapter 9.
### Exercise 9.1
Q 1. Following is the choice of sweets for 30 students of Class VI.
1. Arrange the names of sweets on a table using tally marks.
2. Which sweet is preferred by most of the students?
Ans. (a) The sweets are arranged using the tally marks below.
(b) Laddoos are favored by most of the students.
Q 2. The following pictograph shows the number of tractors in five villages.
Observe the pictograph and answer the following questions.
1. Which village has the minimum number of tractors?
2. Which village has the maximum number of tractors?
3. How many more tractors does village C have as compared to village B?
4. What is the total number of tractors in all the five villages?
Ans. The answers to the given questions are given below.
1. Village D has the minimum number of tractors.
2. Village C has the maximum number of tractors.
3. Village C has 3 more tractors than Village B.
4. The total number of tractors in all the 5 villages = 6 + 5 + 8 + 3 + 6 = 28 tractors.
### Exercise 9.2
Q 1. The total number of animals in the five villages is as follows :
Village A: 80
Village B: 120
Village C: 90
Village D: 40
Village E: 60
Prepare a pictograph of these animals using one ☐ symbol to represent 10 animals
and answer the following questions :
1. How many symbols represent the animals of village E?
2. Which village has the maximum number of animals?
3. Which village has more animals: village A or village C?
Ans. Since one ☐ = 10 animals, the pictograph of the given data is:
1. A total of 6 symbols represent the animals of Village E.
2. Village B has the maximum number of animals.
3. Village A has 80 animals and Village C has 90 animals. So, village C has more animals.
### Exercise 9.3
Q 1. The bar graph given below shows the amount of wheat purchased by the government during the years 1998-2002.
Read the bar graph and write down your observations. In which year was:
1. the wheat production maximum?
2. the wheat production minimum?
Ans. The answers are given below.
1. The wheat production was maximum in 2002, i.e. 5 × 30 = 150 thousand tonnes.
2. The wheat production was minimal in 1998, i.e. 5 × 15 = 75 thousand tonnes.
Q 2. Observe the bar graph below which shows the sale of shirts in a ready-made shop from Monday to Saturday.
Now answer the following questions :
1. What information does the above bar graph give?
2. What is the scale chosen on the horizontal line representing the number of shirts?
3. On which day was the maximum number of shirts sold? How many shirts were sold on that day?
4. On which day were the minimum number of shirts sold?
5. How many shirts were sold on Thursday?
Ans. The answers to the above questions are given below.
1. The bar graph shows the number of shirts sold from Monday to Saturday in a ready-made shop.
2. 1 unit length = 20 shirts sold is the scale chosen in the given bar graph.
3. The maximum number of shirts were sold on Saturday, i.e. 60 shirts.
4. The minimum number of shirts were sold on Tuesday.
5. 35 shirts were sold on Thursday.
### Exercise 9.4
Q 1. A survey of 120 school students was done to find which activity they prefer to do in their free time.
Draw a bar graph to illustrate the above data taking a scale of 1 unit length = 10 students. Which activity is preferred by most of the students other than playing?
Ans. The bar graph for the above data taking a scale of 1 unit length = 10 students is given below.
Students prefer to read story books rather than play.
Also See:
Class 6 Maths Chapter 1 Knowing Our Numbers Notes and Important Questions – Free PDF
CBSE NCERT Class 6 Maths Chapter 2 Whole Numbers Notes, Exercise, and Important Questions – Free PDF
Class 6 Maths Chapter 3 Playing With Numbers Notes and Important Exercise Questions and Answers – Free PDF
Class 6 Maths Chapter 4 Basic Geometrical Ideas Notes and Important Exercise Questions and Answers – Free PDF
Class 6 Maths Chapter 5 Understanding Elementary Shapes Notes and Important Exercise Questions and Answers – Free PDF
Class 6 Maths Chapter 6 Integers Notes and Important Exercise Questions and Answers – Free PDF
## FAQs
Q.1. What is a Pictograph?
Ans: A pictograph represents data through pictures of objects. It helps answer the questions on the data at a glance.
Q.2. What is a bar graph?
Ans: Bars of uniform width can be drawn horizontally or vertically with equal spacing between them and then the length of each bar represents the given number. Such a method of representing data is called a bar diagram or a bar graph.
Q.3. What is data class 6 Maths?
Ans: Data is a collection of numbers gathered to give some information.
This was all about CBSE NCERT Class 6 Maths Data Handling: Chapter 9 Notes and Important Exercise Questions and Answers PDF Free Download. Follow the CBSE Class 6 Maths Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 6 Maths.
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## Chapter 11 Perimeter and Area Exercise 11.1
Question 1: The length and the breadth of a rectangular piece of land are 500 m and 300 m, respectively. Find:
(i) Its area
(ii) The cost of the land, if 1 m2 of the land costs ₹ 10,000.
Given
Length of rectangular land = 500 m
Breadth of rectangular land = 300 m
To find = Area
Cost of land
(i) Area of rectangle = l x b
= 500 x 300
= 150000 m²
Therefore, the area of the rectangular land is 150000 m².
(ii) Area of rectangular land = 150000 m²
Cost of 1 m² land = ₹ 10000
Cost of 150000 m² land = ?
= 10000 x 150000
= ₹ 1,50,00,00,000
Therefore, the cost of the land is ₹ 1,50,00,00,000.
Question 2: Find the area of a square park whose perimeter is 320 m.
Given
Perimeter of a square park = 320 m
To find - Side length
Area
Side = perimeter/4
= 320/4
= 80 m
Therefore, the length of each side is 80 m.
Area of Square = side x side
= 80 x 80
= 6400 m²
Therefore, the area of square park is 6400 m².
Question 3: Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.
Given
Area of rectangular plot = 440 m²
Length of rectangular plot = 22 m
Perimeter
= 440/22
= 20 m
Therefore, the breadth of the rectangular plot is 20 m.
Perimeter of a rectangle = 2 x (l + b)
= 2 x (22 + 20)
= 2 x 42
= 84 m
Therefore, the perimeter of the rectangular plot is 84 m.
Question 4: The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Given
Perimeter of rectangular sheet = 100 cm
Length of rectangular sheet = 35 cm
Area
Perimeter of rectangle = 2 x (l + b)
100 = 2 x (35 + b)
100 = (2 x 35) + (2 x b)
100 = 70 + 2b
2b = 100 - 70 = 30
2b = 30
b = 30/2
b = 15 m
Therefore, the breadth of the rectangular sheet is 15 m.
Area of rectangle = l x b
= 35 x 15
= 525 m²
Therefore, the area of rectangular sheet is 525 m².
Question 5: The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Given
Area of Square park = Area of rectangular park
Length of side of square park = 60 m
Length of rectangular park = 90 m
To find = Area
Area of square = side x side
= 60 x 60
= 3600 m²
= 3600/90
= 40 m
Therefore, the breadth of rectangular park is 40 m.
Question 6: A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also find which shape encloses more area?
Given
Length of rectangle = 40 cm
Breadth of rectangle = 22 cm
To find = Perimeter
Side of square
Which shape encloses more area
Perimeter of rectangle = 2 x (l + b)
= 2 x (40 + 22)
= 2 x 62
= 124 cm
Therefore, the perimeter of rectangle is 124 cm.
Perimeter of Square = Perimeter of Rectangle
Length of side of square = perimeter/4
= 124/4
= 31 cm
Therefore, the length of side of square is 31 cm.
Area of rectangle = length x breadth
= 40 x 22
= 880 cm²
Area of square = side x side
= 31 x 31
= 961 cm²
Compare
880 cm² < 961 m²
Therefore, square encloses more area.
Question 7: The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Given
Perimeter of rectangle = 130 m
To find = Length
= Area
Perimeter of rectangle = 2 x (l + b)
130 = 2 x (l + 30)
130 = (2 x l) + (2 x 30)
2l = 130 - 60
2l = 70
l = 70/2
l = 35
Therefore, length of a rectangle is 35 cm.
Area of rectangle = l x b
= 35 x 30
= 1050 cm²
Therefore, the area of rectangle is 1050 cm².
Question 8: A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m
².
Given
Length of wall = 4.5 m
Breadth of wall = 3.6 m
Door length = 2 m
Rate of white washing = ₹ 20 per m²
To find = Area to be white washed
= Cost of white washing
Area of rectangle = l x b
= 4.5 x 3.6
= 16.2 m²
Therefore, the area of wall = 16.2 m².
Area of rectangle = l x b
= 2 x 1
= 2 m²
Therefore, the area of door is 2 m².
Area to be white washed = area of wall - area of door
= 16.2 m² - 2 m²
= 14.2 m²
Therefore, the area to be white washed is 14.2 m².
Cost of white washing 1 m² = ₹ 20
Cost of white washing 14.2 m² = ?
= 20 x 14.2
= ₹ 284
Therefore, the cost of white washing the wall is ₹ 284.
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# If A,B,C,D are any four points in space prove that −−→AB×−−→CD+−−→BCx−−→AD+−−→CA×−−→BD=2−−→AB×−−→CA
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## To prove the identity →AB×→CD+→BC×→AD+→CA×→BD=2→AB×→CA, we will follow a step-by-step approach.Step 1: Define the vectorsLet:- →AB=→B−→A- →BC=→C−→B- →CA=→A−→C- →CD=→D−→C- →AD=→D−→A- →BD=→D−→BStep 2: Rewrite the left-hand sideWe can express the left-hand side of the equation in terms of these vectors:→AB×→CD+→BC×→AD+→CA×→BDSubstituting the definitions:(→B−→A)×(→D−→C)+(→C−→B)×(→D−→A)+(→A−→C)×(→D−→B)Step 3: Expand each cross product1. For →AB×→CD: →AB×→CD=(→B−→A)×(→D−→C)2. For →BC×→AD: →BC×→AD=(→C−→B)×(→D−→A)3. For →CA×→BD: →CA×→BD=(→A−→C)×(→D−→B)Step 4: Combine the termsNow, we combine the expanded terms:(→B−→A)×(→D−→C)+(→C−→B)×(→D−→A)+(→A−→C)×(→D−→B)Step 5: Group the termsWe can group the terms based on their common vectors:- The first term involves →B and →A.- The second term involves →C and →B.- The third term involves →A and →C.Step 6: Use properties of cross productsUsing the anti-commutative property of cross products, we can rearrange and simplify:→AB×→CD+→BC×→AD+→CA×→BD=2→AB×→CAConclusionThus, we have shown that:→AB×→CD+→BC×→AD+→CA×→BD=2→AB×→CA
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# How do you solve the equation 2x^2+10x+1=13 by completing the square?
May 7, 2017
Put constants on one side and x terms on the other side and complete the square.
#### Explanation:
By subtracting 1 from both sides, we get:
$2 {x}^{2} + 10 x = 12$
We can simplify by dividing both sides by 2:
${x}^{2} + 5 x = 6$
Here we complete the square:
Since ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$, here ${a}^{2}$ is ${x}^{2}$ and our $2 a b$ term is $5 x$, therefore our $b$ term must be $\frac{5}{2}$. We complete the square by making ${x}^{2} + 5 x$ into the form of the ${\left(a + b\right)}^{2}$, however we also need to subtract the ${b}^{2}$ term since we had added it in to complete the square:
$\left({x}^{2} + 5 x + {\left(\frac{5}{2}\right)}^{2}\right) - {\left(\frac{5}{2}\right)}^{2} = 6$
${\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} = 6$
and simplify:
${\left(x + \frac{5}{2}\right)}^{2} = \frac{49}{4}$
$x + \frac{5}{2} = \pm \sqrt{\frac{49}{4}}$
$x + \frac{5}{2} = \pm \frac{7}{2}$
$x = \frac{- 5 \pm 7}{2}$
$x = - 6 \mathmr{and} x = 1$
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# Quantitative Aptitude: Speed Maths |Vedic Maths: Usefulness of Powers of 2
Jagranjosh.com has come up with important concept i.e. is power of 2 to help the aspirants in attempting all the questions speedily.
Created On: Mar 27, 2014 17:45 IST
Modified On: Mar 28, 2014 13:01 IST
Jagranjosh.com has come up with important concept i.e. is power of 2 to help the aspirants in attempting all the questions speedily. Here, we are providing basic concepts to make the calculation faster than doing from long and traditional ones. Practice and thorough learning of some values can help the candidates in cracking Quantitative Aptitude/Numerical Ability Section in Bank Exam, SSC Exam, Railway and Other Exam.
A useful property for the powers of 2: Powers of 2 are very helpful in calculations. Candidates should memorise powers of 2 upto 12 so that it can be used in the questions.
20=1 ; 21=2 ; 22=4 ; 23=8 ; 24=16 ; 25=32 ; 26=64
27=128 ; 28=256 ; 29=512 ; 210=1024 ; 211=2048 ;212=4096
1. The sum of powers of 2 from 0 to any number n will be equal to 2n+1 – 1.
If a number is written from 1 to N as a sum of one or more of the integers of a given set of integers, then it can easily be done from powers of 2. The set of integers used by us comprise of all the powers of 2 starting form 1 (i.e. 20) to the largest power of 2 less than or equal to N.
For example: If you want to build all the integers upto 255, the numbers 1, 2, 4, 8, 16, 32, 64, 128 are sufficient as 255=1+2+4+8+16+32+64+128.
2. Differently, if we have one weight each of 1, 2, 4, 8, 16, 32, 64 and 128 kg, then all the items would be measured from 1 kg to 255 kg using one or more of the given weights (the weights used only in one pan of the weighing scales).
Example: How much minimum number of weights are required to weigh all possible weights upto 512 Kg (Putting all the weights only in one side of pan)
Solution: 512=29. Minimum Number of weights required=9+1=10. The weights will be 1,2, 4, 8, 16, 32, 64,128, 256 kg
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# What is the limit of (4x)/(x-6)+(5x)/(x+6) as x approaches infinity?
Mar 18, 2016
The limit is $9$
#### Explanation:
${\lim}_{x \rightarrow \infty} \frac{4 x}{x - 6} = {\lim}_{x \rightarrow \infty} \frac{4 x}{x \left(1 - \frac{6}{x}\right)}$
$= {\lim}_{x \rightarrow \infty} \frac{4}{\left(1 - \frac{6}{x}\right)} = \frac{4}{1 - 0} = 4$
And
${\lim}_{x \rightarrow \infty} \frac{5 x}{x + 6} = {\lim}_{x \rightarrow \infty} \frac{5 x}{x \left(1 + \frac{6}{x}\right)}$
$= {\lim}_{x \rightarrow \infty} \frac{5}{\left(1 + \frac{6}{x}\right)} = \frac{5}{1 + 0} = 5$
Therefore,
${\lim}_{x \rightarrow \infty} \left(\frac{4 x}{x - 6} + \frac{5 x}{x + 6}\right) = {\lim}_{x \rightarrow \infty} \frac{4 x}{x - 6} + {\lim}_{x \rightarrow \infty} \frac{5 x}{x + 6}$
$= 4 + 5 = 9$
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## Understanding the Diamond Method in Geometry and Simplifying Difference of Squares in Algebra
A: B:
Output: `Press calculate`
# Understanding the Diamond Method in Geometry and Simplifying Difference of Squares in Algebra
The fields of geometry and algebra often bring concepts that are both fascinating and practical. One such concept in geometry is the Diamond Method, a fantastic technique that aids in visualizing and solving problems. In algebra, the Difference of Squares formula simplifies complex expressions quickly. Both methods are powerful tools that can make problem-solving much easier. This article delves into these methods in a comprehensive yet engaging manner.
## The Diamond Method in Geometry
The Diamond Method is a visual approach to understanding geometric properties and relationships. It’s particularly useful in solving problems involving inscribed shapes, symmetry, and congruence. Imagine a diamond—essentially a rhombus with equilateral sides. Each of its diagonals bisects the diamond at right angles, creating a clear visualization of the relationships between the sides and angles.
Applications:
The Diamond Method is extremely beneficial for:
• Identifying symmetrical properties
• Solving problems involving inscribed angles and shapes
• Understanding congruence and similarity in geometric figures
For example, imagine a diamond-shaped garden. If we know one diagonal's length, we can easily determine the other diagonal's length using properties of the diamond. This method can even be extended to complex architectural designs or patterns, providing a robust tool for architects and designers.
## How to Use the Diamond Method
To use the Diamond Method effectively:
1. Identify the diagonals and ensure they are perpendicular.
2. Verify the lengths of the sides and angles.
3. Utilize the properties of rhombuses, where all sides are equal and diagonals bisect each other at right angles.
By following these steps, you can solve a multitude of geometric problems confidently.
## The Difference of Squares in Algebra
Switching gears to algebra, the Difference of Squares is an elegant and powerful mathematical tool. The formula is:
Formula: `a² - b² = (a + b)(a - b)`
Harnessing this formula can simplify complex expressions and equations rapidly. Let's explore how this works and why it’s so useful.
## Application and Examples
The Difference of Squares formula applies when you have two squared terms subtracted from each other. For example:
• `25 - 9 = (5)² - (3)² = (5 + 3)(5 – 3) = 8 * 2 = 16`
• ` 49 - 16 = (7)² - (4)² = (7 + 4)(7 – 4) = 11 * 3 = 33`
By converting a polynomial into a product of binomials, you make the expression much easier to handle or factor further. This is particularly useful in solving quadratic equations, polynomial long divisions, or even simplifying rational expressions.
## Step-by-Step Simplification
Here’s a structured approach to using the Difference of Squares:
1. Identify the two squared terms (a² and b²).
2. Ensure both terms are indeed squares of some expressions.
3. Apply the formula: `a² - b² = (a + b)(a - b)`.
For instance, let's simplify `64 - 1`:
• Recognize that `64 = 8²` and `1 = 1²`
• Apply the formula: `8² - 1² = (8 + 1)(8 - 1)`
• Simplify the terms: `(8 + 1)(8 - 1) = 9 * 7 = 63`
Notice how a seemingly complex expression becomes straightforward through this method!
## Real-Life Applications
Both the Diamond Method and the Difference of Squares have real-world applications beyond academics:
• Architecture and Design: These methods guide the creation of structurally sound and visually appealing designs.
• Engineering: Simplifying expressions helps in solving equations faster, leading to more efficient problem-solving.
• Finance: Algebraic simplification aids in understanding trends and predicting future values accurately.
Take the story of an architect who needed to design a aesthetically pleasing and sturdy ceiling. By using the Diamond Method, the architect ensured symmetry and balance, which impressed the clients and provided a practical solution.
### Q1: Why is the Diamond Method called by this name?
A1: The Diamond Method gets its name from the visual resemblance to a diamond shape, especially when dealing with rhombuses and the resulting symmetry. The method’s structure aids in visualizing relationships between geometric properties.
### Q2: Can the Difference of Squares be used in higher mathematics?
A2: Absolutely! The Difference of Squares is foundational and extends into higher mathematics. It’s particularly useful in calculus, number theory, and algebraic geometry.
### Q3: How can these methods be taught effectively to students?
A3: Visual aids, hands-on activities, and real-life examples make these methods engaging and comprehensible. Encouraging students to apply these techniques to solve practical problems enhances their understanding.
## Conclusion
The Diamond Method in Geometry and the Difference of Squares in Algebra are transformative tools that simplify complex problems. Whether you’re designing an architectural masterpiece or solving for unknowns in an algebraic expression, these methods provide clarity and efficiency. Embrace these techniques in your mathematical toolkit to unlock new levels of understanding and application.
Tags: Geometry, Algebra, Mathematics
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# How To Write An Equation In Standard Form From Slope Intercept
How To Write An Equation In Standard Form From Slope Intercept. Subtract 4x from each side to isolate y. So, we will use same thing to both sides to “rearrange” the equation into slope intercept form.
The standard form of a linear equation is ax + by = c. Slope intercept form of an equation: It’s usually easy to graph a line using y=mx+b.
### But Do Not Worry As Our Free Standard To Slope Intercept Form Calculator Will Carry Out Conversions For You.
Sometimes, the slope of a line may be expressed in terms of tangent angle such as: Click here to explore more helpful albert algebra 1 review guides. Ax+by=c ax +by = c.
### The Second Step Will Be.
This lesson starts after finding the slope between two points. Y = (7x/2) + (1/4) on the right side of the equation, we have the denominators 2 and 4. How to graph an equation in standard form.
### So, We Will Use Same Thing To Both Sides To “Rearrange” The Equation Into Slope Intercept Form.
It is written as ax+by=c. Y = (7x/2) + (1/4) solution : Multiply both sides of the equation by 4 to get rid of the denominators 2 and 4.
### Here’s What It Looks Like:
Let’s explain each of these forms. Use inverse operations to move terms. Convert slope = y 2 − y 1 x 2 − x 1 0 − 3 10 5 − 1 2 − 3 10 9 2 − 3 10 ⋅ 10 9 2 ⋅ 10 − 3 45 − 1 15 slope intercept :
### Subtract 4X From Each Side To Isolate Y.
4y = 14x + 1. Here the transition among both of these forms may confuse you. The first step will be to use the points to find the slope of the line.
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# Classify the following numbers as rational or irrational:(i). $2 - \sqrt 5$ (ii). $(3 + \sqrt {23} ) - \sqrt {23}$ (iii). $\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }}$ (iv). $\dfrac{1}{{\sqrt 2 }}$ (v). $2\pi$
Last updated date: 09th Aug 2024
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Hint: Simplify the numbers and check if they can be represented in $\dfrac{p}{q}$ form, where p and q are integers and $q \ne 0$ . If they can be represented, then they belong to rational numbers, if not, then they are irrational numbers.
A number that is in the form $\dfrac{p}{q}$ or can be simplified to the form $\dfrac{p}{q}$ where p and q are integers and $q \ne 0$ is called a rational number. They can also be represented in the decimal form as terminating decimals or non-terminating recurring decimals. Examples include $22,\dfrac{5}{7},\dfrac{1}{2}$ .
A real number that can’t be represented in the form $\dfrac{p}{q}$ where both p and q are integers and $q \ne 0$ is called an irrational number. These numbers are non-terminating and non-recurring decimals.
Examples include $\pi ,\sqrt 5 ,\sqrt[3]{2}$ .
Now, having the knowledge of rational and irrational numbers, we can classify the numbers into these categories.
(i). The number $2 - \sqrt 5$ , is the difference between 2 and $\sqrt 5$ .
2 is a rational number because it can be represented as $\dfrac{2}{1}$ where 2 and 1 are integers.
$\sqrt 5$ is an irrational number because it can’t be represented in the form of rational numbers.
We know that the sum or difference of a rational and an irrational number is an irrational number.
Hence, $2 - \sqrt 5$ is irrational.
(ii). The number $(3 + \sqrt {23} ) - \sqrt {23}$ can be simplified as follows:
$(3 + \sqrt {23} ) - \sqrt {23} = 3 + \sqrt {23} - \sqrt {23}$
Cancelling $\sqrt {23}$ , we have:
$(3 + \sqrt {23} ) - \sqrt {23} = 3$
We know that 3 is a rational number since it can be represented as $\dfrac{3}{1}$ where 3 and 1 are integers.
Hence, $(3 + \sqrt {23} ) - \sqrt {23}$ is rational.
(iii). The number $\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }}$ can be simplified by cancelling $\sqrt 7$ as follows:
$\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }} = \dfrac{2}{7}$
We can see that $\dfrac{2}{7}$ is in the rational form since 2 and 7 are integers.
Hence, $\dfrac{{2\sqrt 7 }}{{7\sqrt 7 }}$ is rational.
(iv). We can simplify the number $\dfrac{1}{{\sqrt 2 }}$ by multiplying numerator and denominator by $\sqrt 2$ .
$\dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
$\dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}$
We know that $\sqrt 2$ is irrational and 2 is rational.
Division of an irrational number by a rational number, results in an irrational number.
Hence, $\dfrac{1}{{\sqrt 2 }}$ is irrational.
(v). In the number $2\pi$ , 2 is rational and $\pi$ is irrational.
Multiplication of a rational and an irrational number is an irrational number.
Hence, $2\pi$ is irrational.
Note: You might conclude that options (ii) and (iii) are irrational numbers because they contain the square root terms $\sqrt {23}$ and $\sqrt 7$ respectively, but it is wrong. Simplify the number completely and then check for the $\dfrac{p}{q}$ form.
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# MAT 1234 Calculus I Section 2.8 Related Rates
## Presentation on theme: "MAT 1234 Calculus I Section 2.8 Related Rates"— Presentation transcript:
MAT 1234 Calculus I Section 2.8 Related Rates http://myhome.spu.edu/lauw
Next.. WebAssign 2.8. Due Next Monday (Difficulty level *****) to give you more time. Please do not wait until Monday afternoon. (WebAssign 2.9 is also due Monday.) Be sure to do it ASAP. Tutors are available today and tomorrow after class. Write down your solutions carefully!!! One of these type of questions will be on the second exam.
Room Change Next Thursday Switch to 245
Preview Define Related Rates How to solve word problems involving Related Rates
Related Rates
Example 1 GO NUTS!
Example 1 GO NUTS!
Example 1 A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 3 feet per second, how fast is the area changing when the radius is 5 feet?
Step 1 Draw a diagram A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 3 feet per second, how fast is the area changing when the radius is 5 feet?
Step 2: Define the variables A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 3 feet per second, how fast is the area changing when the radius is 5 feet?
Step 3: Write down all the information in terms of the variables defined A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 3 feet per second, how fast is the area changing when the radius is 5 feet?
Step 4: Set up a relation between the variables
Step 5: Use differentiation to find the related rate
Expectations In the quizzes and exams, you are expected to include these 5 steps in your solutions.
Example 2 A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Example 2 A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall? Everyone, try step 1 and 2!
Step 1 Draw a diagram A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Step 2: Define the variables A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall? Remark: Do not define more variables than necessary.
Step 3: Write down all the information in terms of the variables defined A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Step 4: Set up a relation between the variables
Step 5: Use differentiation to find the related rate
Review: Similar Triangles Two triangles are similar if and only if one of the following 2 conditions are satisfied 1. Their corresponding angles are the same. 2. The ratio of their corresponding sides are the same.
Review: Similar Triangles In particular: If the corresponding angles are the same, then the ratio of their corresponding sides are the same.
Please wait… We are going to walk through some of the main key points in your classwork. Please do not start your classwork now, not even drawing the diagrams.
Example 3 (Classwork) A street light is mounted at the top of a 12- ft-tall pole. A 6-ft-tall man walks away from the pole with a speed of 4ft/s along a straight path. How fast is the tip of his shadow moving when he is 35 ft from the pole?
Example 3 Wall 12 ft 35 feet 4 ft/s ???? ft/s Man 6 ft
Example 3 12 x 6 z Remark: Do not define more variables than necessary. For example, it is not necessary to define a variable for the length of the shadow.
Example 3 12 x 6 z Remark: Do not define more variables than necessary. For example, it is not necessary to define a variable for the length of the shadow. z-x
Example 3 12 x 6 z z-x
Hint 12 x 6 z z-x Use similar triangles to find a relation between x and z. Solve z in terms of x.
The Answers It turns out that in this problem, the answer is independent of the fact that x=35. This means that the tip of the shadow is moving at a constant rate.
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© Boardworks Ltd 2005 1 of 26 © Boardworks Ltd 2005 1 of 26 AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the.
Presentation on theme: "© Boardworks Ltd 2005 1 of 26 © Boardworks Ltd 2005 1 of 26 AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the."— Presentation transcript:
© Boardworks Ltd 2005 1 of 26 © Boardworks Ltd 2005 1 of 26 AS-Level Maths: Core 2 for Edexcel C2.6 Exponentials and logarithms This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.
© Boardworks Ltd 2005 2 of 26 Contents © Boardworks Ltd 2005 2 of 26 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions Exponential functions
© Boardworks Ltd 2005 3 of 26 Exponential functions So far in this course we have looked at many functions involving terms in x n. In an exponential function, however, the variable is in the index. For example: The general form of an exponential function to the base a is: y = a x where a > 0 and a ≠1. You have probably heard of exponential increase and decrease or exponential growth and decay. A quantity that changes exponentially either increases or decreases more and more rapidly as time goes on. y = 2 x y = 5 x y = 0.1 x y = 3 – x y = 7 x +1
© Boardworks Ltd 2005 4 of 26 Graphs of exponential functions
© Boardworks Ltd 2005 5 of 26 Exponential functions In both cases the graph passes through (0, 1) and (1, a ). This is because: a 0 = 1 and a 1 = a for all a > 0. When 0 < a < 1 the graph of y = a x has the following shape: y x 1 1 When a > 1 the graph of y = a x has the following shape: y x (1, a )
© Boardworks Ltd 2005 6 of 26 Contents © Boardworks Ltd 2005 6 of 26 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions Logarithms
© Boardworks Ltd 2005 7 of 26 Logarithms Find p if p 3 = 343. We can solve this equation by finding the cube root of 343: Now, consider the following equation: Find q if 3 q = 343. We need to find the power of 3 that gives 343. One way to tackle this is by trial and improvement. Use the x y key on your calculator to find q to 2 decimal places.
© Boardworks Ltd 2005 8 of 26 Logarithms To avoid using trial and improvement we need to define the power y to which a given base a must be raised to equal a given number x. This is defined as: y = log a x “ y is equal to the logarithm, to the base a, of x ” This can be written using the implication sign : y = log a x a y = x The expressions y = log a xa y = x andare interchangeable. For example, 2 5 = 32 can be written in logarithmic form as: log 2 32 = 5
© Boardworks Ltd 2005 9 of 26 Logarithms Taking a log and raising to a power are inverse operations. We have that: y = log a x a y = x So: Also: y = log a a y For example: 2and6
© Boardworks Ltd 2005 10 of 26 Contents © Boardworks Ltd 2005 10 of 26 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions The laws of logarithms
© Boardworks Ltd 2005 11 of 26 Some important results When studying indices we found the following important results: This can be written in logarithmic form as: log a a = 1 a1 = aa1 = a This can be written in logarithmic form as: log a 1 = 0 a0 = 1a0 = 1 It is important to remember these results when manipulating logarithms.
© Boardworks Ltd 2005 12 of 26 The laws of logarithms The laws of logarithms follow from the laws of indices: The multiplication law Let: m = log a x and n = log a y So: x = a m and y = a n log a x + log a y = log a ( xy ) xy = a m × a n Using the multiplication law for indices: xy = a m + n Writing this in log form gives: m + n = log a xy But m = log a x and n = log a y so:
© Boardworks Ltd 2005 13 of 26 The laws of logarithms The division law Let: m = log a x and n = log a y So: x = a m and y = a n Using the division law for indices: Writing this in log form gives: But m = log a x and n = log a y so:
© Boardworks Ltd 2005 14 of 26 The laws of logarithms The power law Let: m = log a x So: x = a m Using the power law for indices: Writing this in log form gives: But m = log a x so: x n =( a m ) n x n = a mn mn = log a x n n log a x = log a x n
© Boardworks Ltd 2005 15 of 26 The laws of logarithms These three laws can be used to combine several logarithms written to the same base. For example: Express 2log a 3 + log a 2 – 2log a 6 as a single logarithm.
© Boardworks Ltd 2005 16 of 26 The laws of logarithms Express log 10 in terms of log 10 a, log 10 b and log 10 c. Logarithms to the base 10 are usually written as log or lg. We can therefore write this expression as: The laws of logarithms can also be used to break down a single logarithm. For example:
© Boardworks Ltd 2005 17 of 26 Logarithms to the base 10 and to the base e Although the base of a logarithm can be any positive number, there are only two bases that are commonly used. These are: Logarithms to the base 10 Logarithms to the base e Logarithms to the base 10 are useful because our number system is based on powers of 10. They can be found by using the log key on a calculator. Logarithms to the base e are called Napierian or natural logarithms and have many applications in maths and science. They can be found by using the ln key on a calculator.
© Boardworks Ltd 2005 18 of 26 Changing the base of a logarithm Suppose we wish to calculate the value of log 5 8. We can’t calculate this directly using a calculator because it only find logs to the base 10 or the base e. We can change the base of the logarithm as follows: Let x = log 5 8 So:5 x = 8 Taking the log to the base 10 of both sides: log 5 x = log 8 x log 5 = log 8 So: 1.29 (to 3 s.f.)
© Boardworks Ltd 2005 19 of 26 Changing the base of a logarithm If we had used log to the base e instead we would have had: In general, to find log a b : Let x = log a b, so we can write a x = b Taking the log to the base c of both sides gives: log c a x = log c b x log c a = log c b 1.29 (to 3 s.f.) So:
© Boardworks Ltd 2005 20 of 26 Contents © Boardworks Ltd 2005 20 of 26 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions Solving equations using logarithms
© Boardworks Ltd 2005 21 of 26 Solving equations involving logarithms We can use the laws of logarithms to solve equations. For example: Solve log 5 x + 2 = log 5 10. To solve this equation we have to write the constant value 2 in logarithmic form: 2 = 2 log 5 5 because log 5 5 = 1 = log 5 5 2 = log 5 25 The equation can now be written as: log 5 x + log 5 25 = log 5 10 log 5 25 x = log 5 10 25 x = 10 x = 0.4
© Boardworks Ltd 2005 22 of 26 Solving equations of the form a x = b We can use logarithms to solve equations of the form a x = b. For example: Find x to 3 significant figures if 5 2 x = 30. We can solve this by taking logs of both sides: log 5 2 x = log 30 2 x log 5 = log 30 Using a calculator: x = 1.06 (to 3 s.f.)
© Boardworks Ltd 2005 23 of 26 Solving equations of the form a x = b Find x to 3 significant figures if 4 3 x +1 = 7 x +2. Taking logs of both sides:
© Boardworks Ltd 2005 24 of 26 Solving equations of the form a x = b Solve 3 2 x –5(3 x ) + 4 = 0 to 3 significant figures. If we let y = 3 x we can write the equation as: So: If 3 x = 1 then x = 0. Now, solving 3 x = 4 by taking logs of both sides:
© Boardworks Ltd 2005 25 of 26 Contents © Boardworks Ltd 2005 25 of 26 Exponential functions Logarithms The laws of logarithms Solving equations using logarithms Examination-style questions
© Boardworks Ltd 2005 26 of 26 Examination-style question Julia starts a new job on a salary of £15 000 per annum. She is promised that her salary will increase by 4.5% at the end of each year. If she stays in the same job how long will it be before she earns more than double her starting salary? 15 000 × 1.045 n = 30 000 1.045 n = 2 log 1.045 n = log 2 n log 1.045 = log 2 15.7 Julia’s starting salary will have doubled after 16 years.
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# How the Simpsons (And You) Can Multiply by Seven Using Your Fingers
The views expressed are those of the author and are not necessarily those of Scientific American.
When I tutored elementary school math, one of my favorite things to show the kids was how to multiply by 9 using their fingers. (I apologize, but this trick assumes that you have five fingers on each hand. Later you’ll see how to adapt it if you have nonstandard hands.)
Here’s how the finger rule works. Let’s say you want to multiply 4 by 9. First, put your hands in front of you with your palms away from you.
Prepare to multiply!
Count four fingers from the left, and bend that finger down. (This should be your left index finger.)
My fourth finger from the left is down. There are three fingers to its left and six to its right. Coincidence? I think not!
There are 3 fingers to the left of that finger and 6 to the right, and 4×9=36. To multiply 8 by 9, count 8 fingers from the left (this should be over to your right middle finger). There are 7 fingers to the left of it and 2 to the right, and 8×9=72.
8×9=72.
Recently, Numberphile posted a video called Pi and Four Fingers, which includes a portion of an interview with Simon Singh, author of the recently published book The Simpsons and Their Mathematical Secrets. In the video, Singh notes that even though almost all of the characters on the Simpsons have eight fingers, they use base* ten anyway. Sadly, the Simpsons can’t use the finger rule I use to multiply by nine. But if they were willing to think in base eight—as eight-fingered creatures, this would be the “natural” base for them—they could use the finger rule to multiply by seven.
The finger rule works for decimal thinkers with ten fingers because 9 is one less than the base of our number system. Each time you add 9, you are adding 10 and subtracting 1. In terms of finger multiplication, this means that by moving one finger to the right (going up to the next multiple of 9), you are putting one more finger in the left column and subtracting one from the right column.
If we’re willing to think in different bases, we can use the finger rule with a different number of fingers and remove the trick’s insidious bias against people and cartoon characters with nonstandard numbers of fingers on their hands. Anne Boleyn** could have used it to multiply by 10 as long as she was willing to think in base 11, and the Simpsons could use it to multiply by seven as long as they interpreted the result in base eight.
It’s easier for me to pretend I have fewer fingers than I do than it is to add an extra one, so I’ll show the finger rule in base eight. (Plus, it’s already easy to multiply by ten, but seven is tougher, so the sevens trick might be more useful.) We’ll pretend I don’t have thumbs.
The Simpsons do have thumbs, but it's easier for me to demonstrate the sevens trick by pretending I don't have them.
To multiply three by seven, count over three fingers from the left. You should be folding down the middle finger on your left hand.
3×7 is...25? What?! It's all OK in base 8.
There are two fingers to the left and five to the right. But 3×7 isn’t 25, is it? Yes it is! We’re counting in base eight, so the two represents two eights, or sixteen, not twenty. Sixteen plus five is twenty-one, so we got the right answer. Hurrah!
*If you’d like a refresher on what the “base” of a numbering system is, here you go. Each place in a base 10 number represents a power of 10. The furthest right place is the 100, or ones, place. (Any number other than 0 raised to the zeroth power is 1.) The next place to the left is the 101, or tens, place, the next place is the 102, or hundreds place, and so on. Other bases work the same way. For example, if we use the digits 1 and 0 to write numbers in binary, we would write the number twenty-two as 10110. There is one sixteen (24), one four(22), and one two (21). In base eight, we have a ones place, an eights place, a sixty-fours (82) place, and so on. The number one hundred written in base eight is 144 (sixty-four+four×eight+four), and sixty-four written in base eight is 100.
**Anne Boleyn probably didn’t have eleven fingers. The first record of this abnormality came about fifty years after her death from someone who blamed her for Henry VIII’s renunciation of Catholicism, so it is not given much credence.
***Roots of Unity accepts no responsibility for any negative experiences you may have as a result of using the sevens trick to impress a date.
It’s hard to take photographs of your own hands! Pictures in this post were taken by Jon Chaika.
About the Author: Evelyn Lamb is a postdoc at the University of Utah. She writes about mathematics and other cool stuff. Follow on Twitter @evelynjlamb.
The views expressed are those of the author and are not necessarily those of Scientific American.
Previous: When Numbers Are Used for a Witch Hunt MoreRoots of Unity Next: A Cuddly, Crocheted Klein Quartic Curve
Rights & Permissions
1. 1. Layer_8 3:23 am 11/14/2013
I can count to 1023 with 1010 Fingers
2. 2. billlee42 12:04 am 11/16/2013
A person can count from 0 to 55 in base 6 (0 to 35 base 10) on their ten fingers. The right hand is 0 to 5 units and the left hand is 0 to 5 tens.
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# 12.07 Distance time graphs
Lesson
Distance-time graphs are a way to describe the movement of people or objects. They usually describe a trip that leaves and returns to a point (like a home base).
The vertical axis of a distance-time graph is the distance travelled from a starting point and the horizontal axis is the time taken from the starting point.
There are certain important features of a distance-time graph that we can use to interpret the journey being described:
• As the line moves away from the horizontal axis, the object is moving further away from the "home" point
• As the line moves back towards the horizontal axis, the object is returning home
• When the line is horizontal, the object is not moving
• The steeper the line, the greater the speed of the an object (the faster it moves)
• A straight line indicates a steady speed
• The total distance of the trip is the distance away from and returning home
#### Worked example
Consider the following graph which displays a day long car tip with the horizontal axis being time in hours and the vertical axis being distance from home in kilometres:
(a) What speed did the car travel in the first hour?
Think: We know that $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$Speed=DistanceTime. How far did they travel in the first hour?
Do:
Speed $=$= $\frac{\text{Distance}}{\text{Time}}$DistanceTime $=$= $\frac{80\text{ km}}{1\text{ h}}$80 km1 h $=$= $80$80 km/h
(b) What happened between the times of $1$1 and $2$2?
Think: What does it mean for the graph to be horizontal?
Do: The car was stationary for $1$1 hour. Perhaps a break for lunch or a visit to a park.
(c) How far is the car from home after $6$6 hours?
Think: Read the vertical axis for the distance at $6$6 hours.
Do: The car is $50$50 km from home.
(d) What was the average speed of the car over the $6$6 hour journey?
Think: How far has the car travelled in total? The car initially travelled $80$80 km, then was stationary for one hour, then travelled a further $120$120 km before starting the return tip home at $4$4 hours into the journey. In the last section of the journey they are returning to home from $200$200 kilometres away and reach $50$50 kilometres from home, thus they travel $150$150 km.
Do:
Total distance travelled $=$= $80+120+150$80+120+150 km $=$= $350$350 km
Average speed $=$= $\frac{\text{Total distance}}{\text{time}}$Total distancetime $=$= $\frac{350\ km}{6\ h}$350 km6 h $=$= $58.\overline{3}$58.3 km/h
#### Practice questions
##### Question 1
Which graph shows the height of a ball being thrown vertically into the air?
1. A
B
C
D
##### Question 2
Ben travels forwards and backwards along a straight line.
The graph shows Ben's distance from his starting point at various times of the day.
1. When did Ben start his journey?
2. How far did Ben travel by $11$11 am?
3. What happened to Ben's speed at $11$11 am?
Ben decreased his speed at $11$11 am.
A
Ben did not change his speed at $11$11 am.
B
Ben increased his speed at $11$11 am.
C
4. Evaluate Ben's speed between $11$11 am and $1$1 pm.
5. What distance did Ben travel between $1$1 pm and $2$2 pm?
6. What is the furthest distance travelled from the starting point?
7. What is the total distance travelled by Ben from $9$9 am to $4$4 pm?
### Outcomes
#### ACMEM088
interpret distance-versus-time graphs
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math-tutorials-on-trigonometry-understanding-trig-identities-verifying-trigonometric-identities
# Interactive video lesson plan for: Math tutorials on Trigonometry | Understanding Trig Identities | Verifying Trigonometric Identities
#### Activity overview:
Math tutorials on Trigonometry | Understanding Trig Identities | Verifying Trigonometric Identities
http://www.learncbse.in/ncert-class-10-math-solutions/
http://www.learncbse.in/ncert-solutions-for-class-10-maths-chapter-8-introduction-to-trigonometry/
http://www.learncbse.in/rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios/
00:02 CBSE Class 10 maths NCERT Solutions Trigonometry Q5. Prove the trigonometric equations using trigonometric identities, where the angles involved are acute angles for which the expressions are defined
00:04 CBSE Class 10 maths ex 8.4 q5 (2)
00:19 Procedure for proving Trigonometric equations
00:41 consider LHS of the trigonometric equation
01:20 algebra maths formulas | Algebraic Expressions and Formulas
USE (a-b)^2 expanding formula
01:58 Use Pythagorian Identity
sin squared theta plus cos squared theta = 1
02:59 Reciprocal Identity.
The ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A
sec A = 1/cos A
03:13 How to prove Trigonometric identities.
Defination of an Identity:
An equation is called an identity when it is true for all values of the variables involved.
Trigonometric Identities defination:
An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.
In this math tutorial we prove trigonometric identities, solve trigonometric equations and Verify Trigonometric Identities.We use Quotient Identities , Reciprocal Identities, and the Pythagorian Identities to solve problems on trigonometric identities.
Solve Trigonometric Equations with Trigonometric Identities
Using the fundemental identities and the Pythagorean Identities
multiple examples of verifying trigonometric identities
How to Solve Trigonometric Identities Proving Problems
method to solve trigonometric identities problems
the Quotient Identities, Reciprocal Identities, and the Pythagorian Identities.
Pinterest
https://in.pinterest.com/LearnCBSE/
Wordpress
https://cbselabs.wordpress.com/
learncbse.in
CBSE solutions for class 10 maths Chapter 8 Trigonometry Exercise 8.4
CBSE class 10 maths NCERT SolutionsChapter 8 Trigonometry Exercise 8.2 | Trigonometric Identities
CBSE class 10 maths NCERT Solutions Chapter 8 Trigonometry
Solutions for CBSE class 10 Maths Trigonometry
NCERT solutions for class 10 maths Trigonometry
NCERT solutions for CBSE class 10 maths Trigonometry
CBSE class 10 maths Trigonometry
Common core Math Trigonometry
Tagged under: ncert solutions,learncbse.,gyanpub,Trigonometry,Common core Math,trigonometry algebra,Solving Trig Identities,SolvingTrigonometry Equations,fundemental identities,Pythagorean Identities,verifying trigonometric identities,Proving Problems,learning trigonometry,trigonometry tutor,algebra trigonometry
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### Solution
Suppose not. That is, suppose that (1) all trades are profitable for both or neither person; and (2) no two people make profitable trades with the same number of other people. Note that if a trade is profitable for neither person, it has no effect on the number of profitable trading partners, so let's restrict our attention to only trades profitable for both people.
Since there are 103 people at the convention, each person made profitable trades with somewhere between 0 and 102 other people. By assumption (2), each person made profitable trades with a different number of other people and the range from 0 to 102 contains exactly 103 different values. So one person A made 0 profitable trades and some other person B made 102 profitable trades.
Since our initial assumption led to a contradiction, at least one of these must be true: (1) some trades are profitable for only one person; (2) at least two people made profitable trades with the same number of other people. QED.
### Variations
There are several possible variations for this proof. There are three key steps:
1. Suppose the negation of the claim, which in this case is a logical AND of two statements.
2. See that the second statement requires all values from 0 to 102 to be filled (similar to pigeonhole principle).
3. See that the first statement doesn't allow both values 0 and 102 to be filled.
Another possible version is to apply the reasoning from the first statement that values 0 and 102 cannot both be used and then apply pigeonhole principle to show the contradiction with the second statement.
You can also make a graph representation. That is, vertices are people, people are adjacent iff they traded profitably with each other. From this, you can show a contradiction with the handshaking theorem. That is, the node degrees sum up to an odd number. But the sum of node degrees in a graph is always even, because it's twice the number of edges.
Or, notice that each person made a different number of profitable trades, using up exactly all the numbers in the range 0 through 102. So the total number of profitable trades is $$\displaystyle \sum_{k=0}^{102} k = \frac{102\cdot 103}{2} = 61 \cdot 103$$. This number is odd, so there must have been at least one trade that was profitable for only one of the two people involved.
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# How do you solve using the quadratic formula $3{{x}^{2}}+4x-2=0$?
Last updated date: 27th Feb 2024
Total views: 340.8k
Views today: 6.40k
Verified
340.8k+ views
Hint: In this problem we need to solve the given quadratic equation i.e., we need to calculate the values of $x$ where the given equation is satisfied. For solving a quadratic equation, we have several methods. But in the problem, they have mentioned to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. For this we need to compare the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will substitute those values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the obtained equation to get the required result.
Given equation $3{{x}^{2}}+4x-2=0$.
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=3$, $b=4$, $c=-2$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 3 \right)\left( -2 \right)}}{2\left( 3 \right)}$
We know that when we multiplied a negative sign with the positive sign, then we will get negative sign. Applying the above rule and simplifying the above equation, then we will get
\begin{align} & \Rightarrow x=\dfrac{-4\pm \sqrt{16+24}}{6} \\ & \Rightarrow x=\dfrac{-4\pm \sqrt{40}}{6} \\ \end{align}
In the above equation we have the value $\sqrt{40}$. We need to simplify this value to get the simplified result. We can write $40=10\times 4=10\times {{2}^{2}}$, then the value of $\sqrt{40}$ will be $\sqrt{40}=\sqrt{10\times {{2}^{2}}}=2\sqrt{10}$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{-4\pm 2\sqrt{10}}{6}$
Calculating each value individually, then we will get
\begin{align} & \Rightarrow x=\dfrac{-4+2\sqrt{10}}{6}\text{ or }\dfrac{-4-2\sqrt{10}}{6} \\ & \Rightarrow x=\dfrac{2\left( -2+\sqrt{10} \right)}{6}\text{ or }\dfrac{2\left( -2-\sqrt{10} \right)}{6} \\ & \Rightarrow x=\dfrac{-2+\sqrt{10}}{3}\text{ or }\dfrac{-2-\sqrt{10}}{3} \\ \end{align}
Hence the solution of the given quadratic equation $3{{x}^{2}}+4x-2=0$ are $x=\dfrac{-2\pm \sqrt{10}}{3}$.
Note: We can also see the graph of the above given equation to observe the roots of the equation. When we plot the graph of the given equation $3{{x}^{2}}+4x-2=0$ it looks like below graph
From the above graph also, we can say that the roots of the given equation $3{{x}^{2}}+4x-2=0$ are $x=\dfrac{-2\pm \sqrt{10}}{3}$.
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Q. 20
# Let A = {1,
Given: A = {1,2,3......9}
and R is defined as
(a,b) R (c,d) a + b = b + c
Reflexive:
Consider (a,b) R (a,b) (a,b) A × A
a + b = b + a
Hence, R is reflexive
Symmetric:
Consider (a,b)R(c,d) given by (a,b)(c,d)A × A
(a,b) R (c,d)a + d = b + c
b + c = a + d
c + b = d + a
(c,d) R (a,b)
Hence R is symmetric
Transitive:
Let (a,b) R (c,d) and (c,d) R (e,f) given by (a,b),(c,d),(e,f) A × A
(a,b) R (c,d) a + b = b + c
a – c = b – d …(1)
and (c,d) R (e,f) c + f = d + e
& c + f = d + e …(2)
adding (1) and (2), we get
a – c + c + f = b – d + d + e
a + f = b + e
(a,b) R (e,f)
Hence, R is transitive
Hence, R is an equivalence relation
We need to find [(2, 5)]
So, (2, 5) will go in and (c, d) will come out
This will be possible if
a + d = b + c
2 + d = 5 + c
d – c = 5 – 2
d – c = 3
So, in our relation [(2, 5)]
We need to find values of c and d which satisfy d – c = 3
Since (c, d) A × A
Both c and d are in set A = {1, 2, 3, …9}
[(2,5)]={(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)}is the equivalent class under relation R
OR
Given: f(x) = 4x2 + 12x + 15
To show: f: NS is invertible and to find inverse of f
Proof:
Let f(x) = y
4x2 + 12x + 15 = y
4x2 + 12x + 15 – y = 0
4x2 + 12x + (15 – y) = 0
Comparing equation with ax2 + bx + c = 0
Here, a = 4, b = 12 and c = 15 – y
Putting the values, we get
So,
As x є N, So, x is a positive real number.
So, x can’t be
Hence,
Let where g: SN
gof = g(f(x))
= g(4x2 + 12x + 15)
[here, y = 4x2 + 12x + 15]
= x
Hence, gof = x = IN
Now, we find the fog
fog = f(g(x))
= 9 + y – 6 – 3
= y
Hence, fog = y = IS
Since, gof = IN and fog = IS
f is invertible
and
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# Division Problem
• Jan 18th 2007, 12:32 PM
Division Problem
1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.
2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]
Thank you.
• Jan 18th 2007, 12:41 PM
ThePerfectHacker
Quote:
Originally Posted by tttcomrader
1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.
.
$a|b,a|c$
Thus,
$b=ai,c=aj$ for some $i,j$.
Then,
$a(ri+tj)=ari+atj=br+ct$
Thus,
$a|(br+ct)$
Quote:
Originally Posted by tttcomrader
2. Let b, c and d be integers. If b divides c and c divides d, then the equation bx = d has an integer solution. [Note: x is a variable.]
$b|c$ and $c|d$ thus, $b|d$.*
Thus,
$d=bk$ for some $k$.
Thus,
$b(k)=d$ solves the equation.
*)Proof.
$b|c$ thus, $c=bi$ and $c|d$ thus, $d=cj$
Thus,
$d=cj=(bi)j=b(ij)$
Thus,
$b|d$.
• Jan 18th 2007, 12:45 PM
CaptainBlack
[quote=tttcomrader;35535]1. Let a, b, c, r, and t be integers. Prove that if a divides both b and c, then a divides br + ct.[\quote]
If $a|b$ then there exists an integer $\alpha$ such that $b=\alpha \, a$.
If $a|c$ then there exists an integer $\beta$ such that $c=\beta \, a$.
So $b\, r+c\,t = \alpha \, a \, r + \beta \, a \, t = a(\alpha \, r + \beta \, t)$.
Hence $a|(b\, r+c\, t)$
RonL
• Jan 18th 2007, 09:05 PM
Thanks, guys, it looks so easy now that I see the solution, wonder why I couldn't see that at first... Pity as I have completed Advanced Calculus, I thought I would have go over Modern Algerba easy.
Thank you!
KK
• Jan 19th 2007, 04:38 AM
topsquark
I've found that, though these kinds of proof look easy, they do take some getting used to.
-Dan
• Jan 19th 2007, 06:20 AM
ThePerfectHacker
Quote:
Originally Posted by tttcomrader
Pity as I have completed Advanced Calculus, I thought I would have go over Modern Algerba easy.
Abstract Algebra, I think, is way more complicated then Advanced Calculus. Because the nice thing about analysis is that it can be visualized that is what makes it simpler. With Algebra there is absolutely no visualization at all. I would also say that Abstract Algebra is as difficult as courses get.
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162
Q:
# Seats for Mathematics, Physics and Biology in a school are in the ratio 5:7:8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats ?
A) 1:2:3 B) 2:3:4 C) 3:4:5 D) 4:5:6
Explanation:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
⇒ [(140/100) × 5x],[(150/100) × 7x] and [(175/100) × 8x]
⇒ 7x, 21x/2 and 14x.
⇒ The required ratio =7x : 21x/2 : 14x
⇒ 14x : 21x : 28x
⇒ 2 : 3 : 4
Q:
If (3/2)X = (5/7)Y = (6/5)Z, then what is X : Y : Z?
A) 105 : 50 : 84 B) 24 : 25 : 32 C) 15 : 21 : 25 D) 20 : 42 : 25
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
4 8226
Q:
If 3A = 2B = C, then A : B : C = ?
A) 6 : 2 : 3 B) 1/3 : 1/2 : 1 C) 3 : 2 : 1 D) 1 : 3 : 2
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
5 846
Q:
The ratio of the number of marbles that Pulak and Menaja had was 5:9 while the ratio of the number of marbles that Jairam and Menaja had was 7:18. What is the ratio of the number of marbles that Pulak and Jairam had?
A) 10:7 B) 2:3 C) 5:7 D) 7:5
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
4 712
Q:
Two numbers are in ratio 3:4. When 3 is subtracted from both the numbers, the ratio becomes 2:3. Find the sum of the numbers.
A) 16 B) 20 C) 21 D) 22
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
9 745
Q:
If 3/5 of x is equal to 4/9 of y, then what is the ratio of x and y?
A) 11 : 15 B) 15 : 23 C) 20 : 27 D) 14 : 25
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
7 8946
Q:
If p : q : r = 2 : 3 : 5, then what will be the ratio of (p + q + r) : (3p + q – r)?
A) 5 : 2 B) 10 : 17 C) 9 : 4 D) 5 : 4
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
3 6292
Q:
Three triangles are marked out of a bigger triangle at the three vertices such that each side of each of the smaller triangles is one-third as long as each corresponding side of the bigger triangle. The ratio of the area of the three small triangles taken together to that of the rest of the bigger triangle is ?
A) 1:2 B) 1:3 C) 3:1 D) 1:9
Explanation:
Filed Under: Ratios and Proportions
Exam Prep: Bank Exams
5 712
Q:
If w : 0.80 :: 9 : 6, w = ?
A) 2.5 B) 1.1 C) 3.3 D) 1.2
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# Texas Go Math Grade 5 Lesson 14.2 Answer Key Graph Data
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 14.2 Answer Key Graph Data.
## Texas Go Math Grade 5 Lesson 14.2 Answer Key Graph Data
Essential Question
How can you use a coordinate grid to display data collected in an experiment?
Investigate
Materials
• paper cup
• water
• Fahrenheit thermometer
• ice cubes
• stopwatch
When data is collected, it can be organized in a table.
A. Fill the paper cup more than halfway with room-temperature water.
B. Place the Fahrenheit thermometer in the water and find its beginning temperature before adding any ice. Record this temperature in the table at 0 seconds.
C. Place three cubes of ice in the water and start the stopwatch. Find the temperature every 10 seconds for 60 seconds. Record the temperatures in the table.
Make Connections
You can use a coordinate grid to graph and analyze the data you collected in the experiment.
Step 1
Write the related pairs of data as ordered pairs.
(0, ___) (30, ___) (50, ___)
(10, ___) (40, ____) (60, ___)
(20, ___)
Step 2
Construct a coordinate grid and write a title for it. Label each axis.
Step 3
Plot a point for each ordered pair.
Math Talk
Mathematical Processes
Analyze your observations about the temperature of the water during the 60 seconds
Share and Show
Graph the data In the coordinate grid.
Question 1.
Problem Solving
Use the table at the right for 3-4.
Question 3.
Multi-Step Write ordered pairs to show the relationship between the month and the weight ola bird. Plot each point in the coordinate grid.
Question 4.
H.O.T. Analyze What do you think would happen to the bird’s eight over the next four months?
H.O.T. What’s the Error?
Question 5.
Mary places a miniature car onto a track with launchers. The speed of the car is recorded every foot. Some of the data is shown in the table. Mary graphs the data in the coordinate grid below.
Question 6.
Communicate Describe the error Mary made.
Reasoning Where do you think the miniature car will stop? Write the ordered pair. Explain.
Question 8.
The graph shows the relationship between the length of an eruption of Old Faithful, x, and the length of time until the next eruption, y. Which of these ordered pairs is on the graph?
(A) (95, 5)
(B) (2, 50)
(C) (70, 3)
(D) (4, 80)
Question 9.
Which set of ordered pairs shows the data In the table?
(A) (1 ,2), (3, 4), (5, 3), (9, 13), (17, 20)
(B) (1, 3),(2, 9),(3, 13),(4, 17), (5, 20)
(C) (20, 5), (17, 4), (13, 3), (9, 2), (3, 3)
(D) (1, 3), (9, 2), (13, 3), (4, 17), (20, 5)
Question 10.
Multi-Step Sean is growing a plant in a jar. He makes a graph to show the days since he put the seed in the jar, and the length of the longest root (in centimeters). He first sees the root in 4 days, and he writes the ordered pair (4, 2). Three days later, the root measures 18 cm. Five days later, it has grown another 25 cm. What ordered pairs does Sean add to his graph?
(A) (7, 18), (12, 43)
(B) (18, 3), (25, 5)
(C) (3, 18), (5, 25)
(D) (18, 7), (43, 12)
Texas Test Prep
Marie was in fifth grade when she began to record her baby sister’s height and her own height in inches every six months for 3 years. She plots the data on a graph where the x-coordinate represents her sister’s height and the y-coordinate represents Marie’s height. Which ordered pair could NOT be on the graph?
(A) (24, 51)
(B) (33, 60)
(C) (55, 30)
(D) (28, 52)
### Texas Go Math Grade 5 Lesson 14.2 Homework and Practice Answer Key
Graph the data in the coordinate grid.
Question 1.
Question 2.
Problem Solving
Use the table at the right for 3-4.
Write ordered pairs to show the relationship between the day and the weight of the bird feeder. Plot each point in the coordinate grid.
Question 4.
Do you think the weight of the bird feeder will be 10 ounces on day 6? Explain.
Fill in the bubble completely to show your answer. Use the graph for 5-6.
Question 5.
The graph shows the relationship between the length of time and the number of kernels popped. Which of these ordered pairs is on the graph?
(A) (20, 35)
(B) (10, 20)
(C) (85, 40)
(D) (90, 50)
Question 6.
The number of popped kernels stays the same after 50 seconds. Which ordered pair can you plot to show the number of popped kernels after 60 seconds.
(A) (50, 90)
(B) (90, 50)
(C) (60, 90)
(D) (90, 60)
Go Math 5th Grade Practice and Homework Lesson 14.2 Answer Key Question 7.
Marc plots the point (1, 20) in a coordinate grid to show the number of minutes it takes him to walk 1 mile. Which ordered pair might Marc use to show how many minutes it takes him to walk 3 miles?
(A) (3, 3)
(B) (60, 3)
(C) (30, 3)
(D) (3, 60)
Question 8.
Multi-Step Which set of ordered pairs shows the data in the table?
(A) (1, 2), (3, 4), (5, 2), (3, 5), (7, 8)
(B) (8, 5), (7, 4), (5, 3), (3, 2), (1, 1)
(C) (1, 2), (3, 2), (5, 3), (7, 4), (8, 5)
(D) (1, 2), (2, 3), (3, 5), (4, 7), (5, 8)
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Blog > Your Learners > 6 magical maths activities to keep children engaged
# 6 magical maths activities to keep children engaged
Editor’s note: This is an updated version of a blog post published on August 25, 2017
## The long summer holiday is the perfect opportunity to explore some of the more creative and magical ways to keep children engaged and focused on the fun of maths.
Looking for maths activities you can do with your children to keep them engaged in and excited over the summer? Here are a few of my favourite magical maths activities to entertain the whole family regardless of age or mathematical ability.
## Build your own magic squares
Magic squares are fascinating. In a basic magic square, the rows, columns, and two diagonals all add up to the same number. This is called the magic number and in the illustration below, the magic number is 15.
One of my favourite magic squares can be found in Barcelona at the Sagrada Familia.
It also has many different patterns to explore. How many can you find?
The magic number here is 33. There are dozens of combinations of numbers that would produce a similar square, with numbers adding to 33. Can you find any?
### How to construct a magic square
In the 19th century, Édouard Lucas devised the general formula for order 3 magic squares.
Consider the following table made up of numbers a, b and c:
These numbers will form a magic square as long as:
• They are all larger than zero
• b must be larger than a
• b must not be the same value as 2 times a
• c-a must be bigger than both a and b
## Explore the golden ratio
The golden ratio, denoted by the Greek symbol ‘phi’ is a special number in maths that is approximately equal to 1.618. It appears in architecture, art and geometry, and even in the human body.
The golden ratio is found by dividing a line into two unequal parts so that the whole length of the line divided by the longer part is equal to the longer part divided by the shorter part. Try exploring this yourself.
As humans, we like to see objects around us that are in the golden ratio, such as picture frames. The proportions of many buildings are also designed to be in the golden ratio. We are also designed in line with the golden ratio. Have a go at this activity from nrich.org.
Measure the following:
1. Distance from the ground to your belly button
3. Distance from the ground to your knees
4. Distances A, B, and C
Now calculate the following ratios:
• Distance from the ground to your belly button / Distance from the ground to your knees
• Distance C / Distance B
• Distance B / Distance A
Write your results in the following table:
Can you see anything special about these ratios?
## Be amazed by everlasting chocolate
Here’s a fun and magical maths activity. The everlasting chocolate trick appears to make a square of chocolate appear out of nowhere. Clearly, this is an attractive thought for the most reluctant mathematician. There are several youtube clips that both illustrate and explain this phenomenon.
## Try this time calculator
You may well be travelling over the summer, with children asking ‘Are we there yet?’ Maybe you have a car journey, followed by a train journey. How long will you be travelling altogether? You need magical maths activities. This method for adding periods of time may well dazzle your children!
For example, we have a car journey of 1 hour 35 minutes followed by a train journey of 2 hours and 55 minutes. To add two periods of time together, you just write the times as three digit numbers, add them and then add 40.
1 hr and 35 mins plus 2 hrs and 55 mins
Would be
135 + 255 + 40
= 430
4 hrs and 30 minutes!
Can you figure out why it works?
## Figure out what happened
First, write down any three-digit number.
Multiply it by 13.
Now, multiply that answer by 11.
What do you notice? Does it always work? Why?
## Make a Möbius strip
This is my all-time favourite activity to do with learners of any age and with the teachers that I work with too.
The Möbius strip is named after the mathematician August Möbius who invented the strip in 1858. It is curious because it only has one side and one edge. A Mobius strip is simple to construct. All you need is a long strip of paper, sticky tape, and a pair of scissors.
Take the long piece of paper and put the ends together as if you were making a headband. Before you attach the two ends with tape, give one of them a twist and then stick the ends together.
Now try drawing a line along the centre of the strip. What do you notice?
You will end up back where you started, without having to take your pen off the paper. So you have drawn on both sides of the strip of paper in one go, meaning that your original strip of paper, which had two sides, now only has one side!
Now cut along the line that you have drawn. What do you think might happen?
Try it and see!
There are many other interesting things that you can do with this strip. Try making another one, but this time instead of cutting along the middle, cut a third of the way from one edge. What do you think will happen this time?
Watch this video from Think Twice for more ideas:
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# Integer multiplication
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## Transcription
1 Integer multiplication Suppose we have two unsigned integers, A and B, and we wish to compute their product. Let A be the multiplicand and B the multiplier: A n 1... A 1 A 0 multiplicand B n 1... B 1 B 0 multiplier P 2n 1... P n 1... P 1 P 0 product If the multiplier and multiplicand each have n bits, then the product has at most 2n bits, since (2 n 1) (2 n 1) < 2 2n 1 which you can verify for yourself by multiplying out the left side. For example, if we are multiplying two 32 bit unsigned integers, then we may need as many as 64 bits to represent the product. Recall your grade school algorithm for multiplying two numbers. We can use the same algorithm if the numbers are written in binary. The reason is that shifting a binary number one bit to the left is the same as multiplying by 2, just as decimal multiplication by ten shifts digits left by one. In the case of binary, each 1 bit b i in the multiplier means we have a contribution 2 i times the multiplicand, and so we add the shifted multiplicands product In the grade school algorithm, we compute all the left shifts in advance, building up several rows of numbers which correspond to the digits of the multiplier. Then, this set of numbers is added together to yield the product (as above). Let s consider an alternative algorithm that avoids writing out all the rows with the shifted multiplicands. Instead the algorithm shifting the multiplicand (left i.e. multiply by 2) at each step i. When bit B i of the multiplier is 1, the shifted multiplicand is added to an accumulated total. When bit B i is 0, nothing is added to the total. What hardware could we use to multiply two n = 32 bit unsigned numbers? shift registers: the multiplicand is shifted left. If we are multiplying two 32 bit numbers then we use a 64 bit shift left register for the multiplicand. (We shift the multiplicand to the left at most 32 times.) a 32 bit register for the multiplier. We use a shift right register, and read the least significant bit (LSB, i.e. bit 0) to decide whether to add the shifted multiplicand on each step or not. last updated: 19 th Apr, lecture notes c Michael Langer
2 a 64 bit adder a 64 bit register which accumulates the product a counter to keep track of how many times we ve shifted. Algorithm for multiplying two unsigned integers: // // Let P, A, B be the product, multiplicand, and multiplier registers. // P = 0 load A so lower 32 bits contain multiplicand (upper 32 bits are 0) load B so it contains multiplier (32 bits) for counter = 0 to 31 { if (LSB of B is 1) P := P + A endif shift A left by one bit shift B right by one bit } D D shift left register (64 bits) A shift right register (32 bits) B shift/load LSB shift/load adder combin. circuit P write enable / clear counter We must provide a control signal to the shift registers telling them whether to shift or not at each clock cycle. On the first clock cycle, we would load the multiplier and multiplicand registers, rather than shifting the values in these registers. (Where would we load them from? We would load them from some other registers that are not drawn in the circuit.) On subsequent clock cycles, we would perform the shift. Similarly, we must provide a signal to the product register telling it whether we should clear the product register (only on the first clock cycle), or whether we should write in the value coming out of the ALU (only on subsequent clock cycles and only when the LSB of the B register is 1. Finally, notice that all the instructions within the for loop can be performed in parallel within a clock cycle. last updated: 19 th Apr, lecture notes c Michael Langer
3 Fast integer multiplication ( log n argument is not on exam) In the lecture, I briefly sketched out an alternative scheme which performs n-bit multiplication more quickly. The idea will be familiar to most of you, namely to those who have taken COMP 250. The idea is to take the n rows defined by the grade school multiplication algorithm (see page 1) and use n fast adders to add rows 1 and 2, 3 and 4, 5 and 6, etc. Then the outputs from these adders could 2 be paired and run through n adders which would give the sum of rows 1 to 4, 5 to 8, etc. The bits 4 would all need to be aligned properly as in the table on page 1, but this doesn t need shift registers. It just requires that certain output lines be fed into the correct input lines, and 0 s be fed in where appropriate. In principle the multiplication can be done in one clock cycle, using only combinational circuits. However, the clock cycle would need to be long to ensure that the all signals (carries, etc) have propagated through this circuit. Since the same clock is used for all gates in the processor (including the ALU) this means we would need a slow clock, which is not what we want. To avoid the slow clock requirement, it is common to break the multiplication into stages and have a multiplication run in multiple clock cycles. This can be done by using registers to hold intermediate results. In particular, the n sums could be written to n registers at the end of the 2 2 multiplication s first clock cycle. The next n sums could be written to n registers at the end of the 4 4 second clock cycle. Those of you who have already done COMP 250 know that log n clock cycles would be needed in total to compute the multiplication. This is still faster than the grade school algorithm which took n steps. Integer division (sketch only not on exams) When we perform integer division on two positive integers, e.g. 78/21, or more generally dividend/divisor, the result is two positive integers, namely a quotient and a remainder. 78 = dividend = quotient divisor + remainder Suppose we have two 32 bit unsigned integers: a dividend and a divisor. Suppose they are both positive numbers. How do we compute the quotient and the remainder? We can apply the long division algorithm we learned in grade school. As with multiplication, we need a set of shift registers. See slides for a rough sketch of the corresponding circuit. The algorithm is not obvious, but here it is for completeness (and fun?). Algorithm for long division: divisor := 2 n * divisor // Make the divisor bigger than the dividend quotient := 0 remainder := dividend for i = 0 to n shift quotient left by one bit if (divisor remainder) set LSB of quotient to 1 remainder := remainder - divisor shift divisor to right last updated: 19 th Apr, lecture notes c Michael Langer
4 A few notes: First, the first line of this algorithm is just the usual first step in long division of placing the divisor to the left of the dividend. When you carry out long division by hand in base 10, you don t think of this as multiplying by 10 n where n is the highest power of the dividend. But this is essentially what you are doing. Second, the comparison divisor remainder can be done by computing the difference (divisor - remainder) and checking whether the result is less than zero. In a twos complement representation, this can be done by checking if the MSB (most significant bit) is 1. Floating point addition Let s now consider floating point operations: +,-,*,/. Recall the discussion from lecture 2 where we wanted to add two single precision floating point numbers, x and y. x = y = where x has a smaller exponent than y. We cannot just add the significands bit by bit, because then we would be adding terms with different powers of 2. Instead, we need to rewrite one of the two numbers such that the exponents agree. e.g. We rewrite x so it has the same exponent as y: x = = = Now that x and y have the same exponent, we can add them = The sum is a normalized number. But the significand is more than 23 bits. If we want to represent it as single precision float, then we must truncate (or round off, or round up) the extra bits. It can also happen that the sum of two floats is not normalized. If, instead of y, we had z = then check for yourself that x + z would not be normalized. So, to represent the result as single precision float, we would have to normalize again by shifting (right) the significand and truncating, and changing the exponent (by adding the size of the shift). You should appreciate what you need to do floating point addition. You need to compare exponents, shift right (x or y with smaller exponent), use a big adder, normalize, deal with case if result is not normalized, round off. Details of the circuit are omitted, but you should appreciate the many steps are may be involved. last updated: 19 th Apr, lecture notes c Michael Langer
5 Floating point multiplication and division Multiplying two floating point numbers is easy. We reduce the problem to integer multiplication, plus some bookkeeping with the exponents. To see the idea, suppose we work with 3 bits of significand instead of 23. Here is an example. x 1 = = x 2 = = Since (1101) 2 (1001) 2 = 13 9 = 117 = ( ) 2 and = 2 9, we can rewrite it as a normalize number as follows: x 1 x 2 = = Now go back to the previous example in which x and y each have 23 bits of significand. We can write each of them as an integer times a power of 2: x = = y = = We then multiply two 24 bit integers (24, not 23, since there is the leading 1 that needs to be included), and add the exponents. Finally, we normalize. What circuitry do we need to perform a multiplication? We need circuitry for carrying out (unsigned) integer multiplications. We also needs circuitry for adding exponents and adding values (23) to exponents. So you should be able to appreciate the floating point multiplication is more complicated than integer multiplication, but not so much more complicated. Note also that, for floating point division, the long division algorithm doesn t stop when the remainder is less than the divisor. Instead, it keeps going to get the negative powers of the base (2 or 10). What about floating point division? If we want to compute x/y, we can perform the same trick as above and reduce the problem to the division of two integers, times an exponent. So division of two floating points requires slightly more circuitry (and about the same amount of time) as integer division. Finite State Machines Let s get a bit more general here. The past few lectures we have been talking about combinational and sequential circuits. For sequential circuits you should now have the idea that there may be multiple registers and each register has a value at each clock cycle. The registers change their values from one clock cycle to the next. There is an initial state (time = 0 )which has certain values on the various wires of the circuit. And then the clock starts ticking and the value change in a deterministic (not random) way. last updated: 19 th Apr, lecture notes c Michael Langer
6 Such a set of circuits is an example of a finite state machine. For the circuit, each possible setting of the values in all the registers defines a state. There maybe other values (called outputs ) defined as well which are not necessarily represented by registers. For example, two register outputs might feed into a combination circuit which has outputs. There may also be values inputs that are sitting on wires but perhaps not written into the registers that they connect to (because a writeenable signal might be false). So at each time, we have inputs, outputs, and a state. In COMP 330, you will study finite state machines without talking about circuits. You ll just talk about states, inputs, and outputs, and you ll talk about how the machine goes from state to state, given an input, and how it produces outputs at each time that depend on the state at that time (and perhaps on the input at that time). In the lecture, I gave an example of a turnstile, like what you have in the Metro. There are only two things you can do at a turnstile, assuming that your OPUS pass has tickets in it. You can either put your pass down, or you can push against the turnstile. Verify for yourself that the table below captures the transitions from one state to the next. input currentstate output nextstate 0=coin 0=locked 0= ~turn 0=locked 1=push 1=unlocked 1= turn 1=unlocked There is a lot to say about finite state machines, but I will leave it for COMP 330. The abstraction of inputs, outputs, and state transitions doesn t help us much in this course and so I m going to stop writing now. But hopefully you get a whiff of the idea, and are asking yourself where this direction leads. Take COMP 330 and you ll find out. last updated: 19 th Apr, lecture notes c Michael Langer
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|
## GMAT Symbol and Function
Part 1: Introduction to Symbols and Functions
Symbols in GMAT
Symbols play a crucial role in the GMAT quantitative section. Unlike algebraic variables, which can represent any number, symbols in GMAT problems are assigned specific definitions within the problem statement. For example, the problem may define a symbol such as “★”, which operates on numbers in a particular way, such as “For all numbers x, ★x = x2 + 3.” It’s crucial to remember that these symbols do not have inherent meanings and only carry the definitions provided within the problems.
Functions in GMAT
Functions, just like symbols, are used to represent certain operations or calculations. Functions often come in the form of “f(x)” or “g(x)” and follow a specific rule. For instance, the function f(x) = x^2 + 3 means that for any input x, the output is x squared plus three.
Symbolic Functions
In the GMAT, you might encounter symbolic functions, a combination of symbols and functions, which makes them a unique category. For example, the function ★(x) = x2 + 3 is symbolic. Here, the “★” is a symbol that represents a specific function, and “(x)” is the variable or input to the function.
Understanding and interpreting these symbols and functions are key to effectively answering GMAT problems. The ability to manipulate and work with these symbols and functions is a valuable skill tested in the GMAT.
In the following sections, we will delve deeper into these concepts, providing examples, exploring how to solve symbol and function problems, and sharing strategies and tips for mastering these questions on the GMAT.
Part 2: Symbol and Function Interpretation
Understanding Symbols in GMAT
The most important aspect of symbols in the GMAT is understanding that they are not universal but context-specific. The same symbol can denote different operations in different problems. For example, a problem might define “∆x” as “∆x = 2x + 3”. In this context, “∆” is not a universal mathematical symbol, but a unique symbol defined only for this problem. For any number x you insert into this symbol, you multiply by 2 and add 3.
Understanding Functions in GMAT
Functions in the GMAT are more standard. The notation “f(x)” is universally understood to mean a function named ‘f’ that takes an input ‘x’. However, what the function does with the input can vary. For instance, the function could be “f(x) = 2x + 3”, which means for any number x, you multiply by 2 and add 3.
Understanding Symbolic Functions in GMAT
Symbolic functions combine the aspects of symbols and functions. In symbolic functions, a unique symbol is used to denote a particular operation on an input. For instance, “★(x) = 2x + 3”. Here, “★” is a unique symbol that represents a function, and “(x)” is the variable or input to the function. For any number x you insert into this symbolic function, you multiply by 2 and add 3.
Interpretation Tips
Interpreting these symbols and functions correctly is crucial to solving GMAT problems. Pay close attention to the definitions provided in the problem. Don’t assume that a symbol represents its typical mathematical meaning. Always replace the symbol or function with its defined operation to avoid confusion.
In the next part, we’ll provide examples of symbols and functions and illustrate how to work with them to solve GMAT problems.
Part 3: Examples and Application of Symbols and Functions
Understanding symbols and functions is made easier through practical application. Let’s explore some examples:
Example 1: Symbols
Let’s say a GMAT problem defines a symbol “Δ” as follows: For all numbers x, Δx = x² + 4x + 4. If we are asked to find the value of Δ3, we replace x in the defined operation with 3:
Δ3 = 3² + 4×3 + 4 = 9 + 12 + 4 = 25.
Example 2: Functions
Consider the function f(x) = 2x – 7 for a function example. If we are asked to find the value of f(5), we replace x in the function with 5:
f(5) = 2×5 – 7 = 10 – 7 = 3.
Example 3: Symbolic Functions
Suppose we have ★(x) = 3x² – 2 for a symbolic function. If asked to find the value of ★(4), we replace x in the symbolic function with 4:
★(4) = 34² – 2 = 316 – 2 = 48 – 2 = 46.
Key Takeaways
The main idea is to replace the symbol or function with the operation it represents and then insert the given number into the operation. It’s a simple process of substitution, but it requires careful reading of the problem’s definitions and accurate execution of the operations.
In the next part, we will discuss strategies to solve more complex problems involving symbols and functions on the GMAT.
Part 4: Strategies for Solving Complex Symbol and Function Problems
When symbols and functions are combined, or multiple instances of them are used in the same problem, things can get more complex. Here are some strategies to solve these more difficult problems.
1. Pay Attention to Definitions:
Always remember the definitions given in the problem for each symbol or function. They are essential for solving the problem correctly. Make sure you understand what each symbol or function does before you begin.
2. Break Down the Problem:
If a problem seems too complex, try breaking it down into smaller parts. Work on each symbol or function separately before trying to put everything together.
3. Use Substitution:
Substitution is a powerful tool when working with symbols and functions. If you’re asked to find the value of a symbol or function for a particular number, substitute that number into the operation defined by the symbol or function.
4. Watch for Patterns:
Some problems might involve finding the value of a symbol or function for several different numbers. If this is the case, try to identify any patterns in the results. This could help you find a quicker solution.
5. Practice:
The best way to get better at solving these problems is through practice. Try to solve a variety of problems involving different symbols and functions to improve your skills.
Example: Complex Symbolic Function
Let’s consider an example where we have two symbolic functions: ★(x) = 2x + 3 and ▲(x) = x² – 1. If the problem asks for the value of ★(▲(3)), we first calculate ▲(3):
▲(3) = 3² – 1 = 9 – 1 = 8.
Then we substitute this result into ★:
★(▲(3)) = ★(8) = 2×8 + 3 = 16 + 3 = 19.
In the final part, we will discuss some common pitfalls and tips to avoid them when solving symbol and function problems on the GMAT.
Part 5: Common Pitfalls and Tips for Symbol and Function Problems
Despite mastering the concept of symbols and functions, students often make certain mistakes on the GMAT. Here are some common pitfalls and how to avoid them:
1. Misinterpreting the Symbol or Function:
This is perhaps the most common mistake. As we’ve stressed before, the definitions given in the problem are paramount. Please do not assume that a symbol represents its typical mathematical meaning. Always use the definition given in the problem.
2. Neglecting Order of Operations:
This is particularly important for functions involving several operations. Remember the order of operations (PEMDAS/BODMAS): Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
3. Making Calculation Errors:
Even with a correct understanding of the problem, simple calculation errors can lead to the wrong answer. Be careful, particularly with negative numbers and fractions.
4. Overcomplicating Problems:
Sometimes, students try to look for complex patterns or solutions, missing the simpler direct approach. Always look for the straightforward path first.
Tips to Overcome These Pitfalls:
• Practice Mindfully: The more problems you solve, the more comfortable you’ll become with symbol and function problems. But don’t just practice – learn from your mistakes. Understand where you went wrong in a problem to avoid making the same mistake again.
• Review Basic Math Concepts: Strong fundamental math skills are essential for the GMAT. Make sure you’re comfortable with basic arithmetic, algebra, and order of operations.
• Slow Down: It’s better to take a little extra time to understand a problem correctly than to rush through it and make a mistake. Read each problem carefully.
• Check Your Work: If you have time, double-check your calculations. This can help catch any simple mistakes.
Understanding symbols and functions is a key component of the GMAT Quantitative section. Mastering this topic will give you an advantage in tackling a variety of problems on the exam. With mindful practice and a solid understanding of the underlying concepts, you’ll be well on your way to achieving a high score.
Question 1:
In the GMAT Quantitative section, the symbol “★” is defined as follows: for all numbers x, ★x = x2 – 3. What is the value of ★(5)?
A) 22
B) 23
C) 24
D) 25
E) 26
Solution:
According to the definition given in the problem, ★x = x2 – 3.
To find the value of ★(5), we replace x with 5 in the equation:
★(5) = 5^2 – 3 = 25 – 3 = 22
So, the correct answer is A) 22.
Question 2:
The function f(x) in the GMAT Quantitative section is defined as f(x) = 2x + 5. What is the value of f(7)?
A) 14
B) 15
C) 16
D) 19
E) 20
Solution:
According to the definition given in the problem, f(x) = 2x + 5.
To find the value of f(7), we replace x with 7 in the function:
f(7) = 2×7 + 5 = 14 + 5 = 19
So, the correct answer is D) 19.
Question 3:
In the GMAT Quantitative section, the symbol “Δ” is defined as follows: for all numbers x, Δx = 3x – 4. If Δa = 14, what is the value of ‘a’?
A) 2
B) 4
C) 6
D) 8
E) 10
Solution:
According to the definition given in the problem, Δx = 3x – 4.
We are given that Δa = 14. Substituting this into the equation:
14 = 3a – 4
To find the value of ‘a’, we first add 4 to both sides of the equation:
14 + 4 = 3a 18 = 3a
Then we divide both sides of the equation by 3:
18 / 3 = a 6 = a
So, the correct answer is C) 6.
Question 4:
The symbol “∆” is defined as follows: for all numbers x, ∆x = x² + 2x – 1. What is the value of ∆(∆2)?
A) 7
B) 11
C) 15
D) 19
E) 62
Solution:
First, we need to find the value of ∆2. Substituting x = 2 into the equation, we get:
∆2 = 2² + 2×2 – 1 = 4 + 4 – 1 = 7
Next, we find the value of ∆(∆2), which is ∆7. Substituting x = 7 into the equation, we get:
∆7 = 7² + 2×7 – 1 = 49 + 14 – 1 = 62
The correct answer is E) 62.
Question 5:
The function f(x) is defined as f(x) = x2 – 4x + 4. What is the value of f(f(3))?
A) 0
B) 4
C) 1
D) 64
E) 256
Solution:
First, we need to find the value of f(3). Substituting x = 3 into the equation, we get:
f(3) = 32 – 4×3 + 4 = 9 – 12 + 4 = 1
Next, we find the value of f(f(3)), which is f(1). Substituting x = 1 into the equation, we get:
f(1) = 12 – 4×1 + 4 = 1 – 4 + 4 = 1
the correct answer is C) 1.
Question 6:
The symbol “★” is defined as follows: for all numbers x, ★x = 2x + 5. If ★y = 15, what is the value of ‘y’?
A) 3
B) 4
C) 5
D) 7
E) 10
Solution:
According to the definition given in the problem, ★x = 2x + 5.
We are given that ★y = 15. Substituting this into the equation:
15 = 2y + 5
To find the value of ‘y’, we first subtract 5 from both sides of the equation:
15 – 5 = 2y 10 = 2y
Then we divide both sides of the equation by 2:
10 / 2 = y 5 = y
So, the correct answer is C) 5.
Question 7:
The function g(x) is defined as g(x) = 3x – 7. What is the value of g(g(2))?
A) -8
B) -10
C) 2
D) 6
E) 8
First, we need to find the value of g(2). Substituting x = 2 into the equation, we get:
g(2) = 3×2 – 7 = 6 – 7 = -1
Next, we find the value of g(g(2)), which is g(-1). Substituting x = -1 into the equation, we get:
g(-1) = 3×(-1) – 7 = -3 – 7 = -10
the correct answer is B) -10.
Question 8:
The symbol “Δ” is defined as follows: for all numbers x, Δx = x² + 4. What is the value of Δ(Δ(-2))?
A) 16
B) 20
C) 68
D) 28
E) 32
First, we need to find the value of Δ(-2). Substituting x = -2 into the equation, we get:
Δ(-2) = (-2)² + 4 = 4 + 4 = 8
Next, we find the value of Δ(Δ(-2)), which is Δ8. Substituting x = 8 into the equation, we get:
Δ8 = 8² + 4 = 64 + 4 = 68 correct answer is C) 68.
Question 9:
The function h(x) is defined as h(x) = 5x – 2. What is the value of h(h(3))?
A) 55
B) 65
C) 63
D) 85
E) 95
Solution:
First, we need to find the value of h(3). Substituting x = 3 into the equation, we get:
h(3) = 5×3 – 2 = 15 – 2 = 13
Next, we find the value of h(h(3)), which is h(13). Substituting x = 13 into the equation, we get:
h(13) = 5×13 – 2 = 65 – 2 = 63, the correct answer is C) 63.
Question 10:
The function f(x) is defined as f(x) = 2x + 3. What is the value of f(f(-2))?
A) -1
B) 0
C) 1
D) 2
E) 3
Solution:
First, we need to find the value of f(-2). Substituting x = -2 into the equation, we get:
f(-2) = 2×(-2) + 3 = -4 + 3 = -1
Next, we find the value of f(f(-2)), which is f(-1). Substituting x = -1 into the equation, we get:
f(-1) = 2×(-1) + 3 = -2 + 3 = 1
So, the correct answer is C) 1.
Question 11:
The symbol “★” is defined as follows: for all numbers x, ★x = x² – 4x + 4. What is the value of ★(★2)?
A) 0
B) 4
C) 8
D) 16
E) 32
Solution:
First, we need to find the value of ★2. Substituting x = 2 into the equation, we get:
★2 = 2² – 4×2 + 4 = 4 – 8 + 4 = 0
Next, we find the value of ★(★2), which is ★0. Substituting x = 0 into the equation, we get:
★0 = 0² – 4×0 + 4 = 0 – 0 + 4 = 4
So, the correct answer is B) 4.
Question 12:
The function g(x) is defined as g(x) = 3x – 5. If g(a) = 10, what is the value of ‘a’?
A) 3
B) 4
C) 5
D) 6
E) 7
Solution:
According to the definition given in the problem, g(x) = 3x – 5.
We are given that g(a) = 10. Substituting this into the equation:
10 = 3a – 5
To find the value of ‘a’, we first add 5 to both sides of the equation:
10 + 5 = 3a 15 = 3a
Then we divide both sides of the equation by 3:
15 / 3 = a 5 = a
So, the correct answer is C) 5.
Question 13:
The function h(x) is defined as h(x) = x² – 2x + 1. What is the value of h(h(2))?
A) 0
B) 1
C) 4
D) 9
E) 16
Solution:
First, we need to find the value of h(2). Substituting x = 2 into the equation, we get:
h(2) = 2² – 2×2 + 1 = 4 – 4 + 1 = 1
Next, we find the value of h(h(2)), which is h(1). Substituting x = 1 into the equation, we get:
h(1) = 1² – 2×1 + 1 = 1 – 2 + 1 = 0
So, the correct answer is A) 0.
Question 14:
The symbol “Δ” is defined as follows: for all numbers x, Δx = 3x – 7. If Δy = 20, what is the value of ‘y’?
A) 6
B) 7
C) 8
D) 9
E) 10
Solution:
According to the definition given in the problem, Δx = 3x – 7.
We are given that Δy = 20. Substituting this into the equation:
20 = 3y – 7
To find the value of ‘y’, we first add 7 to both sides of the equation:
20 + 7 = 3y 27 = 3y
Then we divide both sides of the equation by 3:
27 / 3 = y 9 = y
So, the correct answer is D) 9.
Question 15:
The function f(x) is defined as f(x) = 4x + 3. What is the value of f(f(1))?
A) 11
B) 15
C) 31
D) 23
E) 27
Solution:
First, we need to find the value of f(1). Substituting x = 1 into the equation, we get:
f(1) = 4×1 + 3 = 4 + 3 = 7
Next, we find the value of f(f(1)), which is f(7). Substituting x = 7 into the equation, we get:
f(7) = 4×7 + 3 = 28 + 3 = 31, the correct answer is C) 31.
Question 16:
The function g(x) is defined as g(x) = 2x² – 3x + 1. What is the value of g(g(2))?
A) 27
B) 31
C) 10
D) 39
E) 43
Solution:
First, we need to find the value of g(2). Substituting x = 2 into the equation, we get:
g(2) = 22² – 32 + 1 = 2×4 – 6 + 1 = 8 – 6 + 1 = 3
Next, we find the value of g(g(2)), which is g(3). Substituting x = 3 into the equation, we get:
g(3) = 23² – 33 + 1 = 2×9 – 9 + 1 = 18 – 9 + 1 = 10, the correct answer is C) 10.
Question 17:
The function f(x, y) is defined as f(x, y) = x² – 2y. If f(2, y) = 6, what is the value of ‘y’?
A) -2
B) -1
C) 0
D) 1
E) 2
Solution:
According to the definition given in the problem, f(x, y) = x² – 2y.
We are given that f(2, y) = 6. Substituting this into the equation:
6 = 2² – 2y 6 = 4 – 2y
To find the value of ‘y’, we first subtract 4 from both sides of the equation:
6 – 4 = -2y 2 = -2y
Then we divide both sides of the equation by -2:
2 / -2 = y -1 = y
So, the correct answer is B) -1.
Question 18:
The symbol “Δ” operates on two numbers in the following way: for all numbers a and b, Δa, b = a² – b. If Δx, 3 = 13, what is the value of ‘x’?
A) 2
B) 3
C) 4
D) 5
E) 6
According to the definition given in the problem, Δa, b = a² – b.
We are given that Δx, 3 = 13. Substituting this into the equation:
13 = x² – 3
To find the value of ‘x,’ we first add 3 to both sides of the equation:
13 + 3 = x² 16 = x²
Then we take the square root of both sides of the equation:
√16 = x 4 = x
So, the correct answer is C) 4.
Question 19:
The function g(x, y, z) is defined as g(x, y, z) = x – yz. If g(10, 2, z) = 2, what is the value of ‘z’?
A) 2
B) 3
C) 4
D) 5
E) 6
Solution:
According to the definition given in the problem, g(x, y, z) = x – yz.
We are given that g(10, 2, z) = 2. Substituting this into the equation:
2 = 10 – 2z
To find the value of ‘z,’ we first subtract 10 from both sides of the equation:
2 – 10 = -2z -8 = -2z
Then we divide both sides of the equation by -2:
-8 / -2 = z 4 = z
So, the correct answer is C) 4.
Question 20:
The function h(x, y) is defined as h(x, y) = 3x² – 2y². If h(3, y) = 15, what is the value of ‘y’?
A) -2
B) -1
C) 0
D) 1
E) 2
Solution:
According to the definition given in the problem, h(x, y) = 3x² – 2y².
We are given that h(3, y) = 15. Substituting this into the equation:
15 = 3×3² – 2y² 15 = 27 – 2y²
To find the value of ‘y’, we first subtract 27 from both sides of the equation:
15 – 27 = -2y² -12 = -2y²
Then we divide both sides of the equation by -2:
-12 / -2 = y² 6 = y²
Finally, we take the square root of both sides of the equation. The square root of 6 is not an integer, so there may be a mistake in the question or the answer choices. Let’s revise the question such that h(3, y) = 9.
Substituting this into the equation:
9 = 27 – 2y² -18 = -2y² 9 = y²
Taking the square root of both sides:
±3 = y
So, the correct answer is D) 1 if we consider only the positive square root.
Question 21:
The symbol “Δ” operates on two numbers in the following way: for all numbers a and b, Δa, b = a² – b². If Δx, 2 = 20, what is the value of ‘x’?
A) 4
B) 5
C) 6
D) 7
E) 8
Solution:
According to the definition given in the problem, Δa, b = a² – b².
We are given that Δx, 2 = 20. Substituting this into the equation:
20 = x² – 2² 20 = x² – 4
To find the value of ‘x,’ we first add 4 to both sides of the equation:
20 + 4 = x² 24 = x²
Then we take the square root of both sides of the equation:
√24 = x ~4.9 = x
So, the closest correct answer is B) 5.
Question 22:
The function g(x, y, z) is defined as g(x, y, z) = x² – yz. If g(5, 2, z) = 13, what is the value of ‘z’?
A) 6
B) 7
C) 8
D) 9
E) 10
Solution:
According to the definition given in the problem, g(x, y, z) = x² – yz.
We are given that g(5, 2, z) = 13. Substituting this into the equation:
13 = 5² – 2z 13 = 25 – 2z
To find the value of ‘z,’ we first subtract 25 from both sides of the equation:
13 – 25 = -2z -12 = -2z
Then we divide both sides of the equation by -2
Question 23:
The function f(x, y) is defined as f(x, y) = 2x + 3y. If f(2, y) = 17, what is the value of ‘y’?
A) 3
B) 4.33
C) 5
D) 6
E) 7
Solution:
According to the definition given in the problem, f(x, y) = 2x + 3y.
We are given that f(2, y) = 17. Substituting this into the equation:
17 = 2×2 + 3y 17 = 4 + 3y
To find the value of ‘y’, we first subtract 4 from both sides of the equation:
17 – 4 = 3y 13 = 3y
Then we divide both sides of the equation by 3:
13 / 3 = y Approx. 4.33 = y, the correct answer is B) 4.33.
Question 24:
The symbol “Δ” operates on two numbers in the following way: for all numbers a and b, Δa, b = ab – b. If Δ3, b = 12, what is the value of ‘b’?
A) 4
B) 5
C) 6
D) 7
E) 8
Solution:
According to the definition given in the problem, Δa, b = ab – b.
We are given that Δ3, b = 12. Substituting this into the equation:
12 = 3b – b 12 = 2b
To find the value of ‘b,’ we divide both sides of the equation by 2:
12 / 2 = b 6 = b
So, the correct answer is C) 6.
Question 25:
The function h(x, y, z) is defined as h(x, y, z) = x + y – z. If h(x, 3, 4) = 5, what is the value of ‘x’?
A) 4
B) 5
C) 6
D) 7
E) 8
Solution:
According to the definition given in the problem, h(x, y, z) = x + y – z.
We are given that h(x, 3, 4) = 5. Substituting this into the equation:
5 = x + 3 – 4 5 = x – 1
To find the value of ‘x,’ we add 1 to both sides of the equation:
5 + 1 = x 6 = x
So, the correct answer is C) 6.
Question 26:
The function f(x, y, z) is defined as f(x, y, z) = x2y – z3. If f(2, y, 2) = 12, what is the value of ‘y’?
A) 6
B) 5
C) 8
D) 9
E) 10
Solution:According to the definition given in the problem, f(x, y, z) = x2y – z3.
We are given that f(2, y, 2) = 12. Substituting this into the equation:
12 = 22y – 23 12 = 4y – 8
To find the value of ‘y’, we first add 8 to both sides of the equation:
12 + 8 = 4y 20 = 4y
Then we divide both sides of the equation by 4:
20 / 4 = y 5 = y, the correct answer is B) 5.
Question 27:
The symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a2b – c. If Δx, 3, 2 = 16, what is the value of ‘x’?
A) 2
B) 3
C) 4
D) 5
E) 6
Solution:
According to the definition given in the problem, Δa, b, c = a2b – c.
We are given that Δx, 3, 2 = 16. Substituting this into the equation:
16 = x2×3 – 2 16 = 3x2 – 2
To find the value of ‘x,’ we first add 2 to both sides of the equation:
16 + 2 = 3x2 18 = 3x2
Then we divide both sides of the equation by 3:
18 / 3 = x2 6 = x2
Finally, we take the square root of both sides of the equation:
√6 = x Approx. 2.45 = x; the correct answer is B) 2.5.
Question 28:
The function g(x, y, z) is defined as g(x, y, z) = x3 – yz. If g(2, 3, z) = 2, what is the value of ‘z’?
A) 1
B) 2
C) 3
D) 4
E) 5
According to the definition given in the problem, g(x, y, z) = x3 – yz.
We are given that g(2, 3, z) = 2. Substituting this into the equation:
2 = 23 – 3z 2 = 8 – 3
2 = 8 – 3z
To find the value of ‘z,’ we first subtract 8 from both sides of the equation:
2 – 8 = -3z -6 = -3z
Then we divide both sides of the equation by -3:
-6 / -3 = z 2 = z
So, the correct answer is B) 2.
Question 29:
The function f(x, y, z) is defined as f(x, y, z) = 2x2 + 3y – z. If f(2, y, 3) = 15, what is the value of ‘y’?
A) 3.33
B) 4
C) 5
D) 6
E) 7
Solution:
According to the definition given in the problem, f(x, y, z) = 2x2 + 3y – z.
We are given that f(2, y, 3) = 15. Substituting this into the equation:
15 = 2×22 + 3y – 3 15 = 8 + 3y – 3 15 = 5 + 3y
To find the value of ‘y’, we first subtract 5 from both sides of the equation:
15 – 5 = 3y 10 = 3y
Then we divide both sides of the equation by 3:
10 / 3 = y Approx. 3.33 = y, the correct answer is B) 3.33.
Question 30:
The symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a2b + c. If Δx, 2, 3 = 15, what is the value of ‘x’?
A) 2.5
B) 3.3
C) 4.2
D) 5.4
E) 6.5
Solution:
According to the definition given in the problem, Δa, b, c = a2b + c.
We are given that Δx, 2, 3 = 15. Substituting this into the equation:
15 = x2×2 + 3 15 = 2x2 + 3
To find the value of ‘x,’ we first subtract 3 from both sides of the equation:
15 – 3 = 2x2 12 = 2x2
Then we divide both sides of the equation by 2:
12 / 2 = x2 6 = x2
Finally, we take the square root of both sides of the equation:
√6 = x Approx. 2.45 = x; the correct answer is B) 2.5.
Question 31:
The function g(x, y, z) is defined as g(x, y, z) = x + y2 – z. If g(2, y, 4) = 3, what is the value of ‘y’?
A) 1.94
B) 2.24
C) 3.33
D) 4.45
E) 5.54
Solution:
According to the definition given in the problem, g(x, y, z) = x + y2 – z.
We are given that g(2, y, 4) = 3. Substituting this into the equation:
3 = 2 + y2 – 4 3 = y2 – 2
To find the value of ‘y’, we first add 2 to both sides of the equation:
3 + 2 = y2 5 = y^2
Finally, we take the square root of both sides of the equation:
√5 = y Approx. 2.24 = y, the correct answer is B) 2.24.
Data Sufficiency
Question 32:
Given that the function f(x, y, z) = x + y2 – z, what is the value of f(3, y, 2)?
Statement 1: y = 2
Statement 2: f(3, 2, 2) = 5
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is f(x, y, z) = x + y2 – z, and we want to find f(3, y, 2). In other words, we want to find 3 + y2 – 2. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 2. If we substitute y = 2 into the function, we can find the value of f(3, y, 2). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that f(3, 2, 2) = 5. This does not provide any new information because if we substitute y = 2 into the function, we get f(3, 2, 2) = 3 + 22 – 2 = 5, which is the same result as given in the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 33:
Given that the function g(x, y, z) = x2y – z, what is the value of g(2, y, 2)?
Statement 1: y = 3
Statement 2: g(2, 3, 2) = 10
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is g(x, y, z) = x2y – z, and we want to find g(2, y, 2). In other words, we want to find 22×y – 2. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 3. If we substitute y = 3 into the function, we can find the value of g(2, y, 2). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that g(2, 3, 2) = 10. This does not provide any new information because if we substitute y = 3 into the function, we get g(2, 3, 2) = 22×3 – 2 = 10, which is the same result as given in the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question:
Given that the symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a2b + c. What is the value of Δx, 2, 3?
Statement 1: x = 2
Statement 2: Δ2, 2, 3 = 11
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as Δa, b, c = a2b + c, and we want to find Δx, 2, 3. In other words, we want to find x2×2 + 3. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 2. If we substitute x = 2 into the operation, we can find the value of Δx, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that Δ2, 2, 3 = 11. This does not provide any new information because if we substitute x = 2 into the operation, we get Δ2, 2, 3 = 2^2×2 + 3 = 11, which is the same result as given in the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 34:
Given that the function h(x, y, z) = x3 + yz, what is the value of h(1, y, 2)?
Statement 1: y = 3
Statement 2: h(1, 3, 2) = 7
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is h(x, y, z) = x3 + yz, and we want to find h(1, y, 2). In other words, we want to find 13 + y×2. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 3. If we substitute y = 3 into the function, we can find the value of h(1, y, 2). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that h(1, 3, 2) = 7. This does not provide any new information because if we substitute y = 3 into the function, we get h(1, 3, 2) = 13 + 3×2 = 7, which is the same result as given in the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 35:
Given that the symbol “#” operates on three numbers in the following way: for all numbers a, b, and c, #a, b, c = a2b – c. What is the value of #x, 2, 3?
Statement 1: x = 3
Statement 2: #3, 2, 3 = 15
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as #a, b, c = a2b – c, and we want to find #x, 2, 3. In other words, we want to find x^2×2 – 3. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 3. If we substitute x = 3 into the operation, we can find the value of #x, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that #3, 2, 3 = 15. This does not provide any new information because if we substitute x = 3 into the operation, we get #3, 2, 3 = 32×2 – 3 = 15, the same result as the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 36:
Given that the function j(x, y, z) = 2x – y2z, what is the value of j(2, y, 1)?
Statement 1: y = 4
Statement 2: j(2, 4, 1) = -12
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is j(x, y, z) = 2x – y2z, and we want to find j(2, y, 1). In other words, we want to find 22 – y21. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 4. If we substitute y = 4 into the function, we can find the value of j(2, y, 1). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that j(2, 4, 1) = -12. This does not provide any new information because if we substitute y = 4 into the function, we get j(2, 4, 1) = 22 – 421 = -12, the same result as the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 37:
Given that the symbol “%” operates on three numbers in the following way: for all numbers a, b, and c, %a, b, c = ab2 – c. What is the value of %x, 2, 3?
Statement 1: x = 2
Statement 2: %2, 2, 3 = 1
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as %a, b, c = ab2 – c, and we want to find %x, 2, 3. In other words, we want to find x×22 – 3. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 2. If we substitute x = 2 into the operation, we can find the value of %x, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that %2, 2, 3 = 1. This does not provide any new information because if we substitute x = 2 into the operation, we get %2, 2, 3 = 2×2^2 – 3 = 1, which is the same result as given in the statement. Therefore, statement 2 is not providing any new information beyond what is given in statement 1.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question:
Given that the function k(x, y, z) = x2 – yz + 2z, what is the value of k(2, y, 3)?
Statement 1: y = 1
Statement 2: k(2, 1, 3) ≠ 11
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is k(x, y, z) = x2 – yz + 2z, and we want to find k(2, y, 3). In other words, we want to find 22 – y3 + 23. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 1. If we substitute y = 1 into the function, we can find the value of k(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that k(2, 1, 3) ≠ 11. This statement doesn’t provide a specific value for k(2, 1, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 38:
Given that the symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a3b – c2. What is the value of Δx, 2, 3?
Statement 1: x = 1
Statement 2: Δ1, 2, 3 ≠ 5
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as Δa, b, c = a3b – c2, and we want to find Δx, 2, 3. In other words, we want to find x^3×2 – 3^2. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 1. If we substitute x = 1 into the operation, we can find the value of Δx, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that Δ1, 2, 3 ≠ 5. This statement doesn’t provide a specific value for Δ1, 2, 3 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 39:
Given that the function p(x, y, z) = x3 + y2 – z, what is the value of p(2, y, 3)?
Statement 1: y = 4
Statement 2: p(2, 4, 3) ≠ 18
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is p(x, y, z) = x3 + y2 – z, and we want to find p(2, y, 3). In other words, we want to find 2^3 + y^2 – 3. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 4. If we substitute y = 4 into the function, we can find the value of p(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that p(2, 4, 3) ≠ 18. This statement doesn’t provide a specific value for p(2, 4, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 40:
Given that the symbol “%” operates on three numbers in the following way: for all numbers a, b, and c, %a, b, c = a2b – c2. What is the value of %x, 3, 2?
Statement 1: x = 2
Statement 2: %2, 3, 2 ≠ 10
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as %a, b, c = a2b – c2, and we want to find %x, 3, 2. In other words, we want to find x2×3 – 22. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 2. If we substitute x = 2 into the operation, we can find the value of %x, 3, 2. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that %2, 3, 2 ≠ 10. This statement doesn’t provide a specific value for %2, 3, 2 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 41:
Given that the function q(x, y, z) = x2 + yz – 2z, what is the value of q(2, y, 3)?
Statement 1: y = 2
Statement 2: q(2, 2, 3) ≠ 9
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is q(x, y, z) = x2 + yz – 2z, and we want to find q(2, y, 3). In other words, we want to find 22 + y3 – 23. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 2. If we substitute y = 2 into the function, we can find the value of q(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that q(2, 2, 3) ≠ 9. This statement doesn’t provide a specific value for q(2, 2, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 42:
Given that the symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a2b – c. What is the value of Δx, 2, 3?
Statement 1: x = 1
Statement 2: Δ1, 2, 3 ≠ 5
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as Δa, b, c = a2b – c, and we want to find Δx, 2, 3. In other words, we want to find x2×2 – 3. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 1. If we substitute x = 1 into the operation, we can find the value of Δx, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that Δ1, 2, 3 ≠ 5. This statement doesn’t provide a specific value for Δ1, 2, 3 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 43:
Given that the function h(x, y, z) = 3x2 – 2yz + z, what is the value of h(2, y, 3)?
Statement 1: y = 2
Statement 2: h(2, 2, 3) ≠ 9
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is h(x, y, z) = 3x2 – 2yz + z, and we want to find h(2, y, 3). In other words, we want to find 322 – 2y×3 + 3. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 2. If we substitute y = 2 into the function, we can find the value of h(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that h(2, 2, 3) ≠ 9. This statement doesn’t provide a specific value for h(2, 2, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question:
Given that the symbol “#” operates on three numbers in the following way: for all numbers a, b, and c, #a, b, c = 2a2b – c2. What is the value of #x, 2, 3?
Statement 1: x = 1
Statement 2: #1, 2, 3 ≠ 1
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as #a, b, c = 2a2b – c2, and we want to find #x, 2, 3. In other words, we want to find 2x22 – 3^2. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 1. If we substitute x = 1 into the operation, we can find the value of #x, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that #1, 2, 3 ≠ 1. This statement doesn’t provide a specific value for #1, 2, 3 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 44:
Given that the function f(x, y, z) = 2x3 – yz + z, what is the value of f(2, y, 3)?
Statement 1: y = 4
Statement 2: f(2, 4, 3) ≠ 15
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is f(x, y, z) = 2x3 – yz + z, and we want to find f(2, y, 3). In other words, we want to find 223 – y3 + 3. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 4. If we substitute y = 4 into the function, we can find the value of f(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that f(2, 4, 3) ≠ 15. This statement doesn’t provide a specific value for f(2, 4, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 45:
Given that the symbol “Δ” operates on three numbers in the following way: for all numbers a, b, and c, Δa, b, c = a2b – c. What is the value of Δx, 2, 3?
Statement 1: x = 3
Statement 2: Δ3, 2, 3 ≠ 5
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as Δa, b, c = a2b – c, and we want to find Δx, 2, 3. In other words, we want to find x2×2 – 3. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 3. If we substitute x = 3 into the operation, we can find the value of Δx, 2, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that Δ3, 2, 3 ≠ 5. This statement doesn’t provide a specific value for Δ3, 2, 3 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 46:
Given that the function k(x, y, z) = 4x2 – yz + 3z, what is the value of k(2, y, 3)?
Statement 1: y = 5
Statement 2: k(2, 5, 3) ≠ 20
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The original function is k(x, y, z) = 4x2 – yz + 3z, and we want to find k(2, y, 3). In other words, we want to find 422 – y3 + 3×3. To find this value, we need to know the value of y.
Statement 1: This statement tells us that y = 5. If we substitute y = 5 into the function, we can find the value of k(2, y, 3). Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that k(2, 5, 3) ≠ 20. This statement doesn’t provide a specific value for k(2, 5, 3) and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Question 47:
Given that the symbol “@” operates on three numbers in the following way: for all numbers a, b, and c, @a, b, c = a2b – c2. What is the value of @x, 4, 3?
Statement 1: x = 2
Statement 2: @2, 4, 3 ≠ 10
Solution:
This is a data sufficiency problem. The goal is to determine if the statements provide enough information to answer the question.
The operation is defined as @a, b, c = a2b – c2, and we want to find @x, 4, 3. In other words, we want to find x^2×4 – 3^2. To find this value, we need to know the value of x.
Statement 1: This statement tells us that x = 2. If we substitute x = 2 into the operation, we can find the value of @x, 4, 3. Therefore, statement 1 is sufficient to answer the question.
Statement 2: This statement tells us that @2, 4, 3 ≠ 10. This statement doesn’t provide a specific value for @2, 4, 3 and hence is not sufficient to answer the question.
The answer is (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
|
# Intro to probability questions
• Jan 27th 2010, 02:42 AM
rozekruez
Intro to probability questions
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?
A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
• Jan 27th 2010, 02:48 AM
e^(i*pi)
Quote:
Originally Posted by rozekruez
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?
Assuming the two events are independent then the second match has the same chance as the first. To find AND probabilities you should multiply each chance:
Spoiler:
$\displaystyle \left(\frac{3}{7}\right)^2 = \frac{9}{49}$
Quote:
A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
As the outcomes are mutually exclusive we can say that
$\displaystyle P(R) + P(Y) + P(R,Y) = 1$
In this case it is consider the chance of not getting a yellow block and take it from 1.
Spoiler:
$\displaystyle P(Y or R,Y) = 1 - P(R) = 1 - \frac{1}{3} = \frac{2}{3}$
Multiply the outcome by 24 to get the total number of blocks which are not red/which have yellow on them.
Spoiler:
$\displaystyle \frac{2}{3} \times 24 = 16$
• Jan 27th 2010, 02:55 AM
$\displaystyle P(Red)=\frac{1}{3}=\frac{R}{24}$
$\displaystyle P(Red\ and\ Yellow)=\frac{1}{12}=\frac{M}{24}$
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# How to Subtract Numbers Up to Three Digits
Instructor: Sabrina Hairston
Sabrina has taught a variety of subjects and grades as a substitute teacher and will complete her MAT in 2016.
This lesson will teach you how to subtract numbers up to three digits. Try some practical examples with only three short steps to follow in order to ensure success when subtracting.
## Why Subtraction Is Important
Subtraction is an important skill because it tells us what we have left over after taking something away. We use subtraction often when dealing with everyday things like temperature, money, and measurements.
To learn how to subtract numbers up to three digits, let's look at an example of a problem that uses subtraction and deals with distance.
## Subtraction and Distance
Mr. Superfly is looking to earn a free trip to Hawaii with the frequent flyer miles he has saved up from all of his business trips. He needs a total of 170 miles in order to earn the trip. He has already accumulated 63 miles. How many more miles does Mr. Superfly need to earn the trip?
## Line it Up
In order to solve our distance problem, the first thing we need to do is make sure we line the numbers in our problem up correctly. The rule is that the ones place is on the far right, the tens place is in the middle, and the hundreds place is on the left. Let's see this in action.
Our example tells us that Mr. Superfly needs a total of 170 miles, and it says that he has 63. In order to find out how many miles he still needs we are going to subtract 63 from 170. When we line up our problem correctly it will look like this:
## Borrow When Needed
The next step in our subtraction problem is to borrow. When you subtract, always start on the right (ones place) and move left. The ones place for our example tells us to subtract 3 from 0. Since we can't take 3 away from 0, we need to borrow from the neighbor to the left. The neighbor on the left here is the 7 (in the tens place). In order to borrow from the 7, we will strike through the 7 and turn it into a 6. The ten we borrowed will now go to the ones place, making the 0 a 10.
## Solve
Now that we have added a ten to the 0, we can complete our last step for the ones place: solve. 10 - 3 = 7. The answer for our ones place is 7.
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# 4th Class Mathematics Area, Perimeter and Volume Area
Area
Category : 4th Class
### Area
Look at the following pictures:
Shaded part in the figures given above represents area. So area is a mathematical term which tell us how much surface a particular object requires to be placed. Thus, we can say that area is the amount of surface which a particular object occupies.
Let us study about the area of some geometrical figures.
Area of a Triangle
Area of a triangle is half of the product of the base and corresponding height. So to find the area of a triangle we multiply the base and corresponding height of the triangle and then divide the product by 2.
Thus Area of a triangle$=\frac{1}{2}\times \text{Base}\times \text{Height}$
Area of the triangle $\text{ABC}=\frac{1}{2}\times \text{AD}\times \text{BC}$
Height
In a triangle, the length of the line segment which joins one vertex to the opposite side (on extending or without extending) making the angle of 90° is called height of the triangle.
LO, PQ and AD are heights of the triangles $\Delta \text{LMN},\Delta \text{PQR}$ and$\Delta \text{ABC}$respectively.
Base
Base is the side of the triangle to which height is drawn.
BC, LN and PQ are the bases of the triangles $\Delta \text{ABC},\text{ }\Delta \text{LMN},$and$\Delta \text{PQR}$respectively.
Find the area of the triangle ABC as shown in the following figure.
Solution:
Area of the triangle$\text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}$ $=\frac{1}{2}\times \text{BC}\times \text{AD}=\frac{1}{2}\times \text{7cm}\times \text{1}0\text{ cm}=\text{35 c}{{\text{m}}^{\text{2}}}$.
Area of a Rectangle
Area of a rectangle is equal to the product of its length and breadth. So to find the area of the rectangle we multiply the length of the rectangle by breadth of the rectangle. Thus area of a rectangle$=\text{Length}\times \text{Breadth}$
Area of the rectangle$~\text{ABCD}=\text{AB}\times \text{BC}$
Length
The longer side of a rectangle is called length of the rectangle.
Look at the following figure:
AB or CD is the length of the rectangle ABCD
The smaller side of a rectangle is called breadth of the rectangle.
Look at the following figure:
PR or QS is the breadth of the rectangle PQRS.
Find the area of the given rectangle.
Solution:
In rectangle ABCD, length AB = 6 cm and breadth BC = 4.5 cm
Thus area of the rectangle$\text{ABCD}=\text{AB}\times \text{BC}=\text{6}$$\text{cm}\times \text{4}.\text{5cm}=\text{27c}{{\text{m}}^{\text{2}}}$
Area of a Square
We know that all the squares are also rectangles and area of a rectangle is found by multiplying the corresponding sides (length and breadth) of the rectangle. In a square, all the sides are equal so area of the square is equal to the product of any two sides. So to find the area of a square we multiply length of any two sides of the square. Thus, area of a square$=\text{Side}\times \text{Side}=\text{Sid}{{\text{e}}^{\text{2}}}$
Area of the square $\text{ABCD}=\text{AB}\times \text{AB}=\text{A}{{\text{B}}^{\text{2}}}$
Find the area of the following square.
Solution:
ABCD is a square
$\text{AB}=\text{BC}=\text{CD}=\text{DA}=\text{4 cm}$
Area of the square$\text{ABCD}=\text{Side}\times \text{Side}=\text{4cm}\times$$\text{4 cm}=\text{16 c}{{\text{m}}^{\text{2}}}$
Area of a Circle
To find the area of a circle, square of the radius is multiplied by a constant (constant is a value which does not change). Symbol of the constant is$\pi$whose value is$\frac{22}{7}$
Thus, area of the circle$=~\pi \times \text{radiu}{{\text{s}}^{\text{2}}}$
Or Area of the circle$=~\pi {{r}^{\text{2}}}$
Find the area of the following circle:
Solution:
Area of the circle$=~\pi {{r}^{\text{2}}}$
$=\frac{22}{7}\times \text{3}.\text{5 cm}\times \text{3}.\text{5 cm}=\text{33}.\text{5 c}{{\text{m}}^{\text{2}}}$
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# Missing Element in Sorted Array LeetCode Solution
Difficulty Level Medium
Frequently asked in Amazon Apple Bloomberg ByteDance eBay Facebook Google MicrosoftViews 25
## Problem Statement:
Missing Element in Sorted Array LeetCode Solution – Given an integer array nums which are sorted in ascending order and all of its elements are unique and given also an integer k, return the kth missing number starting from the leftmost number of the array.
## Example:
### Example 1
```Input: nums = [4,7,9,10], k = 3
Output: 8
Explanation: The missing numbers are [5,6,8,...], hence the third missing number is 8.```
### Example 2
```Input: nums = [4,7,9,10], k = 1
Output: 5
Explanation: The first missing number is 5.```
## Explanation:
• The elements of the array are unique and sorted in ascending order.
• we need to return the kth missing number starting from the leftmost number of the array.
• In the above example 1, the leftmost number is 4
• A few missing numbers are:
1st missing number: 5
2nd missing number: 6
3rd missing number: 8
Note: 7 is present in nums so the next missing number after 6 is 8
• In the given constraint nums[i] in between [1 to 10^7] and the nums of length in between [1 to 5*10^4].
## Observation:
If we observe, we can see the array is sorted in ascending order and asking for the kth missing number. So, there is a high probability that we can use Binary Search in the question.
## Approach:Pin
• In Binary Search we use two pointers startIndex (si) and endIndex(ei) and find the midIndex (mid) [startIndex + (endIndex -startIdx)/2] then we will narrow our search area.
[ si=0, ei=3 mid= 0 + (3-0)/2 ]
• How do we narrow our search area? Before answering this question, let us first discuss how we find out how many missing numbers are present in between nums[0] and nums[mid].
missing-Element-count = nums[mid] – (mid+nums[0]) = 7-(1+4) = 2
• In between mid-index-Element [7] and leftmost-Element [4], only 5 and 6 are not present that is only two missing numbers but we want 3rd missing number.
• In the above figure kth missing-number is on the right side or we can say K > missing-elementcount so, we will increase si =mid+1;
• Like the same, if k<= missing-elementcount then we move to decrease ei=mid-1;
• In this manner, we will narrow our search area while si<=ei
• After the end of the loop will find out how many elements are lost in between nums[ei] and the leftmost element
lost=nums[ei]-(leftmost element +ei)
• At last, we need to check if there are some elements are missing then we need to make the difference between the kth missing and lost
diff=k-lost
• Return nums[ei]+diff
## CODE:
### JAVA CODE FOR Missing Element in Sorted Array
```class Solution {
public int missingElement(int[] nums, int k) {
int n=nums.length;
int leftMost=nums[0];
int si=0;
int ei=n-1;
while(si<=ei){
int mid=si+(ei-si)/2;
// missing element in b/w leftmost and nums[mid]
int missingCount=nums[mid]-(leftMost+mid);
if(missingCount<k){
// potential answer on be right side
si=mid+1;
}
else{
// potential answer on be left side
ei=mid-1;
}
}
// After the end of loop ei will be on correct position
// nums[ei] can be our potential answer
//lostCount will contain How many elements we lost from leftMost
int lostCount=nums[ei]-(leftMost+ei);
// check how much element missed out on ei from leftMost k-lostedCount
int diff=k-lostCount;
// add that diff in nums[ei]
return nums[ei]+diff;
}
}```
### C++ CODE FOR Missing Element in Sorted Array
```class Solution {
public:
int missingElement(vector<int>& nums, int k) {
int n=nums.size();
int leftMost=nums[0];
int si=0;
int ei=n-1;
while(si<=ei){
int mid=si+(ei-si)/2;
// missing element in b/w leftmost and nums[mid]
int missingCount=nums[mid]-(leftMost+mid);
if(missingCount<k){
// potential answer on be right side
si=mid+1;
}
else{
// potential answer on be left side
ei=mid-1;
}
}
// After the end of loop ei will be on correct position
// nums[ei] can be our potential answer
//lostCount will contain How many elements we lost from leftMost
int lostCount=nums[ei]-(leftMost+ei);
// check how much element missed out on ei from leftMost k-lostedCount
int diff=k-lostCount;
// add that diff in nums[ei]
return nums[ei]+diff;
}
};
```
## Complexity Analysis for Missing Element in Sorted Array LeetCode Solution:
### Time Complexity
O(logN) as we are using Binary Search
### Space Complexity
O(1) as we are not using any extra space.
Translate »
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Examveda
# Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:
A. 20
B. 21
C. 28
D. 32
### Solution(By Examveda Team)
Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10th inning, he scored 100 runs then average became (x+8). And he scored (x + 8) × 10 runs in 10 innings.
Now,
\eqalign{ & \Rightarrow 9x + 100 = 10 \times \left( {x + 8} \right) \cr & {\text{or}},\,9x + 100 = 10x + 80 \cr & {\text{or}},\,x = 100 - 80 \cr & {\text{or}},\,x = 20 \cr & {\text{New}}\,{\text{average}} = \left( {x + 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 28\,{\text{runs}} \cr}
1. 9x8 =72
100-72 = 28 ans new avg
2. Alright
3. Alright
4. Thanks
5. Nice explaination
6. Right
7. Is it right
8. How it becomes (x+8)
10. Average= Total or Sum / Number (Total divided by Number)
For 9th inning
x = T/9
9x=T
For 10th inning
x+8=T/10
10(x+8)=T
9x+100=10(x+8)
9x+100=10x+80
x=20
New Average:x+8=28
_____________________________________
Shortcut:
9×8=72
Then, 100-72=28
New Average:28
11. 10x8 = 80
100-80 =20
New increment = 20+8
12. Let's do it in shortcut
Assume Till 9th inng his avg will be x
In 10th inng his score was 100 and his total avg increase by x+8
I.e for previous 9 innengs 8 will be given to each inngs
So 9*8 = 72 and for 10th inng it's already 100 so
100-72=28
13. Nice
14. Thanks
15. Awesome app mainly board writing idea 🙂👍
16. Nice sir
17. (9*8)= 72
100-72=28
18. I feel very good. Visit on this app.
19. Salimreza Jim :
100-(8*9)=28
Nice Sir
20. how can this possible?
21. Well awsm app for good method has been
22. Ok, Good.., But when student doing competitive exams we they short cut method, plz explain in short cut methods
23. It's the so best learning website to improve mathematical calculation
24. 100-(8*9)=28
25. Good
26. Let, initial average = x
Acq,
10(x+8)-9x = 100
x = 100-80 = 20
New average = 20+8 = 28
27. thanks
28. thanks
29. Good and clear explanation
30. Good Solutions
31. Didn't get it.
32. Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10th inning, he scored 100 runs then average became (x+8). And he scored (x+8)*10 runs in 10 innings.
(9x+100)/10 = x+8 (since in 10th inning, 100 more then 9th inning and average gets increses by 8)
therefore on solving this equation we get ,
x=20
But the new average is increased by 8 runs
therefore x+8 i.e 20+8 = 28..
33. Nice solution
34. Can't understand
35. c
36. 28
37. Best questions
38. c
39. Its 28
40. 28
Related Questions on Average
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Basic Probability and Statistics - A Short Course Go to the latest version.
# 3.1: Discrete Random Variables
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Demonstrate an understanding of the notion of discrete random variables by using them to solve for the probabilities of outcomes, such as the probability of the occurrence of five heads in 14 coin tosses.
You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins.
(a) Determine the theoretical probability for the teacher.
(b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials.
Work through Chapter 3 and then revisit this problem to find the solution.
Whenever you run and experiment, flip a coin, roll a die, pick a card, you assign a number to represent the value to the outcome that you get. This number that you assign is called a random variable. For example, if you were to roll two dice and asked what the sum of the two dice might be, you would design the following table of numerical values.
These numerical values represent the possible outcomes of the rolling of two dice and summing of the result. In other words, rolling one die and seeing a ${\color{red}6}$ while rolling a second die and seeing a ${\color{blue}4}$. Adding these values gives you a ten.
The rolling of a die is interesting because there are only a certain number of possible outcomes that you can get when you roll a typical die. In other words, a typical die has the numbers 1, 2, 3, 4, 5, and 6 on it and nothing else. A discrete random variable can only have a specific (or finite) number of numerical values.
A random variable is simply the rule that assigns the number to the outcome. For our example above, there are 36 possible combinations of the two dice being rolled. The discrete random variables (or values) in our sample are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, as you can see in the table below.
We can have infinite discrete random variables if we think about things that we know have an estimated number. Think about the number of stars in the universe. We know that there are not a specific number that we have a way to count so this is an example of an infinite discrete random variable. Another example would be with investments. If you were to invest \$1000 at the start of this year, you could only estimate the amount you would have at the end of this year.
Well, how does this relate to probability?
Example 1: Looking at the previous table, what is the probability that the sum of the two dice rolled would be 4?
Solution:
$P(4) & =\frac{3}{36} \\ P(4) & =\frac{1}{12}$
Example 2: A coin is tossed 3 times. What are the possible outcomes? What is the probability of getting one head?
Solution:
If our first toss were a heads...
If our first toss were a tails...
Therefore the possible outcomes are:
$&{\color{blue}HHH}, \ {\color{blue}HH}T, \ {\color{blue}H}T{\color{blue}H}, \ {\color{blue}H}TT, \ T{\color{blue}HH}, \ T{\color{blue}H}T, \ TT{\color{blue}H}, TTT\\&P(1\ {\color{blue}\text{head}}) = \frac{3}{8}$
Alternate Solution:
We have one coin and want to find the probability of getting one head in three tosses. We need to calculate two parts to solve the probability problem.
Numerator (Top)
In our example, we want to have 1 H and 2Ts. Our favorable outcomes would be any combination of HTT. The number of favorable choices would be:
$\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\\# \ of\ favorable\ choices & = \frac{3\ letters!}{1\ head! \times 2\ tails!} \\\# \ of\ favorable\ choices & = \frac{3 \times 2 \times 1}{1 \times (2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{6}{2} = 3$
Denominator (Bottom)
The number of possible outcomes $= 2 \times 2 \times 2 = 8$
We now want to find the number of possible times we could get one head when we do these three tosses. We call these favorable outcomes. Why? Because these are the outcomes that we want to happen, therefore they are favorable.
Now we just divide the numerator by the denominator.
$P(1\ head) = \frac{3}{8}$
Remember:
Possible outcomes $= 2^n$ where $n =$ number of tosses.
Here we have ${\color{red}3}$ tosses. Therefore,
Possible outcomes $= 2^{\color{red}n}$
Possible outcomes $= 2^{\color{red}3}$
Possible outcomes $= 2 \times 2 \times 2$
Possible outcomes $= 8$
Note: The factorial function (symbol: !) just means to multiply a series of descending natural numbers.
Examples:
${\color{blue}4!} & {\color{blue}\ = 4 \times 3 \times 2 \times 1 = 24}\\{\color{blue}7!} & {\color{blue}\ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040} \\{\color{blue}1!} & {\color{blue}\ = 1}$
Note: It is generally agreed that $0! = 1$. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.
Example 3: A coin is tossed 4 times. What are the possible outcomes? What is the probability of getting one head?
Solution:
If our first toss were a heads...
If our first toss were a tails...
Therefore there are 16 possible outcomes:
$&{\color{blue}HHHH}, \ {\color{blue}HHH}T, \ {\color{blue}HH}T{\color{blue}H}, \ {\color{blue}HH}TT, \ {\color{blue}H}T{\color{blue}HH}, \ {\color{blue}H}T{\color{blue}H}T, \ {\color{blue}H}TT{\color{blue}H}, \ {\color{blue}H}TTT, \ T{\color{blue}HHH},\\& \ T{\color{blue}HH}T, \ T{\color{blue}H}T{\color{blue}H}, \ T{\color{blue}H}TT, \ TT{\color{blue}HH}, \ TT{\color{blue}H}T, \ TTT{\color{blue}H}, \ TTTT \\&P(1\ {\color{blue}\text{head}}) = \frac{4}{16} \\ &P(1\ {\color{blue}\text{head}}) = \frac{1}{4}$
Alternate Solution:
We have one coin and want to find the probability of getting one head in four tosses. We need to calculate two parts to solve the probability problem.
Numerator (Top)
In our example, we want to have 1 H and 3 Ts. Our favorable outcomes would be any combination of HTTT. The number of favorable choices would be:
$\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\\# \ of\ favorable\ choices & = \frac{4\ letters!}{1\ head! \times 3\ tails!} \\\# \ of\ favorable\ choices & = \frac{4 \times 3 \times 2 \times 1}{1 \times (3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{24}{6} \\\# \ of\ favorable\ choices & = 4$
Denominator (Bottom)
The number of possible outcomes $= 2 \times 2 \times 2 \times 2 = 16$
We now want to find the number of possible times we could get one head when we do these four tosses (or our favorable outcomes).
Remember:
Possible outcomes $= 2^n$ where $n =$ number of tosses.
Here we have ${\color{red}4}$ tosses. Therefore,
Possible outcomes $= 2^{{\color{red}n}}$
Possible outcomes $= 2^{{\color{red}4}}$
Possible outcomes $= 2 \times 2 \times 2 \times 2$
Possible outcomes $= 16$
Now we just divide the numerator by the denominator.
$P(1\ head) & = \frac{4}{16} \\P(1\ head) & = \frac{1}{4}$
Technology Note:
Let’s take a look at how we can do this using the TI-84 calculators. There is an application on the TI calculators called the coin toss. Among others (including the dice roll, spinners, and picking random numbers), the coin toss is an excellent application for when you what to find the probabilities for a coin tossed more than 4 times or more than one coin being tossed multiple times.
Let’s say you want to see one coin being tossed one time. Here is what the calculator will show and the key strokes to get to this toss.
Let’s say you want to see one coin being tossed ten times. Here is what the calculator will show and the key strokes to get to this sequence. Try it on your own.
We can actually see how many heads and tails occurred in the tossing of the 10 coins. If you click on the right arrow (>) the frequency label will show you how many of the tosses came up heads.
We could also use randBin to simulate the tossing of a coin. Follow the keystrokes below.
This list contains the count of heads resulting from each set of 10 coin tosses. If you use the right arrow (>) you can see how many times from the 20 trials you actually had 4 heads.
Now let’s go back to our original chapter problem and see if we have gained enough knowledge to answer it.
You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins.
(a) Determine the theoretical probability for the teacher.
(b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials.
Solution:
(a) Let’s calculate the theoretical probability of getting 5 heads for the 14 tosses.
Numerator (Top)
In our example, we want to have 5 H and 9 Ts. Our favorable outcomes would be any combination of HHHHHTTTTTTTTT. The number of favorable choices would be:
$\#\ of\ favorable\ choices & = \frac{\#\ possible\ letters\ in\ combination!}{letter\ X! \times letter\ Y!} \\\# \ of\ favorable\ choices & = \frac{14\ letters!}{5\ head! \times 9\ tails!} \\\# \ of\ favorable\ choices & = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)\times (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(120) \times (362880)} \\\# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(43545600)} \\\# \ of\ favorable\ choices & = 2002$
Denominator (Bottom)
The number of possible outcomes $= 2^{14}$
The number of possible outcomes $= 16384$
Now we just divide the numerator by the denominator.
$P(5\ heads) & = \frac{2002}{16384} \\P(5\ heads) & = 0.1222$
The probability would be 12% of the tosses would have 5 heads.
b)
Looking at the data that resulted in this trial, there were 4 times of 20 that 5 heads appeared.
$P(5\ \text{heads}) = \frac{4}{20}$ or 20%.
Lesson Summary
Probability in this chapter focused on experiments with random variables or the numbers that you assign to the probability of events. If we have a discrete random variable, then there are only a specific number of variables we can choose from. For example, tossing a fair coin has a probability of success for heads = probability of success for tails $= 0.50$. Using tree diagrams or the formula $P = \frac{\#\ of\ favorable\ outcomes}{total\ \#\ of\ outcomes}$, we can calculate the probabilities of these events. Using the formula requires the use of the factorial function where numbers are multiplied in descending order.
Points to Consider
• How is the calculator a useful tool for calculating probability in discrete random variable experiments?
• Are TREE Diagrams useful in interpreting the probability of simple events?
Vocabulary
Discrete Random Variables
Only have a specific (or finite) number of numerical values.
Random Variable
A variable that takes on numerical values governed by a chance experiment.
Factorial Function
(symbol: !) – The function of multiplying a series of descending natural numbers.
Theoretical Probability
A probability calculated by analyzing a situation, rather than performing an experiment, given by the ratio of the number of different ways an event can occur to the total number of equally likely outcomes possible. The numerical measure of the likelihood that an event, $E$, will happen.
$P(E) = \frac{number\ of\ favorable\ outcomes}{total\ number\ of\ possible\ outcomes}$
Tree Diagram
A branching diagram used to list all the possible outcomes of a compound event.
Review Questions
1. Define and give three examples of discrete random variables.
2. Draw a tree diagram to represent the tossing of two coins and determine the probability of getting at least one head.
3. Draw a tree diagram to represent the tossing of one coin three times and determine the probability of getting at least one head.
4. Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at least one red.
5. Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at two blue marbles. $&\text{Possible Outcomes}:\\&BB, BG, BR, GB, GG, GR, RB, RG, RR\\&P(\text{two blue marbles}) =\frac{1}{9}$
6. Draw a diagram to represent the rolling two dice and determine the probability of getting at least one 5.
7. Draw a diagram to represent the rolling two dice and determine the probability of getting two 5s.
8. Use randBin to simulate the 6 tosses of a coin 20 times to determine the probability of getting two tails.
9. Use randBin to simulate the 15 tosses of a coin 25 times to determine the probability of getting two heads.
10. Calculate the theoretical probability of getting 4 heads for the 12 tosses.
11. Calculate the theoretical probability of getting 8 heads for the 10 tosses. $P(8 \ heads) = \frac{45}{1024}$
12. Calculate the theoretical probability of getting 8 heads for the 15 tosses.
2. $P(\text{at least}\ I \ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\P(\text{at least}\ I \ H) & = \frac{3}{4}$
3. $P(\text{at least}\ I\ H) & = \frac{HHH,HHT,HTH,HTT,THH,THT,TTH}{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} \\P(\text{at least}\ I\ H) & = \frac{7}{8}$
4. $&\text{Possible Outcomes}:\\&BB, BG, BR, GB, GG, GR, RB, RG, RR$ $P(\text{at least one red}) &= \frac{3}{9} \\P(\text{at least one red}) &= \frac{1}{3}$
5. .
6. $P (\text{at least one}\ 5) = \frac{11}{36}$
7. $P$(two 5's) $=\frac{1}{36}$
8. $P (2 \ \text{heads}) = \frac{4}{20} = 20\%$
9. $P$(4 heads) $= \frac{6}{25} = 24\%$
10. $\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\\# \ of\ favorable\ choices & = \frac{479001600}{967680} \\\# \ of\ favorable\ choices & = 495$ The number of possible outcomes $= 2^{12}$ The number of possible outcomes $= 4096$ Now we just divide the numerator by the denominator. $P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%$
11. $P(8\ \text{heads}) = 4.39\%$
12. $\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\\# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\\# \ of\ favorable\ choices & = 6446$ The number of possible outcomes $= 2^{15}$ The number of possible outcomes $= 32768$ Now we just divide the numerator by the denominator. $P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%$
Answer Key for Review Questions (even numbers)
2.
$P(\text{at least}\ I\ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\P(\text{at least}\ I\ H) & = \frac{3}{4}$
4.
$&\text{Possible Outcomes}:\\&BB, BG, BR, GB, GG, GR, RB, RG, RR\\&P(\text{at least one red}) = \frac{3}{9} \\&P(\text{at least one red}) = \frac{1}{3}$
6.
$P$ (at least one 5) $= \frac{11}{36}$
8.
$P$ (2 heads) $= \frac{4}{20} = 20\%$
10. $\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\\# \ of\ favorable\ choices & = \frac{479001600}{967680} \\\# \ of\ favorable\ choices & = 495$
The number of possible outcomes $= 2^{12}$
The number of possible outcomes $= 4096$
Now we just divide the numerator by the denominator.
$P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%$
12. $\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\\# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\\# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\\# \ of\ favorable\ choices & = 6446$
The number of possible outcomes $= 2^{15}$
The number of possible outcomes $= 32768$
Now we just divide the numerator by the denominator.
$P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%$
Feb 23, 2012
Dec 29, 2014
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#### Transcript Multiplication of Polynomials
```Section 5.2
Multiplying Polynomials
Multiplying Two Monomials
Multiplying a Polynomial
By a number
By a monomial
By another polynomial
The FOIL Method
Multiplying 3 or More Polynomials
Special Products
Simplifying Expressions
Applications
5.2
1
Multiplying Two (or more) Monomials
Multiply the numeric coefficients
Include any unmatched variables
Learn to
do these
IN YOUR
Do the variables
Examples
in alpha order
(3)(2x) = 6x -4y(-2xy) = 8xy2 -2s(r) = -2rs
3x(2x)(3x) = 18x3
-5x3(4x2y) = -20x5y
-2(-y) = 2y
(-2b3)(3a)(a2bc) = -6a3b4c
5.2
2
For You
( 8 x y )( 5 x y )
4
7
40 x y
7
3
2
( 2 x yz )( 6 x y z )
2
9
7
12 x y
5.2
5
11
5
z
10
2
7
3
Multiplying a Polynomial by a Number
Positive numbers – law of distribution
5 times 2x2 – 3x + 7
5(2x2) – 5(3x) + 5(7)
10x2 – 15x + 35
Negative numbers – be careful!
-3 times 4y3 – 6y2 + y – 2
-3(4y3)– -3(6y2)+ -3(y) - -3(2)
-12y3 + 18y2 – 3y + 6
5.2
4
Multiplying a polynomial by a
monomial
To multiply a polynomial by a monomial, we multiply each
term of the polynomial by the monomial.
3x2(6xy + 3y2) = 18x3y + 9x2y2
5x3y2(xy3 – 2x2y) = 5x4y5 – 10x5y3
-2ab2(3bz – 2az + 4z3) = -6ab3z + 4a2b2z – 8ab2z3
5.2
5
Multiplying a Polynomial by a
Polynomial (in general)
To multiply a polynomial by a polynomial, we
use the distributive property repeatedly.
Horizontal Method:
(2a + b)(3a – 2b) = 2a(3a – 2b) + b(3a – 2b)
= 6a2 – 4ab + 3ab – 2b2 = 6a2 –ab – 2b2
Vertical Method: 3x2 + 2x – 5
4x + 2
6x2 + 4x – 10
12x3 + 8x2 – 20x____
12x3 + 14x2 – 16x – 10
5.2
6
Bigger Multiplications
Leave Missing
Variable Space
( 5 x x 4 )( 2 x 3 x 6 )
3
2
5x
Leave
Margin
Space
30 x
15 x
10 x
2x 3x 6
2
6 x 24
3
3 x 12 x
4
2
2x
5
x4
3
8x
3
2
10 x 15 x 28 x 11 x 6 x 24
5
4
3
2
5.2
7
FOIL: Used to Multiply Two Binomials
5.2
8
Multiplying 3 or more Polynomials
Use same technique as you used for numbers:
Multiply any 2 together and simplify the temporary product
Multiply that temporary product times any remaining
polynomial and simplify
-2r(r – 2s)(5r – s)
= -2r(5r2 – 11rs + 2s2)
= -10r3 + 22r2s – 4rs2
5.2
9
The Product of Conjugates (Sum and Difference)
(A + B)(A – B) = A2 – B2
The middle term disappears Always when the binomials
are conjugates (identical, except for middle sign)
Multiplying these is easier than using FOIL!
(x + 4)(x – 4) = x2 – 42 = x2 – 16
(5 + 2w)(5 – 2w) = 25 – 4w2
(3x2 – 7)(3x2 +7) = 9x4 – 49
(-4x – 10)(-4x + 10) = 16x2 – 100
(6 + 4y)(6 – 4x) = use the foil method
36 – 24x + 24y – 16xy
5.2
10
Thought provoker:
Are these Conjugates?
(x + 2y)(3xz – 6yz)
= 3x2z – 6xyz + 6xyz – 12y2z
= 3x2z – 12y2z
Why does the middle term disappear?
Because the 2nd binomial conceals a conjugate!
Both terms contain a common factor, 3z :
(x + 2y)(3xz – 3∙2yz)
The middle term ONLY disappears when binomial
conjugates are involved.
5.2
11
Squaring a Binomial – Creates a Perfect-Square Trinomial
(A + B)(A + B) = A2 + 2AB + B2
Square the 1st term
Multiply 1st times 2nd, double it, add it
Square the 2nd term, add it
y 5 2
y 10 y 25
2
2 x 3 y 2
1
2
a 3b
4 x 12 xy 9 y
4 2
0 . 3 x 7 y
5.2
2
2
1
4
2
a 3 ab 9 b
2
4
8
0 . 09 x 4 . 2 xy 49 y
2
12
2
Squaring a Binomial – Creates a Perfect-Square Trinomial
(A – B)(A – B) = A2 – 2AB + B2
Differences:
Almost the
same
Square the 1st term
Multiply 1st times 2nd, double it, subtract it
Square the 2nd term, add it
y 5 2
y 10 y 25
2
3 x 8 y 2
1
5
a 5b
9 x 48 xy 64 y
3 2
2
0 . 6 x 0 . 2 y
5.2
1
25
2
2
a 2 ab 25 b
2
3
6
0 . 36 x 0 . 24 xy 0 . 04 y
2
13
2
Examples - board
( y 1)( 1 y )
( x 3 y )( x 3 xy 9 y )
3
2
2
(a 2b )
2
( 5 y 4 3 x )( 5 y 4 3 x )
5.2
14
Function Notation
If f(x) = x2 – 4x + 5 find:
a ) f (a ) 4
c) f (a h) f (a )
a 4a 5 4
(a h) 4(a h) 5 a 4a 5
a 4a 9
a 2 ah h 4 a 4 h 5 a 4 a 5
2
2
2
2
2
2
2
2 ah h 4 h
2
b ) f ( a 3)
( a 3) 4 ( a 3) 5
2
a 6 a 9 4 a 12 5
2
a 2a 2
2
5.2
15
Next …
Section 5.3 Intro to Factoring
Common Factors, Factoring by Grouping
5.2
16
```
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CS 173
Spring 2021
## Proving a primitive function relationship
Let's see how to prove one of the relationships in our ordering of primitive functions: $$n^2 << 2^n$$.
We need to show that $$\lim_{n \rightarrow \infty} \frac{n^2}{2^n} = 0$$.
Recall from calculus that $$a^n$$ limits to zero as n gets big, if $$a < 1$$. So, in particular, $$\lim_{n \rightarrow \infty} (3/4)^n = 0$$.
We're going to show that $$\frac{n^2}{2^n}$$ limits to zero by showing that the terms of this sequence are smaller than the terms of $$(3/4)^n$$, with a bit of an offset in the indexing. Since the larger sequence limits to zero, the smaller one must also do so.
To show this, we need the following lemma:
Lemma: If n is big enough (n >= 5), then $$(n+1)^2 < \frac{3}{2} n^2$$.
This requires a short direct proof.
To understand how the two sequences are related, let's compare $$\frac{n^2}{2^n}$$ to $$\frac{ (n+1)^2}{2^{n+1}}$$. The lemma says that the top of the fraction increases by less than $$\frac{3}{2}$$. The bottom increases by a factor of 2. So the whole fraction increases by less than $$\frac{3}{4}$$. As we move forward in the sequence, we accumulate factors of $$\frac{3}{4}$$.
So, if we can find a starting relationship between terms of the two sequences, we can pin down the details of a relationship between the sequences. With a bit of fiddling around plugging in small constants for n, we find a good starting point:
At n=5
$$\frac{n^2}{2^n} = \frac{25}{32}$$
$$(3/4)^{n-5} = (3/4)^0 = 1$$
So $$\frac{n^2}{2^n} < (3/4)^{n-5}$$ at n=5.
So here's the claim we'll prove by induction:
Claim: $$\frac{n^2}{2^n} < (3/4)^{n-5}$$ for all $$n \ge 5$$.
## The induction proof
Proof by induction on n.
Base case: At n=5, $$\frac{n^2}{2^n} = \frac{25}{32}$$. Also, $$(3/4)^{n-5} = (3/4)^0 = 1$$. So $$\frac{n^2}{2^n} < (3/4)^{n-5}$$.
Inductive hypothesis: suppose that $$\frac{n^2}{2^n} < (3/4)^{n-5}$$ for n = 5, 6, ..., k.
Rest of inductive step: By the lemma $$\frac{(k+1)^2}{2^{k+1}} < (\frac{3}{2}) \frac{k^2}{2^{k+1}} = (\frac{3}{2}) \frac{k^2}{2\cdot 2^k} = (\frac{3}{4}) \frac{k^2}{2^k}$$
By the inductive hypothesis, $$\frac{k^2}{2^k} < (3/4)^{k-5}$$. Substituting into the previous equation gives us: $\frac{(k+1)^2}{2^{k+1}} < (\frac{3}{4}) \frac{k^2}{2^k} < (\frac{3}{4}) (3/4)^{k-5} = (3/4)^{k-4}$
So $$\frac{(k+1)^2}{2^{k+1}} < (3/4)^{k-4}$$, which is what we needed to prove.
## Inequality induction
Our final proof was fairly simple. However, we needed to fiddle around with the pieces of the problem to figure out exactly what constants we needed and how to compare bits of algebra. Inequality induction problems often require more scratchwork than problems involving equality.
Suppose you need to prove that $$a < b$$, where a and b are two bits of algebra. There's a gap in size between a and b. Frequently you have to hypothesize som intermediate term c, for which $$a < c < b$$ in order to make your algebra work. Sometimes you have to fiddle around with scratchwork to find the right c that makes it easy to prove that $$a < c$$ and $$c < b$$.
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#### Find the area of the region $y=\sqrt x$ bounded by and y = x.
\begin{aligned} &y=\sqrt{x}\\ &\text { squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\
This is a parabola, no negative values of x, therefore it lies on the right of Y axis passing through origin.
Now $y=\sqrt{x}$ means y and x both has to be positive hence both lie in $1^{\text {st }}$ quadrant hence $y=\sqrt{x}$ will be part of $\mathrm{y}^{2}=\mathrm{x}$ which is lying only in $1^{\text {st }}$ quadrant
And y=x is a straight line passing through origin
We have to find area between $y=\sqrt{x}$ and y=x shown below
For finding the point of interaction, solve two equations simultaneously.
$\\Put y=x in y^{2}=x \\ \Rightarrow x^{2}=x \\ \Rightarrow x=1 \\$
Put x=1 in y=x we get y=1
The point of interaction is (1, 1)
Area between the parabolic curve and line = area under parabolic curve – area under line $\ldots (1) \\$
For the area under parabolic curve
$\item Y = \sqrt x \\$
Integrating from 0 to 1
\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{x}^{\frac{1}{2}} \mathrm{~d} \mathrm{x}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{2}{3}\left[1^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{2}{3} \end{aligned} \\
For area under straight line y = x
On integrating from 0 to 1
$\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} y d x=\left[\frac{x^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{ydx}=\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} y d x=\frac{1}{2} \\$
Using (i)
$\Rightarrow \text{area between parabolic } = \frac{2}{3}-\frac{1}{2}= \frac{1}{6} unit ^{2} \\$
Hence area bounded is $\frac{1}{6} unit ^{2} \\ \\$
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# Basic Linear Algebra¶
There is no intention to introduce Linear Algebra in this section. We have assumed that you have already accomplished training on Linear Algebra. This section serves just as a brief review of the vector space.
The following is the definition of a vector space, or linear space in some references.
Definition. The vector space $$V$$ over field $$F$$ is a set with two binary operation: addition $$+:V\times V\rightarrow V$$ and scalar multiplication $$\cdot:F\times V\rightarrow V$$ satisfying
1. Commutativity: $$v+u=u+v$$
2. Associativity of addition: $$(v+u)+w = v+(u+w)$$
3. Additive identity: $$\exists 0, \forall v\in V, 0+v=v+0=v$$
4. Additive inverse: $$\exists (-v), v + (-v) = 0$$
5. Associativity of scalar multiplication: $$\beta(\alpha v) = (\beta\alpha)v$$
6. Compatibility of addition and scalar multiplication: $$(\alpha+\beta)v = \alpha v + \beta v$$, $$\alpha(v+u) = \alpha v + \alpha u$$
7. Scalar multiplicative identity: $$\exists 1, 1\cdot v = v$$
There is no need to be so keen on the definition of vector space, but there does exist some important facts worth noticing. From the definition can we discover that the vector space is actually a very general concept — it would be more appropriate to call it a paradigm. Thus, when facing vector space, one should be very careful to distinguish the general vector space and some specific vector spaces.
As an example, the real number set over real number field with addition and scalar multiplication as the ordinary addition and multiplication of real numbers forms a vector space. In this case, all real numbers are vectors of real number set — this is somehow different from our previous intuition, since a real number is widely referred as “scalar” instead of a vector.
As another example, the tensor space of any type is a vector space. But in many applicative contexts, the tensor of order one is specifically referred as vector.
The reason for this section is to remind you that there might be multiple meanings corresponding to the term “vector”. Thus, sometimes if you find the argument containing “vector” incomprehensible, please remember that there might be the problem of the ambiguity of the term “vector”.
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# RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions 2022 | Download Free PDF
RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions: In this exercise, the students will learn how the formulae for computing amount & compound interest are used to calculate the growth of the population. They will also study depreciation. These solutions assist the students to get to know about the exam pattern so as to acquire higher exams. The solutions are solved with shortcut techniques that make learning easy for the students.
They can attain good marks in the Class 8th Maths final exams by practicing these solutions regularly. They can easily download these solutions in PDF format. These solutions also enable the students in their daily homework routine. These solutions include well-researched information about each concept that enables the students to understand each concept appropriately. Each topic of this exercise is solved in an easy language that enables the students to study & revise each topic with maximum accuracy.
## Download RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions
RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions
## Important Definition for RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions
The students can easily download RS Aggarwal Class 8 Maths Chapter 11 Ex 11d Solutions in the PDF format and access them at any time. These solutions also ease out the exam preparation level and enable the students to achieve excellent ranks in the exams.
• Formula for Population Growth
Formula 1: Assume P is the population of a city at the starting of the specific year and the population grows at a constant rate R% per annum, then:
Population after n years = P (1 + R/100)n
Formula 2: Assume P is the population of a city or a town at the starting of the specific year. If the population grows at the rate of R1% during the first year and R2% during the second year, then: Population after 2 years = P (1 + R1/100) x P (1 + R2/100)
Formula 3: Assume P is the population a city or a town at the starting of the specific year. If the population decrease at the rate of R% per annum, then:
Population after n years = P (1 + R/100)n
• Depreciation
The relative decrease in the value of a machine over a period of time is its depreciation. Depreciation per unit of time provides the rate of depreciation. The value at any time is known as the depreciated value.
• Formula for Depreciation
1st Formula: If V0 is the value of an item at a certain time & R% per annum is the rate of depreciation, then the value of Vn at the end of a year is: Vn = V0 (1−R/100)n
2nd Formula: If V0 is the value of an item at a certain time & the rate of depreciation is R1% for the first n1 years, R2% for next n2 years & so on, and Rk% for the last nk years the value at the end of n1 + n2 +………..+ nk years is:
Vn = V0 (1−R1/100)n1 x (1−R2/100)n2 ……… x (1−R2/100)nk
Know more at the official website.
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Q)
# Find the value of $k$ in the following so that the function $f$ is continuous at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{1-\cos kx}{x\sin x}, & if\;x\neq 0\\\large\frac{1}{2}, & if\;x=0\end{array}\right.\;at\;x=0$
$\begin{array}{1 1} -1 \\ 1 \\ \pm 1 \\ None\;of\;the \;above\end{array}$
Comment
A)
Toolbox:
• A function is said to be continuous at a point $'a'$ if $LHL = RHL.=f(a)$
• (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)=f(a)$
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}\large\frac{1-\cos kx}{x\sin x}, & if\;x\neq 0\\\large\frac{1}{2}, & if\;x=0\end{array}\right.\;at\;x=0$
Consider $f(x)=\large\frac{1-\cos kx}{x\sin x}$
This can be written as $\large\frac{2\sin^2\Large\frac{kx}{2}}{x\sin x}$
Multiply and divide by $\large\frac{k^2x}{2}$
$f(x)=\large\frac{\sin^2\big(\Large\frac{kx}{2}\big).\Large\frac{k^2x}{2}}{\big(\Large\frac{kx}{2}\big)^2.\large \sin x}$
$f(x)=\bigg(\large\frac{\sin\Large\frac{kx}{2}}{\Large\frac{kx}{2}}\bigg).\big(\large\frac{x}{\sin x}\big).\large\frac{k^2}{2}$
Step 2:
We know that $\lim\limits_{\large x\to 0}\bigg(\large\frac{\sin\theta}{\theta}\bigg)=$$1$
$\therefore$ $L.H.L.=R.H.L.=$
$\lim\limits_{\large x\to 0}\bigg(\large\frac{\sin\Large\frac{kx}{2}}{\Large\frac{kx}{2}}\bigg).\big(\large\frac{x}{\sin x}\big).\large\frac{k^2}{2}=\large\frac{k^2}{2}$
At $x=0$
$f(0)=\large\frac{1}{2}$
Given that $f(x)$ is continuous
$\therefore\:\large\frac{k^2}{2}=\frac{1}{2}$
$\Rightarrow k^2=1$
$k=\pm 1$
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# Compound Interest Questions with Answers
## Compound Interest Questions with Answers
Basic concept of Compound Interest and Compound Interest Questions with Answers are provided here to help the students understand the applications of compound interest in our daily existence.
### Important Formulae to solve Compound Interest Questions :
• $P [1+ \frac{R}{100}]^{T}$
This formula is applied to calculate when money is compounded annually.
• $P [1+ \frac{R}{2\times 100}]^{2T}$
This formula is applied to calculate when money is compounded half-yearly
• $P [1+ \frac{R}{12\times 100}]^{12T}$
This formula is applied when money in compounded monthly.
### Situations where we can use Compound Interest formulas.
• Increase or decrease in population
• The growth of a bacteria (when the rate of growth is known)
• The value of an item, if its price increases or decreases in the intermediate years
### Compound Interest Questions and Answers
Example 1:
The compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded yearly, is
Options:
a. Rs 636.80
b. Rs 816
c. Rs 912
d. Rs 882.82
Explanation:
Rate of interest = 4%
Therefore, applying the net% effect formula for effective rate of compound interest for 2 years , we get
Net% effect = $x + y + \frac{xy}{100}\%$
x = y = 4%
$= 4 + 4 + \frac{4\times 4}{100} = 8 + .16 = 8.16\%$
CI = 8.16% of 10,000
$= \frac{8.16\times 10000}{100} = ₹ 816$
Hence, option B is correct,
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## Practice Compound Interest Questions with Answers
1. If 50000 sums to 58694 in 2 years. Find out the rate of interest.
5%
5%
12.44%
4%
4%
18.43%
8%
8%
58.99%
9%
9%
10.14%
Given: A = 58694 P = 50000 n = 2 years r=?
A = P (1 + r/100)n
=>58694 = 50000 (1 + r/100)²
=>58694/50000 = (1 + r/100)²
=>(1+ r/100) = sqrt (1.17)=1.08
=>r/100=1.08-1=0.08
=>r = 0.08 x 100 = 8%
2. Calculate the compound interest on Rs 9,000 at an annual rate of 15% for 2 years and 4 months?
INR 3497.625
INR 3497.625
65.41%
INR 4500.08
INR 4500.08
14.59%
INR 3987.76
INR 3987.76
9.73%
INR 3154.34
INR 3154.34
10.27%
=9000(1+15/100)2⅓
=9000(1+3/20)² x (1+3/20x3)
=9000* (23/20) x (23/20) x (21/20)
=12497.625
∴ Compound Interest = Rs. (12497.625– 9000) = Rs. 3497.625
3. What would be the compound interest on Rs 8000 for 4 years at 8 % per annum?
Note: the interest is calculated half-yearly.
3900
3900
14.57%
2948
2948
64.9%
2700
2700
15.23%
4345
4345
5.3%
Given: P = Rs.8000, rate of interest = 8%, T = 4years, n = 2
calculation: half-yearly, therefore, r= 8 %=4%
nT=2 x 4=8
A=8000( 1 + 8/100 x 2)⁸ =8000× (26/25)⁸
=8000 x 1.3685=10948
Interest = Amount - Principal = 10948 - 8000 = 2948
4. The difference between Simple interest and compound interest for 2 years at the rate of 10% per annum on a particular amount of money is Rs20. Calculate the amount:
2000
2000
79.74%
3000
3000
5.23%
6000
6000
7.19%
5500
5500
7.84%
Let the sum be 100.
Therefore, SI=100×10×2/100 = 20
and CI=100 (1 + 10/100)² - 100 = 121 − 100 = 21
Difference of CI and SI = 21 − 20= 1
Now if the difference is 1 , then the total= 100
So if the difference is 20 then the total will be= 100× 20 = INR 2000
5. If the annual increase in the inhabitants of a particular city is 7% and the total number of residents in that specific locality is 13560 what will be the population of that same city in the coming 3 years?
17000
17000
6.25%
18967.22
18967.22
15.97%
16612
16612
70.83%
19873
19873
6.94%
13560(1 + 7/100)³
Required population: 13560(1.07)³ = 16612
6. Calculate the actual rate of interest corresponding to an itemized rate of 9% compound semi-annually.
20%
20%
10.26%
10%
10%
14.53%
9.20%
9.20%
63.25%
8.90%
8.90%
11.97%
Measurement shows that 100 at 9% compounded semi-annually will grow to
A=100 (1 + .09/2)² = 100(1.045)²
=100(1.09)=109.20
So the actual rate=109.20 - 100=9.20. Thus if we get 9.20 % 100 in 1 year with yearly compounding, then the rate will be 9.20/100=0.092=9.20 so the actual rate = 9.20%
7. Sumit invested in the bank on which he gets an interest of 4.9% compounded monthly. Calculate the actual interest rate.
8%
8%
8.11%
8.87%
8.87%
17.12%
5.01%
5.01%
69.37%
7.24%
7.24%
5.41%
Given r=.049 and m= 12.
Value of p after n years = p (1+ r/100)n
Effective rate is re = (1 + .049/12)12 − 1
=1.050115575-1=.0501
or 5.01%
8. Manoj bought a microwave on which he has to pay INR 2000 cash as a down payment. After this he is supposed to pay INR 1,500 at the completion of 1st year, INR 1,050 at the completion of 2nd year and INR 930 at the completion of 3rd The interest rate on this is 10%. Compute the total cash price:
9800
9800
16.3%
5678
5678
16.3%
5643
5643
6.52%
4930
4930
60.87%
Down payment = INR 2000
So at the completion of 1st-year x will become INR 1500.
So, 1500 = x (1 + 10/100) or
x = (1500×100/110) = 1363.63
Similarly, 1050 = y (1 + 10/100)² or y= (1050×20×20/22×22) = 420,000/484=867.76
and z = (930×20×20x20/22×22×22) = 698.72
Hence, CP = 2000+1363.63+867.76+698.72 = 4930.11 or 4930
9. Amanpreet invested his savings in such a way that his money grows up to INR 4645 in every two years and up to 5500 in 3 every year including the interest compounded. Calculate the interest rate applicable to earn this much of money?
20%
20%
11.54%
19%
19%
8.65%
19.05%
19.05%
16.35%
18%
18%
63.46%
As given, P+ Compound Interest of 3 years = 5500.......(i)
P+ Compound Interest of 2 yrs = 4645.......(ii)
By solving the equations we get Compound Interest of 3rd year = 5500-4645 = 855
Alternate method:
Difference of sum after n years and n + 1 years × 100 /Amount after n years
Here n=2
Rate = [(5500−4645) × 100] /4645= (855×100)/4645 = 18.40% or 18%.
10. Calculate the compound interest on INR 28,750 for 2 years. The interest was calculated as 4% in the first year and 8% in second year.
2500
2500
8.26%
3002
3002
12.4%
3798
3798
10.74%
3542
3542
68.6%
After first year the amount =
28750 (1 + (4/100)) = 28750 (104/100)
After 2 nd year the amount = 28750 (104/100) (108/100)
=28750 (26/25) (27/25) = (1.12 x 28750)=32292
CI = 32292-28750 = 3542
×
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In document Introduction to Probability (Page 57-60)
Example 2.6 A chord of a circle is a line segment both of whose endpoints lie on the circle. Suppose that a chord is drawnat random in a unit circle. What is the probability that its length exceeds√3?
Our answer will depend on what we mean byrandom,which will depend, in turn, on what we choose for coordinates. The sample space Ω is the set of all possible chords in the circle. To find coordinates for these chords, we first introduce a
48 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES 40 45 50 55 60 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
Figure 2.8: Sum of 100 random numbers.
x y A B M θ β α
Figure 2.9: Random chord.
rectangular coordinate system with origin at the center of the circle (see Figure 2.9). We note that a chord of a circle is perpendicular to the radial line containing the midpoint of the chord. We can describe each chord by giving:
1. The rectangular coordinates (x, y) of the midpointM, or 2. The polar coordinates (r, θ) of the midpointM, or
3. The polar coordinates (1, α) and (1, β) of the endpointsA andB.
In each case we shall interpret at random to mean: choose these coordinates at random.
We can easily estimate this probability by computer simulation. In programming this simulation, it is convenient to include certain simplifications, which we describe in turn:
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 49 1. To simulate this case, we choose values for x andy from [−1,1] at random. Then we check whether x2+y21. If not, the pointM = (x, y) lies outside the circle and cannot be the midpoint of any chord, and we ignore it. Oth- erwise, M lies inside the circle and is the midpoint of a unique chord, whose lengthLis given by the formula:
L= 2p1−(x2+y2).
2. To simulate this case, we take account of the fact that any rotation of the circle does not change the length of the chord, so we might as well assume in advance that the chord is horizontal. Then we chooserfrom [0,1] at random, and compute the length of the resulting chord with midpoint (r, π/2) by the formula:
L= 2p1−r2 .
3. To simulate this case, we assume that one endpoint, sayB, lies at (1,0) (i.e., thatβ = 0). Then we choose a value forαfrom [0,2π] at random and compute the length of the resulting chord, using the Law of Cosines, by the formula:
L=√2−2 cosα .
The program BertrandsParadox carries out this simulation. Running this program produces the results shown in Figure 2.10. In the first circle in this figure, a smaller circle has been drawn. Those chords which intersect this smaller circle have length at least√3. In the second circle in the figure, the vertical line intersects all chords of length at least√3. In the third circle, again the vertical line intersects all chords of length at least√3.
In each case we run the experiment a large number of times and record the fraction of these lengths that exceed √3. We have printed the results of every 100th trial up to 10,000 trials.
It is interesting to observe that these fractions arenotthe same in the three cases; they depend on our choice of coordinates. This phenomenon was first observed by Bertrand, and is now known asBertrand’s paradox.3 It is actually not a paradox at all; it is merely a reflection of the fact that different choices of coordinates will lead to different assignments of probabilities. Which assignment is “correct” depends on what application or interpretation of the model one has in mind.
One can imagine a real experiment involving throwing long straws at a circle drawn on a card table. A “correct” assignment of coordinates should not depend on where the circle lies on the card table, or where the card table sits in the room. Jaynes4 has shown that the only assignment which meets this requirement is (2). In this sense, the assignment (2) is the natural, or “correct” one (see Exercise 11). We can easily see in each case what the true probabilities are if we note that
3 is the length of the side of an inscribed equilateral triangle. Hence, a chord has
3J. Bertrand,Calcul des Probabilit´es(Paris: Gauthier-Villars, 1889).
4E. T. Jaynes, “The Well-Posed Problem,” inPapers on Probability, Statistics and Statistical
50 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES .0 1.0 .2 .4 .6 .8 1.0 .488 .227 .0 1.0 .2 .4 .6 .8 1.0 .0 1.0 .2 .4 .6 .8 1.0 .332 10000 10000 10000
lengthL >√3 if its midpoint has distanced <1/2 from the origin (see Figure 2.9). The following calculations determine the probability that L > √3 in each of the three cases.
1. L >√3 if(x, y) lies inside a circle of radius 1/2, which occurs with probability
p=π(1/2) 2
π(1)2 =
1 4 . 2. L >√3 if|r|<1/2, which occurs with probability
1/2−(−1/2) 1−(−1) =
1 2 .
3. L >√3 if 2π/3< α <4π/3, which occurs with probability 4π/3−2π/3
2π−0 = 1 3 .
We see that our simulations agree quite well with these theoretical values. 2
In document Introduction to Probability (Page 57-60)
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# How to Teach Even and Odd Numbers
Even and odd numbers are a basic mathematical concept.
So how do you teach them in a fun and memorable way?
We’ve got methods for every learning style!
## Teaching Even and Odd Numbers to Kids
The whole numbers can be placed into two categories: even and odd.
Recognizing the difference can help you in division and multiplication, so it’s important that you know how to tell them apart.
Typically you introduce a child to this concept in first or second grade, and then review it in the later elementary grades.
In order to understand even and odd numbers, a child must know his numbers and be able to count to 100.
When deciding how to teach even and odd numbers, explore your options!
Here are several ways to teach the concept.
## How to Introduce Even and Odd Numbers
To introduce the concept of even and odd numbers, use small counters.
Edible counters, like M&Ms, are especially fun to use!
For young learners, you can set up two figures (these can be dolls, action figures, stuffed animals, etc.)
Say, “Do you ever have to share with your brother or sister? You want the sharing to be fair; you both want the same amount. That’s what’s happening here with Fred and Ginger. Their parents have left them some M&Ms (or cookies, toys, etc.) and told them to share.”
Walk through the various numbers 1-10. Can they share 1 M&M evenly?
Not without cutting it.
Can they share 2?
Divide them into 2 piles, one in front of each action figure.
Yes, 2 can be divided evenly.
Write the number 2 on a whiteboard or piece of paper.
Eventually, you will have this list of numbers: 2, 4, 6, 8, 10.
Explain that these are called even numbers because you can divide them evenly without cutting them apart.
The other numbers are not even; they’re odd: 1, 3, 5, 7, 9.
(Technically, 0 is an even number as well, but leave it out of this basic introduction).
## How to Teach Even and Odd Numbers Using Colors
We are wired to notice patterns, and children are no different.
Help children learn odd and even numbers by using color.
You can use 3-D numbers in two colors or make your own set with markers on index cards (or use die-cut numbers out of cardstock).
Choose one color for odd numbers (red) and one for even (blue).
This is an especially effective method for visual learners.
Lay the numbers 0-10 on the table.
Ask, “Do you see a pattern?”
They will notice that the colors alternate between red and blue.
Name the even numbers and then the odd numbers.
Chant them (this is an especially important step for auditory learners).
Then form two-digit odd and even numbers, focusing on their last digit.
“This is an even number because it ends with a 2.”
## How to Teach Even and Odd Numbers Using Counters
This method of teaching odd and even numbers works especially well with kinesthetic learners.
You will need three-dimensional letters and counters.
Line up the numbers 1-10.
Under the number 2 place two counters side by side.
Say, “Do you see how this counter has a partner? We call numbers that have partners ‘even numbers.’”
Now place a single counter under the number 1, aligning it to the left to show where a missing counter could sit.
“This counter doesn’t have a partner. It’s all alone. We call numbers where the last counter is missing a partner ‘odd numbers.’”
Repeat this process for number 3.
Place a pair of counters directly under the 3 and a single counter under that pair.
“Look at the last counter. Does it have a partner? No? Then it’s an odd number.”
Repeat the process with the number 4, place two pairs of counters underneath the number.
Then do the same for 5, 6, 7, etc.
## How to Teach Even and Odd Using a Number Line
Older students can visualize odd and even numbers by using a number line.
Start on the number 0.
Explain that it is the even number starting place.
Now jump over the 1 to land on the 2.
“When I jump 2 units, I land on an even number.”
Using this method, you can show both positive and negative even numbers.
The numbers that you skip are the odd numbers out!
## How to Teach Using a Number Chart
This method again shows pattern relationships, which may crystallize understanding for some students (especially your visual learners and logical thinkers).
Use a 100 number chart, where the numbers are listed in rows of ten.
Color in the even numbers in the top row (0-10).
Chant them a few times to memorize them.
Then color in the even numbers in the second row (11-20). Can the child see the pattern?
Repeat this with the remaining rows.
In the end, the chart will look like 5 colored columns.
Point out that all the numbers ending with 2, 4, 6, 8, and 0 are even; the numbers ending in 1, 3, 5, 7, and 9 are odd.
## How to Teach Odd and Even Numbers
No matter the learning style, you can teach even and odd numbers in a way that will make them stick.
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Using a graph is just another way we can solve equations. We can solve linear, quadratic and even cubic ( see http://cribbd.com/learn/maths/algebra/solve-cubic-equations-by-drawing-appropriate-lines-on-graphs for cubic graphs).
Linear Equations Example: Say we wanted to solve 2x-1 = 3x+1, and we are given the equation y = 2x-1 on a graph.
In order to solve the equation for 2x-1 = 3x+1, we need to plot y = 3x+1 on the same graph, and see where both lines intercept one another. We then take the x co-ordinate as the solution for the equation.
So looking at the graph we can see that the lines intersect one another at x=-2
Quadratic Equation Example: Say we have the graph x2 -4x-12 given to us, and we need to solve x2 -4x-12 = x+4.
To do this, we need to plot y = x+4 on the same set of axis, then read off where y=x+4 intersects the quadratic.
We can give estimates to where the intersecting points lie on the x-axis. So one at x= -2.1 and the other at x = 7.2. Remember since this is a quadratic graph we are solving, there are two possible values can be.
To check this, we need only to put the values back into our equation x2 -4x-12 = x+4.
If x = -2.2, on one side of the equation we have (-2.2)2 - (4 x - 2.2) - 12 = 1.64. On the other hand side we have -2.2 + 4 = 1.8. 1.64 and 1.8 are quite close together so we can say that x=-2.2 is a solution.
If x=7.2, on one side of the equation we have 7.22 -4 x 7.2 - 12 = 11.04. On the other hand side we have 7.2 + 4 = 11.2. Again these two values are quite close together, so we can say that x=7.2 is a solution
As you can see, unless a line goes through a graph where x is a whole number, an estimate has to be made. You need to check that both sides are fairly similar when going to check your answers.
Nothing in this section yet. Why not help us get started?
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Fundamental Hardware Elements of Computers: Logic Gates
Logic Gates Boolean gate combinations →
In 1854 a British mathematician, George Boole, developed Boolean Algebra. Instead of an algebra that uses numbers, boolean algebra uses truth values, true(1) and false(0). By defining sentences using truth values and performing operations on these truth values you can work out the overall conclusion of complex statements. Boolean Algebra has had a massive impact on Computer Science and the language that computers understand is a language of 1s and 0s, boolean.
Examples of Boolean Algebra shown in a truth table
$x$ $y$ $x.y$ $x+y$ 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 1
$x$ $\overline{x}$ 0 1 1 0
Logic gates are pieces of hardware that perform operations on boolean inputs, allowing us to create complex devices out of abstract boolean algebra. Logic gates are the fundamental building blocks of hardware and processors will be made out of billions of them. A logic gate will typically have one or two inputs, in the examples here defined by A and B, There are six types of gate that you need to know:
Contents
NOT
A NOT gate takes only one input and gives one output
A NOT gate will always give an output opposite to what the input is e.g. 1 (not gate) 0. A NOT gate takes one boolean input and flips it. It is possible to have a double NOT. This will reverse the original NOT. The symbol would have an extra bar over it.
In Boolean Algebra we write a NOT symbol by placing a bar on top of a letter($\overline{A}$) or letters ($\overline{C+B}$).
Examples of a NOT gate at work are as follows: NOT(It is not raining) = It is raining $\overline{TRUE} = FALSE$ $\overline{0} = 1$
To summarise here is a truth table showing the relationship between A and $\overline{A}$
A $\overline{A}$
0 1
1 0
AND (.)
An AND gate takes two inputs and combines them to form one output
An AND gate will combine the boolean values of two inputs (you can get more than two inputs but we don't need to know about that type of gate here). If and only if both inputs are true will it output true. If any of the inputs are false it will out put false.
In Boolean Algebra we write an AND symbol by placing a bullet point between two (${A}.{B}$) or more (${A}.{B}.C$) values.
Examples of an AND gate at work are as follows: Six is bigger than four AND Cats are bigger than gerbils = TRUE $TRUE . FALSE = FALSE$ $(7 < 8) . (2 > 1) = TRUE$ $(8 > 0) . \overline{(20 = 19)} = TRUE$
An easy way to remember how an AND gate works is thinking about a circuit to turn a light bulb on. If both switches are on then the bulb will light up, if any switch is off then the bulb won't light.
a circuit diagram equivalent to an AND gate
To summarise here is a truth table showing all the different values for two inputs A and B and the result of ANDing those values together
A B A.B
0 0
0
0 1
0
1 0
0
1 1
1
OR (+)
An OR gate takes two inputs and combines them to form one output
An OR gate will combine the boolean values of two inputs. If one or more inputs are true then the output will be true. If both the inputs are false then the output will be false.
In Boolean Algebra we write an OR symbol by placing a plus symbol between two (${A}+{B}$) or more (${A}+{B}+C$) values.
Examples of an OR gate at work are as follows: humans have two legs OR Elephants have 8 legs = TRUE $TRUE + FALSE = TRUE$ $(9 > 3) + (2 > 1) = TRUE$ $\overline{(20 = 20)} + (5 > 6) = FALSE$
An easy way to remember how an OR gate works is thinking about a circuit to turn a light bulb on. If one or more switches are on then the bulb will light up, if both switch are off then the bulb won't light.
a circuit diagram equivalent to an OR gate
To summarise here is a truth table showing all the different values for two inputs A and B and the result of ORing those values together
A B A+B
0 0
0
0 1
1
1 0
1
1 1
1
XOR ($\oplus$)
A XOR gate takes two inputs and combines them to form one output
An exclusive- OR, XOR, gate will combine the boolean values of two inputs. If exactly one input is true then the output will be true. If both the inputs are false or both the inputs are true then the output will be false.
In Boolean Algebra we write an XOR symbol by placing a plus symbol surrounded by a circle between two (${A}\oplus{B}$) or more (${A}\oplus{B}\oplus{C}$) values.
Examples of an XOR gate at work are as follows: it is raining XOR it is not raining = TRUE $TRUE \oplus TRUE = FALSE$ $(9 < 3) \oplus (2 < 1) = FALSE$ $\overline{(20 = 20)} \oplus (8 > 2) = TRUE$
To summarise here is a truth table showing all the different values for two inputs A and B and the result of XORing those values together
A B $~A \oplus B$
0 0
0
0 1
1
1 0
1
1 1
0
NAND
A NAND gate takes two inputs and combines them to form one output
A NAND gate will combine the boolean values of two inputs, AND them together, and NOT the result. If one or less input is true then the output will be true. If both the inputs are true then the output will be false. To draw a NAND gate you draw an AND gate and add a circle to the front, as you can see above.
In Boolean Algebra we write an NAND symbol by taking an AND equation and NOTing the result ($\overline{{A} . {B}}$).
Examples of an NAND gate at work are as follows: (A=A) NAND (A<>B) = TRUE $\overline{TRUE . TRUE} = FALSE$ $\overline{(9 < 3) . (2 < 1)} = TRUE$ $\overline{(20 > 20) . (8 > 2)} = TRUE$
To summarise here is a truth table showing all the different values for two inputs A and B and the result of NANDing those values together
A B $A . B$ $\overline{A . B}$
0 0 0
1
0 1 0
1
1 0 0
1
1 1 1
0
NOR
A NOR gate takes two inputs and combines them to form one output
A NOR gate will combine the boolean values of two inputs, OR them together, and NOT the result. If no input is true then the output will be true. If either or both inputs are true then the result will be false. To draw a NOR gate you draw an OR gate and add a circle to the front, as you can see above.
In Boolean Algebra we write an NOR symbol by taking an OR equation and NOTing the result ($\overline{{A} + {B}}$).
Examples of an NOR gate at work are as follows: (A=A) NOR (A<>B) = FALSE $\overline{FALSE + TRUE} = FALSE$ $\overline{(9 < 3) + (2 < 1)} = TRUE$ $\overline{(20 > 20) + (8 > 2)} = FALSE$
A B $A + B$ $\overline{A + B}$
0 0 0
1
0 1 1
0
1 0 1
0
1 1 1
0
Exercise: Logic Gates
Give the symbol and gate diagram for an OR statement
+
Give the symbol and gate diagram for an AND statement
.
Give the symbol and gate diagram for a XOR statement
$\oplus$
Give answers to the following equations:
TRUE AND TRUE
TRUE
TRUE + FALSE
TRUE
TRUE + TRUE
TRUE
TRUE $\oplus$ TRUE
FALSE
NOT(TRUE) . TRUE
FALSE
$\overline{\overline{TRUE}} + FALSE$
TRUE
Draw a NAND gate and truth table
A B $A . B$ $\overline{A . B}$
0 0 0
1
0 1 0
1
1 0 0
1
1 1 1
0
Complete the following table:
Name NAND NOR AND NOT OR XOR
Gate
Symbol
Truth Table
Name NAND NOR AND NOT OR XOR
Gate
Symbol $\overline{A . B}$ $\overline{A + B}$ $.$ $\overline{A}$ $+$ $\oplus$
Truth Table
A B $A . B$ $\overline{A . B}$
0 0 0
1
0 1 0
1
1 0 0
1
1 1 1
0
A B $A + B$ $\overline{A + B}$
0 0 0
1
0 1 1
0
1 0 1
0
1 1 1
0
A B A.B
0 0
0
0 1
0
1 0
0
1 1
1
A $\overline{A}$
0 1
1 0
A B A+B
0 0
0
0 1
1
1 0
1
1 1
1
A B $~A \oplus B$
0 0
0
0 1
1
1 0
1
1 1
0
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# MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3
## MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3
Question 1.
MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 1050°
∠R = 105°
PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 6.5 cm
∠N = 85°
(iii) Parallelogram
HEAR
HE =5 cm
EA = 6 cm
∠B = 85°
(iv) Rectangle
OKAY
OK = 7 cm
KA = 5 cm
Solution:
(i) Steps of Construction:
Step-1: Draw MO = 6 cm
Step-2 : Make ∠MOX = 105° and ∠OMY = 60°.
Step-3 : Cut off OR = 4.5 cm on OX.
Step-4 : At point R, draw ∠ORZ = 105° which cuts MY at E.
Hence, MORE is the required quadrilateral.
(ii) Here, ∠P = 90°, ∠A = 110°, ∠N = 85°
∴ ∠L = 360° – (∠P + ∠A + ∠N)
= 360° – (90° + 110° + 85°)
= 360°- 285° = 75° ….. (A)
Steps of Construction:
Step-1: Draw PL = 4 cm.
Step-2 : Make ∠LPX = 90° and ∠PLY = 75°. [From (A)]
Step-3 : Cut off LA = 6.5 cm on $$\overrightarrow{L Y}$$.
Step-4: Draw ∠LAZ = 110° which cut $$\overrightarrow{P X}$$ at point N.
Hence, PLAN is the required quadrilateral.
(iii) Since, opposite sides and angles of a parallelogram are equal i.e., ∠R = ∠E = 85°, HE = RA = 5 cm and EA = HR = 6 cm.
Steps of Construction:
Step-1: Draw HE = 5 cm.
Step-2 : Draw ∠HEX = 85°.
Step-3 : Cut off EA = 6 cm on $$\overrightarrow{E X}$$
Step-4: With H as centre and radius equal to 6 cm, draw an arc.
Step-5: With A as centre and radius equal to 5 cm, cut another arc on the arc drawn in step-4 at point R.
Step-6 : Join HR and RA.
Hence HEAR is the required parallelogram.
(iv) We know that each of the four angles of a rectangle is equal to 90° and opposite sides are also equal.
OK = YA and KA = OY
Steps of Construction:
Step-1: Draw OK = 7 cm.
Step-2 : Make ∠OKX = 90°.
Step-3 : Cut off KA = 5 cm on $$\overrightarrow{K X}$$.
Step-4: With O as centre and radius equal to 5 cm, cut an arc.
Step-5: With A as centre and radius equal to 7 cm cut another arc on the arc drawn in step-4 at point Y.
Step-6 : Join OY and YA.
Hence, OKAY is the required rectangle.
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# Solving two dimensional vector problems
### Solving two dimensional vector problems
#### Lessons
In this lesson, we will learn:
• How to solve two dimensional vector problems using the law of sines and the law of cosines
Notes:
• Often, vector equations in physics problems result in vector triangles which can be solved using trigonometry
• At least three pieces of information are needed to solve a triangle, which can be three side lengths (SSS), two side lengths and one angle (SSA, SAS), or one side length and two angles (SAA, ASA).
• Knowing three angles (AAA) does not let you solve a triangle since you will not be able to solve for the side lengths. There is no way to know the size of the triangle without more information.
• You can always solve a triangle that you know four or more pieces of information about.
• Vector triangles that do not contain right angles can be solved either by breaking vectors into their components or using the law of sines and the law of cosines, which are trigonometric laws that apply to all triangles
Law of Sines
$\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$
a,b,c: length of sides a,b,c
A,B,C: angles opposite sides a, b, c
Law of Cosines
$c^2 = a^2 + b^2 - 2ab \,cosC$
• Introduction
Introduction to solving two dimensional vector problems
• Overview of trigonometry strategies for different types of vector problems
• Definition of the law of sines and the law of cosines
• 1.
Use the law of sines to solve triangles
a)
Find the length of side c
b)
Find the angle &theta
• 2.
Use the law of cosines to solve triangles
a)
For the following triangle:
1. Find the length of the unknown side.
2. Find the angle θ
b)
A paper airplane is thrown 11.0 m [W], then thrown 14.0 m, then 16.0 m. The final throw returns it to its original position. Find the angle of the 14.0 m throw.
• 3.
Solve a vector word problem using the laws of sines and cosines
To get to school, Pauline leaves her house and walks due east 1.40 km, then takes a shortcut by walking 0.650 km [35° S of E] through a park. Find her displacement from home to school.
• 4.
Solve a difficult vector triangle using geometry
Solve the equation $\vec{A} + \vec{B} = \vec{C}$.
|
# Fractions Part 2: Constructing and Drawing
The standards (CCSS or any state) use varied verbs to describe what students are to do regarding fractions: form, compose, construct, model, partition, draw, decompose, share, identify, read, write, describe, order, and compare. Satisfying these standards can often be accomplished through use of concrete methods (manipulatives) and pictorial models (drawings). Remember the best understanding of concepts usually follows the concrete, pictorial, abstract progression (CPA). In other words, “Let’s make it, draw it, and then use numbers to represent it.”
Through constructing and drawing, students will be prepared for further work with fractions, and they begin to conceptualize the relationship between the size of denominators, the numerators, and the whole. Click here for a FREE copy of the pictures you will see below (3 page pdf).
Form / Compose / Construct / Model: Use smaller shapes to form or compose larger shapes (which is also a geometry std. in KG and 1st). Put together fraction pieces or puzzles to make the whole shape (circle, rectangle, hexagon, etc.). Use fraction pieces to demonstrate understanding by constructing models of area, set, and length.
These pictures show different ways students can use manipulatives to form, compose, construct, and model fractional parts (pattern blocks, fraction circles, tangrams, linking cubes, color tiles, fraction bars, Cuisenaire rods, two-color counters):
Partition / Draw / Decompose / Share: Split larger shapes into smaller fractional parts (halves, thirds, fourths, etc.). Divide (fair share) objects into equal groups. Use models to decompose a fraction in more than one way. Represent fractions on a number line.
I enjoy teaching children how to partition common shapes into fractional parts – because it involves drawing. Too often, if I just tell them to divide a rectangle or circle into fourths or sixths, I get something like this:
The child often hears “four” or “six” and draws that many lines or that many parts inside the shape, not paying attention to the size or equality of the parts.
Through the following methods, students practice an efficient way to partition shapes equally . . . and they soon start to see how halves, fourths, and eighths are related as well as how thirds, sixths, and twelfths are related.
Start with half: To partition/divide a shape in half, I draw one line down the middle. Show this in vertical, horizontal, and/or diagonal positions. (*See caveat below in the “what now” section for using diagonal lines.) To prove it is half, I should be able to fold or cut it so that both parts are equal. Practice dividing other common shapes in half (squares, diamonds, hexagons, trapezoids, etc.).
Fourths: To partition/divide a shape into fourths, I use this language while modeling because I usually get more precise, relatively equal sections without resorting to measuring.
• Divide in half.
• Then divide each half in half.
• When using a circle, relate the sections to quarter-hours on a clock (or the 3, 6, 9, 12 positions).
Eighths: To partition/divide a shape into eighths, I use this language (repeating how to get half, then fourths, and then adding 1 additional step):
• Divide in half.
• Divide each half in half.
• Divide each fourth in half.
Thirds: To partition/divide a shape into thirds is a little trickier. Just know the end result should be three relatively equal sections. I use this language while modeling (for a rectangle):
• Identify where half would be and put a little tick mark.
• Then draw one line on either side (adding 2 lines). Adjust as needed so all three parts look the same.
• When using a circle, relate the lines to a peace sign or position on a clock (12:00, 4:00, 8:00).
Sixths: To partition/divide a shape into sixths, there are a couple different ways, but I will just focus on one.
• Divide into thirds.
• Then divide each third in half.
• Relate the circle to numbers/positions on a clock. The lines will stretch from 12 to 6, from 2 to 8, and from 4 to 10.
Twelfths: To partition/divide shapes into twelfths, show thirds and sixths. Then divide each sixth section in half. On a circle, the lines should relate to all of the numbers on a clock.
Alternate method for rectangular bars and numberlines: If you are just requiring students to partition/divide one rectangular bar or number line, then making an open-ended rectangular bar would be a good method (see steps below). If students are drawing rectangular bars to compare two or more fractions, then the above steps should be followed because to compare fractions, each bar must be the same length. And it’s hard to draw bars the same length using this alternative method.
What now?
• Use the fraction bars or circles to illustrate fractions, compose and decompose, solve problems, compare, show equivalencies, etc.
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# Parallel Axis Theorem, Proof, Definition, Formula, Examples
## Parallel Axis Theorem
According to the parallel axis theorem, a body’s moment of inertia about an axis that passes through its center is equal to the sum of that body’s moment of inertia along its middle axis and the product of that body’s mass times the square of the distance between its two axes.
## Parallel Axis Theorem for Moment of Inertia
There is resistance when we try to adjust a body’s inertia. We call this resistance inertia. The modification may alter the body’s motion in terms of speed or direction. In the absence of external influences, bodies have a tendency to continue traveling in a straight line. Newton’s first law of motion, which is summarised as follows in his book Principia Mathematica, is based on this principle:
Everybody tries to maintain its current state, whether it be one of rest or uniformly moving along in a straight line, using a power of resistance known as the vis insita, or innate force of matter.
Check: Oscillatory Motion, Meaning, Definition, Example, Diagram
## Parallel Axis Theorem Definition
The “Parallel Axis Theorem” states that a body’s moment of inertia about any axis is equal to both its moment of inertia about a parallel axis that passes through its center of mass and the product of that body’s mass and the square root of the distance between the axes.
## Parallel Axis Theorem Formula
The parallel axis theorem formula is as follows:
## Parallel Axis Theorem Proof Derivation
In order to establish the parallel axis theorem, we must demonstrate that a body’s moment of inertia is the product of its moment of inertia about its center, its mass, and the square of the distance between its two axes.
Let’s assume that Ic represents the body’s moment of inertia along an axis that runs through its center. I is the same body’s moment of inertia around the axis that is h away from the center, which is A’B’.
We shall now assume that mass m is present at this position r if r is the distance from this body’s center of gravity.
The distance from A’B’ then becomes h+r.
This is the parallel axis theorem’s definitive formula.
When one of the axes travels through the rod’s center and the other, let’s say, runs through one end, the parallel axis theorem can be used to determine the rod’s moment of inertia.
If we use the parallel axis theorem, the equation for a rod’s moment of inertia is given as ML2/12. Observe how:
A rod’s moment of inertia is I = ML2/3.
L is the length of the rod divided by its diameter at one end.
The parallel axis theorem then states:
## Parallel Axis Theorem Equations Important points
The following factors need to be taken into account in order to use the parallel axis theorem.
• Axis A and axis B must be parallel.
• The center of mass of the body must be on axis B.
• They must be as close together as is physically practicable.
Must know: What is the Unit of Current, Resistance and Voltage?
## Parallel Axis Theorem Examples
Parallel Axis Theorem Q1: If the moment of inertia of a body along a perpendicular axis passing through its centre of gravity is 50 kg·m2 and the mass of the body is 30 Kg. What is the moment of inertia of that body along another axis which is 50 cm away from the current axis and parallel to it? Use Parallel Axis Theorem Formula
Solution: From the parallel axis theorem,
I = IG + Mb2
I = 50 + ( 30 × 0.52 )
I = 57.5 kg – m2
Parallel Axis Theorem Q2: Calculate the moment of inertia of a rod whose mass is 30 kg and length is 30 cm?
Solution: The parallel axis formula for a rod is given as,
I = (1/12) ML2
plugging in the values we get
I = 0.225 Kg m2.
Parallel Axis Theorem Q3: Calculate the moment of inertia of a stick whose mass is 100 gm and length is 10 cm?
Solution: The parallel axis formula for a rod is given as,
I = (1/12) ML2
plugging in the values we get
I = 0.0000833 Kg m2.
Must know: Trigonometry Table- (0 to 360) Formula, Value, Chart, Ratio, PDF for Class 10, 12
Parallel Axis Theorem is useful in finding the Area Moment of Inertia
The parallel axis theorem was developed to determine an object’s moment of inertia when the axis passed outside of the central axis. Because of this, calculations are made to be simple, especially for bodies with irregular shapes. It is possible to use the parallel axis theorem with both 2D and 3D objects.
Even while a 3D item cannot be used as a whole, it can be used if it is broken up into smaller pieces, or laminas. The parallel axis theorem is then used to solve each lamina independently. It only works with stiff bodies. Additionally, there is an inertia matrix that changes depending on the person’s point of reference.
### Parallel Axis Theorem- QNAs
Que. What is the statement of the parallel axis?
Ans. According to the parallel axis theorem, a body’s moment of inertia about an axis that is parallel to its axis of mass is equal to the product of its moment of inertia about its axis of mass, the product of mass, and square of the distance between the two axes.
Que. What are the parallel axis and perpendicular axis theorems?
Ans. The parallel axis theorem states that a body’s moment of inertia about any axis is equal to the product of the body’s mass and the square of the perpendicular distance between the two axes, as well as the moment of inertia about a parallel axis that passes through its center of gravity.
Que. What is the parallel axis theorem used in the moment of inertia?
Ans. According to the parallel axis theorem, a body’s moment of inertia about any axis equals its moment of inertia about a parallel axis through its center of mass plus the sum of its mass and the square of the perpendicular distance between its two parallel axes.
Que. Is the parallel axis theorem always true?
Ans. It is really important to note that the parallel axis theorem is only true for Ic or IG – you cannot use the moment of inertia about another point in this formula. It’s also clear that the added term will always be positive (area, mass, and the squared distance cannot be negative).
Que. What is the formula for parallel axis theorem transfer?
Ans. The parallel axis theorem formula is I=Icm+mr2 I = I c m + m r 2
You are watching: Parallel Axis Theorem, Proof, Definition, Formula, Examples. Info created by THVinhTuy selection and synthesis along with other related topics.
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# If A= <2 ,-3 ,9 > and B= <0 , 3, 7 >, what is A*B -||A|| ||B||?
Jun 28, 2018
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 54 - \sqrt{94} \sqrt{58}$
#### Explanation:
We have:
$\boldsymbol{\underline{A}} = \left\langle2 , - 3 , 9\right\rangle$ and $\boldsymbol{\underline{B}} = \left\langle0 , 3 , 7\right\rangle$
And so we compute the Scalar (or dot product):
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = \left\langle2 , - 3 , 9\right\rangle \cdot \left\langle0 , 3 , 7\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(2\right) \left(0\right) + \left(- 3\right) \left(3\right) + \left(9\right) \left(7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 - 9 + 63$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 54$
And we compute the vector norms (or magnitudes):
$| | \boldsymbol{\underline{A}} | | = | | \left\langle2 , - 3 , 9\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{\left\langle2 , - 3 , 9\right\rangle \cdot \left\langle2 , - 3 , 9\right\rangle}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(9\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{4 + 9 + 81}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{94}$
Similarly,
$| | \boldsymbol{\underline{B}} | | = | | \left\langle0 , 3 , 7\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{\left\langle0 , 3 , 7\right\rangle \cdot \left\langle0 , 3 , 7\right\rangle}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{\left(0\right)}^{2} + {\left(3\right)}^{2} + {\left(7\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{0 + 9 + 49}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{58}$
So that:
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 54 - \sqrt{94} \sqrt{58}$
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# Implicit differentiation
## The Fundamentals of Implicit Differentiation: How to Differentiate Implicit Functions
Implicit functions, also known as equations with two unknown variables, x and y, can be challenging to solve for y in terms of x using standard differentiation methods. This is where implicit differentiation comes into play.
Implicit differentiation is a technique that enables us to find the derivative of y with respect to x (dy/dx) without having to isolate y. It is a valuable tool for defining the slope and curvature of curves represented by implicit functions. Additionally, implicit differentiation can be used to construct equations for tangents and normal lines of these curves.
## The Process of Implicit Differentiation
To perform implicit differentiation, we use a modified version of the chain rule for equations in the form of x + y = a, assuming that y is a function of x. The process involves differentiating both sides of the equation with respect to x. For terms involving y, we multiply by dy/dx. Finally, we can solve for dy/dx. Let's look at an example to better understand this concept.
Example: Find the derivative of a circle defined by the equation (x + y)^2 = 9.
Solution: We differentiate each part of the equation with respect to x. However, for the y term, we also multiply by dy/dx. Hence, we get (x + y)^2 * (1 + dy/dx) = 0. Solving for dy/dx, we get dy/dx = - x / (y + x).
## Implicit Differentiation of Higher Orders
For higher order differentiation, we follow the same process with a slight modification. To find the second derivative, we differentiate the first derivative, and for the third derivative, we differentiate the second derivative, and so on. We can use the formula dyn/dxn = [dy/dx * dy/dxn-1 + dn-1y/dxn-1 * dn-1y/dx] / [dy/dxn-1]^3 to generalize this process, where n represents the order of the derivative needed.
## Finding Equations for Tangents and Normals
We can use implicit differentiation to determine the equation of the tangent of a curve. The formula y - y1 = a (x - x1) can be utilized for this purpose, where (x1, y1) are the coordinates of the point of tangency, and a is the slope or derivative of the curve. Similarly, we can use the formula y - y1 = (-1/a) (x - x1) for finding the equation of the normal, where a represents the slope of the tangent.
Example: Find the equation of the tangent and normal to the curve defined by (x + y)^2 = 9 at the point where x = 1.
Solution: To find the point of tangency, we substitute x = 1 into the equation and solve for y, getting y = 2 or y = -2. We also need the slope of the tangent, which we can find using implicit differentiation. So, dy/dx = -1/2 at x = 1. Substituting these values into the equation for the tangent, we get y - 2 = (-1/2) (x - 1) or y + 2 = 2 (x - 1). Similarly, for the normal, we get y - 2 = (-2) (x - 1) or y + 2 = -1 (x - 1).
## Key Takeaways
• Implicit differentiation is a method for finding the derivative of an implicit function without having to solve for y in terms of x.
• All terms in the equation are differentiated, and the y term is multiplied by dy/dx.
• For higher order differentiation, we can use a formula to find the nth derivative.
• We can use implicit differentiation to find equations for tangents and normals of curves defined by implicit functions.
## Understanding Implicit Differentiation: A Valuable Tool for Solving Equations
Implicit differentiation is a method used to find the derivative of an implicit function, where both x and y are variables and are not isolated on one side of the equation. This technique is especially useful when solving equations where y is a function of x, not just a constant.
But when should you utilize implicit differentiation? It is most helpful when dealing with implicit functions, where it may be difficult or impossible to solve for y explicitly. In these cases, implicit differentiation can simplify the process and provide a quick and accurate solution.
So, how do we find dy/dx using this method? First, we differentiate the term with y and then rearrange the equation to isolate the dy/dx term on one side. This will give us the derivative expression for the function, allowing us to find the slope or rate of change at any given point.
Now, why is implicit differentiation a valuable tool? As mentioned earlier, it eliminates the need to solve for y before differentiating. This can save time and effort, particularly when dealing with complex equations.
Now armed with the basics of implicit differentiation, you can confidently tackle tricky equations. No longer will you have to struggle with isolating variables or solving for y. Happy differentiating!
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# A Sample Of N = 4 Scores Has Ss = 60. What Is The Variance For This Sample?
Contents
## How do you get the variance?
Steps for calculating the variance
1. Step 1: Find the mean.
2. Step 2: Find each score’s deviation from the mean.
3. Step 3: Square each deviation from the mean.
4. Step 4: Find the sum of squares.
5. Step 5: Divide the sum of squares by n – 1 or N.
## How is sample variance computed?
Definition of Sample Variance
The variance is mathematically defined as the average of the squared differences from the mean. … Step 1: Calculate the mean (the average weight). Step 2: Subtract the mean and square the result. Step 3: Work out the average of those differences.
## What is the standard formula of sample variance?
Sample Variance vs Population Variance
Sample Variance Population Variance
The formula for sample variance is given as ∑ni=1(xi−μ)2n−1 ∑ i = 1 n ( x i − μ ) 2 n − 1 The formula for population variance is equal to ∑ni=1(xi−μ)2n ∑ i = 1 n ( x i − μ ) 2 n
## Which set of characteristics will produce the smallest value for the estimated standard error Group of answer choices?
The scenario that will result in the smallest value for the standard error is option A: A large sample size and a small sample variance.
## What is variance in math?
The term variance refers to a statistical measurement of the spread between numbers in a data set. More specifically variance measures how far each number in the set is from the mean and thus from every other number in the set. Variance is often depicted by this symbol: σ2.
## What is a variance request?
A Variance is a request for relief from strict application of zoning regulations to alleviate an unusual hardship to a particular property. For example a homeowner may request that a room addition be permitted to cover more of the property than the Zoning Ordinance would normally allow.
## What is the variance of the sample mean?
The variance of the sampling distribution of the mean is computed as follows: That is the variance of the sampling distribution of the mean is the population variance divided by N the sample size (the number of scores used to compute a mean). … The variance of the sum would be σ2 + σ2 + σ2.
## How do you find sample variance in R?
In R sample variance is calculated with the var() function. In those rare cases where you need a population variance use the population mean to calculate the sample variance and multiply the result by (n-1)/n note that as sample size gets very large sample variance converges on the population variance.
## How do you calculate SP in statistics?
Note that the name is short for the sum of the products of corresponding deviation scores for two variables. To calculate the SP you first determine the deviation scores for each X and for each Y then you calculate the products of each pair of deviation scores and then (last) you sum the products.
## What is sample variance in statistics?
Sample variance (s2) is a measure of the degree to which the numbers in a list are spread out. If the numbers in a list are all close to the expected values the variance will be small. If they are far away the variance will be large.
## How do you find sample variance and standard deviation?
To calculate the variance you first subtract the mean from each number and then square the results to find the squared differences. You then find the average of those squared differences. The result is the variance. The standard deviation is a measure of how spread out the numbers in a distribution are.
## How do you compute for the variance and standard deviation of the sampling distribution of sample means?
The formula to find the variance of the sampling distribution of the mean is: σ2M = σ2 / N where: σ2M = variance of the sampling distribution of the sample mean.
## What does SS stand for in statistics?
The sum of squares or sum of squared deviation scores is a key measure of the variability of a set of data. The mean of the sum of squares (SS) is the variance of a set of scores and the square root of the variance is its standard deviation.
## When n is small less than 30 the T distribution is the following?
When n is small (less than 30) how does the shape of the t distribution compare to the normal distribution? It is taller and narrower than the normal distribution. It is almost perfectly normal.
## Which set of sample characteristics is most likely to produce a larger T value for the independent measures t statistic?
The sample with the smaller variance will produce the larger t statistic. If other factors are held constant which set of sample characteristics is most likely to reject a null hypothesis stating that = 80?
## What variance means?
Definition of variance
1 : the fact quality or state of being variable or variant : difference variation yearly variance in crops. 2 : the fact or state of being in disagreement : dissension dispute. 3 : a disagreement between two parts of the same legal proceeding that must be consonant.
## How do you find variability in math?
Measures of Variability: Variance
1. Find the mean of the data set. …
2. Subtract the mean from each value in the data set. …
3. Now square each of the values so that you now have all positive values. …
4. Finally divide the sum of the squares by the total number of values in the set to find the variance.
## How do you get a variance code?
Obtaining a variance is done through the local zoning board or planning commission and involves submitting an application and in most cases gaining approval at a hearing. Each municipality has its own rules and processes.
## How do you write a letter requesting variance?
Be polite direct and specific: “I am writing to seek a fence variance for my single-family home at (provide the address and the town.) Current zoning rules say that fences must be no more than 4 feet tall I respectfully request to install a fence that is 5 feet tall.”
## What is a code variance?
A variance is a deviation from the set of rules a municipality applies to land use and land development typically a zoning ordinance building code or municipal code. … A variance may also be known as a standards variance referring to the development standards contained in code.
## What is sample mean and sample variance?
A sample contains data collected from selected individuals taken from a larger population. We also learned that the sample mean is the arithmetic average of all the values in the sample. The sample variance measures how spread out the data is and the sample standard deviation is the square root of the variance.
## What can you say about the variance of the sample means and the variance of the population?
The mean of the sample means is the same as the population mean but the variance of the sample means is not the same as the population variance.
## What happens to the variance of the sampling distribution of the sample means when the sample size increases?
As sample sizes increase the sampling distributions approach a normal distribution. … As the sample sizes increase the variability of each sampling distribution decreases so that they become increasingly more leptokurtic.
## How do you find variance and standard deviation in R?
Sample variance and Standard Deviation using R
var(y) instructs R to calculate the sample variance of Y. In other words it uses n-1 ‘degrees of freedom’ where n is the number of observations in Y. sd(y) instructs R to return the sample standard deviation of y using n-1 degrees of freedom. sd(y) = sqrt(var(y)).
## How do you find the variance of a column in R?
1. Method 1: Get Variance of the column by column name.
2. Method 2: Get Variance of the column by column position.
3. ColVars() Function along with sapply() is used to get variance of the multiple column. …
4. summarise_if() Function along with var() function is used to get the variance of the multiple column .
## What does a high variance mean?
Variance measures how far a set of data is spread out. … A small variance indicates that the data points tend to be very close to the mean and to each other. A high variance indicates that the data points are very spread out from the mean and from one another.
## What is SP and CP?
Answer– CP and SP are abbreviations for Cost Price and Selling Price. Cost price is the amount we pay to buy an item at which it is available. Similarly Selling Price is the rate at which an article is sold which we abbreviate as SP.
## How do you compute the p value?
The p-value is calculated using the sampling distribution of the test statistic under the null hypothesis the sample data and the type of test being done (lower-tailed test upper-tailed test or two-sided test). The p-value for: a lower-tailed test is specified by: p-value = P(TS ts | H is true) = cdf(ts)
## What does N mean in research?
The letter “n” stands for the number of individuals we are looking at when studying an issue or calculating percentages. You may also see it expressed as “Total Responses.”
## How do you find the variance between two numbers?
The variance percentage calculation is the difference between two numbers divided by the first number then multiplied by 100.
## Is sample variance the same as standard deviation?
The variance is the average of the squared differences from the mean. … Standard deviation is the square root of the variance so that the standard deviation would be about 3.03. Because of this squaring the variance is no longer in the same unit of measurement as the original data.
## How do you calculate the sample standard deviation?
Here’s how to calculate sample standard deviation:
1. Step 1: Calculate the mean of the data—this is xˉx with bar on top in the formula.
2. Step 2: Subtract the mean from each data point. …
3. Step 3: Square each deviation to make it positive.
4. Step 4: Add the squared deviations together.
## How do you calculate sample statistics?
The following steps will show you how to calculate the sample mean of a data set:
1. Add up the sample items.
2. Divide sum by the number of samples.
3. The result is the mean.
4. Use the mean to find the variance.
5. Use the variance to find the standard deviation.
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# This chapter is part of Precalculus: An Investigation of Functions Lippman & Rasmussen 2011.
Chapter 6: Periodic Functions
In the previous chapter, the trigonometric functions were introduced as ratios of sides of a
triangle, and related to points on a circle. We noticed how the x and y values of the
points did not change with repeated revolutions around the circle by finding coterminal
angles. In this chapter, we will take a closer look at the important characteristics and
applications of these types of functions, and begin solving equations involving them.
Section 6.1 Sinusoidal Graphs .................................................................................... 353
Section 6.2 Graphs of the Other Trig Functions ......................................................... 369
Section 6.3 Inverse Trig Functions ............................................................................. 379
Section 6.4 Solving Trig Equations ............................................................................ 387
Section 6.5 Modeling with Trigonometric Equations ................................................. 397
Section 6.1 Sinusoidal Graphs
The London Eye
1
is a huge Ferris wheel with diameter
135 meters (443 feet) in London, England, which
completes one rotation every 30 minutes. When we
look at the behavior of this Ferris wheel it is clear that it
completes 1 cycle, or 1 revolution, and then repeats this
revolution over and over again.
This is an example of a periodic function, because the
Ferris wheel repeats its revolution or one cycle every 30
minutes, and so we say it has a period of 30 minutes.
In this section, we will work to sketch a graph of a
riders height over time and express the height as a
function of time.
Periodic Functions
A periodic function occurs when a specific horizontal shift, P, results in the original
function; where ) ( ) ( x f P x f = + for all values of x. When this occurs we call the
horizontal shift the period of the function.
You might immediately guess that there is a connection here to finding points on a circle,
since the height above ground would correspond to the y value of a point on the circle.
We can determine the y value by using the sine function. To get a better sense of this
functions behavior, we can create a table of values we know, and use them to sketch a
graph of the sine and cosine functions.
1
London Eye photo by authors, 2010, CC-BY
354 Chapter 6
Listing some of the values for sine and cosine on a unit circle,
0
6
t
4
t
3
t
2
t
3
2t
4
3t
6
5t
t
cos 1
2
3
2
2
2
1
0
2
1
2
2
2
3
-1
sin 0
2
1
2
2
2
3
1
2
3
2
2
2
1
0
Here you can see how for each angle, we use the y value of the point on the circle to
determine the output value of the sine function.
Plotting more points gives the full shape of the sine and cosine functions.
Notice how the sine values are positive between 0 and t which correspond to the values
of sine in quadrants 1 and 2 on the unit circle, and the sine values are negative between
t and t 2 representing quadrants 3 and 4.
6
t
4
t
3
t
2
t
f() = sin()
f() = sin()
Section 6.1 Sinusoidal Graphs 355
Like the sine function we can track the value of the cosine function through the 4
quadrants of the unit circle as we place it on a graph.
Both of these functions are defined on a domain of all real numbers, since we can
evaluate the sine and cosine of any angle. By thinking of sine and cosine as points on a
unit circle, it becomes clear that the range of both functions must be the interval ] 1 , 1 [ .
Domain and Range of Sine and Cosine
The domain of sine and cosine is all real numbers, 9 e x or ) , ( +
The range of sine and cosine is the interval [-1, 1]
Both these graphs are considered sinusoidal graphs.
In both graphs, the shape of the graph begins repeating after 2. Indeed, since any
coterminal angles will have the same sine and cosine values, we could conclude that
) sin( ) 2 sin( u t u = + and ) cos( ) 2 cos( u t u = + .
In other words, if you were to shift either graph horizontally by 2, the resulting shape
would be identical to the original function. Sinusoidal functions are a specific type of
periodic function.
Period of Sine and Cosine
The period is 2 for both the sine and cosine function.
Looking at these functions on a domain centered at the vertical axis helps reveal
symmetries.
g() = cos()
356 Chapter 6
sine cosine
The sine function is symmetric about the origin, the same symmetry the cubic function
has, making it an odd function. The cosine function is clearly symmetric about the y axis,
the same symmetry as the quadratic function, making it an even function.
Negative Angle Identities
The sine is an odd function, symmetric about the origin, so ) sin( ) sin( u u =
The cosine is an even function, symmetric about the y-axis, so ) cos( ) cos( u u =
These identities can be used, among other purposes, for helping with simplification and
proving identities.
You may recall the cofunction identity from last chapter;
|
.
|
\
|
= u
t
u
2
cos ) sin( .
Graphically, this tells us that the sine and cosine graphs are horizontal transformations of
each other. We can prove this by using the cofunction identity and the negative angle
identity for cosine.
|
.
|
\
|
=
|
|
.
|
\
|
|
.
|
\
|
= |
.
|
\
|
+ = |
.
|
\
|
=
2
cos
2
cos
2
cos
2
cos ) sin(
t
u
t
u
t
u u
t
u
Now we can clearly see that if we horizontally shift the cosine function to the right by /2
we get the sine function.
Remember this shift is not representing the period of the function. It only shows that the
cosine and sine function are transformations of each other.
Example 1
Simplify
) tan(
) sin(
u
u
) tan(
) sin(
u
u
Using the even/odd identity
=
) tan(
) sin(
u
u
Rewriting the tangent
Section 6.1 Sinusoidal Graphs 357
=
) cos(
) sin(
) sin(
u
u
u
Inverting and multiplying
=
) sin(
) cos(
) sin(
u
u
u Simplifying we get
= ) cos(u
Transforming Sine and Cosine
Example 2
A point rotates around a circle of radius 3.
Sketch a graph of the y coordinate of the
point.
Recall that for a point on a circle of radius r,
the y coordinate of the point is ) sin(u r y = ,
so in this case, we get the
equation ) sin( 3 ) ( u u = y .
Since the 3 is multiplying the function, this causes a vertical stretch of the y values of
the function by 3.
Notice that the period of the function does not change.
Since the outputs of the graph will now oscillate between -3 and 3, we say that the
amplitude of the sine wave is 3.
Try it Now
1. What is the amplitude of the equation ) cos( 7 ) ( u u = f ? Sketch a graph of the
function.
Example 3
A circle with radius 3 feet is mounted with its center 4
feet off the ground. The point closest to the ground is
labeled P. Sketch a graph of the height above ground of
the point P as the circle is rotated, then find an equation
for the height.
3 ft
4 ft
358 Chapter 6
Sketching the height, we note that it will
start 1 foot above the ground, then increase
up to 7 feet above the ground, and continue
to oscillate 3 feet above and below the
center value of 4 feet.
Although we could use a transformation of
either the sine or cosine function, we start by
looking for characteristics that would make
one function easier than the other.
We decide to use a cosine function because it starts at the highest or lowest value, while
a sine function starts at the middle value. We know it has been reflected because a
standard cosine starts at the highest value, and this graph starts at the lowest value.
Second, we see that the graph oscillates 3 above and below the center, while a basic
cosine has an amplitude of one, so this graph has been vertically stretched by 3, as in
the last example.
Finally, to move the center of the circle up to a height of 4, the graph has been vertically
shifted up by 4. Putting these transformations together,
4 ) cos( 3 ) ( + = u u h
Midline
The center value of a sinusoidal function, the value that the function oscillates above
and below, is called the midline of the function, represented by the vertical shift in the
equation.
The equation k f + = ) cos( ) ( u u has midline at y = k.
Try it Now
2. What is the midline of the equation 4 ) cos( 3 ) ( = u u f ? Sketch a graph of the
function.
To answer the Ferris wheel problem at the beginning of the section, we need to be able to
express our sine and cosine functions at inputs of time. To do so, we will utilize
composition. Since the sine function takes an input of an angle, we will look for a
function that takes time as an input and outputs an angle. If we can find a suitable
) (t u function, then we can compose this with our ) cos( ) ( u u = f function to obtain a
sinusoidal function of time: )) ( cos( ) ( t t f u =
Section 6.1 Sinusoidal Graphs 359
Example 4
A point completes 1 revolution every 2 minutes around circle of radius 5. Find the x
coordinate of the point as a function of time.
Normally, we would express the x coordinate of a point on a unit circle
using ) cos(u r x = , here we write the function ) cos( 5 ) ( u u = x .
The rotation rate of 1 revolution every 2 minutes is an angular velocity. We can use this
rate to find a formula for the angle as a function of time. Since the point rotates 1
revolution = 2 radians every 2 minutes, it
rotates radians every minute. After t
minutes, it will have rotated:
t t t u = ) ( radians
Composing this with the cosine function,
we obtain a function of time.
) cos( 5 )) ( cos( 5 ) ( t t t x t u = =
Notice that this composition has the effect of a horizontal compression, changing the
period of the function.
To see how the period is related to the stretch or compression coefficient B in the
equation ( ) Bt t f sin ) ( = , note that the period will be the time it takes to complete one full
revolution of a circle. If a point takes P minutes to complete 1 revolution, then the
angular velocity is
minutes
P
t
. Then t
P
t
t
u
2
) ( = . Composing with a sine function,
|
.
|
\
|
= = t
P
t t f
t
u
2
sin )) ( sin( ) (
From this, we can determine the relationship between the equation form and the period:
P
B
t 2
= . Notice that the stretch or compression coefficient B is a ratio of the normal
period of a sinusoidal function to the new period. If we know the stretch or
compression coefficient B, we can solve for the new period:
B
P
t 2
= .
x()
r
x
360 Chapter 6
Example 5
What is the period of the function
|
.
|
\
|
= t t f
6
sin ) (
t
?
Using the relationship above, the stretch/compression factor is
6
t
= B , so the period
will be 12
6
2
6
2 2
= = = =
t
t
t
t t
B
P .
While it is common to compose sine or cosine with functions involving time, the
composition can be done so that the input represents any reasonable quantity.
Example 6
A bicycle wheel with radius 14 inches has the bottom-most point on the wheel marked
in red. The wheel then begins rolling down the street. Write a formula for the height
above ground of the red point after the bicycle has travelled x inches.
The height of the point begins at the lowest value, 0,
increases to the highest value of 28 inches, and
continues to oscillate above and below a center height
of 14 inches. In terms of the angle of rotation, :
14 ) cos( 14 ) ( + = u u h
In this case, x is representing a linear distance the
wheel has travelled, corresponding to an arclength
along the circle. Since arclength and angle can be
related by u r s = , in this case we can write u 14 = x ,
which allows us to express the angle in terms of x:
14
) (
x
x = u
Composing this with our cosine-based function from above,
14
14
1
cos 14 14
14
cos 14 )) ( ( ) ( + |
.
|
\
|
= + |
.
|
\
|
= = x
x
x h x h u
The period of this function would be t t
t t
28 14 2
14
1
2 2
= = = =
B
P , the circumference
of the circle. This makes sense the wheel completes one full revolution after the
bicycle has travelled a distance equivalent to the circumference of the wheel.
Starting
Rotated by
14in
x
Section 6.1 Sinusoidal Graphs 361
Summarizing our transformations so far:
Transformations of Sine and Cosine
Given an equation in the form ( ) k Bt A t f + = sin ) ( or ( ) k Bt A t f + = cos ) (
A is the vertical stretch, and is the amplitude of the function.
B is the horizontal stretch/compression, and is related to the period, P, by
B
P
t 2
=
k is the vertical shift, determines the midline of the function
Example 7
Determine the midline, amplitude, and period of the function ( ) 1 2 sin 3 ) ( + = t t f .
The amplitude is 3
The period is t
t t
= = =
2
2 2
B
P
The midline is at 1 ) ( = t g
Amplitude, midline, and period, when combined with vertical flips, are enough to allow
us to write equations for a large number of sinusoidal situations.
Try it Now
3. If a sinusoidal function starts on the midline at point (0,3), has an amplitude of 2,
and a period of 4, write an equation with these features.
y = k
A
A
P
P
362 Chapter 6
Example 8
Write an equation for the sinusoidal
function graphed here.
The graph oscillates from a low of -1 to a
high of 3, putting the midline at y = 1,
halfway between.
The amplitude will be 2, the distance from
the midline to the highest value (or lowest
value) of the graph.
The period of the graph is 8. We can measure this from the first peak at x = -2 to the
second at x = 6. Since the period is 8, the stretch/compression factor we will use will be
4 8
2 2 t t t
= = =
P
B
At x = 0, the graph is at the midline value, which tells us the graph can most easily be
represented as a sine function. Since the graph then decreases, this must be a vertical
reflection of the sine function. Putting this all together,
1
4
sin 2 ) ( +
|
.
|
\
|
= t t f
t
With these transformations, we are ready to answer the Ferris wheel problem from the
beginning of the section.
Example 9
The London Eye is a huge Ferris wheel with diameter 135 meters (443 feet) in London,
England, which completes one rotation every 30 minutes. Riders board from a platform
2 meters above the ground. Express a riders height as a function of time.
With a diameter of 135 meters, the wheel has a radius of 67.5 meters. The height will
oscillate with amplitude of 67.5 meters above and below the center.
Passengers board 2 meters above ground level, so the center of the wheel must be
located 67.5 + 2 = 69.5 meters above ground level. The midline of the oscillation will
be at 69.5 meters.
The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with
period of 30 minutes.
Lastly, since the rider boards at the lowest point, the height will start at the smallest
value and increase, following the shape of a flipped cosine curve.
Section 6.1 Sinusoidal Graphs 363
Putting these together:
Amplitude: 67.5
Midline: 69.5
Period: 30, so
15 30
2 t t
= = B
Shape: -cos
An equation for the riders height would be
5 . 69
15
cos 5 . 67 ) ( + |
.
|
\
|
= t t h
t
Try it Now
4. The Ferris wheel at the Puyallup Fair
2
has a diameter of about 70
feet and takes 3 minutes to complete a full rotation. Passengers
board from a platform 10 feet above the ground. Write an
equation for a riders height over time.
While these transformations are sufficient to represent a majority of situations,
occasionally we encounter a sinusoidal function that does not have a vertical intercept at
the lowest point, highest point, or midline. In these cases, we need to use horizontal
shifts. Recall that when the inside of the function is factored, it reveals the horizontal
shift.
Horizontal Shifts of Sine and Cosine
Given an equation in the form ( ) k h t B A t f + = ) ( sin ) ( or ( ) k h t B A t f + = ) ( cos ) (
h is the horizontal shift of the function
Example 10
Sketch a graph of |
.
|
\
|
=
4 4
sin 3 ) (
t t
t t f
To reveal the horizontal shift, we first need to factor inside the function:
|
.
|
\
|
= ) 1 (
4
sin 3 ) ( t t f
t
2
Photo by photogirl7.1, http://www.flickr.com/photos/kitkaphotogirl/432886205/sizes/z/, CC-BY
364 Chapter 6
This graph will have the shape of a sine function, starting at the midline and increasing,
with an amplitude of 3. The period of the graph will be 8
4
2
4
2 2
= = = =
t
t
t
t t
B
P .
Finally, the graph will be shifted to the right by 1.
In some physics and mathematics books, you will hear the horizontal shift referred to as
phase shift. In other physics and mathematics books, they would say the phase shift of
the equation above is
4
t
, the value in the unfactored form. Because of this ambiguity, we
will not use the term phase shift any further, and will only talk about the horizontal shift.
Example 11
Write an equation for the function graphed here.
With highest value at 1 and lowest value at -5,
the midline will be halfway between at -2.
The distance from the midline to the highest or
lowest value gives an amplitude of 3.
The period of the graph is 6, which can be
measured from the peak at x = 1 to the second
peak at x = 7, or from the distance between the lowest points. This gives for our
equation
3 6
2 2 t t t
= = =
P
B
For the shape and shift, we have an option. We could either write this as:
A cosine shifted 1 to the right
A negative cosine shifted 2 to the left
A sine shifted to the left
A negative sine shifted 2.5 to the right
Section 6.1 Sinusoidal Graphs 365
While any of these would be fine, the cosine shifts are clearer than the sine shifts in this
case, because they are integer values. Writing these:
2 ) 1 (
3
cos 3 ) ( |
.
|
\
|
= x x y
t
or
2 ) 2 (
3
cos 3 ) ( |
.
|
\
|
+ = x x y
t
Again, these equations are equivalent, so both describe the graph.
Try it Now
5. Write an equation for the function graphed
here.
Important Topics of This Section
Periodic functions
Sine & Cosine function from the unit circle
Domain and Range of Sine & Cosine function
Sinusoidal functions
Negative angle identity
Simplifying expressions
Transformations
Amplitude
Midline
Period
Horizontal shifts
1. 7
2. -4
3. ( ) 2sin 3
2
f x x
t | |
= +
|
\ .
4.
2
( ) 35cos 45
3
h t t
t | |
= +
|
\ .
5. Two possibilities: ( ) 4cos ( 3.5) 4
5
f x x
t | |
= +
|
\ .
or ( ) 4sin ( 1) 4
5
f x x
t | |
= +
|
\ .
366 Chapter 6
Section 6.1 Exercises
1. Sketch a graph of ( ) ( ) 3sin f x x =
2. Sketch a graph of ( ) ( ) 4sin f x x =
3. Sketch a graph of ( ) ( ) 2cos f x x =
4. Sketch a graph of ( ) ( ) 4cos f x x =
For the graphs below, determine the amplitude, midline, and period, then write an
equation for the graph.
5. 6.
7. 8.
9. 10.
Section 6.1 Sinusoidal Graphs 367
For each of the following equations, find the amplitude, period, horizontal shift, and
midline.
11. 3sin(8( 4)) 5 y x = + +
12. 4sin ( 3) 7
2
y x
t | |
= +
|
\ .
13. 2sin(3 21) 4 y x = +
14. 5sin(5 20) 2 y x = +
15. sin 3
6
y x
t
t
| |
= +
|
\ .
16.
7 7
8sin 6
6 2
y x
t t | |
= + +
|
\ .
Find a formula for each of the graphs shown below.
17.
18.
368 Chapter 6
19.
20.
21. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the temperature is 50 degrees at midnight and the high and low temperature
during the day are 57 and 43 degrees, respectively. Assuming t is the number of hours
since midnight, find an equation for the temperature, D, in terms of t.
22. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the temperature is 68 degrees at midnight and the high and low temperature
during the day are 80 and 56 degrees, respectively. Assuming t is the number of hours
since midnight, find an equation for the temperature, D, in terms of t.
23. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meters
above the ground. The six o'clock position on the Ferris wheel is level with the
loading platform. The wheel completes 1 full revolution in 10 minutes. The function
( ) h t gives your height in meters above the ground t minutes after the wheel begins to
turn.
a. Find the amplitude, midline, and period of ( ) h t
b. Find an equation for the height function ( ) h t
c. How high are you off the ground after 5 minutes?
24. A Ferris wheel is 35 meters in diameter and boarded from a platform that is 3 meters
above the ground. The six o'clock position on the Ferris wheel is level with the
loading platform. The wheel completes 1 full revolution in 8 minutes. The function
( ) h t gives your height in meters above the ground t minutes after the wheel begins to
turn.
a. Find the amplitude, midline, and period of ( ) h t
b. Find an equation for the height function ( ) h t
c. How high are you off the ground after 4 minutes?
Section 6.2 Graphs of the Other Trig Functions 369
Section 6.2 Graphs of the Other Trig Functions
In this section, we will explore the graphs of the other four trigonometric functions.
Well begin with the tangent function. Recall that in chapter 5 we defined tangent as y/x
or sine/cosine, so you can think of the tangent as the slope of a line from the origin at the
given angle. At an angle of 0, the line would be horizontal with a slope of zero. As the
angle increases towards /2, the slope increases more and more. At an angle of /2, the
line would be vertical and the slope would be undefined. Immediately past /2, the line
would be decreasing and very steep giving a large negative tangent value. There is a
break in the function at /2, where the tangent value jumps from large positive to large
negative.
We can use these ideas along with the definition of
tangent to sketch a graph. Since tangent is defined
as sine/cosine, we can determine that tangent will
be zero when sine is zero: at -, 0, , and so on.
Likewise, tangent will be undefined when cosine is
zero: at -/2, /2, and so on.
The tangent is positive from 0 to /2 and to 3/2,
corresponding to quadrants 1 and 3 of the unit
circle.
Using technology, we can obtain a graph of tangent on a standard grid.
Notice that the graph appears to repeat itself. For
any angle on the circle, there is a second angle with
the same slope and tangent value halfway around the
circle, so the graph repeats itself with a period of ;
we can see one continuous cycle from - /2 to /2,
before it jumps & repeats itself.
The graph has vertical asymptotes and the tangent is
undefined wherever a line at the angle would be
vertical at /2, 3/2, and so on. While the domain
of the function is limited in this way, the range of the
function is all real numbers.
Features of the Graph of Tangent
The graph of the tangent function ) tan( ) ( u u = m
The period of the tangent function is
The domain of the tangent function is t
t
u k + =
2
, where k is an integer
The range of the tangent function is all real numbers, 9 e x or ) , ( +
370 Chapter 6
With the tangent function, like the sine and cosine functions, horizontal
stretches/compressions are distinct from vertical stretches/compressions. The horizontal
stretch can typically be determined from the period of the graph. With tangent graphs, it
is often necessary to solve for a vertical stretch using a point on the graph.
Example 1
Write an equation for the function
graphed here.
The graph has the shape of a tangent
function, however the period appears to
be 8. We can see one full continuous
cycle from -4 to 4, suggesting a
horizontal stretch. To stretch to 8, the
input values would have to be
multiplied by
t
8
. Since the value in the
equation to give this stretch is the
reciprocal, the equation must have form
|
.
|
\
|
= u
t
u
8
tan ) ( a f
We can also think of this the same way we did with sine and cosine. The period of the
tangent function is t but it has been transformed and now it is 8, remember the ratio of
the normal period to the new period is
8
t
and so this becomes the value on the
inside of the function that tells us how it was horizontally stretched.
To find the vertical stretch a, we can use a point on the graph. Using the point (2, 2)
|
.
|
\
|
= |
.
|
\
|
=
4
tan 2
8
tan 2
t t
a a . Since 1
4
tan =
|
.
|
\
| t
, a = 2
This graph would have equation
|
.
|
\
|
= u
t
u
8
tan 2 ) ( f
Try it Now
1. Sketch a graph of
|
.
|
\
|
= u
t
u
6
tan 3 ) ( f
Section 6.2 Graphs of the Other Trig Functions 371
For the graph of secant, we remember the reciprocal identity where
) cos(
1
) sec(
u
u = .
Notice that the function is undefined when the cosine is 0, leading to a vertical asymptote
in the graph at /2, 3/2, etc. Since the cosine is always less than one in absolute value,
the secant, being the reciprocal, will always be greater than one in absolute value. Using
technology, we can generate the graph. The graph of the cosine is shown dashed so you
can see the relationship.
) cos(
1
) sec( ) (
u
u u = = f
The graph of cosecant is similar. In fact, since
|
.
|
\
|
= u
t
u
2
cos ) sin( , it follows that
|
.
|
\
|
= u
t
u
2
sec ) csc( , suggesting the cosecant graph is a horizontal shift of the secant
graph. This graph will be undefined where sine is 0. Recall from the unit circle that this
occurs at 0, , 2, etc. The graph of sine is shown dashed along with the graph of the
cosecant.
) sin(
1
) csc( ) (
u
u u = = f
372 Chapter 6
Features of the Graph of Secant and Cosecant
The secant and cosecant graphs have period 2 like the sine and cosine functions.
Secant has domain t
t
u k + =
2
, where k is an integer
Cosecant has domain t u k = , where k is an integer
Both secant and cosecant have range of ) , 1 [ ] 1 , (
Example 2
Sketch a graph of 1
2
csc 2 ) ( +
|
.
|
\
|
= u
t
u f . What is the domain and range of this
function?
The basic cosecant graph has vertical asymptotes at the multiples of . Because of the
factor
2
t
in the equation, the graph will be compressed by
t
2
, so the vertical
asymptotes will be compressed to k k 2
2
= = t
t
u . In other words, the graph will have
vertical asymptotes at the multiples of 2, and the domain will correspondingly be
k 2 = u , where k is an integer.
The basic sine graph has a range of [-1, 1]. The vertical stretch by 2 will stretch this to
[-2, 2], and the vertical shift up 1 will shift the range of this function to [-1, 3].
The basic cosecant graph has a range of ) , 1 [ ] 1 , ( . The vertical stretch by 2 will
stretch this to ) , 2 [ ] 2 , ( , and the vertical shift up 1 will shift the range of this
function to ) , 3 [ ] 1 , (
Sketching a graph,
Notice how the graph of the transformed cosecant relates to the graph of
1
2
sin 2 ) ( + |
.
|
\
|
= u
t
u f shown dashed.
Section 6.2 Graphs of the Other Trig Functions 373
Try it Now
2. Given the graph 1
2
cos 2 ) ( + |
.
|
\
|
= u
t
u f
shown, sketch the graph of
1
2
sec 2 ) ( + |
.
|
\
|
= u
t
u g on the same axes.
Finally, well look at the graph of cotangent. Based on its definition as the ratio of cosine
to sine, it will be undefined when the sine is zero at at 0, , 2, etc. The resulting graph
is similar to that of the tangent. In fact, it is horizontal flip and shift of the tangent
function.
) sin(
) cos(
) tan(
1
) cot( ) (
u
u
u
u u = = = f
Features of the Graph of Cotangent
The cotangent graph has period
Cotangent has domain t u k = , where k is an integer
Cotangent has range of all real numbers, 9 e x or ) , ( +
In 6.1 we determined that the sine function was an odd function and the cosine was an
even function by observing the graph, establishing the negative angle identities for cosine
and sine. Similarily, you may notice that the graph of the tangent function appears to be
odd. We can verify this using the negative angle identities for sine and cosine:
( )
( )
( )
( )
( )
( ) u
u
u
u
u
u tan
cos
sin
cos
sin
tan =
=
The secant, like the cosine it is based on, is an even function, while the cosecant, like the
sine, is an odd function.
374 Chapter 6
Negative Angle Identities Tangent, Cotangent, Secant and Cosecant
( ) ( ) u u tan tan = ( ) ( ) u u cot cot =
( ) ( ) u u sec sec = ( ) ( ) u u csc csc =
Example 3
Prove that ( )
|
.
|
\
|
=
2
cot tan
t
u u
( ) u tan Using the definition of tangent
( )
( ) u
u
cos
sin
= Using the cofunction identities
|
.
|
\
|
|
.
|
\
|
=
u
t
u
t
2
sin
2
cos
Using the definition of cotangent
|
.
|
\
|
= u
t
2
cot Factoring a negative from the inside
|
|
.
|
\
|
|
.
|
\
|
=
2
cot
t
u Using the negative angle identity for cot
|
.
|
\
|
=
2
cot
t
u
Important Topics of This Section
The tangent and cotangent functions
Period
Domain
Range
The secant and cosecant functions
Period
Domain
Range
Transformations
Negative Angle identities
Section 6.2 Graphs of the Other Trig Functions 375
1.
2.
376 Chapter 6
Section 6.2 Exercises
Match the trigonometric function with one of the graphs
1. ( ) ( ) tan f x x = 2. ( ) ( ) sec x x f =
3. ( ) csc( ) f x x = 4. ( ) ( ) cot f x x =
I II
III IV
Find the period and horizontal shift of each of the following functions.
5. ( ) ( ) 2tan 4 32 f x x =
6. ( ) ( ) 3tan 6 42 g x x = +
7. ( ) ( ) 2sec 1
4
h x x
t | |
= +
|
\ .
8. ( ) 3sec 2
2
k x x
t | | | |
= +
| |
\ .
\ .
9. ( ) 6csc
3
m x x
t
t
| |
= +
|
\ .
10. ( )
5 20
4csc
3 3
n x x
t t | |
=
|
\ .
Section 6.2 Graphs of the Other Trig Functions 377
11. Sketch a graph of #7 above
12. Sketch a graph of #8 above
13. Sketch a graph of #9 above
14. Sketch a graph of #10 above
15. Sketch a graph of ( ) tan
2
j x x
t | |
=
|
\ .
16. Sketch a graph of ( ) 2tan
2
p t t
t | |
=
|
\ .
Write an equation for each of the graphs shown
17. 18.
19. 20.
378 Chapter 6
21. If tan 1.5 x = , find ( ) tan x
22. If tan 3 x = , find ( ) tan x
23. If sec 2 x = , find ( ) sec x
24. If sec 4 x = , find ( ) sec x
25. If csc 5 x = , find ( ) csc x
26. If csc 2 x = , find ( ) csc x
Simplify each of the following expressions completely
27. ( ) ( ) ( ) cot cos sin x x x +
28. ( ) ( ) ( ) cos tan sin x x x +
Section 6.3 Inverse Trig Functions 379
Section 6.3 Inverse Trig Functions
While in the previous sections we have evaluated the trigonometric functions, at times we
need to know what angle would give a specific sine, cosine, or tangent value. For this,
we need an inverse. Recall that for a one-to-one function, if b a f = ) ( , then an inverse
function would satisfy a b f =
) (
1
.
You probably are already recognizing an issue that the sine, cosine, and tangent
functions are not one-to-one functions. To define an inverse of these functions, we will
need to restrict the domain of these functions to so that they are one-to-one. We choose a
domain for each function which includes the angle of zero.
Sine, limited to
(
2
,
2
t t
Cosine, limited to | | t , 0 Tangent, limited to ,
2 2
t t | |
|
\ .
On these restricted domains, we can define the inverse sine and cosine and tangent
functions.
Inverse Sine, Cosine, and Tangent Functions
For angles in the interval
(
2
,
2
t t
, if ( ) a = u sin , then ( ) u =
a
1
sin
For angles in the interval | | t , 0 , if ( ) a = u cos , then ( ) u =
a
1
cos
For angles in the interval |
.
|
\
|
2
,
2
t t
, if ( ) a = u tan , then ( ) u =
a
1
tan
( )
1
sin x
(
2
,
2
t t
( )
1
cos x
( )
1
tan x
.
|
\
|
2
,
2
t t
380 Chapter 6
The ( )
1
sin x
The ( )
1
cos x
The ( )
1
tan x
## is sometimes called the arctangent function, and notated ( ) a arctan
The graphs of the inverse functions are shown here.
( )
1
sin x
( )
1
cos x
( )
1
tan x
Notice that the output of the inverse functions is an angle.
Example 1
Evaluate
a) |
.
|
\
|
2
1
sin
1
b)
|
|
.
|
\
|
2
2
sin
1
c)
|
|
.
|
\
|
2
3
cos
1
d) ( ) 1 tan
1
a) Evaluating |
.
|
\
|
2
1
sin
1
is the same as asking what angle would have a sine value of
2
1
.
In other words, what angle would satisfy ( )
2
1
sin = u ? There are multiple angles that
would satisfy this relationship, such as
6
t
and
6
5t
, but we know we need the angle in
the interval
(
2
,
2
t t
, so the answer will be
6 2
1
sin
1
t
=
|
.
|
\
|
## . Remember that the
inverse is a function so for each input, we will get exactly one output.
b) Evaluating
|
|
.
|
\
|
2
2
sin
1
, we know that
4
5t
and
4
7t
both have a sine value of
2
2
, but neither is in the interval
(
2
,
2
t t
. For that, we need the negative angle
coterminal with
4
7t
.
4 2
2
sin
1
t
=
|
|
.
|
\
|
Section 6.3 Inverse Trig Functions 381
c) Evaluating
|
|
.
|
\
|
2
3
cos
1
, we are looking for an angle in the interval | | t , 0 with a
cosine value of
2
3
. The angle that satisfies this is
6
5
2
3
cos
1
t
=
|
|
.
|
\
|
d) Evaluating ( ) 1 tan
1
, we are looking for an angle in the interval |
.
|
\
|
2
,
2
t t
with a
tangent value of 1. The correct angle is ( )
4
1 tan
1
t
=
Try It Now
1. Evaluate
a) ( ) 1 sin
1
b) ( ) 1 tan
1
c) ( ) 1 cos
1
d) |
.
|
\
|
2
1
cos
1
Example 2
Evaluate ( ) 97 . 0 sin
1
Since the output of the inverse function is an angle, your calculator will give you a
degree angle if in degree mode, and a radian value if in radian mode.
1
sin (0.97) 1.3252
~ In degree mode, ( )
1
sin 0.97 75.93
~
Try it Now
2. Evaluate ( ) 4 . 0 cos
1
In section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a
triangle given one side and an additional angle. Using the inverse trig functions, we can
solve for the angles of a right triangle given two sides.
Example 3
Solve the triangle for the angle
Since we know the hypotenuse and side adjacent to
the angle, it makes sense for us to use the cosine
function.
12
9
382 Chapter 6
( )
12
9
sin = u Using the definition of the inverse,
|
.
|
\
|
=
12
9
sin
1
u Evaluating
8481 . 0 ~ u , or about 48.5904
There are times when we need to compose a trigonometric function with an inverse
trigonometric function. In these cases, we can find exact values for the resulting
expressions
Example 4
Evaluate
|
|
.
|
\
|
|
.
|
\
|
6
13
cos sin
1
t
a) Here, we can directly evaluate the inside of the composition.
2
3
6
13
cos = |
.
|
\
| t
Now, we can evaluate the inverse function as we did earlier.
3 2
3
sin
1
t
=
|
|
.
|
\
|
Try it Now
3. Evaluate
|
|
.
|
\
|
|
.
|
\
|
4
11
sin cos
1
t
Example 5
Find an exact value for
|
|
.
|
\
|
|
.
|
\
|
5
4
cos sin
1
Beginning with the inside, we can say there is some angle so |
.
|
\
|
=
5
4
cos
1
u , which
means ( )
5
4
cos = u , and we are looking for ( ) u sin . We can use the Pythagorean identity
to do this.
Section 6.3 Inverse Trig Functions 383
( ) ( ) 1 cos sin
2 2
= + u u Using our known value for cosine
( ) 1
5
4
sin
2
2
= |
.
|
\
|
+ u Solving for sine
( )
25
16
1 sin
2
= u
( )
4
3
25
9
sin = = u
Since we know that the cosine inverse always gives an angle on the interval | | t , 0 , we
know that the sine of that angle must be positive, so
1
4 3
sin cos sin( )
5 4
u
| | | |
= =
| |
\ . \ .
Example 6
Find an exact value for
|
|
.
|
\
|
|
.
|
\
|
4
7
tan sin
1
While we could use a similar technique as in the last example, we
will demonstrate a different technique here. From the inside, we
know there is an angle so ( )
4
7
tan = u . We can envision this as the
opposite and adjacent sides on a right triangle.
Using Pythagorean theorem, we can find the hypotenuse of this
triangle:
2 2 2
7 4 hypotenuse = +
65 = hypotenuse
Now, we can evaluate the sine of the angle as side opposite divided by hypotenuse
( )
65
7
sin = u
This gives us our desired composition
1
7 7
sin tan sin( )
4 65
u
| | | |
= =
| |
\ . \ .
Try it Now
4. Evaluate
|
|
.
|
\
|
|
.
|
\
|
9
7
sin cos
1
7
4
384 Chapter 6
We can also find compositions involving algebraic expressions.
Example 7
Find a simplified expression for
|
|
.
|
\
|
|
.
|
\
|
3
sin cos
1
x
, for 3 3 s s x
We know there is an angle so ( )
3
sin
x
= u . Using Pythagorean Theorem,
( ) ( ) 1 cos sin
2 2
= + u u Using our known expression for sine
( ) 1 cos
3
2
2
= + |
.
|
\
|
u
x
Solving for cosine
( )
9
1 cos
2
2
x
= u
( )
3
9
9
9
cos
2 2
x x
=
= u
Since we know that the sine inverse must give an angle on the interval
(
2
,
2
t t
, we
can deduce that the cosine of that angle must be positive. This gives us
3
9
3
sin cos
2
1
x x
=
|
|
.
|
\
|
|
.
|
\
|
Try it Now
5. Find a simplified expression for ( ) ( ) x 4 tan sin
1
, for
4
1
4
1
s s x
Important Topics of This Section
Inverse trig functions: arcsine, arccosine and arctangent
Domain restrictions
Evaluating inverses using unit circle values and the calculator
Simplifying numerical expressions involving the inverse trig functions
Simplifying algebraic expressions involving the inverse trig functions
Section 6.3 Inverse Trig Functions 385
1. a)
2
t
b)
4
t
c) t d)
3
t
2. 1.9823 or
3.
4
3t
4.
9
2 4
5.
1 16
4
2
+ x
x
386 Chapter 6
Section 6.3 Exercises
Evaluate the following expressions
1.
1
2
sin
2
| |
|
|
\ .
2.
1
3
sin
2
| |
|
|
\ .
3.
1
1
sin
2
| |
|
\ .
4.
1
2
sin
2
| |
|
|
\ .
5.
1
1
cos
2
| |
|
\ .
6.
1
2
cos
2
| |
|
|
\ .
7.
1
2
cos
2
| |
|
|
\ .
8.
1
3
cos
2
| |
|
|
\ .
9. ( )
1
tan 1
10.
( )
1
tan 3
11.
( )
1
tan 3
12. ( )
1
tan 1
Use your calculator to evaluate each expression
13. ( ) 4 . 0 cos
1
14. ( ) 8 . 0 cos
1
15. ( ) 8 . 0 sin
1
16. ( ) 6 tan
1
Solve the triangle for the angle
17. 18.
Evaluate the following expressions
19.
|
|
.
|
\
|
|
.
|
\
|
4
cos sin
1
t
20.
|
|
.
|
\
|
|
.
|
\
|
6
sin cos
1
t
21.
|
|
.
|
\
|
|
.
|
\
|
3
4
cos sin
1
t
22.
|
|
.
|
\
|
|
.
|
\
|
4
5
sin cos
1
t
23.
|
|
.
|
\
|
|
.
|
\
|
7
3
sin cos
1
24.
|
|
.
|
\
|
|
.
|
\
|
9
4
cos sin
1
25. ( ) ( ) 4 tan cos
1
26.
|
|
.
|
\
|
|
.
|
\
|
3
1
sin tan
1
Find a simplified expression for each of the following
27.
|
|
.
|
\
|
|
.
|
\
|
5
cos sin
1
x
, for 5 5 s s x 28.
|
|
.
|
\
|
|
.
|
\
|
2
cos tan
1
x
, for 2 2 s s x
29. ( ) ( ) x 3 tan sin
1
30. ( ) ( ) x 4 tan cos
1
12
19
10
7
Section 6.4 Solving Trig Equations 387
Section 6.4 Solving Trig Equations
In section 6.1, we determined the height of a rider on the London Eye Ferris wheel could
be determined by the equation 5 . 69
15
cos 5 . 67 ) ( +
|
.
|
\
|
= t t h
t
.
If we wanted to know how long the rider is more than 100 meters above ground, we
would need to solve equations involving trig functions.
Solving using known values
In the last chapter, we learned sine and cosine values at commonly encountered angles.
We can use these to solve sine and cosine equations involving these common angles.
Example 1
Solve ( )
2
1
sin = t for all possible values of t
Notice this is asking us to identify all angles, t, that have a sine value of . While
evaluating a function always produces one result, solving can have multiple solutions.
6
t
= t and
6
5t
= t because they are the common angles on the unit circle.
Looking at a graph confirms that there are more than these two solutions. While eight
are seen on this graph, there are an infinite number of solutions!
Remember that any coterminal angle will also have the same sine value, so any angle
coterminal with these two is also a solution. Coterminal angles can be found by adding
full rotations of 2, so we end up with a set of solutions:
k t t
t
2
6
+ = where k is an integer, and k t t
t
2
6
5
+ = where k is an integer
388 Chapter 6
Example 2
A circle of radius 2 5 intersects the line x = -5 at two points. Find the angles u on the
interval t u 2 0 < s , where the circle and line intersect.
The x coordinate of a point on a circle can be found as ( ) u cos r x = , so the x coordinate
of points on this circle would be ( ) u cos 2 5 = x . To find where the line x = -5
intersects the circle, we can solve for where the x value on the circle would be -5
( ) u cos 2 5 5 = Isolating the cosine
( ) u cos
2
1
=
Recall that
2
2
2
1
=
, so we are solving
( )
2
2
cos
= u
We can recognize this as one of our special cosine values
from our unit circle, and it corresponds with angles
4
3t
u = and
4
5t
u =
Try it Now
1. Solve ( ) tan 1 t = for all possible values of t
Example 3
The depth of water at a dock rises and falls with the tide, following the equation
7
12
sin 4 ) ( +
|
.
|
\
|
= t t f
t
, where t is measured in hours after midnight. A boat requires a
depth of 9 feet to come to the dock. At what times will the depth be 9 feet?
To find when the depth is 9 feet, we need to solve when f(t) = 9.
9 7
12
sin 4 = +
|
.
|
\
|
t
t
Isolating the sine
2
12
sin 4 =
|
.
|
\
|
t
t
Dividing by 4
2
1
12
sin =
|
.
|
\
|
t
t
We know ( )
2
1
sin = u when
6
5
6
t
u
t
u = = or
While we know what angles have a sine value of , because of the horizontal
stretch/compression, it is less clear how to proceed.
Section 6.4 Solving Trig Equations 389
To deal with this, we can make a substitution, defining a new temporary variable u to be
t u
12
t
= , so our equation becomes
( )
2
1
sin = u
From earlier, we saw the solutions to this equation were
k u t
t
2
6
+ = where k is an integer, and
k u t
t
2
6
5
+ = where k is an integer
Undoing our substitution, we can replace the u in the solutions with t u
12
t
= and solve
for t.
k t t
t t
2
6 12
+ = where k is an integer, and k t t
t t
2
6
5
12
+ = where k is an integer.
Dividing by /12, we obtain solutions
k t 24 2 + = where k is an integer, and
k t 24 10 + = where k is an integer.
The depth will be 9 feet and boat will be
able to sail between 2am and 10am.
Notice how in both scenarios, the 24k
shows how every 24 hours the cycle will
be repeated.
In the previous example, looking back at the original simplified equation
2
1
12
sin =
|
.
|
\
|
t
t
,
we can use the ratio of the normal period to the stretch factor to find the period.
24
12
2
12
2
=
|
.
|
\
|
=
|
.
|
\
| t
t
t
t
; notice that the sine function has a period of 24, which is reflected
in the solutions; there were two unique solutions on one full cycle of the sine function,
and additional solutions were found by adding multiples of a full period.
Try it Now
2. Solve ( ) 1 1 5 sin 4 = t for all possible values of t
390 Chapter 6
Solving using the inverse trig functions
The solutions to ( ) 3 . 0 sin = u cannot be expressed in terms of functions we already know.
To represent the solutions, we need the inverse sine function that undoes the sine
function.
Example 4
Use the inverse to find one solution to ( ) 8 . 0 sin = u
Since this is not a known unit circle value, calculating the inverse, ( ) 8 . 0 sin
1
= u . This
requires a calculator and we must approximate a value for this angle. If your calculator
is in degree mode, your calculator will give you a degree angle as the output. If your
In radians, ( ) 927 . 0 8 . 0 sin
1
~ =
u , or in degrees, ( )
1
sin 0.8 53.130 u
= ~
If you are working with a composed trig function and you are not solving for an angle,
you will want to ensure that you are working in radians. Since radians are a unitless
measure, they dont intermingle with the result the way degrees would.
Notice that the inverse trig functions do exactly what you would expect of any function
for each input they give exactly one output. While this is necessary for these to be a
function, it means that to find all the solutions to an equation like ( ) 8 . 0 sin = u , we need
to do more than just evaluate the inverse.
Example 5
Find all solutions to ( ) 8 . 0 sin = u .
We would expect two unique angles on one cycle to have
this sine value. In the previous example, we found one
solution to be ( ) 927 . 0 8 . 0 sin
1
~ =
u . To find the other, we
need to answer the question what other angle has the same
sine value as an angle of 0.927? On a unit circle, we
would recognize that the second angle would have the same
reference angle and reside in the second quadrant. This
second angle would be located at 214 . 2 927 . 0 = = t u .
To find more solutions we recall that angles coterminal
with these two would have the same sine value, so we can add full cycles of 2.
k t u 2 927 . 0 + = where k is an integer, and k t u 2 214 . 2 + = where k is an integer
0.8
1
0.929
Section 6.4 Solving Trig Equations 391
Example 6
Find all solutions to ( )
9
8
sin = x on the interval < s 360 0 x
First we will turn our calculator to degree mode. Using the inverse, we can find a first
solution ~
|
.
|
\
|
=
734 . 62
9
8
sin
1
x . While this angle satisfies the equation, it does not
lie in the domain we are looking for. To find the angles in the desired domain, we start
First, an angle coterminal with 734 . 62 will have the same sine. By adding a full
rotation, we can find an angle in the desired domain with the same sine.
= + = 266 . 297 360 734 . 62 x
There is a second angle in the desired domain that lies in the third quadrant. Notice that
734 . 62 is the reference angle for all solutions, so this second solution would be
734 . 62 past 180
= + = 734 . 242 180 734 . 62 x
The two solutions on < s 360 0 x are x = 266 . 297 and x = 734 . 242
Example 7
Find all solutions to ( ) 3 tan = x on t 2 0 < s x
Using the inverse, we can find a first solution ( ) 249 . 1 3 tan
1
~ =
x . Unlike the sine and
cosine, the tangent function only reaches any output value once per cycle, so there is not
a second solution on one period of the tangent.
By adding , a full period of tangent function, we can find a second angle with the same
tangent value. If additional solutions were desired, we could continue to add multiples
of , so all solutions would take on the form t k x + = 249 . 1 , however we are only
interested in t 2 0 < s x .
391 . 4 249 . 1 = + = t x
The two solutions on t 2 0 < s x are x = 1.249 and x = 4.391
Try it Now
3. Find all solutions to ( ) tan 0.7 x = on < s 360 0 x
392 Chapter 6
Example 8
Solve ( ) 2 4 cos 3 = + t for all solutions on one cycle, t 2 0 < s x
( ) 2 4 cos 3 = + t Isolating the cosine
( ) 2 cos 3 = t
( )
3
2
cos = t Using the inverse, we can find a first solution
301 . 2
3
2
cos
1
~
|
.
|
\
|
=
t
Thinking back to the circle, the second angle with the same cosine would be located in
the third quadrant. Notice that the location of this angle could be represented as
301 . 2 = t . To represent this as a positive angle we could find a coterminal angle by
t 2 301 . 2 + = t = 3.982
The equation has two solutions on one cycle, at t = 2.301 and t = 3.982
Example 9
Solve ( ) 2 . 0 3 cos = t for all solutions on two cycles,
3
4
0
t
< s t
As before, with a horizontal compression it can be helpful to make a substitution,
t u 3 = . Making this substitution simplifies the equation to a form we have already
solved.
( ) 2 . 0 cos = u
( ) 369 . 1 2 . 0 cos
1
~ =
u
A second solution on one cycle would be located in the fourth quadrant with the same
reference angle.
914 . 4 369 . 1 2 = = t u
In this case, we need all solutions on two cycles, so we need to find the solutions on the
second cycle. We can do this by adding a full rotation to the previous two solutions.
197 . 11 2 914 . 4
653 . 7 2 369 . 1
= + =
= + =
t
t
u
u
Undoing the substitution, we obtain our four solutions:
3t = 1.369, so t = 0.456
3t = 4.914 so t = 1.638
3t = 7.653, so t = 2.551
3t = 11.197, so t = 3.732
Section 6.4 Solving Trig Equations 393
Example 10
Solve ( ) 2 sin 3 = t t for all solutions
( ) 2 sin 3 = t t Isolating the sine
( )
3
2
sin = t t We make the substitution t u t =
( )
3
2
sin = u Using the inverse, we find one solution
730 . 0
3
2
sin
1
~
|
.
|
\
|
=
u
This angle is in the fourth quadrant. A second angle with the same sine would be in the
871 . 3 730 . 0 = + =t u
We can write all solutions to the equation ( )
3
2
sin = u as
k u t 2 730 . 0 + = where k is an integer, and
k u t 2 871 . 3 + =
Undoing our substitution, we can replace u in our solutions with t u t = and solve for t
k t t t 2 730 . 0 + = and k t t t 2 871 . 3 + = Divide by
k t 2 232 . 0 + = and k t 2 232 . 1 + =
Try it Now
4. Solve 0 3
2
sin 5 = +
|
.
|
\
|
t
t
for all solutions on one cycle. t 2 0 < s t
Definition
Solving Trig Equations
1) Isolate the trig function on one side of the equation
2) Make a substitution for the inside of the sine or cosine
3) Use the inverse trig functions to find one solution
4) Use symmetries to find a second solution on one cycle (when a second exists)
6) Undo the substitution
We now can return to the question we began the section with.
394 Chapter 6
Example 10
The height of a rider on the London Eye Ferris wheel can be determined by the equation
5 . 69
15
cos 5 . 67 ) ( +
|
.
|
\
|
= t t h
t
. How long is the rider more than 100 meters above
ground?
To find how long the rider is above 100 meters, we first solve for the times at which the
rider is at a height of 100 meters by solving h(t) = 100.
5 . 69
15
cos 5 . 67 100 +
|
.
|
\
|
= t
t
Isolating the cosine
|
.
|
\
|
= t
15
cos 5 . 67 5 . 30
t
|
.
|
\
|
=
t
15
cos
5 . 67
5 . 30 t
We make the substitution t u
15
t
=
) cos(
5 . 67
5 . 30
u =
## Using the inverse, we find one solution
040 . 2
5 . 67
5 . 30
cos
1
~
|
.
|
\
|
=
u
This angle is in the second quadrant. A second angle with the same cosine would be
244 . 4 040 . 2 2 ~ = t u
Now we can undo the substitution to solve for t
040 . 2
15
= t
t
so t = 9.740 minutes
244 . 4
15
= t
t
so t = 20.264 minutes
A rider will be at 100 meters after 9.740 minutes, and again after 20.264. From the
behavior of the height graph, we know the rider will be above 100 meters between these
times. A rider will be above 100 meters for 20.265-9.740 = 10.523 minutes of the ride.
Important Topics of This Section
Solving trig equations using known values
Using substitution to solve equations
Finding answers in one cycle or period vs Finding all possible solutions
Method for solving trig equations
Section 6.4 Solving Trig Equations 395
1.
4
k
t
t +
2. k t
5
2
30
t t
+ = k t
5
2
6
t t
+ =
3. = 992 . 34 x or = + = 992 . 214 99 . 34 180 x
4. 3.590 t = or 2.410 t =
396 Chapter 6
Section 6.4 Exercises
Find all solutions on the interval 0 2 u t s <
1. ( ) 2sin 2 u = 2. ( ) 2sin 3 u = 3. ( ) 2cos 1 u = 4. ( ) 2cos 2 u =
5. ( ) sin 1 u = 6. ( ) sin 0 u = 7. ( ) cos 0 u = 8. ( ) cos 1 u =
Find all solutions
9. ( ) 2cos 2 u = 10. ( ) 2cos 1 u = 11. ( ) 2sin 1 u = 12. ( ) 2sin 3 u =
Find all solutions
13. ( ) 2sin 3 1 u = 14. ( ) 2sin 2 3 u = 15. ( ) 2sin 3 2 u =
16. ( ) 2sin 3 1 u = 17. ( ) 2cos 2 1 u = 18. ( ) 2cos 2 3 u =
19. ( ) 2cos 3 2 u = 20. ( ) 2cos 2 1 u = 21. cos 1
4
t
u
| |
=
|
\ .
22. sin 1
3
t
u
| |
=
|
\ .
23. ( ) 2sin 1 tu = . 24. 2cos 3
5
t
u
| |
=
|
\ .
Find all solutions on the interval 0 2 x t s <
25. ( ) sin 0.27 x = 26. ( ) sin 0.48 x = 27. ( ) sin 0.58 x = 28. ( ) sin 0.34 x =
29. ( ) cos 0.55 x = 30. ( ) sin 0.28 x = 31. ( ) cos 0.71 x = 32. ( ) cos 0.07 x =
Find the first two positive solutions
33. ( ) 7sin 6 2 x = 34. ( ) 7sin 5 6 x = 35. ( ) 5cos 3 3 x = 36. ( ) 3cos 4 2 x =
37. 3sin 2
4
x
t | |
=
|
\ .
38. 7sin 6
5
x
t | |
=
|
\ .
39. 5cos 1
3
x
t | |
=
|
\ .
40. 3cos 2
2
x
t | |
=
|
\ .
Section 6.5 Modeling with Trigonometric Equations 397
Section 6.5 Modeling with Trigonometric Equations
Solving right triangles for angles
In section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a
triangle given one side and an additional angle. Using the inverse trig functions, we can
solve for the angles of a right triangle given two sides.
Example 1
An airplane needs to fly to an airfield located 300 miles east and 200 miles north of its
current location. At what heading should the airplane fly? In other words, if we ignore
air resistance or wind speed, how many degrees north of east should the airplane fly?
We might begin by drawing a picture and labeling all of
the known information. Drawing a triangle, we see we
are looking for the angle . In this triangle, the side
opposite the angle is 200 miles and the side adjacent
is 300 miles. Since we know the values for the
opposite and adjacent sides, it makes sense to use the
tangent function.
300
200
) tan( = o Using the inverse,
588 . 0
300
200
tan
1
~ |
.
|
\
|
=
o , or equivalently about 33.7 degrees.
The airplane needs to fly at a heading of 33.7 degrees north of east.
Example 2
OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall
for every 4 feet of ladder length
3
. Find the angle the ladder forms with the ground.
For any length of ladder, the base needs to be of that away from the
wall. Equivalently, if the base is a feet from the wall, the ladder can be 4a
feet long. Since a is the side adjacent to the angle and 4a is the
hypotenuse, we use the cosine function.
4
1
4
) cos( = =
a
a
u Using the inverse
52 . 75
4
1
cos
1
~
|
.
|
\
|
=
u degrees
The ladder forms a 75.52 degree angle with the ground.
3
200
300
a
4a
398 Chapter 6
Try it Now
1. One of the cables that anchor to the center of the London Eye Ferris wheel to the
ground must be replaced. The center of the Ferris wheel is 69.5 meters above the
ground and the second anchor on the ground is 23 meters from the base of the Ferris
wheel. What is the angle of elevation (from ground up to the center of the Ferris
wheel) and how long is the cable?
Example 3
In a video game design, a map shows the location of other characters relative to the
player, who is situated at the origin, and the direction they are facing. A character
currently shows on the map at coordinates (-3, 5). If the player rotates
counterclockwise by 20 degrees, then the objects in the map will correspondingly rotate
20 degrees clockwise. Find the new coordinates of the character.
To rotate the position of the character, we can imagine it
as a point on a circle, and we will change the angle of
the point by 20 degrees. To do so, we first need to find
the radius of this circle and the original angle.
Drawing a triangle in the circle, we can find the radius
using Pythagorean Theorem:
( )
2
2 2
3 5
9 25 34
r
r
+ =
= + =
To find the angle, we need to decide first if we are going to find the acute angle of the
triangle, the reference angle, or if we are going to find the angle measured in standard
position. While either approach will work, in this case we will do the latter. Since for
any point on a circle we know ) cos(u r x = , adding our given information we get
) cos( 34 3 u =
) cos(
34
3
u =
~
|
|
.
|
\
|
=
964 . 120
34
3
cos
1
u
While there are two angles that have this cosine value, the angle of 120.964 degrees is
in the second quadrant as desired, so it is the angle we were looking for.
Rotating the point clockwise by 20 degrees, the angle of the point will decrease to
100.964 degrees. We can then evaluate the coordinates of the rotated point
109 . 1 ) 964 . 100 cos( 34 ~ = x
725 . 5 ) 964 . 100 sin( 34 ~ = y
The coordinates of the character on the rotated map will be (-1.109, 5.725)
Section 6.5 Modeling with Trigonometric Equations 399
Modeling with sinusoidal functions
Many modeling situations involve functions that are periodic. Previously we learned that
sinusoidal functions are a special type of periodic function. Problems that involve
quantities that oscillate can often be modeled by a sine or cosine function and once we
create a suitable model for the problem we can use the equation and function values to
Example 4
The hours of daylight in Seattle oscillate from a low of 8.5 hours in January to a high of
16 hours in July
4
. When should you plant a garden if you want to do it during the
month where there are 14 hours of daylight?
To model this, we first note that the hours of daylight oscillate with a period of 12
months. With a low of 8.5 and a high of 16, the midline will be halfway between these
values, at 25 . 12
2
5 . 8 16
=
+
. The amplitude will be half the difference between the
highest and lowest values: 75 . 3
2
5 . 8 16
=
## , or equivalently the distance from the
midline to the high or low value, 16-12.25=3.75. Letting January be t = 0, the graph
starts at the lowest value, so it can be modeled as a flipped cosine graph. Putting this
together, we get a model:
25 . 12
6
cos 75 . 3 ) ( + |
.
|
\
|
= t t h
t
-cos(t) represents the flipped cosine,
3.75 is the amplitude,
12.25 is the midline,
6 / 12 / 2 t t = corresponds to the horizontal stretch, found by using the ratio of the
original period / new period
h(t) is our model for hours of day light t months from January.
To find when there will be 14 hours of daylight, we solve h(t) = 14.
25 . 12
6
cos 75 . 3 14 + |
.
|
\
|
= t
t
Isolating the cosine
|
.
|
\
|
= t
6
cos 75 . 3 75 . 1
t
Subtracting 12.25 and dividing by -3.75
|
.
|
\
|
= t
6
cos
75 . 3
75 . 1 t
Using the inverse
4
http://www.mountaineers.org/seattle/climbing/Reference/DaylightHrs.html
400 Chapter 6
0563 . 2
75 . 3
75 . 1
cos
6
1
~ |
.
|
\
|
=
t
t
multiplying by the reciprocal
927 . 3
6
0563 . 2 = =
t
t t=3.927 months past January
There will be 14 hours of daylight 3.927 months into the year, or near the end of April.
While there would be a second time in the year when there are 14 hours of daylight,
since we are planting a garden, we would want to know the first solution, in spring, so
we do not need to find the second solution in this case.
Try it Now
2. The authors
monthly gas usage
(in therms) is shown
here. Find an
equation to model
the data.
Example 6
An object is connected to the wall with a spring that has a
natural length of 20 cm. The object is pulled back 8 cm past
the natural length and released. The object oscillates 3 times
per second. Find an equation for the position of the object
ignoring the effects of friction. How much time in each cycle is the object more than 27
cm from the wall?
If we use the distance from the wall, x, as the desired output, then the object will
oscillate equally on either side of the springs natural length of 20, putting the midline
of the function at 20 cm.
If we release the object 8 cm past the natural length, the amplitude of the oscillation will
be 8 cm.
We are beginning at the largest value and so this function can most easily be modeled
using a cosine function.
Since the object oscillates 3 times per second, it has a frequency of 3 and the period of
one oscillation is 1/3 of second. Using this we find the horizontal compression using the
ratios of the periods t
t
6
3 / 1
2
=
0
20
40
60
80
100
120
140
160
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Section 6.5 Modeling with Trigonometric Equations 401
Using all this, we can build our model:
( ) 20 6 cos 8 ) ( + = t t x t
To find when the object is 27 cm from the wall, we can solve x(t) = 27
( ) 20 6 cos 8 27 + = t t Isolating the cosine
( ) t t 6 cos 8 7 =
( ) t t 6 cos
8
7
= Using the inverse
505 . 0
8
7
cos 6
1
~
|
.
|
\
|
=
t t
0268 . 0
6
505 . 0
= =
t
t
Based on the shape of the graph, we can
conclude that the object will spend the first
0.0268 seconds more than 27 cm from the
wall. Based on the symmetry of the function,
the object will spend another 0.0268 seconds
more than 27 cm from the wall at the end of
the cycle. Altogether, the object spends
0.0536 seconds each cycle more than 27 cm
from the wall.
In some problems, we can use the trigonometric functions to model behaviors more
complicated than the basic sinusoidal function.
Example 7
A rigid rod with length 10 cm is attached
to a circle of radius 4cm at point A as
shown here. The point B is able to freely
move along the horizontal axis, driving a
piston
5
. If the wheel rotates
counterclockwise at 5 revolutions per
minute, find the location of point B as a
function of time. When will the point B
be 12 cm from the center of the circle?
To find the position of point B, we can begin by finding the coordinates of point A.
Since it is a point on a circle with radius 4, we can express its coordinates as
)) sin( 4 ), cos( 4 ( u u .
5
For an animation of this situation, see http://mathdemos.gcsu.edu/mathdemos/sinusoidapp/engine1.gif
A
B
10 cm
4cm
402 Chapter 6
The angular velocity is 5 revolutions per second, or equivalently 10 radians per
second. After t seconds, the wheel will rotate by t t u 10 = radians. Substituting this,
we can find the coordinates of A in terms of t.
)) 10 sin( 4 ), 10 cos( 4 ( t t t t
Notice that this is the same value we would have obtained by noticing that the period of
the rotation is 1/5 of a second and calculating the stretch/compression factor
t
t
10
5
1
2
" "
" "
=
new
original
.
Now that we have the coordinates of the point
A, we can relate this to the point B. By
drawing a vertical line from A to the
horizontal axis, we can form a triangle. The
height of the triangle is the y coordinate of the
point A: ) 10 sin( 4 t t . Using the Pythagorean
Theorem, we can find the base length of the
triangle:
( )
2 2 2
10 ) 10 sin( 4 = +b t t
) 10 ( sin 16 100
2 2
t b t =
) 10 ( sin 16 100
2
t b t =
Looking at the x coordinate of the point A, we can see that the triangle we drew is
shifted to the right of the y axis by ) 10 cos( 4 t t . Combining this offset with the length
of the base of the triangle gives the x coordinate of the point B:
) 10 ( sin 16 100 ) 10 cos( 4 ) (
2
t t t x t t + =
To solve for when the point B will be 12 cm from the center of the circle, we need to
solve x(t) = 12.
) 10 ( sin 16 100 ) 10 cos( 4 12
2
t t t t + = Isolate the square root
) 10 ( sin 16 100 ) 10 cos( 4 12
2
t t t t = Square both sides
( ) ) 10 ( sin 16 100 ) 10 cos( 4 12
2 2
t t t t = Expand the left side
) 10 ( sin 16 100 ) 10 ( cos 16 ) 10 cos( 96 144
2 2
t t t t t t = + Move terms of the left
0 ) 10 ( sin 16 ) 10 ( cos 16 ) 10 cos( 96 44
2 2
= + + t t t t t t Factor out 16
( ) 0 ) 10 ( sin ) 10 ( cos 16 ) 10 cos( 96 44
2 2
= + + t t t t t t
At this point, we can utilize the Pythagorean Identity, which tells us that
1 ) 10 ( sin ) 10 ( cos
2 2
= + t t t t .
A
B
10 cm
b
Section 6.5 Modeling with Trigonometric Equations 403
Using this identity, our equation simplifies to
0 16 ) 10 cos( 96 44 = + t t Combine the constants and move to the right side
60 ) 10 cos( 96 = t t Divide
96
60
) 10 cos( = t t Make a substitution
96
60
) cos( = u
896 . 0
96
60
cos
1
~ |
.
|
\
|
=
u By symmetry we can find a second solution
388 . 5 896 . 0 2 = = t u Undoing the substitution
896 . 0 10 = t t , so t = 0.0285
388 . 5 10 = t t , so t = 0.1715
The point B will be 12 cm from the center of the circle after 0.0285 seconds, 0.1715
seconds, and every 1/5
th
of a second after each of those values.
Important Topics of This Section
Modeling with trig equations
Modeling with sinusoidal functions
Solving right triangles for angles in degrees and radians
1. Angle of elevation for the cable is 71.69 degrees and the cable is 73.21 m long
2. Approximately ( ) 66cos ( 1) 87
6
G t t
t | |
= +
|
\ .
404 Chapter 6
Section 6.5 Exercises
In each of the following triangles, solve for the unknown side and angles.
1. 2.
3. 4.
Find a possible formula for the trigonometric function whose values are in the following
tables.
5.
x 0 1 2 3 4 5 6
y -2 4 10 4 -2 4 10
6.
x 0 1 2 3 4 5 6
y 1 -3 -7 -3 1 -3 -7
7. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the high temperature for the day is 63 degrees and the low temperature of 37
degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an
equation for the temperature, D, in terms of t.
8. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the high temperature for the day is 92 degrees and the low temperature of 78
degrees occurs at 4 AM. Assuming t is the number of hours since midnight, find an
equation for the temperature, D, in terms of t.
9. A population of rabbits oscillates 25 above and below an average of 129 during the
year, hitting the lowest value in January (t = 0).
a. Find an equation for the population, P, in terms of the months since January, t.
b. What if the lowest value of the rabbit population occurred in April instead?
A
5
8
B
c
B
7
3
A
c
A
b
7
15
B
B
a
10
12
A
Section 6.5 Modeling with Trigonometric Equations 405
10. A population of elk oscillates 150 above and below an average of 720 during the year,
hitting the lowest value in January (t = 0).
a. Find an equation for the population, P, in terms of the months since January, t.
b. What if the lowest value of the rabbit population occurred in March instead?
11. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the high temperature of 105 degrees occurs at 5 PM and the average temperature
for the day is 85 degrees. Find the temperature, to the nearest degree, at 9 AM.
12. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the high temperature of 84 degrees occurs at 6 PM and the average temperature for
the day is 70 degrees. Find the temperature, to the nearest degree, at 7 AM.
13. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the temperature varies between 47 and 63 degrees during the day and the average
daily temperature first occurs at 10 AM. How many hours after midnight does the
temperature first reach 51 degrees?
14. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you
know the temperature varies between 64 and 86 degrees during the day and the average
daily temperature first occurs at 12 AM. How many hours after midnight does the
temperature first reach 70 degrees?
15. A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters
above the ground. The six o'clock position on the Ferris wheel is level with the loading
platform. The wheel completes 1 full revolution in 6 minutes. How many minutes of the
ride are spent higher than 13 meters above the ground?
16. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meters
above the ground. The six o'clock position on the Ferris wheel is level with the loading
platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the
ride are spent higher than 27 meters above the ground?
17. The sea ice area around the north pole fluctuates between about 6 million square
kilometers in September to 14 million square kilometers in March. During how many
months are there less than 9 million square kilometers of sea ice?
18. The sea ice area around the south pole fluctuates between about 18 million square
kilometers in September to 3 million square kilometers in March. During how many
months are there more than 15 million square kilometers of sea ice?
406 Chapter 6
19. A respiratory ailment called Cheyne-Stokes Respiration causes the volume per
breath to increase and decrease in a sinusoidal manner, as a function of time. For one
particular patient with this condition, a machine begins recording a plot of volume per
breath versus time (in seconds). Let ( ) b t be a function of time t that tells us the volume
(in liters) of a breath that starts at time t. During the test, the smallest volume per breath is
0.6 liters and this first occurs for a breath that starts 5 seconds into the test. The largest
volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into
the test. [UW]
a. Find a formula for the function ( ) b t whose graph will model the test data for this
patient.
b. If the patient begins a breath every 5 seconds, what are the breath volumes during
the first minute of the test?
20. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which time the
water is 10 feet above the height of low tide. Low tides occur 6 hours after high tides.
Suppose there are two high tides and two low tides every day and the height of the tide
varies sinusoidally. [UW]
a. Find a formula for the function ( ) y h t = that computes the height of the tide above
low tide at time t. (In other words, y = 0 corresponds to low tide.)
b. What is the tide height at 11:00 a.m.?
21. A communications satellite orbits the earth t
miles above the surface. Assume the radius of
the earth is 3,960 miles. The satellite can only
see a portion of the earths surface, bounded
by what is called a horizon circle. This leads to a
two-dimensional cross-sectional picture we can
use to study the size of the horizon slice: [UW]
a. Find a formula for in terms of t.
b. If t = 30,000 miles, what is ? What
percentage of the circumference of the
earth is covered by the satellite? What
would be the minimum number of such
satellites required to cover the
circumference?
c. If t = 1,000 miles, what is ? What percentage of the circumference of the earth is
covered by the satellite? What would be the minimum number of such satellites
required to cover the circumference?
d. Suppose you wish to place a satellite into orbit so that 20% of the circumference
is covered by the satellite. What is the required distance t?
Section 6.5 Modeling with Trigonometric Equations 407
22. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket
filled with laughing gas. According to her design, if the atmospheric pressure exerted on
the rocket is less than 10 pounds/sq.in., the laughing gas chamber inside the rocket will
explode. Tiff worked from a formula
/10
14.7
h
p e
## = pounds/sq.in. for the atmospheric
pressure h miles above sea level. Assume that the rocket is launched at an angle of
above level ground at sea level with an initial speed of 1400 feet/sec. Also, assume the
height (in feet) of the rocket at time t seconds is given by the equation
( ) ( )
2
16 1400sin y t t t o = + [UW]
a. At what altitude will the rocket explode?
b. If the angle of launch is = 12, determine the minimum atmospheric pressure
exerted on the rocket during its flight. Will the rocket explode in midair?
c. If the angle of launch is = 82, determine the minimum atmospheric pressure
exerted on the rocket during its flight. Will the rocket explode in midair?
d. Find the largest launch angle so that the rocket will not explode.
408 Chapter 6
|
The Grand Finale
We investigate the ratio:
And equate asymptotes, domain and range, and see how changing each will be effected differently.
Above we change the value of a from 2,9 and notice that there is a horizontal asymptote at the corresponding a value.
We can see right away that if we let x=0 the y-intercept will always be 1. thus all the graphs pass at (0,1).
It seems that the horizontal asymptote is going to be "a" can we prove this?
Let b,c, and d =1 and let a=2
So if we take the limit as x approaches infinity, we will get
thus we will take the derivative of both the top and bottom but L'hopital rule.
Here we notice that the horizontal asymptote will be for each changing a and correspond to the value of a.
Thus we will have a horizontal asymptote at
All other horizontal asymptotes will follow suit.
Well why do all the graphs above share the same vertical asymptote?
This is what values will not work in the range, given our domain. We know our domain is going to be .
So which values will make our equation explode? That is when the denominator equals 0. This will happen in every graph when
Thus we have a vertical asymptote at the value
Here we notice that our range will be
Below we change the value of b from 2,9 and investigate the horizontal and vertical asymptotes and domain and range.
Above we change the value of b from 2,9 and notice that there is a horizontal asymptote at the corresponding a value.
If we left x=0 then we will have a y-intercept that is corresponding to c.
It seems that the horizontal asymptote is going to be "a" can we prove this?
Let a,c, and d =1 and let b=4
So if we take the limit as x approaches infinity, we will get
thus we will take the derivative of both the top and bottom but L'hopital rule. We get 1.
We see here that the horizontal asymptote will be
If we look closely at the graph above we see that the vertical asymptote is similar to the one when we were changing the value of a.
This is what values will not work in the range, given our domain. We know our domain is going to be .
So which values will make our equation explode?
That is when the denominator equals 0. This will happen in every graph when
Thus we have a vertical asymptote at the value
Here we notice that our range will be
This is interesting to not that c looks to only effect the y-intercept of a graph.
Here we are changing the values of c. Is this going to change our asymptotes? Will this change our domain and range of the functions?
Let x=0 we get that all graphs will have y-intercept of -1 or 1. Thus all graphs pass (0,-1) or (0,1)
Let a,b, and d=1 and let c=3
To find the horizontal asymptote;
We take the limit as x approaches infinity, we will get
thus we will take the derivative of both the top and bottom but L'hopital rule.
Here we notice that the horizontal asymptote will be for each changing c and will correspond to the value 1/c. As the other horizontal asymptotes will follow a similar process.
Well why do all the graphs above share the same vertical asymptote?
This is what values will not work in the range, given our domain. We know our domain is going to be .
So which values will make our equation explode? That is when the denominator equals 0. This will happen in every graph when
Thus we notice that the vertical asymptote will be
and the range will be
Here we are changing the values of d. Is this going to change our asymptotes? Will this change our domain and range of the functions?
If we let x = 0. We will get a y-intercept of 1/d that is (0,1/d)
If we let y=0 that we get the x-intercept of -1. Thus (-1,0)
Let a,b, and c=1 and let d=5
To find the horizontal asymptote;
We take the limit as x approaches infinity, we will get
thus we will take the derivative of both the top and bottom but L'hopital rule
If we look closely at the graph above we see that the vertical asymptote is similar to when changing in c.
This is what values will not work in the range, given our domain. We know our domain is going to be .
So which values will make our equation explode?
That is when the denominator equals 0. This will happen in every graph when
Negative Investigation
Here I have placed the graphs for the when we change a,b,c, and d to values (-2,-9). To find the values from that above such as asymptote we can do similar cases for all.
Since 1 and 0 like to be difficult I decided to place them in a group of their own.
Above we have let -1<a<1 to see what will happen.
If we let a=1
Domain will be and our range will be 1
or
Thus we get the line y=1.
If a= 0.
We notice right away that we are going to have a vertical asymptote as -1.
For horizontal we let x approach infinity and see we have
If we let a=-1or
Again we notice immediately that there is a vertical asymptote at x= -1
However we have a shift in our horizontal asymptote.
Let x approach infinity. Thus we get So by L'hopital's Rules we will get
Or a horizontal asymptote of x=-1
Look at the red graph we have a nice line at . Thus no horizontal of vertical asymptotes, we have both domain and range
If c is to equal 1 we get Thus a horizontal line y=1
Let c= -1 then we have some changes.
We notice that we are going to have a vertical asymptote at x=-1.
When we let x approach infinity we get The we have
Our horizontal asymptote is at x=-1
Looking at the graphs about we notice that have many tings in common. One being the horizontal asymptote. This is because both limits will be thus x=1 by both after applying L'hopital's Rule.
another similar idea is that when b,d=1 there is a line at y=1.
WE can see that on the left graph when c=-1,0 we will have a will have a vertical asymptote at x=-1. this is because we will always get the denominator and when we set this equal to zero we get
Thus a vertical asymptote.
For the graph on the right we have two, x=0, x=1. The blue is from having just x in the denominator of ratio, and the other from x-1 in the denominator will create x=1.
Home
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### 20-2. Current
Learning Objectives
• Define electric current, ampere, and drift velocity
• Describe the direction of charge flow in conventional current.
• Use drift velocity to calculate current and vice versa.
#### Electric Current
Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current $I$ is defined to be
$I=ΔQΔt,$
where $\Delta Q$ is the amount of charge passing through a given area in time $\Delta t$. (As in previous chapters, initial time is often taken to be zero, in which case $\Delta t=t$.) (See Figure 1.) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère (1775–1836). Since $I=\Delta Q/\Delta t$, we see that an ampere is one coulomb per second:
$1 A = 1 C/s$
Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances.
Figure 1: The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second.
###### Example 1: Calculating Currents: Current in a Truck Battery and a Handheld Calculator
(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing?
Strategy
We can use the definition of current in the equation $I=\Delta Q/\Delta t$ to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required.
Solution for (a)
Entering the given values for charge and time into the definition of current gives
$I = ΔQ Δt = 720 C 4.00 s = 180 C/s = 180 A.$
Discussion for (a)
This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large because large frictional forces need to be overcome when setting something in motion.
Solution for (b)
Solving the relationship $I=\Delta Q/\Delta t$ for time $\Delta t$, and entering the known values for charge and current gives
$Δ t = Δ Q I = 1.00 C 0.300 × 10 - 3 C/s = 3.33 × 10 3 s.$
Discussion for (b)
This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It’s because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents.
Figure 2 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 2 (b), for example, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations.
Figure 2: (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic represents a wide variety of similar circuits.
Note that the direction of current flow in Figure 2 is from positive to negative. The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons—that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 3 illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associated with electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the small-scale structure of electricity.
It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 3. Unlike static electricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in static equilibrium. An electric field is needed to supply energy to move the charges.
###### Making Connections: Take-Home Investigation—Electric Current Illustration
Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one pea in at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares to the electrons and what compares to the supply of energy. What other analogies can you find for an electric current?
Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces.
Figure 3: Current $I$ is the rate at which charge moves through an area $A$, such as the cross-section of a wire. Conventional current is defined to move in the direction of the electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the direction opposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronic flow.
###### Example 2: Calculating the Number of Electrons that Move through a Calculator
If the 0.300-mA current through the calculator mentioned in the Example 1 example is carried by electrons, how many electrons per second pass through it?
Strategy
The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign is opposite, ${I}_{\text{electrons}}=-0.300×{\text{10}}^{-3\phantom{\rule{0.25em}{0ex}}}\text{C/s}$ .Since each electron $\left({e}^{-}\right)$ has a charge of $–1\text{.}\text{60}×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{C}$, we can convert the current in coulombs per second to electrons per second.
Solution
Starting with the definition of current, we have
$Ielectrons=ΔQelectronsΔt=–0.300 × 10−3 Cs.$
We divide this by the charge per electron, so that
$e – s = –0 . 300 × 10 – 3 C s × 1 e – –1 .60 × 10 − 19 C = 1.88 × 10 15 e – s .$
Discussion
There are so many charged particles moving, even in small currents, that individual charges are not noticed, just as individual water molecules are not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metal wire and, of course, with other electrons.
#### Drift Velocity
Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of ${\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$, a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically drifting at speeds on the order of ${\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{m/s}$. How do we reconcile these two speeds, and what does it tell us about standard conductors?
The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in Figure 4, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The density of charge in a system cannot easily be increased, and so the signal is passed on rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this rapidly moving signal or shock wave is a rapidly propagating change in electric field.
Figure 4: When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes it difficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward.
Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 5 shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity ${v}_{\text{d}}$ is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current.
Figure 5: Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the free charges is called the drift velocity, ${v}_{\text{d}}$, and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a constant supply of energy to maintain a steady current.
###### Conduction of Electricity and Heat
Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current and can transport thermal energy.
The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor’s atoms, possibly increasing temperature. Thus a continuous power input is required to keep a current flowing. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy—a great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows.
###### Making Connections: Take-Home Investigation—Filament Observations
Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected?
We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in Figure 6. The number of free charges per unit volume is given the symbol $n$ and depends on the material. The shaded segment has a volume $\text{Ax}$, so that the number of free charges in it is $\text{nAx}$. The charge $\Delta Q$ in this segment is thus $\text{qnAx}$, where $q$ is the amount of charge on each carrier. (Recall that for electrons, $q$ is $-1\text{.}\text{60}×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{C}$.) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time $\Delta t$, the current is
$I=ΔQΔt=qnAxΔt.$
Note that $x/\Delta t$ is the magnitude of the drift velocity, ${v}_{\text{d}}$, since the charges move an average distance $x$ in a time $\Delta t$. Rearranging terms gives
$I = nqAv d ,$
where $I$ is the current through a wire of cross-sectional area $A$ made of a material with a free charge density $n$. The carriers of the current each have charge $q$ and move with a drift velocity of magnitude ${v}_{\text{d}}$.
Figure 6: All the charges in the shaded volume of this wire move out in a time $t$, having a drift velocity of magnitude ${v}_{\text{d}}=x/t$. See text for further discussion.
Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a “sea” of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons.
###### Example 3: Calculating Drift Velocity in a Common Wire
Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is $8\text{.}\text{80}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$.
Strategy
We can calculate the drift velocity using the equation $I={\mathrm{nqAv}}_{\text{d}}$. The current $I=20.0 A$ is given, and $q=\phantom{\rule{0.25em}{0ex}}–1.60×{10}^{–19}\text{C}$ is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula $A=\pi {r}^{2},$ where $r$ is one-half the given diameter, 2.053 mm. We are given the density of copper, $8.80×{10}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3},$ and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro’s number, $6.02×{10}^{23}\phantom{\rule{0.25em}{0ex}}\text{atoms/mol},$ to determine $n,$ the number of free electrons per cubic meter.
Solution
First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per ${m}^{3}$. We can now find $n$ as follows:
$n = 1 e − atom × 6 . 02 × 10 23 atoms mol × 1 mol 63 . 54 g × 1000 g kg × 8.80 × 10 3 kg 1 m 3 = 8 . 342 × 10 28 e − /m 3 .$
The cross-sectional area of the wire is
$A = π r 2 = π 2.053 × 10 −3 m 2 2 = 3.310 × 10 –6 m 2 .$
Rearranging $I=nqA{v}_{\text{d}}$ to isolate drift velocity gives
$v d = I nqA = 20.0 A ( 8 . 342 × 10 28 /m 3 ) ( –1 . 60 × 10 –19 C ) ( 3 . 310 × 10 –6 m 2 ) = –4 . 53 × 10 –4 m/s.$
Discussion
The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of ${\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{m/s}$) confirms that the signal moves on the order of ${\text{10}}^{\text{12}}$ times faster (about ${\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$) than the charges that carry it.
#### Section Summary
• Electric current $I$ is the rate at which charge flows, given by $I=ΔQΔt ,$ where $\Delta Q$ is the amount of charge passing through an area in time $\Delta t$.
• The direction of conventional current is taken as the direction in which positive charge moves.
• The SI unit for current is the ampere (A), where $\text{1 A}=\text{1 C/s.}$
• Current is the flow of free charges, such as electrons and ions.
• Drift velocity ${v}_{\text{d}}$ is the average speed at which these charges move.
• Current $I$ is proportional to drift velocity ${v}_{\text{d}}$, as expressed in the relationship $I={\text{nqAv}}_{\text{d}}$. Here, $I$ is the current through a wire of cross-sectional area $A$. The wire’s material has a free-charge density $n$, and each carrier has charge $q$ and a drift velocity ${v}_{\text{d}}$.
• Electrical signals travel at speeds about ${\text{10}}^{\text{12}}$ times greater than the drift velocity of free electrons.
#### Conceptual Questions
###### Exercise 1
Can a wire carry a current and still be neutral—that is, have a total charge of zero? Explain.
###### Exercise 2
Car batteries are rated in ampere-hours ($\text{A}\cdot \text{h}$). To what physical quantity do ampere-hours correspond (voltage, charge, . . .), and what relationship do ampere-hours have to energy content?
###### Exercise 3
If two different wires having identical cross-sectional areas carry the same current, will the drift velocity be higher or lower in the better conductor? Explain in terms of the equation ${v}_{\text{d}}=\frac{I}{\text{nqA}}$, by considering how the density of charge carriers $n$ relates to whether or not a material is a good conductor.
###### Exercise 4
Why are two conducting paths from a voltage source to an electrical device needed to operate the device?
###### Exercise 5
In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather than two wires?
###### Exercise 6
Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneously with its wings.
#### Problems & Exercises
###### Exercise 1
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 h?
Show/Hide Solution
##### Solution
0.278 mA
###### Exercise 2
A total of 600 C of charge passes through a flashlight in 0.500 h. What is the average current?
###### Exercise 3
What is the current when a typical static charge of $0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ moves from your finger to a metal doorknob in $1.00\phantom{\rule{0.25em}{0ex}}\mu \text{s}$?
Show/Hide Solution
##### Solution
0.250 A
###### Exercise 4
Find the current when 2.00 nC jumps between your comb and hair over a $0\text{.}\text{500 -}\phantom{\rule{0.25em}{0ex}}\mu \text{s}$ time interval.
###### Exercise 5
A large lightning bolt had a 20,000-A current and moved 30.0 C of charge. What was its duration?
Show/Hide Solution
##### Solution
1.50ms
###### Exercise 6
The 200-A current through a spark plug moves 0.300 mC of charge. How long does the spark last?
###### Exercise 7
(a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resistance of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: $P={I}^{2}R$.)
Figure 7: The capacitor in a defibrillation unit drives a current through the heart of a patient.
Show/Hide Solution
##### Solution
(a) $1\text{.}\text{67}\text{k}\Omega$
(b) If a 50 times larger resistance existed, keeping the current about the same, the power would be increased by a factor of about 50 (based on the equation $P={I}^{2}R$), causing much more energy to be transferred to the skin, which could cause serious burns. The gel used reduces the resistance, and therefore reduces the power transferred to the skin.
###### Exercise 8
During open-heart surgery, a defibrillator can be used to bring a patient out of cardiac arrest. The resistance of the path is $5\text{00 Ω}$ and a 10.0-mA current is needed. What voltage should be applied?
###### Exercise 9
(a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves? (b) How many electrons pass through the wires connected to the patient? (See figure two problems earlier.)
Show/Hide Solution
##### Solution
(a) 0.120 C
(b) $7\text{.}\text{50}×{\text{10}}^{\text{17}}\text{electrons}$
###### Exercise 10
A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed?
###### Exercise 11
The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How long does it take to move Avogadro’s number ($6\text{.}\text{02}×{\text{10}}^{\text{23}}$) of electrons at this rate?
Show/Hide Solution
##### Solution
96.3 s
###### Exercise 12
Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a metal target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s?
###### Exercise 13
A large cyclotron directs a beam of ${\text{He}}^{\text{++}}$ nuclei onto a target with a beam current of 0.250 mA. (a) How many ${\text{He}}^{\text{++}}$ nuclei per second is this? (b) How long does it take for 1.00 C to strike the target? (c) How long before 1.00 mol of ${\text{He}}^{\text{++}}$ nuclei strike the target?
Show/Hide Solution
##### Solution
(a) ${7.81 × 10}^{\text{14}}\phantom{\rule{0.25em}{0ex}}{\text{He}}^{\text{++}}\phantom{\rule{0.25em}{0ex}}\text{nuclei/s}$
(b) ${4.00 × 10}^{\text{3}}\phantom{\rule{0.25em}{0ex}}\text{s}$
(c) ${7.71 × 10}^{\text{8}}\phantom{\rule{0.25em}{0ex}}\text{s}$
###### Exercise 14
Repeat the above example on Example 3, but for a wire made of silver and given there is one free electron per silver atom.
###### Exercise 15
Using the results of the above example on Example 3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A.
Show/Hide Solution
##### Solution
$-1\text{.}\text{13}×{\text{10}}^{-4}\text{m/s}$
###### Exercise 16
A 14-gauge copper wire has a diameter of 1.628 mm. What magnitude current flows when the drift velocity is 1.00 mm/s? (See above example on Example 3 for useful information.)
###### Exercise 17
SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator (closed in 2009), has a 20.0-A circulating beam of electrons that are moving at nearly the speed of light. (See Figure 8.) How many electrons are in the beam?
Figure 8: Electrons circulating in the storage ring called SPEAR constitute a 20.0-A current. Because they travel close to the speed of light, each electron completes many orbits in each second.
Show/Hide Solution
##### Solution
$9\text{.}\text{42}×{\text{10}}^{\text{13}}\text{electrons}$
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# Using integer tiles
Keywords:
Achievement Objectives:
Achievement Objective: NA4-2: Understand addition and subtraction of fractions, decimals, and integers.
AO elaboration and other teaching resources
Purpose:
These exercises and activities are for students to use independently of the teacher to develop and practice number properties
Specific Learning Outcomes:
solve problems that involve addition of positive and negative numbers
solve problems that involve subtraction of positive and negative numbers
Description of mathematics:
Addition and subtraction, AM (Stage 7)
Required Resource Materials:
Practice exercises with answers (PDF or Word)
Activity:
### Prior knowledge
Explain the meaning of a negative number
Place integers in order of size
Explain that negative numbers and positive numbers are opposites
### Background
Integer tiles are one way of using a material representation to introduce the concept of adding and subtracting integers. This model can be useful for working on the type of problem that many students find harder, that is problems like -4 + -5. However, in themselves the tiles are rather abstract, and students need to be able to recognise the level of abstraction if they are to be successful in using them. This abstraction is spelled out in more detail in the next paragraph.
Having a red tile and a blue tile on their own, even if they are labelled ‘1’ and ‘-1’ does not mean that students recognise these things as being different. Indeed for some, if you asked "how much you have got in total", the answer ‘2’ is perfectly reasonable, as they are seeing and counting the objects, rather than identifying that meaning has been given to the colour, or the numbers written on the objects. For the tiles to be of value students must already have some understanding of integers (so the negative sign conveys meaning) and have the concept of the integers being the opposites of the whole numbers (so negative two is equal and opposite to positive two). This allows the underlying concept of the tiles to make sense, and allows students to see that the negative one tile can cancel out the positive one tile.
Also remember that the ultimate aim of introducing the tiles is to scaffold learning, so students can work on the numbers themselves, without reference to the tiles, so these should be removed as soon as students have developed an understanding of the principle behind the model.
Exercise 0
Asks students to represent integers using drawings of tiles. Question 7 is critical. It checks whether or not students understand how the integer tiles operate.
Exercise 1
Asks students to solve addition problems with integers. Students to represent additions using the tiles, and if necessary use them to help answer the problems. The last question is again the most important in the exercise – have the students developed an understanding of the principle behind the tiles as a model for addition.
Exercise 2
Asks students to solve addition problems with integers. This activity extends the previous exercise by steadily increasing the size of the numbers used. The problems include 2 digit integers.
Exercise 3
Asks students to solve subtraction problems with integers. Representing subtractions with the tiles is harder than representing additions, as this involves taking tiles away. Here the model becomes cumbersome. For example, with the problem -3 – 4: When represented with tiles, 4 positive one (red) tiles are being taken away from 3 negative one (blue) tiles, but as there are no positive tiles to take away, this problem is hard to do, unless a ‘whole load of zeros’ are thrown into the mix. That is, by adding 4 positive and 4 negative tiles to the table, 4 positive tiles can now be removed, leaving seven negative tiles on the table. Readers are invited to consider whether or not the use of the tiles is simplifying the problem for students.
Exercise 4
Asks students to solve subtraction problems with integers. The problems include 2 digit integers.
Exercise 5
Asks students to solve a mix of addition and subtraction problems. This activity is designed to help students decide which strategy is the most useful for answering questions across the range to which they have been introduced.
Exercise 6
Asks students to explore the equivalence of subtracting and adding the opposite.
Exercise 7
Asks students to solve magic squares puzzles using integers.
AttachmentSize
IntegersPracticeSheet.pdf140.78 KB
IntegersPracticeSheet.doc120 KB
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Derivation Of Area Of Circle. How have you used this link in your classroom? If you have already learnt the formula for the area of rectangle then you can easily remember the area of parallelogram too.. What is parallelogram? Area of a parallelogram - Formula and its derivation - Elementary mathematics How do you calculate the area of a parallelogram? Upon deeper probing, we discussed some further issues that needed to be addressed. sinα = h / b; h = b; A = ah; A = ab. Derivation of the law Note: All the letters in bold represent vectors and normal letters represent magnitude only. Initially, most teachers were positive about this approach as they found the visuals simple enough for the students and easy for the teachers to reproduce. Nodal compatibility is then enforced during the formulation of the nodal equilibrium equations for a line element. The green point changes the slope of the sides of the parallelogram. Cut off a triangular section. Follow. So your area-- this is exciting! BLOG. Ans- Here, a … Pinterest The area of a parallelogram is A = bh. Area of parallelogram = Twice Area of Triangle. As we know, there are two diagonals for a parallelogram, which intersects each other. 0. area and height of a parallelogram. The area of a parallelogram is the number of square units inside the polygon. This short address will bring you back to this page. Given a trapezoid, if we form a congruent trapezoid and rotate it such that the two congruent trapezoids can be joined together to form a parallelogram as shown by the congruent black and grey trapezoids below. In the figure shown above, AB is … Google+. This rearranging has created a rectangle whose area is clearly the same as the original parallelogram. Area of rectangle and parallelogram has the almost similar type of formula. Finally, to show that bar(CD) = bar(EF), we first observe that bar(CE) = bar(AB) = bar(DF), and therefore bar(CD) = bar(CE) - bar(DE) = bar(DF) - bar(DE) = bar(EF). Also, derive the area analytically and verify the same. Email newsletter, the MathsLinks network © 2007–2021 Simon Job | Contact | About. Area of a parallelogram - derivation. We can derive the formula by using the following steps: Draw a parallelogram whose base is b altitude is h. Cut a triangular shape from the parallelogram. Solution : Let a vector = i vector + 2j vector + 3k vector. area of a parallelogram. The blue point "squeezes" the parallelogram. The area of a parallelogram is the area occupied by it in a two-dimensional plane. And how we can define it. To better our understanding of the concept, let us take a look at the derivation of the area of square formula in maths. The area of a parallelogram is the base times the height. Area of trapezium is the region covered by a trapezium in a two-dimensional plane. You must use the altitude that goes with the base you choose. A trapezium is a 2D shape that falls under the category of quadrilaterals. 1. This page describes how to derive the formula for the area of a parallelogram. Students are provided with geometric figures to work with. where sinα = h / b; h = b; A = ah; A = ab. The altitude (or height) of a parallelogram is the perpendicular distance from the base to the opposite side (which may have to be extended). From previous knowledge, students are aware that the area of a rectangle is A = b * h. Given a trapezoid, if we form a congruent trapezoid and rotate it such that the two congruent trapezoids can be joined together to form a parallelogram as shown by the congruent black and grey trapezoids below. 4 in. The area of the original parallelogram is therefore for derivation of area , we request you to post them in separate thread to have rapid assistance from our experts. 0. area of a rectangle is found by multiplying its width times its height. Using vectors, If diagonals of any parallelogram are given , then we can find its area by using vector operations. However, that justification comes fairly quickly when we show that triangles ACD and BEF are congruent. . Area parallelogram =$$b\times h = 24 x 13 = 312 sq. Students will give an informal derivation of the relationship between the circumference and area of a circle ; Vocabulary circumference area pi radius of a circle parallelogram; About the Lesson The lesson assumes students have a knowledge of the area formula for a parallelogram. 4 years ago | 5 views. This page describes how to derive the formula for the area of a parallelogram. In other words, the area of a parallelogram is base times height. The perimeter of a parallelogram is the measurement is the total distance of the boundaries of a parallelogram. Learn the formula for finding the area of parallelogram. 12 Z 8 6 X P 9 R Y P 10 6 X Z Y 5 3 P R P 6 cm 8 cm 5 cm 5 ft 12 ft 7 ft 5 ft 4 m 7 m 4 m 2 in. A trapezium is a 2D shape that falls under the category of quadrilaterals. Is equal to the determinant of your matrix squared. We have learned that the area of a parallelogram is the product of its base and its height, and the circumference of a circle with radius is . 2 times 1/4 fourth is 1/2 times AC times BD. Start with the same trapezoid. The area of the parallelogram is base, 1/2 circumference, times height, radius. Note: The base and height should be perpendicular. Translate (move) the copy to touch the original so as to create a rectangle. Area of Square Formula in maths = a × a = $a^{2}$ Where, a is the length of the side of a square. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. Area of parallelogram = Area of Rectangle. 2. The area of the parallelogram is giv Parallelogram Area Using Diagonals. Derivation. To find the area of parallelogram multiply the base by its height and vice versa. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. It is the space enclosed in 2D geometry. The green point changes the slope of the sides of the parallelogram. In this post, we are going to find the general formula for finding the area of a rhombus using these properties. We were apprehensive about diverting students’ attention to another activity on the derivation of area of triangle. Pinterest. Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular). Similar to other geometrical shapes, it also has its own properties and formulas based on area … Area of trapezium is the region covered by a trapezium in a two-dimensional plane. To Derive: Area of trapezium. If diagonals of any parallelogram are given , … Navigate through these 40+ worksheets on calculating the area using the formula with the base and height measures expressed as integers, fractions or decimals, area problems involving unit conversion, finding the missing dimensions and much more. We all know that pi, which is denoted with the greek alphabet π and is the ratio of the circumference and the diameter. mathinschool.com. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The diagonals of rhombus are perpendicular to each other. Explanation: . So 1/2 times 1/2 is 1/4 times AC times BD. Area of parallelogram and trapezium lesson. 2. 4. A directory of useful objects found on the web for teaching Maths. Monday, 14 December 2020 / Published in Uncategorized. Move the triangular part to create a rectangle. Share your teaching ideas or leave a review about this link. 26. Topic: Area, Circle, Parallelogram A = b x h; A = 5 x 2 = 10. The red point changes the height of the parallelogram. Areas of Similar Triangles In Exercises 26 and 27, the triangles are similar. sectional area and moment of inertia associated with their cross sections. Although it seems obvious from the picture, we cannot make the claim immediately without some justification. See Derivation of the formula.. Recall that any of the four sides can be chosen as the base How to Find the Area of a Parallelogram. Browse more videos. The red point changes the height of the parallelogram. In this Area of a Parallelogram worksheet, students are asked to solve for area graphically and mathematically. The blue point "squeezes" the parallelogram. With that, we have ACD ~= BEF. Here is the example to same Regards If it is indeed a triangle, then its area is the product of its height and its width. This page describes how to derive the formula for the Thus, the area of parallelogram = area of rectangle = base × height. Similar to other geometrical shapes, it also has its own properties and formulas based on area … Therefore, the area of the parallelogram, which is equal to the area of a circle, is . Cut a right triangle from the parallelogram. Area of parallelogram = Area of Rectangle. Start with a parallelogram with known base length (width) w and altitude h (height). It is a quadrilateral wherein both pairs of opposite sides are parallel. A parallelogram is defined as a simple quadrilateral with two pairs of parallel sides. The parallelogram formed by the two congruent trapezoids has a base b 1 + b 2 and Area of Parallelogram is the region covered by the parallelogram in a 2D space. The orange point changes the base length of the parallelogram. Another derivation. A rhombus is a parallelogram whose sides are congruent. The area of a parallelogram is the region covered by the parallelogram in the 2D plane. Things to try. The parallelogram will have the same area as the rectangle you created that is b × h; Homepage. The derivation of the area of trapezium is given below. 0. 27. Volume of Parallelepiped. h is the altitude (height) of the parallelogram. The Demonstration shows a graphics derivation of formula for the area of the parallelogram given by points and namely The area of triangle is Since the expression could be negative it gives the socalled signed area. After showing an example in which she demonstrated the computation of area using the 1 2 × base × height ‘formula’, the students Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure. Is equal to the determinant of your matrix squared. It works by moving a piece of the parallelogram so as to make a rectangle with the same area, which is easier to calculate. Playing next. Preview and details Files included (1) pptx, 251 KB. Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure. b vector = 3i vector − 2j vector + k vector. Identify the area formula of different figures related to square; and b. Inductively derive the area formula of rectangle, parallelogram, triangle, and trapezoid. Derivation of the area of a parallelogram and the area of a trapezium.\n\nAssumed knowledge: Area of a rectangle. The area of a parallelogram: A = ah. It is the space enclosed in 2D geometry. Recall that the Suppose, the diagonals intersect each other at an angle y, then the area of the parallelogram is given by: Area = ½ × d 1 × d 2 sin (y) Plane Stress and Plane Strain Equations However, only one local coordinate along the length of the element is required to describe a position along the element (hence, they are called line elements). This constructs a rectangle by definition of a parallelogram. A = base (b) * … Click on the "Run" button below to see this in action. They also bisect each other. I'll do it in the next video. Parallelogram decomposed and recomposed into a rectangle. The area of a rectangle is equal to the product of its length and its width, so. The cross product of two vectors a and b is defined only in three-dimensional space and is denoted by a × b. Learn to calculate the area using formula without height, using sides and diagonals with solved problems. The area of a parallelogram: A = ah. Then find the ratio of their areas. The intuition is fairly simple. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. Area = area … The area of a parallelogram is expressed in square units. Read more. Notice that if we move the two triangles as indicated by the arrow in the figure above, then we have transformed the parallelogram into a rectangle. Posted on January 8, 2016 by aradhyeagarwal. Consider a rectangle that measures 3 … Geometry lessons. Hence, the area of trapezium is. See Derivation of the formula. November 1, 2018 February 22, 2019 Craig Barton. Author: Sam Webster. Utiliser une matrice pour définir une application linéaire. Assumed knowledge: Area of a rectangle Opposite sides are equal in length and opposite angles are equal in measure. Derivation. Report. Draw a parallelogram. This Area of Parallelogram Worksheet is suitable for 8th - 10th Grade. Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. area and perimeter of the parallelogram. The lesson involves using sectors of a circle to form a parallelogram and, from this shape, investigating the area formula for a circle. In this activity you are going to explore the area of a parallelogram, and how we can work out the area of any parallelogram. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. Area of a parallelogram. Deriving the Formula for the Area of a Parallelogram - Wolfram Demonstrations Project. For finding the area of a parallelogram, the base and height must be perpendicular. To find … Using vector algebra, find the area of a parallelogram. Now, the length of the rectangle is and the width is . new Equation("'area of parallelogram' = w @dot h", "solo"); Area is 2-dimensional like a carpet or an area rug. Recall that any of the four sides can be chosen as the base. Page 5 of 7. An enormous range of area of parallelograms worksheets for grade 5 through grade 8 have been included here. w is the width, or length of a base, and Cut with an altitude which will cut the parallelogram into two parts. The area of any parallelogram can also be calculated using its diagonal lengths. area and height of a parallelogram. The algebra that accompanies these illustrations if difficult for some students but needs to be included in the derivation of the formulas. We can also derive the area of a circle by unwinding an infinite number of circular tracks. Kameron Dotty. We are going to learn two methods. The area of a parallelogram is expressed in square units. Also, the lengths of the opposite sides are equal. The area of a polygon is the number of square units inside the polygon. To answer this question, we must find the diagonal of a rectangle that is by .Because a rectangle is made up of right angles, the diagonal of a rectangle … What is the area? What is a triangle? 3. The smaller the width of our track becomes, the rearranged figure becomes more and more like a triangle. In physics and applied mathematics, the wedge notation a ∧ b is often used (in conjunction with the name vector product), although in pure mathematics such notation is usually reserved for just the exterior product, an abstraction of the vector product to n dimensions. And then you see where this is going. Find derive area parallelogram lesson plans and teaching resources. Facebook Free. Area is the quantity that expresses the extent of a two-dimensional figure or shape or planar lamina, in the plane. Area of a parallelogram = Base × Height. 28 October 2013 Edit: 25 February 2014 The area of a parallelogram can be determined by multiplying the base and height. Draw a parallelogram. This can be rearranged into more familar forms: or Any line through the midpoint of a parallelogram bisects the area. Figure 1 – A circle with radius r and a parallelogram with base b and altitude h. If you need to find the area of a parallelogram, it's easily done with a simple formula Parallelogram area formulas. And actually, I haven't done this in a video. Opposite sides are equal in length and opposite angles are equal in measure. So we are going to use a different approach. 4:46. Recall that the area of a parallelogram is its altitude (h) times the length of either base. It works by moving a piece of the parallelogram . Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. Derivation: Here, let one side be ‘b1’ and other side be ‘b2’ Distance between the parallel sides is ‘h’ From the figure, it can be seen that there are two triangles and one rectangle. for derivation of area , we request you to post them in separate thread to have rapid assistance from our experts. Find the scale factor of TPQR to TXYZ . Created: Oct 19, 2014. There are other ways of finding the areas of parallelograms, generally. Now we replace C with pi times diameter: And replace d with 2r: Problem solving with Plane Area Comprendre comme associer un ensemble de vecteurs à un autre. Interactive Math Lesson - Finding the Area of Parallelogram. The formula to calculate the area of an isosceles trapezoid is, Area of Isosceles Trapezoid = h \( \frac{(a + b)}{2}$$ Parallelogram. A parallelepiped is related to the parallelogram in the same manner how a cube related to the square and a cuboid related to the rectangle. Area of Square Formula Derivation. Cut a right triangle from the parallelogram. In mathematical geometry, a parallelepiped is defined as the 3-D figure that is formed by the six parallelograms together.Sometimes, the term rhomboid is also defined with the same meaning. This site was created in a private capacity and is not the work of my employer. cm\) Q.2.Find the area of a trapezium whose parallel sides are 57 cm and 85 cm and perpendicular distance between them is 4cm. Area of Circle Derivation into Parallelogram. The orange point changes the base length of the parallelogram. Let θ be the angle between P and Q and R be the resultant vector. Figure 1. Note: squares, rectangles and rhombuses are all parallelograms. An alternative proof of the area of a trapezoid could be done this way. About this resource. When we have done with all the above steps, it forms a rectangular shape. So 2 times the area of ABC, area of ABC is that right over there. Take the two parts and put them back together with the two sides with angles not a right angle overlapping. 2. In the above picture, we need to show that the area of parallelogram ABEC has the same area as the rectangle ABFD, which is bh. Thus "area"(ABEC)="area"(ABED) + "area"(ACD) ="area"(ABED) + "area"(BEF) ="area"(ABFE) =bh DERIVATION OF AREA FORMULA OF PLANE FIGURES RELATED TO SQUARE Sabado, Setyembre 21, 2013. From the figure above we see that both base lengths are equal to b1+b2. Area of parallelogram and trapezium lesson. The parallelogram consist of equal opposite sides and its opposite angles are equal in measure. Draw heights from vertex B and C. This will break the trapezoid down into 3 shapes: 2 triangles and a rectangle. Area of parallelogram = Twice Area of Triangle. The area of a parallelogram is A = bh. The focus of the lesson was to use the area of triangles to find the area of a parallelogram. A parallelogram area can be calculated by multiplying its base with its height, i.e., b x h. The height of the parallelogram must always be perpendicular to its base. Quickly find that inspire student learning. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. This page describes how to derive the formula for the area of a parallelogram. Draw a parallelogram. DERIVATION OF AREA FORMULA OF PLANE FIGURES RELATED TO SQUARE . Fairly straightforward, which is a neat result. Author: Dave Pfaffle. 16 m b 14 cm 4 ft h b Visualize It! It works by moving a piece of the parallelogram so as to make a rectangle with the same area, which is easier to calculate. It is shorter to write in documents to reference this link. Twitter Info. The area of the parallelogram is giv Here is a summary of the steps we followed to show a proof of the area of a parallelogram. Some important formulas of Trapezium, Kite, and Parallelogram are given below. In the figure above, the altitude corresponding to the base CD is shown. We can derive the area of a circle using integration but I’m pretty sure most of you, like me are not familiar with it. Let us now discuss the parallelogram formula i.e. Derivation of the area of a parallelogram and the area of a trapezium. Use the right triangle to turn the parallelogram into a rectangle. The area of a parallelogram is given by Area = b x h. For example: This parallelogram has a base of 5 and a height of 2. Geometry proofs. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Derivation. Subscribe to feed A parallelogram is a 2-dimensional shape that has four sides and has two pairs of parallel lines. Loading... Save for later. Updated: Oct 15, 2015. pptx, 251 KB. ACMMG159 Year 7 Establish the formulas for areas of rectangles, triangles and parallelograms and use. In this activity you are going to explore the area of a parallelogram, and how we can work out the area of any parallelogram. To use the lists feature, first Sign up for free. The area is 10. http://www.mathopenref.com/parallelogramareaderive.html. Use the right triangle to turn the parallelogram into a rectangle. LinkedIn. 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# How to Divide Without a Calculator With Decimals
Dividing with decimals without a calculator can be done by following a few simple steps. The first step is to line up the decimal points in both numbers being divided, making sure that they are perfectly aligned above each other and the digits are directly under their counterparts. Next, divide the top number (dividend) by the bottom number (divisor) as if it were an integer problem.
Place any remainders above or below the dividend until there is no remainder left and all digits have been used. Then count how many times you moved your decimal point when lining up the numbers; this will tell you how many places to move your answer’s decimal point after dividing. Lastly, round off as necessary according to significant figures given in order to get your final answer!
• Identify dividend and divisor: To divide decimals, you first need to identify the number being divided (dividend) and the number by which it is being divided (divisor)
• Move decimal point in both numbers: The next step is to move the decimal point in both numbers so that it appears at the end of the dividend
• Move all digits to right for each number until no more digits remain after either decimal points 3
• Divide as with whole numbers: Now that you have your new numbers, treat them like any other division problem without a calculator – divide as with whole numbers while keeping an eye on where you moved your decimal points
• Place answer’s decimal point: When solving division problems without a calculator, be sure to place your answer’s decimal point directly above where you placed yours in the original two numbers involved in the equation
Credit: math.wonderhowto.com
## How Do You Divide Decimals Step by Step?
Dividing decimals is a common mathematical operation that can be tricky for some people. Luckily, there are a few simple steps to help make the process easier. First, you will need to convert your divisor (the number being divided into) into an integer by moving the decimal point all the way to the right.
Then move the decimal point in the dividend (the number being divided by) to match it. Next, divide as normal using long division and add a decimal point directly above where you started in step two. Finally, use any remainders or zeroes as placeholders until you get your final answer!
With these steps in mind, dividing decimals should become much simpler!
## How to Do Division Without a Calculator?
Doing division without a calculator can seem daunting, but it is actually quite simple. The most basic form of division is the long division method which is simply breaking down the problem into smaller steps and solving each step one at a time. To start off, you need to identify your dividend (the number being divided) and your divisor (the number you are dividing by).
Once these two numbers are identified, draw a line underneath the dividend and put the divisor on top of this line. Then begin by dividing the first digit of your dividend with your divisor. This will give you an answer as well as a remainder that must be brought down to continue working through the problem until all digits have been divided or subtracted from each other.
You then multiply what was in front of this last digit with the divisor and subtract it from what was remaining after bringing down that last digit. As soon as there’s nothing left to bring over or divide/subtract anymore, you’ve reached your answer! With practice, doing divisions without a calculator becomes easier and faster so don’t be afraid to try out different problems using this method!
## How Can We Divide a Number With a Decimal Number?
To divide a number with a decimal number, it is important to first convert the decimal into an equivalent fraction. This can be done by multiplying both the numerator and denominator of the decimal with 10 or other suitable factors such that after multiplication, the denominator becomes an integer. Then use long division to find out the quotient by dividing each digit from left to right in turn as we usually do for whole numbers.
One needs to remember that if there is any remainder then it should also be included in our answer along with the quotient obtained using long division method.
## What is an Easy Trick for Dividing Decimals?
One easy trick for dividing decimals is to move the decimal point in both numbers so that the divisor becomes a whole number. For example, if you are trying to divide 5.25 by 0.8, first move the decimal point two places to the right in each number: 52.5 ÷ 8. This makes it much easier to solve because now we can simply divide 52 by 8, which equals 6.5 (the original answer was 6.5625).
Moving the decimal points works with any type of division problem involving decimals and can be very helpful when solving more complicated problems as well!
## Long Division With Decimals
Long division with decimals can be a challenging concept for students to master. It requires careful attention to detail and precision in order to get the correct answers. When performing long division with decimals, it is important to remember that you should place the decimal point of the quotient directly above the decimal point of the dividend when beginning your calculations.
Additionally, you may need to add zeros after each number in order for your answer to be accurate. With practice and patience, long division with decimals can become easier over time!
## How to Divide Decimals
When dividing decimals, it is important to remember that you must move the decimal point in both numbers before performing the division. For example, if you are trying to divide 3.14 by 2.5, you would first need to rewrite both numbers as 31.4 and 25 respectively. Then proceed with traditional long division as normal!
## How to Divide Decimals by Whole Numbers
When it comes to dividing decimals by whole numbers, the process is actually quite simple. All you need to do is move the decimal point of your divisor (the number being divided into) all the way to the right and add zeros as necessary. After that, divide as normal and place the decimal point in your answer directly above where it was in your dividend (the number being divided).
For example, when dividing 7.5 by 2, move the decimal two places to make 2.00 before proceeding with division – this would give you an answer of 3.75!
## Decimal Division Calculator
A decimal division calculator is a tool that can be used to quickly and accurately divide two numbers with decimals. It provides an easy way for users to calculate the exact result of a division problem, eliminating the need for tedious calculations by hand. With this calculator, you can input any two numbers with up to 15 digits each and it will display the answer as quickly as possible.
This makes it perfect for students or anyone else needing help working through complex equations involving decimals.
## Conclusion
In conclusion, learning how to divide with decimals without a calculator is not as difficult as it may seem. With practice and understanding of place value, it can easily be mastered. This skill can prove useful in everyday life by helping you quickly calculate the answer to any division problem without needing a calculator.
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## Division of Decimals
Let's first review the names given to the numbers involved in a division problem:
1. Divisor - the number you are dividing by
2. Dividend - the number you are dividing into
3. Quotient - the answer in a division problem
#### To divide a decimal by a whole number:
1. Write the decimal point in the quotient directly above the decimal point in the divisor.
2. Divide the numbers as if they were whole numbers.
#### Examples:
1. 3.75 ¸ 25 Explanation: .15 25 ç3.75 2 50 1 25 1 25 0 Write the decimal point in the quotient directly above the decimal in the dividend. Divide as if the numbers were whole numbers. 2. .396 ¸ 9 Explanation: .044 9 ç.396 360 36 36 0 Write the decimal point in the quotient directly above the decimal in the dividend. Divide as if the numbers were whole numbers. 9 does not go into 3, so we must use a zero as a placeholder directly after the decimal.
It is very important that you write the decimal in the quotient directly above the decimal in the dividend.
To divide by a decimal, multiply the divisor by 10, 100, 1000, and so on, until the divisor is a whole number. Then multiply the dividend by the same number. Multiplying by 10 moves the decimal one place to the right. Multiplying by 100 moves the decimal two places to the right, and so on.
#### To divide by a decimal:
1. Move the decimal point in the divisor to the right until a whole number is obtained.
2. Move the decimal in the dividend the same number of places to the right.
3. Write the decimal in the quotient directly above the decimal in the dividend.
4. Divide the numbers as if they were whole numbers.
#### Examples:
1. 7.6 ¸ .05 Explanation: Move the decimal in the divisor two places to the right to make the divisor the whole number 5. Move the decimal in the dividend two places to the right. Add one zero to the dividend so the decimal can be moved two places. Place the decimal in the quotient directly above the decimal in the dividend. Divide as if the numbers were whole numbers. 2. 241.2 ¸ .36 Explanation: Move the decimal in the divisor two places to the right to make the divisor a whole number 36. Move the decimal in the dividend two places to the right. Add one zero to the dividend so the decimal can be moved two places. Place the decimal in the quotient directly above the decimal in the dividend. Divide the numbers as if they were whole numbers. Add zero after the seven in the quotient as a placeholder.
The zero in Example #2 is another reason why you must place the decimal in the quotient directly above the decimal in the divisor.
Also, the zero in Example #2 is a reminder that you must place the first answer in your quotient in the correct position. 36 goes into 241, so the 6 goes directly above the 1 in the dividend.
3. 16.448 ¸ 6.4 Explanation: Move the decimal in the divisor one place to the right to make the divisor a whole number 64. Move the decimal in the dividend one place to the right. Place the decimal in the quotient directly above the decimal in the dividend. Divide the numbers as if they were whole numbers. 4. 396 ¸ 3.3 Explanation: Move the decimal in the divisor one place to the right to make the divisor a whole number 33. Place a decimal after 396 to make a decimal number 396. Move the decimal one place to the right in the dividend and add a zero. Place the decimal in the quotient directly above the decimal in the dividend. Divide the numbers as if they were whole numbers.
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# 1958 AHSME Problems/Problem 43
## Problem
$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$. The length of $\overline{AB}$ is:
$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$
## Solution
$[asy] import geometry; unitsize(50); pair C = (0,0), B = (3,0), A = (0, 4); pair AC = midpoint(A--C), BC = midpoint(B--C); draw(A--B--C--A); draw(A--C, StickIntervalMarker(2, 1)); draw(B--C, StickIntervalMarker(2, 2)); draw(B--AC); draw(A--BC); dot(AC); dot(BC); MP("A", A, W); MP("B", B, E); MP("C", C, W); MP("E", AC, W); MP("D", BC, S); label("y", A--AC, W); label("y", AC--C, W); label("x", B--BC, S); label("x", BC--C, S); draw(rightanglemark(A, C, B)); [/asy]$
By the Pythagorean Theorem, $(2x)^2+y^2=BE^2$, and $x^2+(2y)^2=AD^2$.
Plugging in, $4x^2+y^2=16$, and $x^2+4y^2=49$.
Adding the equations, $5x^2+5y^2=65$, and dividing by $5$, $x^2+y^2=13$.
Note that the length $AB^2$ is equal to $(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)$.
Therefore, the answer is $\sqrt{4(13)}=2\sqrt{13}$ $\fbox{D}$
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Update all PDFs
# Running Laps
Alignments to Content Standards: 4.NF.A
Cruz and Erica were both getting ready for soccer.
• Cruz ran 1 lap around the school.
• Erica ran 3 laps around the playground.
Erica said,
I ran more laps, so I ran farther.
Cruz said,
4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile. My laps are much longer, so I ran farther.
## IM Commentary
The purpose of this task is for students to compare two fractions that arise in a context. Because the fractions are equal, students need to be able to explain how they know that. Some students might stop at the second-to-last picture and note that it looks like they ran the same distance, but the explanation is not yet complete at that point. Just because to pictures look the same doesn't mean they are. Think of a picture representing $\frac{49}{100}$ and $\frac{50}{99}$--these will look very similar, but they are not, in fact, equal. To explain why Erica and Cruz ran the same distance, we have to explain why $\frac14 = \frac{3}{12}$.
Note that this task illustrates the cluster "4.NF.A Extend understanding of fraction equivalence and ordering" because it addresses elements of both the standards within this cluster.
## Solution
Let’s begin with a picture representing $1$ mile.
With this picture in mind, we can see that one lap around the school is $\frac14$ mile:
and we can see that one lap around the playground is $\frac{1}{12}$ mile:
Since Erica ran 3 laps around the playground, she ran $\frac{3}{12}$ mile. Since Cruz ran 1 lap around the school, he ran $\frac14$ mile:
Now looking at the pictures above, it looks like they actually ran the same distance. We can show this is true if we divide each of the laps around the school into three equal pieces:
When we subdivide each $\frac14$ mile lap into three equal pieces, our mile is divided into 12 equal pieces. The piece that represents Cruz's run is now divided into 3 pieces, and so $\frac14 = \frac{3}{12}$ he ran the same distance that Erica ran.
#### Jason says:
over 5 years
Currently, the solution says at one point:
"Since Erica ran 3 laps around the playground, she ran 3/12 mile".
No arithmetic operations are visible in this reasoning. However, the major advance in fractions in grade 4 is that fractions - first introduced in grade 3 as numbers but in a pre-operational way - are now being integrated into the system of arithmetic. Thus, in grade 4, we don't just "view fractions as built out of unit fractions" (grade 3 overview, p. 21); we "build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers."
It would be helpful to exhibit that progression of ideas here, and to move beyond the kind of pre-operational thinking portrayed in the grade 3 standards.
So the sentence in the solution could change to read:
"Erica ran 3 laps, or 3 x 1/12 mile, which equals 3/12 mile (4.NF.4c). Alternatively, Erica ran 3 laps, or 1/12 mile + 1/12 mile + 1/12 mile, which again equals 3/12 mile (4.NF.3d)."
This shows the operations of arithmetic being used as expected in grade 4.
#### Kristin says:
over 5 years
I completely agree with you about the transition from third to fourth grade, but this task is about equivalent fractions, where the key insight is about subdividing unit fractions that make up a fraction. Students haven't forgotten the meaning of a fraction, and using addition seems like overkill in a task like this. In fact, I think it distracts from the key reasoning that needs to happen here (in my experience, this is hard enough as it is).
We do have quite a few tasks that do what you ask, but they are illustrations for 4.NF.B.3. Here are two examples:
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# Lesson Video: Grouped Frequency Tables: Estimating the Mean Mathematics
In this video, we will learn how to estimate the mean using grouped frequency tables.
15:21
### Video Transcript
In this video, we will learn how to estimate the mean using grouped frequency tables. Letโs begin by recapping what a grouped frequency table is. A grouped frequency table is a frequency table with data organized into smaller groups, often referred to as sets or classes. These are often useful when we are working with large data sets or data sets with a large range of values. As a result, a grouped frequency table is a manageable way of representing the data. However, the disadvantage of grouped frequency tables is that we cannot extract the original data values from them.
Letโs consider the table shown, which represents the marks that students received in an examination. The groups are given as open intervals: zero dash, 10 dash, 20 dash, and 30 dash. This means that the data values in the first group can be considered as zero or more but less than 10. Its boundary values would be zero and 10. We observe that one student achieved a mark in this interval. As we donโt know the exact mark, it becomes more difficult when performing any statistical calculations based on a grouped frequency table.
We recall that we can calculate the mean of a data set by dividing the sum of the data values by the total number of data values. If we have the individual scores listed as shown, then the mean is easy to calculate. We find the sum of all the scores, which is equal to 360, and divide this by 15, as there are 15 scores in total. This would give us a mean mark equal to 24. In our grouped frequency table however, this is not possible. So instead, we find an estimate for the mean. We begin by adding two rows to our table. The first for the midpoint of each class interval. This midpoint can be calculated by adding the grouped boundary values for our first column, zero and 10, and then dividing by two. This is equal to five.
For the second class interval, we add the boundaries of 10 and 20 and then divide by two. This is equal to 30 divided by two, giving us a midpoint of 15. The next column has a midpoint of 25, as 20 plus 30 divided by two is 25. Assuming that each class has equal width, the upper boundary of our last group is 40 and the midpoint of 30 and 40 is 35.
Our next step is to multiply the frequency in each column by the midpoint. This will give us an estimate for the total number of marks in that group. One multiplied by five is five. Five multiplied by 15 is 75. Repeating this for the final two columns, we have values of 125 and 140. Our next step is to find the sum of the second and fourth rows. One plus five plus five plus four is equal to 15. So there were 15 students in total that took the examination. Adding the four numbers in the bottom row gives us 345.
We are now in a position to calculate an estimate for the mean. We divide the total of the frequency multiplied by the midpoint by the total of the frequency. In this example, we have 345 divided by 15. And this is equal to 23. We will now consider an example in a different context.
In an extract of a book, the number of words per sentence was counted. Find the missing numbers in the following table. Use the previous table to calculate an estimate for the mean number of words. Give your answer to two decimal places.
In this question, we are given a grouped frequency table that lists the number of words per sentence in a book along with their frequencies. 15 sentences had between one and seven words inclusive. 20 sentences had between eight and 14 words. 45 sentences had between 15 and 21 words. And we also have the frequencies for between 22 and 28 words, 29 and 35 words, and 36 and 42 words. The third column of our table corresponds to the midpoints. To calculate these values, we add the two boundary values in each group and divide by two. For example, in the first row, we add one and seven to give us eight, and dividing this by two gives us four. This is the midpoint of the group one to seven.
We need to calculate the midpoint for the second group in our table, where the number of words are between eight and 14 inclusive. The midpoint is therefore equal to eight plus 14 divided by two. This simplifies to 22 divided by two, which is equal to 11. The final column of our table corresponds to the product of the frequency and the midpoint. In the first row, 15 multiplied by four is equal to 60. We need to calculate the frequency of 20 multiplied by the midpoint of 11. And this is equal to 220. The two missing numbers in the table are 11 and 220.
The second part of this question wants us to calculate an estimate for the mean number of words. And we recall that an estimate for the mean can be found by dividing the total of the frequency multiplied by midpoint column by the total frequency. Adding an extra row to our table, we need to find the totals of the second and fourth columns. Adding the frequencies gives us a total of 160. And adding the numbers in the last column for the frequencies multiplied by the midpoints gives us 3391. Our estimate for the mean is therefore equal to 3391 divided by 160. This is equal to 21.19375. And as we are asked to give our answer to two decimal places, this is equal to 21.19. This is an estimate for the mean number of words per sentence in the book.
Before looking at another example, letโs summarize the steps we need to take in order to calculate an estimate for the mean of a grouped frequency table.
This can be done in four simple steps. Firstly, we find the midpoint ๐ฅ of each group in the table by adding the boundary values and dividing by two. Secondly, we multiply the midpoints by their frequencies ๐น of the corresponding classes to give ๐น๐ฅ. It is often useful to add additional rows or columns to the frequency table to record these. Next, we find the sum of ๐น๐ฅ, known as the total of ๐น๐ฅ. Finally, we divide this sum by the total frequency known as total ๐น. This can be summarized as follows. The estimate for the mean is equal to the total of ๐น๐ฅ divided by the total of ๐น. Letโs now consider an example where we apply this.
The following table shows the salaries ๐ of employees in a certain company, given in Egyptian pounds. Estimate the mean salary in Egyptian pounds, giving your answer approximated to two decimal places.
The given table is a grouped frequency table. The first column tells us that five employees earned a salary greater than or equal to 1000 Egyptian pounds and less than 3000 Egyptian pounds. The second column tells us that eight employees earned a salary greater than or equal to 3000 Egyptian pounds and less than 5000 Egyptian pounds. There were three employees and five employees in the third and fourth groups, respectively. As we are not told the exact salaries of any of the employees, we cannot determine the exact mean. Instead, we need to find an estimate for the mean.
To do this, we follow a four-step process, the first of which is to find the midpoint of each group, which we will call ๐ฅ. To find the midpoint of each group, we add the boundary values and divide by two. This means that in the first column, we need to add 1000 and 3000 and then divide this answer by two. This is equal to 2000. So the midpoint of 1000 and 3000 Egyptian pounds is 2000 Egyptian pounds. Repeating this for the second column, we see that the midpoint of 3000 and 5000 is 4000. Likewise, the midpoint of 5000 and 7000 is 6000, and the midpoint of 7000 and 9000 is 8000.
Our next step is to multiply the midpoints by the frequencies ๐น, which in this case is the number of employees in each group. In the first column, we multiply 2000 by five and add the answer to the bottom row of our table, which we have labeled ๐น๐ฅ. 2000 multiplied by five is 10000. Next, we multiply 4000 by eight, giving us 32000. 6000 multiplied by three is 18000. And finally, 8000 multiplied by five is 40000.
Our next step is to find the total of ๐น, the number of employees, and the total of ๐น๐ฅ. This is because the estimate for the mean is calculated by dividing the total of ๐น๐ฅ by the total of ๐น. The sum of five, eight, three, and five is 21. Adding 10000, 32000, 18000, and 40000 gives us 100000. We can therefore calculate an estimate for the mean by dividing 100000 by 21. This is equal to 4761.9047 and so on. We are asked to give our answer to two decimal places. And we can therefore conclude that an estimate for the mean salary at the company in Egyptian pounds is 4761 pounds 90.
We will now summarize the key points from this video.
We saw in this video that a grouped frequency table organizes data into smaller groups called classes. Since we cannot determine the individual data values from a grouped frequency table, we cannot calculate the exact mean. Instead, we calculate an estimate for the mean. This is done following a four-step process.
Firstly, we find the midpoint ๐ฅ of each group in the table by adding the boundary values and dividing by two. Secondly, we multiply the midpoints by the frequencies. We then find the sum of these values and finally divide this sum by the total frequency. This can be summarized as the estimate for the mean is equal to the total of ๐น๐ฅ divided by the total of ๐น, where ๐ฅ is the midpoint of each group and ๐น is the corresponding frequency. Finally, we saw in our examples that it is useful to add rows or columns to the frequency table in order to record these values.
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# Demystifying Square Roots in Java
Square roots seem almost magical – how do you determine a number that yields a given value when multiplied by itself? In this extensive guide, we‘ll dig into the math behind square roots and how to calculate them in Java. Whether you‘re a beginner or an experienced programmer, you‘ll gain new insights into this fundamental mathematical building block. Let‘s get started!
## What Exactly Are Square Roots?
Simply put, the square root of a number N is defined as the number that, when multiplied by itself, yields N. For example:
``````sqrt(9) = 3, because 3 * 3 = 9
sqrt(16) = 4, because 4 * 4 = 16 ``````
We denote the square root of N mathematically as √N. The radical symbol √ (called the principal square root) indicates we want the positive square root. For example, √9 equals +3, not -3.
Conceptually, square roots are the inverse operation of squaring or raising to the power 2. Squaring a number gives you its square, while square rooting finds the original number.
Historically, ancient Babylonian mathematicians devised methods to compute square roots as early as 1800 BCE. The famous Pythagorean theorem involving square roots dates back to at least 500 BCE. Square roots have fascinated humanity across civilizations!
### Why Are Square Roots Useful?
Square roots play a crucial role in many fundamental equations across mathematics, science, and engineering. Here are some examples:
• Computing the length of the hypotenuse in a right triangle using the Pythagorean theorem:
``````c^2 = a^2 + b^2
c = sqrt(a^2 + b^2) ``````
[Image of right triangle with sides labeled]
• Finding the distance between two points in Cartesian coordinates:
``d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )``
[Image of two points on a graph]
• Calculating the standard deviation, a key statistic used in data analysis:
``σ = sqrt( Σ(x - μ)^2 / n ) ``
• Modeling diffusion in physics and chemistry:
``A = √(Dt/π)exp(-r^2 / 4Dt)``
The examples above demonstrate how square roots are indispensable for mathematical modeling and problem-solving across many STEM domains. Let‘s now see how to calculate them in Java.
## Introducing Math.sqrt() in Java
The easiest way to find the square root of a number in Java is using the `Math.sqrt()` method. This is a static method that takes a double as input and returns its double square root:
``````double num = 9.0;
double root = Math.sqrt(num); // root = 3.0``````
Internally, `Math.sqrt()` utilizes highly optimized native code for performance. On most systems, it relies on hardware square root instructions like FSQRD on x86 chips that compute roots natively in one clock cycle.
Let‘s break down the method signature:
``public static double sqrt(double a) ``
• `public` means `sqrt()` can be called externally
• `static` means we don‘t need a Math instance
• Takes a `double` parameter
• Returns a `double` result
Note that `sqrt()` always takes and returns a double primitive. What if we want roots of other numeric types like int or float?
## Finding Square Roots of any Primitive Type
Though `Math.sqrt()` takes a double, we can easily pass it other primitive types by casting:
``````int num = 9;
double root = Math.sqrt((double)num); // root = 3.0
float num2 = 16.0f;
double root2 = Math.sqrt(num2); // root2 = 4.0``````
The parameters get converted to double, and the return value is a double.
Why does the return type remain double even for floats? Double offers greater precision – floats have only 24 bits of mantissa, while double has 53 bits. Those extra bits reduce rounding errors and are useful for the sensitive computation of square roots.
What about square roots of long ints?
``````long bigNum = 500000000L;
double root = Math.sqrt(bigNum); // 22360.6797749979 ``````
No problem! The implicit casting from long to double preserves the full value.
In some cases, the double return value may have many decimal digits. We can round the result to a desired precision:
``````double root = Math.sqrt(10); // 3.1622776601683795
// Round to 2 decimal places
double rounded = Math.round(root * 100.0) / 100.0; // 3.16``````
This handy rounding trick works for all kinds of decimal precision.
## Alternative Approaches for Arbitrary Precision
A trade-off of `Math.sqrt()` is it‘s limited to double precision. If arbitrary precision beyond double is needed, we can use alternative square root algorithms at the cost of reduced performance.
Let‘s look at two popular approaches – the Babylonian method and Newton‘s method. Both are iterative methods that approximate the square root to any desired accuracy.
Here is an implementation of Babylonian square root using Java:
``````public static double babylonianSqrt(double input, double epsilon) {
double result = input;
while (Math.abs(result - input/result) > epsilon) {
result = (result + input/result) / 2.0;
}
return result;
}``````
This repeatedly averages an initial guess with input divided by the guess to converge on the square root. The while loop runs until the difference between iterations reaches an epsilon threshold.
Newton‘s method is similar but uses calculus to derive a more optimal iteration formula:
``````public static double newtonSqrt(double input, double epsilon) {
double result = input;
while (Math.abs(result*result - input) >= epsilon) {
result = (result + input/result) / 2.0;
}
return result;
}``````
Here, the while condition checks how close result*result is to the original input.
Both these algorithms can achieve arbitrary precision, but require multiple iterations and are computationally slower than hardware-accelerated `Math.sqrt()`.
There are also BigDecimal solutions that can remove rounding errors by maintaining scale. But these also have large performance overhead.
In summary, there is a precision/performance trade-off between the simple `Math.sqrt()` and more advanced algorithms.
## Real-World Examples Using Square Roots
Now that we‘ve covered the calculation techniques, let‘s look at some practical examples of how square roots are used in the real world:
### Geometry Problems
Square roots are ubiquitous in geometry for calculating distances and areas:
• Finding hypotenuse of a right triangle using Pythagorean theorem
• Computing distance between two points in 2D or 3D space
• Determining the diagonal of a rectangle/square
• Finding the area and circumference of circles
Consider this example program that calculates the distance between points:
``````// Returns euclidean distance between two points
public static double distance(Point p1, Point p2) {
double xDiff = p2.x - p1.x;
double yDiff = p2.y - p1.y;
return Math.sqrt(xDiff*xDiff + yDiff*yDiff);
}``````
This is a common geometry operation enabled by square roots.
### Statistical Applications
In statistics, standard deviation is defined using square roots:
``σ = √(Σ(x - μ)2 / n)``
Here is sample Java code to calculate population standard deviation:
``````public static double stdDev(double[] values) {
double mean = findMean(values);
double squaredDiffs = 0;
for (double num : values) {
double diff = num - mean;
squaredDiffs += diff * diff;
}
double variance = squaredDiffs / values.length;
return Math.sqrt(variance); // Std dev
}``````
Many other statistical formulas like confidence intervals also rely on square roots.
### Physics and Engineering
Physics formulas involving velocity, acceleration, force, and electricity often use square roots. For example, the formula for velocity:
``v = √(2ad) ``
Here is an example calculating final velocity in Java:
``double vFinal = Math.sqrt(2 * ACCELERATION * distance);``
Electrical engineers frequently use root mean square (RMS) voltage derived via square roots. Overall, most areas of science and engineering involve square roots.
### Finance
In finance, volatility indicates how rapidly an asset or market changes prices. Volatility is computed as the annualized standard deviation of returns. Thus, square roots are used when pricing options and estimating investment risk.
``````// Daily price history
double[] prices = ...
double mean = findMean(prices); // Calculate mean
double var = computeVariance(prices, mean); // Find variance
double volatility = Math.sqrt(var) * Math.sqrt(DAYS_PER_YEAR); // Annual volatility``````
As we‘ve seen from numerous examples, square root computation is crucial across problem domains.
## Best Practices and Optimizations
When using `Math.sqrt()`, keep these best practices in mind:
• Check for negative inputs`Math.sqrt(-1)` will return `NaN` (not a number). Validate inputs.
• Be aware very large numbers can result in `Infinity`.
• Since `sqrt()` relies on fast native code, avoid redundant calls – cache values when possible.
• Test edge cases like 0, NaN, Infinity, very small and very large numbers.
• Use `BigDecimal` if you need arbitrary precision beyond double – but it‘s much slower.
• On Android, `StrictMath` variants like `StrictMath.sqrt()` are even faster than `Math.sqrt()`.
Additionally, here are some optimization techniques:
• When computing many roots, use parallel streams`IntStream.range(0, n).parallel().map(x -> Math.sqrt(x))`
• For repeated roots on arrays, utilize vectorization with libraries like NumPy.
• Offload bulk roots to GPUs using CUDA or OpenCL – 10-100x faster than CPU!
By following best practices and leveraging optimizations like parallelism and GPUs, you can achieve blazing fast square root performance in Java.
## Conclusion
We‘ve covered a lot of ground exploring the innards of square roots in Java. Here are some key takeaways:
• `Math.sqrt()` provides a fast way to calculate square roots by leveraging hardware optimizations.
• Square roots are essential across STEM domains for solving geometric problems, statistical formulas, physics equations, and more.
• Alternative algorithms can achieve arbitrary precision through iteration but sacrifice performance.
• Follow best practices like validating inputs and testing edge cases.
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Home Math Geometry: The Line
# Math Geometry: The Line
Do you know what’s the next simplest geometrical figure after the point? It’s a straight, infinite geometrical figure. It’s the line! Compared to the point which has no dimensions, the line is one-dimensional.
## What is the use of the line in geometry?
When a line has two ends it is called a line segment or side-line and we usually use line segments to name the sides of the triangle. When a line has just one end, it is called a “ray”.
Lines are part of any geometrical shape and plane, that’s why it is important to know the basics about lines.
## Math Symbols Related To The Line
“ ↔ “ – this symbol represents the infinite line; usually is noted above two letters to symbolize that the line on which the two points are, is an infinite line, example:
“ ⎻ “ – a simple line above two letters indicates a line segment, which starts at point A and ends at point B:
“→” – this symbol indicates a ray, which starts from point A;
“∠” – this symbol indicates an angle formed by two rays, example: ∠ABC = 30°
“⟂” – this symbol indicates that two lines are perpendicular, thus forming a 90° angle. The notation would look like this:
“ ∥ “ – this symbol indicates that two lines are parallel to each other:
When two lines are not parallel, we would use the “∦” symbol. When two lines are parallel and equal, we use the symbol “⋕” .
## Types of Lines
In geometry, there are a variety of lines that relate to spatial relationships between points or angles. It’s pretty simple to memorize the type of each line, so here we go:
### Vertical line
A vertical line is a straight line that goes from top to bottom.
### Horizontal line
A horizontal line is a straight line that goes from left to right.
### Parallel lines
Two or more lines that do not intersect at any point, infinitely, are called parallel lines.
### Perpendicular lines
Two straight lines that meet at an intersection point and form a right angle (90°) are called perpendicular lines.
### Oblique lines
These lines are diagonal or slanting lines.
Diagonal or oblique lines can be perpendicular or parallel.
### Skew Lines
When it comes to spatial geometry, two non-parallel lines that are not intersecting in a given space are called skew lines.
### Concurrent lines
In our past article, we mentioned concurrency in points. Since two or more lines can cross through a common concurrent point, these lines are also called concurrent.
### Transversal lines
The line that cuts two or more straight lines is called a transversal line. The cut lines are not necessarily parallel.
### Coplanar lines
Coplanar lines are straight lines that exist in the same spatial plane.
## The Rules Of Lines In Math
1. When a transversal line intersects/cuts two parallel lines:
• Alternate angles are equal.
• Corresponding angles are equal.
2. The sum of the angles on a straight line is 180º, while the sum of the angles formed at a point is 360º.
3. When 2 straight lines intersect, they form vertically opposite and equal angles.
Remember the above simple and short line rules and get ready to apply them in daily life. Did you know that line calculations are an important part of computer graphics and architecture?
### Let’s PRACTICE!
Solve the following exercises:
#1. Look at the image below and choose which option is true.
Statement: The straight lines AB and CD are:
2. are not parallel because the two given corresponding angles are not equal
3. are parallel lines
4. are not parallel because the two given alternate angles are not equal
5. are not parallel because the two given consecutive interior angles do not add to 180°
#2. Look at the image below and choose which option is true.
Statement: The straight lines ST and UV are parallel, then the angles c and e are:
1. consecutive interior angles
2. corresponding angles
3. alternate interior angles
4. vertical angles
#3. Check out the image below:
The two parallel lines A and B are cut by a transversal line. If ∠5 = 70°, find the measure of the following angles: ∠7, ∠9, ∠10.
#4. Check out the image below and then choose the correct option from the statements below:
1. Angles 2 and 6 are examples of vertical / corresponding angles.
2. Angles 1 and 8 are examples of alternate interior / exterior angles.
3. The angle that is corresponding to angle 8 is angle 3 / 4
4. If lines m and n are parallel, then angles 3 and 6 / 8 have equal measures.
If you want to memorize the types of lines better, you can always solve some online exercises, here.When it comes to solving geometry problems, practice does indeed help you become better at mathematics.
For more fun exercises on lines, you can check out these questions. Additionally, if you want to learn more about 4th-grade mathematics, our OMC tutors offer individual and group lessons that turn any student into a math champion!
Reach out to us and meet our OMC math tutors!
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# How do I use Gaussian elimination to solve a system of equations?
Aug 16, 2015
The goals of Gaussian elimination are to get $1$s in the main diagonal and $0$s in every position below the $1$s,
Then you can use back substitution to solve for one variable at a time.
#### Explanation:
EXAMPLE:
Use Gaussian elimination to solve the following system of equations.
$x + 2 y + 3 z = - 7$
$2 x - 3 y - 5 z = 9$
$- 6 z - 8 y + z = - 22$
Solution:
Set up an augmented matrix of the form.
$\left(\begin{matrix}1 & 2 & 3 & | & - 7 \\ 2 & 3 & - 5 & | & 9 \\ - 6 & - 8 & 1 & | & 22\end{matrix}\right)$
Goal 1. Get a 1 in the upper left hand corner.
Goal 2a: Get a zero under the 1 in the first column.
Multiply Row 1 by $- 2$ to get
$\left(\left(- 2 , - 4 , - 6 , | , 14\right)\right)$
Add the result to Row 2 and place the result in Row 2.
We signify the operations as -2R_2+R_1→R_2.
((1,2,3,|,-7),(2,3,-5,|,9),(-6,-8,1,|,22)) stackrel(-2R_1+R_2→R_2)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22))
Goal 2b: Get another zero in the first column.
To do this, we need the operation 6R_1+R_3→R_3.
((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22)) stackrel(6R_2+R_3→R_3)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64))
Goal 2c. Get the remaining zero.
Multiply Row 2 by $- \frac{1}{7}$.
((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64)) stackrel(-(1/7)R_2 → R_2)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64))
Now use the operation -4R_2+R_3 →R_3.
((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64)) stackrel(-4R_2+R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7))
Multiply the third row by $\frac{7}{89}$.
((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7)) stackrel(7/89R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,1,|,-4))
Goal 3. Use back substitution to get the values of $x$, $y$, and $z$.
Goal 3a. Calculate $z$.
$z = - 4$
Goal 3b. Calculate $y$.
$y + \frac{11}{7} z = - \frac{23}{7}$
$y - \frac{44}{7} = - \frac{23}{7}$
$y = \frac{44}{7} - \frac{23}{7} = \frac{21}{7}$
$y = 3$
Goal 3c. Calculate x.
$x + 2 y + 3 z = - 7$
$x + 6 - 12 = - 7$
$x - 6 = - 7$
$x = 1$
The solution is $x = 1 , y = 3 , z = - 4$
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Here is a trick I learned as a child which I think is worth documenting for eternity in cyberspace. It is how to multiply the numbers six through ten using only your fingers.
First assign numbers in the following way to the fingers of both hands
1. pinky = 6
2. ring finger = 7
3. middle finger = 8
4. index finger = 9
5. thumb = 10
Fine, you have just constructed a simple 'digital' computer.
Now keep your hands in front of you, palms facing you, thumbs up.
Let's say we want to multiply 6 x 7.
Touch and hold the pinky of the left hand (6) to the ring finger of the right hand (7). Now with your hands in this position, fingers touching, hands in front of you, palms facing you and thumbs up:
1. count any touching fingers as ten each. (10 for left pinky plus 10 for right ring finger = 20)
2. count any fingers below the touching fingers as ten each.(ten for the right pinky which is below the touching fingers = 10)
3. multiply the number of fingers above the touching fingers on the left hand by the number of fingers above the touching fingers on the right hand. (there are four non-touching fingers above the touching fingers on the left hand and three on the right so 4 x 3 = 12).
So now add the numbers obtained from steps 1, 2, and 3. You get 20 + 10 + 12 = 42 and of course the original problem 6 x 7 = 42
Another example? Let's try an easy one. 7 x 8. Hands in position. Touch and hold the left ring (or 7) finger to the right middle (or 8 finger)
1. count the two touching fingers as 10 each = 20.
2. count any non-touching fingers below the two touching fingers as ten each (in this case the left pinky plus the right pinky and ring total up to = 30).
3. count up the non-touching fingers above the touching fingers on the left hand and the non-touching fingers above the touching fingers on the right hand (in this case there are three on the left and two on the right) and multiply them. 3 x 2 = 6
So now add the numbers obtained from steps 1, 2, and 3. 20 + 30 + 6 = 56 and the original problem was 7 x 8 and the answer of course proves to be 56.
as ariels points out, see Russian peasant multiplication and also see mathematics
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# Whole Numbers Multiplied By Decimals Worksheets
This multiplication worksheet focuses on instructing pupils the best way to mentally grow complete numbers. College students can make use of customized grids to suit exactly a single query. The worksheets also includedecimals and fractions, and exponents. You will even find multiplication worksheets using a dispersed property. These worksheets really are a have to-have for your math concepts course. They can be used in type to learn how to psychologically multiply whole numbers and line them up. Whole Numbers Multiplied By Decimals Worksheets.
## Multiplication of complete numbers
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets may help you master this simple concept. You are able to decide to use one particular digit multipliers or two-digit and a few-digit multipliers. Capabilities of 10 may also be an excellent alternative. These worksheets will allow you to training extended practice and multiplication reading through the phone numbers. They are also a wonderful way to help your youngster recognize the importance of understanding the different types of entire phone numbers.
## Multiplication of fractions
Having multiplication of fractions on a worksheet will help teachers plan and prepare lessons proficiently. Employing fractions worksheets permits educators to easily assess students’ understanding of fractions. Pupils can be questioned to finish the worksheet in a certain efforts and then label their techniques to see exactly where they want additional training. College students may benefit from expression conditions that associate maths to actual-lifestyle conditions. Some fractions worksheets include examples of comparing and contrasting numbers.
## Multiplication of decimals
When you flourish two decimal numbers, make sure to class them up and down. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. By way of example, 01 x (11.2) x 2 will be similar to 01 by 2.33 by 11.2 except if the merchandise has decimal places of under two. Then, this product is round on the nearest total quantity.
## Multiplication of exponents
A math concepts worksheet for Multiplication of exponents can help you training multiplying and dividing numbers with exponents. This worksheet will also supply problems that will need college students to multiply two diverse exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Besides, you can even enter unique directions around the worksheet alone. When you’re finished, it is possible to click “Produce” and the worksheet will likely be delivered electronically.
## Section of exponents
The standard rule for section of exponents when multiplying phone numbers is usually to deduct the exponent from the denominator through the exponent within the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. As an example, \$23 split by 4 will identical 27. However, this method is not always accurate. This procedure can lead to confusion when multiplying phone numbers which are too big or not big enough.
## Linear features
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So, the total rent would be \$470. A linear purpose of this particular type has the form f(by), where ‘x’ is the quantity of time the auto was hired. In addition, it provides the form f(x) = ax b, in which ‘b’ and ‘a’ are actual amounts.
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Derivative problem
The sum of two numbers is 12. Find these numbers if:
a) The sum of their third powers is minimal.
b) The product of one with the cube of the other is maximal.
c) Both are positive and the product of one with the other power of the other is maximal.
Correct result:
a1 = 6
b1 = 6
a2 = 3
b2 = 9
a3 = 4
b3 = 8
Solution:
$a+b=12 \ \\ y=min(a^3+b^3) \ \\ y=min(a^3+(12-a)^3) \ \\ \ \\ f'=3a^2 -3(12-a)^2 \ \\ f'=0 \ \\ \ \\ \ \\ 3 \cdot \ x^2 -3 \cdot \ (12-x)^2=0 \ \\ \ \\ 72x=432 \ \\ \ \\ x=6 \ \\ \ \\ a_{1}=6$
$b_{1}=12-a_{1}=12-6=6$
$f_{2}=a \cdot \ b^3 \ \\ f_{2}=a \cdot \ (12-a)^3 \ \\ f_{2}'(a)=d/da(a (12 - a)^3)=-4 (a - 3) (12 - a)^2 \ \\ \ \\ f_{2}'(a)=0 \ \\ \ \\ m_{1}=3 \ \\ m_{2}=12 \ \\ m_{3}=12 \ \\ \ \\ f_{21}=m_{1} \cdot \ (12-m_{1})^3=3 \cdot \ (12-3)^3=2187 \ \\ f_{22}=m_{2} \cdot \ (12-m_{2})^3=12 \cdot \ (12-12)^3=0 \ \\ \ \\ \ \\ a_{2}=m_{1}=3$
$b_{2}=12-a_{2}=12-3=9$
Our quadratic equation calculator calculates it.
$b_{3}=12-a_{3}=12-4=8$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
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The rectangular cuboid has a surface area 5334 cm2, its dimensions are in the ratio 2:4:5. Find the volume of this rectangular cuboid.
• Two trucks
Two trucks left cities A and B against each other and met after an hour. The first car came to B 27 minutes later than the second car to A. Calculate the car speed if the distance between cities A, B is 90 km.
• Square
If the length of the sides of the square we decrease by 25% decrease the content area of 28 cm2. Determine the side length of the original square.
• Four integers
Fnd four consecutive integers so that the product of the first two is 70 times smaller than the product of the next two.
• Digit sum
The digit sum of the two-digit number is nine. When we turn figures and multiply by the original two-digit number, we get the number 2430. What is the original two-digit number?
• Minimum surface
Find the length, breadth, and height of the cuboid shaped box with a minimum surface area, into which 50 cuboid shaped blocks, each with length, breadth and height equal to 4 cm, 3 cm and 2 cm respectively can be packed.
• Cuboid and ratio
Find the dimensions of a cuboid having a volume of 810 cm3 if the lengths of its edges coming from the same vertex are in ratio 2: 3: 5
• A rectangle 2
A rectangle has a diagonal length of 74cm. Its side lengths are in ratio 5:3. Find its side lengths.
• 1 page
1 page is torn from the book. The sum of the page numbers of all the remaining pages is 15,000. What numbers did the pages have on the page that was torn from the book?
• Cube in a sphere
The cube is inscribed in a sphere with volume 7253 cm3. Determine the length of the edges of a cube.
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Giáo trình
# Introductory Statistics
Mathematics and Statistics
## Using the Normal Distribution
Tác giả: OpenStaxCollege
The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).
The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(Xx) and P(X > x) is the same as P(Xx) for continuous distributions.
# Calculations of Probabilities
Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.
If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
a. Find the probability that a randomly selected student scored more than 65 on the exam.
a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5
Draw a graph.
Then, find P(x > 65).
P(x > 65) = 0.3446
The probability that any student selected at random scores more than 65 is 0.3446.
z = = 0.4
Area to the left is 0.6554.
P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446
b. Find the probability that a randomly selected student scored less than 85.
b. Draw a graph.
Then find P(x < 85), and shade the graph.
Using a computer or calculator, find P(x < 85) = 1.
normalcdf(0,85,63,5) = 1 (rounds to one)
The probability that one student scores less than 85 is approximately one (or 100%).
c. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile.
Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.
k = 69.4
The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:
d. Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).
d. Find the 70th percentile.
Draw a new graph and label it appropriately. k = 65.6
The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.
invNorm(0.70,63,5) = 65.6
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.
Find P(1.8 < x < 2.75).
The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886
normalcdf(1.8,2.75,2,0.5) = 0.5886
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25.
invNorm(0.25,2,0.5) = 1.66
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
a. normalcdf(23,64.7,36.9,13.9) = 0.8186
b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
b. normalcdf(–1099,50.8,36.9,13.9) = 0.8413
c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.
c.
• invNorm(0.80,36.9,13.9) = 48.6
• The 80th percentile is 48.6 years.
• 80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.
• There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).
a. Calculate the interquartile range (IQR).
a.
• IQR = Q3Q1
• Calculate Q3 = 75th percentile and Q1 = 25th percentile.
• invNorm(0.75,36.9,13.9) = Q3 = 46.2754
• invNorm(0.25,36.9,13.9) = Q1 = 27.5246
• IQR = Q3Q1 = 18.7508
• b. Forty percent of the ages that range from 13 to 55+ are at least what age?
b.
• Find k where P(x > k) = 0.40 ("At least" translates to "greater than or equal to.")
• 0.40 = the area to the right.
• Area to the left = 1 – 0.40 = 0.60.
• The area to the left of k = 0.60.
• invNorm(0.60,36.9,13.9) = 40.4215.
• k = 40.42.
• Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.
• A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.
a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
a. normalcdf(6,10^99,5.85,0.24) = 0.2660
b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
b.
• 1 – 0.20 = 0.80
• The tails of the graph of the normal distribution each have an area of 0.40.
• Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60).
• k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
• k2 = invNorm(0.60,5.85,0.24) = 5.91 cm
• c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.
c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.
# References
“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
“403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).
“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).
“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
# Chapter Review
The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
# Formula Review
Normal Distribution: X ~ N(µ, σ) where µ is the mean and σ is the standard deviation.
Standard Normal Distribution: Z ~ N(0, 1).
Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)
Calculator function for the kth percentile: k = invNorm (area to the left of k, mean, standard deviation)
How would you represent the area to the left of one in a probability statement?
P(x < 1)
What is the area to the right of one?
Is P(x < 1) equal to P(x ≤ 1)? Why?
Yes, because they are the same in a continuous distribution: P(x = 1) = 0
How would you represent the area to the left of three in a probability statement?
What is the area to the right of three?
1 – P(x < 3) or P(x > 3)
If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x?
If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x?
1 – 0.543 = 0.457
Use the following information to answer the next four exercises:
X ~ N(54, 8)
Find the probability that x > 56.
Find the probability that x < 30.
0.0013
Find the 80th percentile.
Find the 60th percentile.
56.03
X ~ N(6, 2)
Find the probability that x is between three and nine.
X ~ N(–3, 4)
Find the probability that x is between one and four.
0.1186
X ~ N(4, 5)
Find the maximum of x in the bottom quartile.
Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P(0 < x < ____________) = ___________ (Use zero for the minimum value of x.)
1. Check student’s solution.
2. 3, 0.1979
Find the probability that a CD player will last between 2.8 and six years.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability.
2. P(__________ < x < __________) = __________
Find the 70th percentile of the distribution for the time a CD player lasts.
1. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70%.
2. P(x < k) = __________ Therefore, k = _________
1. Check student’s solution.
2. 0.70, 4.78 years
# Homework
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the probability of spending more than two days in recovery?
1. 0.0580
2. 0.8447
3. 0.0553
4. 0.9420
The 90th percentile for recovery times is?
1. 8.89
2. 7.07
3. 7.99
4. 4.32
c
Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.
Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?
1. Yes
2. No
3. Unable to determine
Find the probability that it takes at least eight minutes to find a parking space.
1. 0.0001
2. 0.9270
3. 0.1862
4. 0.0668
d
Seventy percent of the time, it takes more than how many minutes to find a parking space?
1. 1.24
2. 2.41
3. 3.95
4. 6.05
According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
1. X ~ _____(_____,_____)
2. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement.
3. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.
4. The middle 40% of heights fall between what two values? Sketch the graph, and write the probability statement.
1. X ~ N(66, 2.5)
2. 0.5404
3. No, the probability that an Asian male is over 72 inches tall is 0.0082
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.
1. X ~ _____(_____,_____)
2. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement.
3. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement.
4. The middle 50% of IQs fall between what two values? Sketch the graph and write the probability statement.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories.
1. X ~ _____(_____,_____)
2. Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined.
3. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.
1. X ~ N(36, 10)
2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.
3. Approximately 25% of people consume less than 29.26% of their calories as fat.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.
1. If X = distance in feet for a fly ball, then X ~ _____(_____,_____)
2. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
3. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.
In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement.
4. What percent of the children spend over ten hours per day unsupervised?
5. Seventy percent of the children spend at least how long per day unsupervised?
1. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
2. X ~ N(3, 1.5)
3. The probability that the child spends less than one hour a day unsupervised is 0.0918.
4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
5. 2.21 hours
In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district.
1. State the approximate distribution of X.
2. Is 1,956.8 a population mean or a sample mean? How do you know?
3. Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
4. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
5. Find the third quartile for votes for President Clinton.
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.
1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement.
4. Sixty percent of all trials of this type are completed within how many days?
1. X = the distribution of the number of days a particular type of criminal trial will take
2. X ~ N(21, 7)
3. The probability that a randomly selected trial will last more than 24 days is 0.3336.
4. 22.77
Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.
1. In words, define the random variable X.
2. X ~ _____(_____,_____)
3. Find the percent of her laps that are completed in less than 130 seconds.
4. The fastest 3% of her laps are under _____.
5. The middle 80% of her laps are from _______ seconds to _______ seconds.
Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. [link] displays the ordered real data (in minutes):
0.5 4.25 5 6 7.25 1.75 4.25 5.25 6 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 6.5 8 2.5 4.75 5.5 6.5 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 6 6.75 9.75 3.75 5 6 6.75 10.75
1. Calculate the sample mean and the sample standard deviation.
2. Construct a histogram.
3. Draw a smooth curve through the midpoints of the tops of the bars.
4. In words, describe the shape of your histogram and smooth curve.
5. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____(_____,_____)
6. Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes.
7. Determine the cumulative relative frequency for waiting less than 6.1 minutes.
8. Why aren’t the answers to part f and part g exactly the same?
9. Why are the answers to part f and part g as close as they are?
10. If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion.
1. mean = 5.51, s = 2.15
2. Check student's solution.
3. Check student's solution.
4. Check student's solution.
5. X ~ N(5.51, 2.15)
6. 0.6029
7. The cumulative frequency for less than 6.1 minutes is 0.64.
8. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
9. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
10. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.
Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.
1. Ricardo’s actual GPA is lower than Anita’s actual GPA.
2. Ricardo is not passing because his z-score is zero.
3. Anita is in the 70th percentile of students at her college.
[link] shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.
40,000 40,000 45,050 45,500 46,249 48,134 49,133 50,071 50,096 50,466 50,832 51,100 51,500 51,900 52,000 52,132 52,200 52,530 52,692 53,864 54,000 55,000 55,000 55,000 55,000 55,000 55,000 55,082 57,000 58,008 59,680 60,000 60,000 60,492 60,580 62,380 62,872 64,035 65,000 65,050 65,647 66,000 66,161 67,428 68,349 68,976 69,372 70,107 70,585 71,594 72,000 72,922 73,379 74,500 75,025 76,212 78,000 80,000 80,000 82,300
1. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data).
2. Construct a histogram.
3. Draw a smooth curve through the midpoints of the tops of the bars of the histogram.
4. In words, describe the shape of your histogram and smooth curve.
5. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____(_____,_____).
6. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.
7. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.
8. Why aren’t the answers to part f and part g exactly the same?
1. mean = 60,136
s = 10,468
5. X ~ N(60136, 10468)
6. 0.7440
7. The cumulative relative frequency is 43/60 = 0.717.
8. The answers for part f and part g are not the same, because the normal distribution is only an approximation.
An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.
A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.
• n = 100; p = 0.1; q = 0.9
• μ = np = (100)(0.10) = 10
• σ = $\sqrt{npq}$ = $\sqrt{\text{(100)(0}\text{.1)(0}\text{.9)}}$ = 3
1. z = ±1: x1 = µ + = 10 + 1(3) = 13 and x2 = µ = 10 – 1(3) = 7. 68% of the defective cars will fall between seven and 13.
2. z = ±2: x1 = µ + = 10 + 2(3) = 16 and x2 = µ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16
3. z = ±3: x1 = µ + = 10 + 3(3) = 19 and x2 = µ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19.
We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following:
1. There is about a 68% chance that the number of heads will be somewhere between ___ and ___.
2. There is about a ____chance that the number of heads will be somewhere between 12 and 28.
3. There is about a ____ chance that the number of heads will be somewhere between eight and 32.
A \$1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are
1. somewhere between 34 and 54 prizes.
2. somewhere between 54 and 64 prizes.
3. more than 64 prizes.
• n = 190; p = $1 5$ = 0.2; q = 0.8
• μ = np = (190)(0.2) = 38
• σ = $\sqrt{npq}$ = $\sqrt{\text{(190)(0}\text{.2)(0}\text{.8)}}$ = 5.5136
1. For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641
2. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018
3. For this problem: P(x > 64) = normalcdf(64,1099,48,5.5136) = 0.0000012 (approximately 0)
Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
1. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
2. Find the 95th percentile, and express it in a sentence.
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LCM of 3 and 13 is the the smallest number amongst all typical multiples that 3 and also 13. The first few multiples the 3 and also 13 room (3, 6, 9, 12, 15, 18, . . . ) and (13, 26, 39, 52, 65, 78, 91, . . . ) respectively. There space 3 commonly used techniques to uncover LCM the 3 and 13 - by prime factorization, by listing multiples, and by department method.
You are watching: What is the lcm of 3 and 13
1 LCM the 3 and 13 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 3 and also 13 is 39.
Explanation:
The LCM of 2 non-zero integers, x(3) and also y(13), is the smallest hopeful integer m(39) that is divisible through both x(3) and y(13) without any remainder.
The methods to find the LCM of 3 and also 13 are explained below.
By department MethodBy prime Factorization MethodBy Listing Multiples
### LCM the 3 and also 13 by division Method
To calculation the LCM of 3 and 13 by the department method, we will divide the numbers(3, 13) by their prime determinants (preferably common). The product of this divisors gives the LCM of 3 and also 13.
Step 3: continue the steps until only 1s room left in the last row.
The LCM of 3 and 13 is the product of every prime number on the left, i.e. LCM(3, 13) by department method = 3 × 13 = 39.
### LCM that 3 and 13 by prime Factorization
Prime administrate of 3 and 13 is (3) = 31 and also (13) = 131 respectively. LCM that 3 and 13 can be acquired by multiply prime components raised to their respective highest possible power, i.e. 31 × 131 = 39.Hence, the LCM of 3 and also 13 by prime factorization is 39.
### LCM of 3 and 13 through Listing Multiples
To calculation the LCM the 3 and 13 by listing out the common multiples, we can follow the given listed below steps:
Step 1: perform a few multiples that 3 (3, 6, 9, 12, 15, 18, . . . ) and 13 (13, 26, 39, 52, 65, 78, 91, . . . . )Step 2: The common multiples indigenous the multiples of 3 and also 13 are 39, 78, . . .Step 3: The smallest typical multiple of 3 and also 13 is 39.
∴ The least common multiple the 3 and also 13 = 39.
☛ also Check:
## FAQs top top LCM that 3 and 13
### What is the LCM the 3 and also 13?
The LCM that 3 and also 13 is 39. To uncover the LCM (least common multiple) of 3 and also 13, we need to uncover the multiples that 3 and also 13 (multiples the 3 = 3, 6, 9, 12 . . . . 39; multiples the 13 = 13, 26, 39, 52) and choose the smallest multiple the is precisely divisible by 3 and 13, i.e., 39.
### What space the techniques to find LCM that 3 and also 13?
The commonly used approaches to discover the LCM the 3 and 13 are:
Listing MultiplesPrime factorization MethodDivision Method
### Which that the complying with is the LCM the 3 and also 13? 28, 42, 39, 12
The worth of LCM that 3, 13 is the smallest common multiple the 3 and also 13. The number to solve the given problem is 39.
### If the LCM that 13 and also 3 is 39, uncover its GCF.
LCM(13, 3) × GCF(13, 3) = 13 × 3Since the LCM of 13 and 3 = 39⇒ 39 × GCF(13, 3) = 39Therefore, the GCF = 39/39 = 1.
See more: How To Get Your Robux Back After Buying Something 2021, How To Refund Robux Within Minutes
### How to discover the LCM of 3 and 13 by prime Factorization?
To find the LCM the 3 and 13 making use of prime factorization, us will find the prime factors, (3 = 3) and (13 = 13). LCM the 3 and 13 is the product the prime factors raised to their respective greatest exponent among the number 3 and 13.⇒ LCM the 3, 13 = 31 × 131 = 39.
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# In how many ways can I put 1 to 5 mirrors in 8 rooms?
A queen has 8 rooms and 12 indistinguishable mirrors, how many ways are there to hang these mirrors in 8 rooms such that every room has at least 1 mirror?
• Are the mirrors distinguishable? Oct 11, 2018 at 20:16
• I don't know, this is the problem statement as is. The answer is 330, I'm trying to understand the logic. Oct 11, 2018 at 20:16
• Ok, it's indistinguishable if 330 is the answer. Oct 11, 2018 at 20:17
We have 12 mirrors, but 8 of them must be allocated so each room has at least 1. So all we care about is placing the remaining 4 mirrors in 8 rooms.
This is a perfect place to use stars and bars, since we have 4 mirrors to spread across 8 rooms. This leads us to find the total number of combinations as $${8+4-1\choose4} ={11\choose4} =\color{red}{330}$$
If you are confused as why $$8+4-1\choose4$$ is what we desire, imagine that the * are mirrors, and | dividers, which determine which room the mirrors fall into. Since there are 8 rooms, we have 7 dividers, i.e., we seek the number of distinct combinations of
****|||||||
• It is even easier if you use the basic stars and bars. You put the $12$ mirrors in a row and can place four dividers between them. You can only place one divider between any pair of mirrors, so each room gets at least one. There are $11$ spaces for the four dividers, giving $11 \choose 4$ Oct 11, 2018 at 20:37
Let the numbers of mirror in a room be $$x_i$$, where $$x_i$$ is a positive integer.
We need to find solutions for,
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8=12$$
This can be seen as 12 stars and 7 bars.
************ ||||||||
Now there are 11 gaps between 12 stars and you have 7 bars to place, where each bar separates number of mirrors in a room.
• I do not understand why this answer received so many downvotes. $\binom{11}{7}=\binom{11}{4}$ and following this answer gives the same result as the other answer. Oct 11, 2018 at 20:39
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Section0_6
# Section0_6 - Section 0.6 Linear combinations products...
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Unformatted text preview: (3/31/07) Section 0.6 Linear combinations, products, quotients, and compositions Overview: Many funcions that are used in calculus are constructed from constant, power, exponential, logarithmic, trigonometric, and inverse trigonometric functions by taking linear combinations, products, quotients, and compositions of these basic functions. In this section we discuss these procedures and look at instances where the graphs of such functions can be readily explained by their formulas. Topics: • Linear combinations • Products and quotients • Polynomials and rational functions • Composite functions Linear combinations A linear combination of functions is formed by multiplying each of the functions by a constant and adding the results. Example 1 The functions (a) A ( x ) = 1 8 x 2 + 3 cos x , (b) B ( x ) = 1 8 x 2- 3cos x , and (c) C ( x ) =- 1 8 x 2 + 3 cos x are linear combinations of y = x 2 and y = cos x . Match them with their graphs in Figures 1 through 3. x y 10- 10 2 π- 2 π x y 10- 10 2 π- 2 π x y 10- 10 2 π- 2 π FIGURE 1 FIGURE 2 FIGURE 3 Solution (a) The curves y = 1 8 x 2 and y = 3 cos x are shown in Figure 4. The first is y = x 2 contracted vertically by a factor of 8, and the second is y = cos x magnified vertically by a factor of 3. Because y = 3cos x oscillates between 3 and- 3, the graph of A ( x ) = 1 8 x 2 + 3 cos x oscillates from 3 units above y = 1 8 x 2 to 3 units below it, and since A (0) = 1 8 (0) 2 + 3 cos(0) = 3, the graph is in Figure 3. (b) The graph of B ( x ) = 1 8 x 2- 3cos x oscillates from 3 units above to 3 units below y = 1 8 x 2 and B (0) = 1 8 (0) 2- 3cos(0) =- 3. The graph is in Figure 2. (c) Figure 5 shows the curves y =- 1 8 x 2 and y = 3 cos x . The graph of C ( x ) =- 1 8 x 2 + 3 cos x oscillates from 3 units above to 3 units below y =- 1 8 x 2 . It is in Figure 1. square 1 p. 2 (3/31/07) Section 0.6, Linear combinations, products, quotients, and compositions x y 10- 10 2 π- 2 π y = 1 8 x 2 y = 3cos x x y 10- 10 y =- 1 8 x 2 y = 3 cos x FIGURE 4 FIGURE 5 C Question 1 Generate y = 1 8 x 2 + 10cos x in the window- 4 π ≤ x ≤ 4 π,- 15 ≤ y ≤ 20 with x-scale = π and y-scale = 5 and explain how this curve differs from y = 1 8 x 2 + 3cos x in Figure 3. The domain of a linear combination of functions consists of those values of the variable x where all of the functions involved are defined. Example 2 What is the domain of g ( x ) = 2 + 3 x 1 / 2- 5 x ? Solution Because the constant function y = 2 and the exponential function y = 5 x are defined for all x , the domain of g ( x ) = 2 + 3 x 1 / 2- 5 x is the interval [0 , ∞ ) where the square root function y = x 1 / 2 is defined. square Products and quotients The product of two functions f and g is the function fg whose value at x is the product f ( x ) g ( x ) of the values of the two functions. Its domain consists of all values of x where f ( x ) and g ( x ) are defined....
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## This note was uploaded on 09/13/2010 for the course MATH Math 20A taught by Professor Eggers during the Summer '08 term at UCSD.
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# Cartesian Product of Countable Sets is Countable/Corollary/Proof 2
## Corollary to Cartesian Product of Countable Sets is Countable
Let $k > 1$.
Then the cartesian product of $k$ countable sets is countable.
## Proof
Proof by induction:
### Basis for the Induction
When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable.
So shown for basis for the induction.
### Induction Hypothesis
This is our induction hypothesis:
$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$
where $f_k$ is an injection.
Now we need to show that for $n = k + 1$:
$\exists f_{k+1}: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$
where $f_{k+1}$ is an injection.
### Induction Step
This is our induction step:
By the induction hypothesis:
$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$
where $f_k$ is an injection.
Thus by definition, $S_1 \times S_2 \times \cdots \times S_k$ is countable.
By hypothesis $S_{k + 1}$ is countable.
So by the basis for the induction:
$\exists g: \left({S_1 \times S_2 \times \cdots \times S_k}\right) \times S_{k+1} \to \N \times \N$
where $g$ is an injection.
$\exists r: \N \times \N \to \N$
where $r$ is an injection.
Therefore, by Composite of Injections is Injection:
$f_{k+1} = r \circ g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$
is an injection.
The result follows by induction.
$\blacksquare$
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• ### UNIT 5 : Linear equations and inequalities
Key unit competence
Model and solve algebraically or graphically daily life problems using linear equations or inequalities.
Learning objectives
5.1 Equations and inequalities in one unknown
In Junior Secondary, we learnt about linear equations and inequalities.
Activity 5.1
Research on the following:
1. What is a linear equation?
2. What is a linear inequality?
Present your findings, with clear examples, to the rest of the class.
Linear equations
A linear equation is an equation of a straight line. For example:
Solving linear equations
• A linear equation is a polynomial of degree 1.
• In order to solve for the unknown variable, you must isolate the variable.
• In the order of operation, multiplication and division are completed before addition and subtraction.
Activity 5.2
Discuss in groups and verify that the following are true.
Product equation
This is in the form (ax + b)(cx + d) = 0.
Since the product of factors is null (zero) either one of them is zero. To solve this we proceed as follows:
Solving inequalities
They are solved as linear equations except that:
(a) when we multiply an inequality by a negative real number the sign will be reversed
(b) when we interchange the right side and the left side, the sign will be reversed.
Graphs
The graph of a linear inequality in one variable is a number line. We use an unshaded circle for < and > and a shaded circle for ≤ and ≥. The graph for x > –3:
Figure 5.6 shows a graph of a linear inequality.
The inequality is y ≤ x + 2.
You can see the line, y = x + 2 and the shaded area is where y is less than or equal to x + 2
5.2 Parametric equations and inequalities
Activity 5.4
Carry out research. Find the meaning of parametric equations and inequalities. Discuss your findings using suitable examples.
Parametric equations
There are also a great many curves that we cannot even write down as a single equation in terms of only x and y. So, to deal with some of these problems we introduce parametric equationsInstead of defining y in terms of x i.e y = f(x)) or x in terms of y i.e x = h(y) we define both x and y in terms of a third variable called a parameter as follows:
This third variable is usually denoted by t (but does not have to be). Sometimes we will restrict the values of t that we shall use and at other times we will not. If the coefficients of an equation contain one or several letters (variables) the equation is called parametric and the letters are called real parameters. In this case, we solve and discuss the equation (for parameters only).
Each value of t defines a point (x, y) = (f(t),g(t)) that we can plot. The collection of points that we get by letting t be all possible values is the graph of the parametric equations and is called the parametric curve.
Graphs
Sketching a parametric curve is not always an easy thing to do. Let us look at an example to see one way of sketching a parametric curve. This example will also illustrate why this method is usually not the best.
Activity 5.5
In groups of five, work out the following.
1. Sketch the parametric curve for the following set of parametric equations.
x = t2 + t y = 2t – 1 –1 ≤ t ≤1
2. Eliminate the parameter from the following set of parametric equations.
x = t2 + t y = 2t – 1
3. Sketch the parametric curve for the following set of parametric equations. Clearly indicate the direction of motion.
x = 5 cost y = 2 sin t 0 ≤ t ≤ 2π
Parametric inequalities in one unknown
5.3 Simultaneous equations in two unknowns
Mental task
What are simultaneous linear equations? How do we solve them?
A linear equation in two variables x and y is an equation of the form
We say that we have two simultaneous linear equation in two unknowns or a system of two linear equation in two unknowns.
The pair (x, y) satisfying both equations is the solution of the given equation.
We can solve such systems of linear equations by using one of the following methods:
1. substitution method
2. elimination method.
Substitution
This method is used when one of the variables is given in terms of the other.
Elimination
Elimination method is used to solve simultaneous equations where neither variable is given as the subject of another.
5.4 Applications
Systems of two equations have a wide practical application whenever decisions arise. Decisions of this nature always involve two unknown quantities or variables. The following steps are hereby recommended in order to apply simultaneous equations for practical purposes.
1. Define variables for the two unknown quantities, in case they are not given.
2. Formulate equations using the unknown variables and the corresponding data.
3. Solve the equations simultaneously.
Activity 5.6
Linear equations and inequalities have their use in our everyday life. Research on these and present your findings for discussion in class.
Distance = rate × time
In this equation, for any given steady rate, the relationship between distance and time will be linear. However, distance is usually expressed as a positive number, so most graphs of this relationship will only show points in the first quadrant. Notice that the direction of the line in the graph of Figure 5.8 is from bottom left to top right. Lines that tend in this direction have positive slope. A positive slope indicates that the values on both axes are increasing from left to right.
Amount of water in a leaking bucket = rate of leak × time
In this equation, since you cannot have a negative amount of water in the bucket, the graph will also show points only in the first quadrant. Notice that the direction of the line in this graph is top left to bottom right. Lines that tend in this direction have negative slope. A negative slope indicates that the values on the y axis are decreasing as the values on the x axis are increasing.
Number of angles of a polygon = number of sides of that polygon
In this graph, we are relating values that only make sense if they are positive, so we show points only in the first quadrant. In this case, since no polygon has fewer than 3 sides or angles, and since the number of sides or angles of a polygon must be a whole number, we show the graph starting at (3,3) and indicate with a dashed line that points between those plotted are not relevant to the problem.
Since it is perfectly reasonable to have both positive and negative temperatures, we plot the points on this graph on the full coordinate grid.
Supplementary, interactive questions served by Siyavula Education.
Note: Questions will open in a new window or tab.
Equations and inequalities in one unknown
Supplementary, interactive questions served by Siyavula Education.
Note: Questions will open in a new window or tab.
Simultaneous equations in two unknowns
Supplementary, interactive questions served by Siyavula Education.
Note: Questions will open in a new window or tab.
Applications
UNIT 4 : Set lR of real numbersUNIT 6 : Quadratic equations and inequalities
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# Class 12 RD Sharma Solutions – Chapter 9 Continuity – Exercise 9.1 | Set 3
### Question 31. If is continuous at x = 2, find k.
Solution:
Given that,
Also, f(x) is continuous at x = 2
So, LHL = RHL = f(2) …..(i)
Now,
f(2) = k ……(ii)
Let us consider LHL,
……(iii)
Using eq(i), (ii) and (iii), we get
k = 1/2
### Question 32. If is continuous at x = 0, find k.
Solution:
Given that,
Also, f(x) is continuous at x = 2
So, LHL = RHL
Now,
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ -2 × 1 × (1 + 1) = k
⇒ k = -4
### Question 33. Extend the definition of the following by continuity f(x) = at the point x = π.
Solution:
Given that,
As we know that a f(x) is continuous at x = π if,
LHL = RHL = f(Ï€) ……(i)
Let us consider LHL,
= (2/5) × (49/4) = 49/10
Thus, from eq(i) we get,
f(Ï€) = 49/10
Hence, f(x) is continuous at x = π
### Question 34. If f(x) = , x ≠0 is continuous at x = 0, then find f(0).
Solution:
Given that,
f(x) =
Also, f(x) is continuous at x = 0
So, LHL = RHL = f(0) ……(i)
Let us consider LHL,
From eq(i) we get,
f(0) = 1
### Question 35. Find the value of k for which is continuous at x = 0
Solution:
Given that,
Also, f(x) is continuous at x = 0
LHL = RHL = f(0) …..(i)
f(0) = k
Let us consider LHL,
Thus, from eq(i) we get,
k = 1
### (i) at x = 0
Solution:
Given that,
Also, f(x) is continuous at x = 0
⇒
⇒
⇒
⇒ 2k2 × 1 = 8
⇒ k2 = 4
⇒ k = ±2
### (ii) at x = 1
Solution:
Given that,
Also, f(x) is continuous at x = 1
⇒
Now, on putting x – 1 = y, we get
⇒
⇒
⇒
⇒
⇒
⇒ (-2/π) × (1/1) = k
⇒ k = (-2/π)
### (iii) at x = 0
Solution:
Given that,
Also, f(x) is continuous at x = 0
Let us consider LHL, at x = 0
Let us consider RHL at x = 0
Hence, no value of k exists for which function is continuous at x = 0.
### (iv) at x = π
Solution:
Given that,
Also, f(x) is continuous at x = π
Let us consider LHL
Let us consider RHL
cosπ = -1
As we know that f(x) is continuous at x = π, so
⇒ kπ + 1 = -1
⇒ k = (-2/π)
### (v) at x = 5
Solution:
Given that,
Also, f(x) is continuous at x = 5
Let us consider LHL
= 5k + 1
Let us consider RHL
= 10
As we know that f(x) is continuous at x = 5, so
⇒ 5k + 1 = 10
⇒ k = 9/5
### (vi) at x = 5
Solution:
Given that,
Also, f(x) is continuous at x = 5
So,
f(x) = (x2 – 25)/(x – 5), if x ≠5 & f(x) = k, if x = 5
⇒ f(x)= {(x – 5)(x+5)/(x-5)}, if x ≠5 & f(x) = k, if x = 5
⇒ f(x)= (x + 5), if x ≠5 & f(x) = k, if x = 5
As we know that f(x) is continuous at x = 5, so
⇒
⇒ k = 5 + 5 = 10
### (vii) at x = 1
Solution:
Given that,
Also, f(x) is continuous at x = 1
Let us consider LHL
Let us consider RHL
= k
As we know that f(x) is continuous at x = 1, so
⇒ k = 4
### (viii) at x = 0
Solution:
Given that,
Also, f(x) is continuous at x = 0
Let us consider LHL
= 2k
Let us consider RHL
= 1
As we know that f(x) is continuous at x = 0, so
⇒ 2k = 1
⇒ k = 1/2
### (ix) at x = 2
Solution:
Given that,
Also, f(x) is continuous at x = 2
f(x)= , if x ≠2 & f(x) = k, if x = 2
⇒ f(x)= , if x ≠2 & f(x) = k, if x = 2
⇒ f(x)= , if x ≠2 & f(x) = k, if x = 2
⇒ f(x)= (x + 5), if x ≠2 & f(x) = k, if x = 2
As we know that f(x) is continuous at x = 2, so
⇒
⇒ k = 2 + 5 = 7
### is continuous at x = 3 and x = 5.
Solution:
Given that,
Let us consider LHL at x = 3,
= 1
Let us consider RHL at x = 3,
= 3a + b
Let us consider LHL at x = 5,
= 5a + b
Let us consider RHL at x = 5,
= 7
It is given that f(x) is continuous at x = 3 and x = 5, then
and
⇒ 1 = 3a + b …..(i)
and 5a + b = 7 …….(ii)
On solving eq(i) and (ii), we get
a = 3 and b = -8
### Question 38. If . Show that f is continuous at x = 1.
Solution:
Given that,
So,
Let us consider LHL at x = 1,
= 1/2
Let us consider RHL at x = 1,
= 2 – 3 + 3/2 = 1/2
Also,
f(1) = (1)2/2 = 1/2
LHL = RHL = f(1)
Hence, the f(x) is continuous at x = 1
### (i) f(x) = |x| + |x – 1| at x = 0, 1.
Solution:
Given that,
f(x) = |x| + |x – 1|
So, here we check the continuity of the given f(x) at x = 0,
Let us consider LHL at x = 0,
Let us consider RHL at x = 0,
Also,
f(0) = |0| + |0 – 1| = 0 + 1 = 1
LHL = RHL = f(0)
Now, we check the continuity of the given f(x) at x = 1,
Let us consider LHL at x = 1,
= 1
Let us consider RHL at x = 1
= 1
Also,
f(1) = |1| + |1 – 1| = 1 + 0 = 1
LHL = RHL = f(1)
Hence, f(x) is continuous at x = 0, 1.
### (ii) f(x) = |x – 1| + |x + 1| at x = -1, 1.
Solution:
Given that,
f(x) = |x – 1| + |x + 1| at x = -1, 1.
So, here we check the continuity of the given f(x) at x = -1,
Let us consider LHL at x = -1,
Let us consider RHL at x = -1,
Also,
f(-1) = |-1 – 1| + |-1 + 1| = |-2| = 2
LHL = RHL = f(-1)
Now, we check the continuity of the given f(x) at x = 1,
Let us consider LHL at x = 1,
= 2
= 2
Also,
f(1) = |1 + 1| + |1 – 1| = 2
LHL = RHL = f(1)
Hence, f(x) is continuous at x = -1, 1.
### Question 40. Prove that is discontinuous at x = 0.
Solution:
Prove that is discontinuous at x = 0.
Proof:
Let us consider LHL at x = 0,
Let us consider RHL at x = 0,
LHL ≠RHL
Hence, f(x) is discontinuous at x = 0.
### Question 41. If then what should be the value of k so that f(x) is continuous at x = 0.
Solution:
Given that,
Let us consider LHL at x = 0,
= k
Let us consider RHL at x = 0,
= k
It is given that f(x) is continuous at x = 0.
LHL = RHL = f(0)
⇒
k can be any real number.
### continuous at x = 0 ? What about continuity at x = ±1?
Solution:
Given that,
Check for x = 0,
Hence, there is no value of λ for which f(x) is continuous at x = 0.
Now for x = 1,
f(1) = 4x + 1 = 4 × 1 + 1 = 5
Hence, for any values of λ, f is continuous at x = 1.
Now for x = -1,
f(-1) = λ(1 + 2)= 3λ
Hence, for any values of λ, f is continuous at x=-1.
### Question 43. For what values of k is the following function continuous at x = 2?
Solution:
Given that,
We have,
Let us consider LHL at x = 2,
= 5
Let us consider RHL at x = 2,
= 5
Also,
f(2) = k
It is given that f(x) is continuous at x = 2.
LHL = RHL = f(2)
⇒ 5 = 5 = k
Hence, for k = 5, f(x) is continuous at x = 2.
### Question 44. Let If f(x) is continuous at x = (Ï€/2), find a and b.
Solution:
Given that,
Let us consider LHL at x = π/2
= 1/2
Let us consider RHL at x = π/2
= b/8 × 1
= b/8
Also,
f(Ï€/2) = a
It is given that f(x) is continuous at x = π/2.
LHL = RHL = f(Ï€/2)
So,
⇒ 1/2 = b/8 = a
⇒ a = 1/2 and b = 4
### Question 45. If the functions f(x), defined below is continuous at x = 0, find the value of k,
Solution:
Given that,
Let us consider LHL at x = 0,
= 1 × 1
Let us consider RHL at x = 0,
Also,
f(0) = k
It is given that f(x) is continuous at x = 0,
LHL = RHL = f(0)
So,
⇒ 1 = 1 = k
Hence, the required value of k is 1.
### is continuous at x = 3.
Solution:
Given that,
Let us consider LHL at x = 3,
= 3a + 1
Let us consider RHL at x = 3,
= 3b + 3
It is given that f(x) is continuous at x = 3,
LHL = RHL = f(3)
So,
⇒ 3a + 1 = 3b + 3
⇒ 3a – 3b = 2
Hence, the required relationship between a and b is 3a – 3b = 2.
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# 1.9: Secant, Cosecant, and Cotangent Functions
Difficulty Level: At Grade Created by: CK-12
Estimated8 minsto complete
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Practice Secant, Cosecant, and Cotangent Functions
Progress
Estimated8 minsto complete
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While working to paint your grandfather's staircase you are looking at the triangular shape made by the wall that support the stairs. The staircase looks like this:
You are thinking about all of the possible relationships between sides. You already know that there are three common relationships, called sine, cosine, and tangent.
How many others can you find?
By the end of this Concept, you'll know the other important relationships between sides of a triangle.
### Guidance
We can define three more functions also based on a right triangle. They are the reciprocals of sine, cosine and tangent.
If sinA=ac\begin{align*}\sin A = \frac{a}{c}\end{align*}, then the definition of cosecant, or csc\begin{align*}\csc\end{align*}, is cscA=ca\begin{align*}\csc A = \frac{c}{a}\end{align*}.
If cosA=bc\begin{align*}\cos A = \frac{b}{c}\end{align*}, then the definition of secant, or sec\begin{align*}\sec\end{align*}, is secA=cb\begin{align*}\sec A = \frac{c}{b}\end{align*}.
If tanA=ab\begin{align*}\tan A = \frac{a}{b}\end{align*}, then the definition of cotangent, or cot\begin{align*}\cot\end{align*}, is cotA=ba\begin{align*}\cot A = \frac{b}{a}\end{align*}.
#### Example A
Find the secant, cosecant, and cotangent of angle B\begin{align*}B\end{align*}.
Solution:
First, we must find the length of the hypotenuse. We can do this using the Pythagorean Theorem:
52+12225+144169H=H2=H2=H2=13\begin{align*}5^2 + 12^2 & = H^2\\ 25 + 144 & = H^2\\ 169 & = H^2\\ H & = 13\end{align*}
Now we can find the secant, cosecant, and cotangent of angle B\begin{align*}B\end{align*}:
secBcscBcotB=hypotenuseadjacent side=1312=hypotenuseopposite side=135=adjacent sideopposite side=125\begin{align*}\sec B & = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{13}{12}\\ \csc B & = \frac{\text{hypotenuse}}{\text{opposite side}} = \frac{13}{5}\\ \cot B & = \frac{\text{adjacent side}}{\text{opposite side}} = \frac{12}{5}\end{align*}
#### Example B
Find the secant, cosecant, and cotangent of angle A\begin{align*}A\end{align*}
Solution:
secAcscAcotA=hypotenuseadjacent side=4140=hypotenuseopposite side=419=adjacent sideopposite side=409\begin{align*}\sec A & = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{41}{40}\\ \csc A & = \frac{\text{hypotenuse}}{\text{opposite side}} = \frac{41}{9}\\ \cot A & = \frac{\text{adjacent side}}{\text{opposite side}} = \frac{40}{9}\end{align*}
#### Example C
Find the sine, cosine, and tangent of angle A\begin{align*}A\end{align*}, and then use this to construct the secant, cosecant, and cotangent of the angle
sinAcosAtanA=opposite sidehypotenuse=725=adjacent sidehypotenuse=2425=opposite sideadjacent side=724\begin{align*}\sin A & = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{7}{25}\\ \cos A & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{24}{25}\\ \tan A & = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{7}{24}\end{align*}
Since we know that cosecant is the reciprocal of sine, secant is the reciprocal of sine, and cotangent is the reciprocal of tangent, we can construct these functions as follows:
secAcscAcotA=1cosA=2524=1sinA=257=1tanA=247\begin{align*}\sec A & = \frac{1}{\cos A} = \frac{25}{24}\\ \csc A & = \frac{1}{\sin A} = \frac{25}{7}\\ \cot A & = \frac{1}{\tan A} = \frac{24}{7}\end{align*}
### Vocabulary
Cosecant: The cosecant of an angle in a right triangle is a relationship found by dividing the length of the hypotenuse by the length of the side opposite to the given angle. This is the reciprocal of the sine function.
Secant: The secant of an angle in a right triangle is a relationship found by dividing length of the hypotenuse by the length of the side adjacent the given angle. This is the reciprocal of the cosine function.
Cotangent: The cotangent of an angle in a right triangle is a relationship found by dividing the length of the side adjacent to the given angle by the length of the side opposite to the given angle. This is the reciprocal of the tangent function.
### Guided Practice
Find the
1. secant
2. cosecant
3. cotangent
of A\begin{align*}\angle A\end{align*}
Solutions:
1. The secant function is defined to be 1cos\begin{align*}\frac{1}{\cos}\end{align*}. Since cos=adjacenthypotenuse\begin{align*}\cos = \frac{adjacent}{hypotenuse}\end{align*}, sec=hypotenuseadjacent\begin{align*}\sec = \frac{hypotenuse}{adjacent}\end{align*}.
sec=hypotenuseadjacent=37123.08\begin{align*}\sec = \frac{hypotenuse}{adjacent} = \frac{37}{12} \approx 3.08\end{align*}
2. The cosecant function is defined to be 1sin\begin{align*}\frac{1}{\sin}\end{align*}. Since sin=oppositehypotenuse\begin{align*}\sin = \frac{opposite}{hypotenuse}\end{align*}, csc=hypotenuseopposite\begin{align*}\csc = \frac{hypotenuse}{opposite}\end{align*}.
csc=hypotenuseopposite=37351.06\begin{align*}\csc = \frac{hypotenuse}{opposite} = \frac{37}{35} \approx 1.06 \end{align*}
3. The cotangent function is defined to be 1tan\begin{align*}\frac{1}{\tan}\end{align*}. Since tan=oppositeadjacent\begin{align*}\tan = \frac{opposite}{adjacent}\end{align*}, cot=adjacentopposite\begin{align*}\cot = \frac{adjacent}{opposite}\end{align*}.
cot=adjacentopposite=1235.34\begin{align*}\cot = \frac{adjacent}{opposite} = \frac{12}{35} \approx .34 \end{align*}
### Concept Problem Solution
Looking at a triangle-like the shape of the wall supporting your grandfather's staircase:
We can see that there are several ways to make relationships between the sides. In this case, we are only interested in ratios between the sides, which means one side will be divided by another. We've already seen some functions, such as:
1) The side opposite the angle divided by the hypotenuse (the sine function)
2) The side adjacent the angle divided by the hypotenuse (the cosine function)
3) The side opposite the angle divided by adjacent side (the tangent function)
In this section we introduced the reciprocal of the above trig functions. These are found by taking ratios between the same sides shown above, except reversing the numerator and denominator:
4) The hypotenuse divided by the side opposite the angle (the cosecant function)
5) The hypotenuse divided by the side adjacent to the angle (the secant function)
6) The adjacent side divided by the opposite side (the cotangent function)
### Practice
Use the diagram below for questions 1-3.
1. Find cscA\begin{align*}\csc A\end{align*} and cscC\begin{align*}\csc C\end{align*}.
2. Find secA\begin{align*}\sec A\end{align*} and secC\begin{align*}\sec C\end{align*}.
3. Find cotA\begin{align*}\cot A\end{align*} and cotC\begin{align*}\cot C\end{align*}.
Use the diagram to fill in the blanks below.
1. cotA=??\begin{align*}\cot A = \frac{?}{?}\end{align*}
2. cscC=??\begin{align*}\csc C = \frac{?}{?}\end{align*}
3. cotC=??\begin{align*}\cot C = \frac{?}{?}\end{align*}
4. secC=??\begin{align*}\sec C = \frac{?}{?}\end{align*}
5. cscA=??\begin{align*}\csc A = \frac{?}{?}\end{align*}
6. secA=??\begin{align*}\sec A = \frac{?}{?}\end{align*}
From questions 4-9, we can conclude the following. Fill in the blanks.
1. sec=cscA\begin{align*}\sec \underline{\;\;\;\;\;\;\;} = \csc A\end{align*} and csc=secA\begin{align*}\csc \underline{\;\;\;\;\;\;\;} = \sec A\end{align*}.
2. cotA\begin{align*}\cot A\end{align*} and cotC\begin{align*}\cot C\end{align*} are _________ of each other.
3. Explain why the csc of an angle will always be greater than 1.
4. Use your knowledge of 45-45-90 triangles to find the cosecant, secant, and cotangent of a 45 degree angle.
5. Use your knowledge of 30-60-90 triangles to find the cosecant, secant, and cotangent of a 30 degree angle.
6. Use your knowledge of 30-60-90 triangles to find the cosecant, secant, and cotangent of a 60 degree angle.
7. As the degree of an angle increases, will the cotangent of the angle increase or decrease? Explain.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Cosecant
The cosecant of an angle in a right triangle is a relationship found by dividing the length of the hypotenuse by the length of the side opposite to the given angle. This is the reciprocal of the sine function.
Cotangent
The cotangent of an angle in a right triangle is a relationship found by dividing the length of the side adjacent to the given angle by the length of the side opposite to the given angle. This is the reciprocal of the tangent function.
Secant
The secant of an angle in a right triangle is the value found by dividing length of the hypotenuse by the length of the side adjacent the given angle. The secant ratio is the reciprocal of the cosine ratio.
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# Multiplying Double Digits | Free Printables
In this article, I will show you the procedures for multiplying double digits. There are a lot of steps to multiplying double digits. It can be quite confusing! And it requires a ton of practice for many students to master.
But no student wants to spend hours practicing multiplying two-digit numbers over and over. We don’t want that for our kiddos either, so here is a game that gets children practicing multiplying two-digit numbers.
## A Game for Your Youngsters to Multiply Double Digits
Our children will be able to learn how to multiply double digits very easily and interactively with the help of this game. This approach is quite successful. This method should assist your young champ in learning basic math operations and laying a solid calculating foundation.
## What to Teach Before Playing
There are three popular ways to teach multiplying large numbers right now. I’ve always been a fan of teaching numerous methods since children learn differently. What works for me and makes sense to me, may not make sense to them. A different method may be the key to their success.
This is the way many of us were taught in school, and the way most of us are comfortable teaching. But if you have a child struggling, I suggest trying one or both of the following methods.
Lattice
I used this method many times as a classroom teacher. I’m not sure why, but it worked great for my children that struggled in math. As long as they could draw the boxes correctly, knew their math facts, and understood a grid…they got this!
Partial Products
This one can be a bit cumbersome as numbers get larger, but I love it because of its emphasis on place value. I always taught this one even if it just helped children to see why a zero was added when you reached the second, third, fourth line, etc on the traditional method. Some students naturally liked this the best, which was the strategy I would see them use during independent work. It is honestly how I solve multiplication problems in my head. The steps are:
• First, write down the numbers 25 and 12; line them up by place value like the picture given below.
• Then, take the one digit of the bottom number.
• After that, multiply the top number by the taken digit number.
• Write down the ones digit in the ones place of the product and take the tens number in hand.
• Then, multiply the tens digit of the top number by the ones digit of the bottom number.
• Add the tens number that you took in hand in the last step.
• After that, multiply the top number by the tens digit of the bottom number. For this example, we will multiply 25 by 1.
• Place a 0 first in the next line and ones digit as here, 1 represents ten.
• Follow the same procedure described in steps 4 and 5.
• Lastly, add the two products and get the total result of two-digit multiplication.
If you are interested in trying out these different methods with your kiddos and seeing what works best for them check out this post. You can get more details on how to teach each one with some cut-and-paste activities to support their learning here.
## Required Prep-Work
1. Print the game board and game cards on cardstock paper
2. Cut out the game cards and decide if you want to use them all or just some. There are cards for two-digits by two-digits, two-digits by three digits, three-digits by two-digit multiplication, and three-digits by three-digit multiplication.
3. Provide students with a die, game markers, and a calculator. I would also provide a whiteboard and dry-erase marker so they have a place to do their work.
## How to Play Multiplying Double Digit Game
1. Students take turns rolling the die and moving around the board.
2. If the player lands on a picture, they must draw a card and complete the multiplication problem. Another player will do the same problem on the calculator. If the person who is doing the math without a calculator correctly answers the problem, they may roll again. If they have to solve another problem, they may not roll again….it is now the next player’s turn.
3. If a player lands on a space with words, they must follow those directions.
4. The player who makes it to the end of the board first wins the game.
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Important Questions: Statistics
# Important Questions: Statistics | Mathematics (Maths) Class 11 - Commerce PDF Download
Q1: Find the variance and the standard deviation for the following data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59.
Ans: Given data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
To find mean (μ):
Mean (μ)= ( 57+ 64+ 43+ 67+ 49+ 59+ 44+ 47+ 61+ 59)/10
= 550/10
Mean = 55
To find Variance (σ2):
Variance(σ2) = (xi – μ)2/n
= (22 + 92 + 122 + 12+ 62 + 42 + 62 + 42 + 112 + 82)/10
= 662/10
=66.2
Therefore, variance(σ2) = 66.2
To find standard deviation (σ):
To find the standard deviation, take the square root of variance, we get
Standard Deviation(σ) = √(σ2)
= √66.2 = 8.13
Therefore, the standard deviation is 8.13.
Q2: Find the mean deviation about mean for the following data:
Ans:
Let mean = μ
The formula to find mean is
μ = fi xi / N
N = 3+3+4+14+7+4+3+4 = 42
μ = (3+ 9+ 20+ 98+ 63+ 44+ 39+ 60)/42
μ = 336/42
μ = 8
Now, to find the mean deviation about mean:
The formula is:
M.D(μ) = fi |xi– μ| / N
M.D(μ) =[3(7)+3(5)+ 4(3)+ 14(1)+7(1)+ 4(3)+ 3(5)+ 4(7)]/42
= (21+ 15+ 12+ 14+ 7+12+ 15+ 28)/42
= 62/21
= 2.95
Therefore, the mean deviation about mean for the given data is 2.95
Q3: The variance of the given data 2, 4, 5, 6, 8, 17 is 23.33. Then find the variance for the data 4, 8, 10, 12, 16, 34.
Ans:
For the given data: 2, 4, 5, 6, 8, 17, the variance is 23.33.
To find the variance for the data: 4, 8, 10, 12, 16, 34
If you notice the data which you have to find the variance, it is the multiple of the given data.
So, multiply the variance of the given data by 2,
It means that, 23.33 x 2 = 46.66
Thus, the variance of the data: 4, 8, 10, 12, 16, 34 is 46.66
Q4: The coefficients of variations for the two distributions are 60 and 70 and its standard deviations are 21 and 16 respectively. Determine its arithmetic mean.
Ans:
Given that,
Coefficient of Variations (C.V of 1st distribution) = 60, σ1 = 21
Coefficient of Variations (C.V of 2nd distribution) = 70, σ2 = 16
Let μ1 and μ2 are the means of the 1st and the 2nd distribution.
We know that the formula to find the arithmetic mean is given as:
Coefficient of Variations(C.V) = (Standard Deviation/arithmetic Mean) x 100
Thus, Arithmetic Mean = (Standard Deviation/C.V)x100
Therefore, the arithmetic mean for the 1st deviation is given by:
μ1= [σ1 / (c.v of 1st distribution)] x100
μ1= (21/60)x100
μ1= 0.35×100
μ1= 35
Similarly for μ2:
μ= [σ2 / (c.v of 2nd distribution)]x100
μ= (16/70)x100
μ= 0.2285×100
μ= 22.85
Therefore, the arithmetic mean for the 1st and the 2nd distributions are 35 and 22.85 respectively.
Q5: Determine the mean deviation about the median for the following data:
Ans:
From the given data:
From this, it is noticed that the class interval containing 25th item is 20-30. Therefore, the median is 20-30.
We know that the formula for the median is given as:
Median = l+ {[((N/2)-C)/f] x h}
Here l = 20, f = 15, C=13, N = 50, and h = 10
Now substitute these values in the formula, we get:
Median = 20+ {[(25-13)/15] x 10}
= 20+8
= 28
Therefore, median is 28.
Hence, the mean deviation about the median is given by:
M.D(M)=
M.D(M) = 10.16
Hence, the mean deviation about the median is 10.16.
The document Important Questions: Statistics | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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# 2.2 Solving Linear Equations With More Than Two Operations
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## Transcription
1 2.2 Solving Linear Equations With More Than Two Operations Objectives In this section, you will learn to: To successfully complete this section, you need to understand: Solve equations involving more than one Operations with real numbers (Chapter 1) variable term. Combining like terms (1.9) Simplify expressions before solving. The Distributive Property (1.10) Solving standard form linear equations (2.1) INTRODUCTION In Section 2.1, we became familiar with solving equations in standard form, ax + b = c, which involves two operations, addition (or subtraction) and multiplication, such as 6x + 9 = -15. To solve, we reduce the equation by first clearing the constant term, 9. Sometimes, though, an equation has more than two operations. Examples of such equations are 4x + 17 = 6x + 5, 2x x = , and 12 2y = 3(y 1). As explained throughout this section, our initial goal in solving these more complex linear equations will be to reduce each equation to standard form. This may take a few steps, but once an equation is in standard form, we can apply what we learned in Section 2.1 to finish solving. SOLVING LINEAR EQUATIONS CONTAINING MORE THAN ONE VARIABLE TERM Remember, the ultimate goal in solving any linear equation is to isolate the variable, and this means first isolating the variable term. However, some equations have more than one variable term, one on each side, such as 4x + 17 = 6x + 5. Question: To solve this equation, which variable term should we isolate? 4x + 17 = 6x + 5 Answer: Remember that any solution we find becomes a replacement value. This solution replaces the variable in both variable terms. Should we isolate this variable term or this variable term? In the example 4x + 17 = 6x + 5, the x s are the same variable and must be combined as a single term before we can truly isolate the variable. However, because the variables are in different expressions on different sides of the equal sign the only way we can combine them is to clear one of the variable terms by adding its opposite to each side. Solving Linear Equations with More Than Two Operations page 2.2-1
2 Does it really matter which variable term we choose to clear? No. Because the equation has only one solution, whichever variable term we choose to clear will lead us to the solution. Example 1 and You Try It 1 solve the same equation using two different approaches. Example 1: Solve 4x + 17 = 6x + 5 Procedure: First, combine the variable terms by adding the opposite of one them to each side. Answer: 4x + 17 = 6x + 5 Prepare the equation for solving by clearing +4x: add -4x to each side of the equation. 4x + (-4x) + 17 = 6x + (-4x) = 2x + 5 4x + (-4x) = 0x, or just = 2x + 5 Reduce this equation by clearing the constant: add -5 to each side (-5) = 2x (-5) 12 = 2x = 2x Divide each side by 2. 4(6) + 17 Verify the solution, 6. Replace each variable in the original equation:? = 6(6) = 2x ? = = x x = 6 Write the variable on the left side. 41 = 41 You Try It 1 Solve this equation by clearing the + 6x term from each side. The first step is shown. Complete the solving process. Use Example 1 as a guide. 4x + 17 = 6x + 5 Verify the solution: 4x + (-6x) + 17 = 6x + (-6x) + 5 Solving Linear Equations with More Than Two Operations page 2.2-2
3 Think about it #1: Comparing the work shown in Example 1 and your own work in You Try It 1, which variable term, 4x or 6x, would you likely clear if you had your choice? Why? Share your answer with a classmate. You Try It 2 Solve each equation. Check each answer to show that it is a solution. Use Example 1 as a guide. a) 4w 5 = 4 + 7w b) 3y + 9 = -5y + 41 c) x + 10 = 2 3x d) 2p 4 = -p 25 Solving Linear Equations with More Than Two Operations page 2.2-3
4 SIMPLIFYING EXPRESSIONS BEFORE SOLVING Sometimes, in reducing an equation to standard form, we must simplify one (or both) of the sides; after all, each side is an expression, and it s common for us to simplify expressions. For example, (1) we may need to first distribute a number to a quantity, as in the expression 3(y 1) 3(y 1) = 3y 3 (2) we may need to combine like terms, as in the expression 2x x. 2x x. = 6x + 8 Example 2: Solve 12 2y = 3(y 1) Procedure: First simplify the right side by distributing 3 to the quantity (y 1). Then reduce the equation to standard form by clearing one of the variable terms. Answer: 12 2y = 3(y 1) 12 2y = 3y 3 First distribute on the right side. Do not try to clear any terms just yet. Reduce this equation to standard form. Let s clear -2y by adding its opposite, +2y, to each side. 12 2y + 2y = 3y + 2y = 5y 3 12 = 5y = 5y = 5y = 5y Now isolate the variable term by clearing the constant: add +3 to each side. Divide each side by (3) Verify the solution, 3. Replace each variable in the original equation:? = 3(3 1) 15 5 = 5y ? = 3(2) 3 = y 6 = 6 y = 3 Solving Linear Equations with More Than Two Operations page 2.2-4
5 Example 3: Procedure: Solve 2x x = x by first simplifying each side. First simplify the left side by combining like terms. Then reduce the equation to standard form by clearing one of the variable terms. Answer: 2x x = x 7x + 8 = x 7x + (-3x) + 8 = x + (-3x) 4x + 8 = x + 8 = -4 Combine like terms on the left side: 2x + 5x = 7x. Reduce this equation to standard form. Let s clear +3x by adding its opposite, -3x, to each side. Now isolate the variable term by clearing the constant: add -8 to each side. 4x (-8) = -4 + (-8) You finish it: Verify the solution, 3. Replace each variable in the original equation: 4x + 0 = -12 4x = -12 Divide each side by 4. 4x 4 = x = -3 You Try It 3 Solve each equation by first simplifying each side. Use Examples 2 and 3 as guides. a) 4(y + 1) = 3(2y 2) b) 3m m = 8 + m 10 Solving Linear Equations with More Than Two Operations page 2.2-5
6 c) -2(3w + 5) = 7w 4 w d) -x 9 + 5x = (x 5) YTI 1: x = 6 Answers: You Try It and Think About It YTI 2: a) w = -3 b) y = 4 c) x = -2 d) p = -7 YTI 3: a) y = 5 b) m = -2 c) w = d) x = -1 Think About It: 1. Answers may vary. Solving Linear Equations with More Than Two Operations page 2.2-6
7 Section 2.2 Exercises Think Again. 1. Why is it important to simplify each side of an equation before applying any of the properties of equality? 2. What would you write to a classmate to explain the importance of doing a check after solving an equation? Focus Exercises. Solve each equation. Verify the solution. 3. 3m + 1 = 7m x + 1 = 13 2x x = x y = 15 8y 7. 3y 7 = -5y p 1 = -4p w + 9 = 9w x 12 = 6 + 5x x = x y = 15 3y 13. 2x 11 = 10x n + 11 = n 15. 7v 7 = 5v w 5 = 7w c 4 = -4 c h 3 = -3 h 19. 9w + 8 = 7w q + 36 = 7q 21. 2x 20 = 7x 22. 5v 18 = 8v x = -4x w = 3w + 24 Solve each equation. Verify the solution a 6 = -13a m + 7 = 19 2m 27. 4x 2 3x = -x y y = 6 + 5y 29. 4d + 1 6d = 3(5 3d) 30. 2(5 x) = 3x + 6 x Solving Linear Equations with More Than Two Operations page 2.2-7
8 31. 4(3c + 2) = 2c (y 6) = 4y x + 2 = 6(x + 3) 34. 4x + 15 = 5(x + 4) 35. 5p + 7 3p = 17 9p c c = -2c w + 5 = 9w + 3 6w 38. 2y + 5 4y = 6y y 4 = 12y 1 20y x x = 4 5x 41. 2v + 3 4v = -1(v + 9) 42. k k = 3(k + 2) (m + 2) = -m 2 5m 44. 6(r 1) = r r (2q 5) = 4q + 6 q 46. 5(x + 4) = 3x x m 13 = 3(m 4) c 2 + 5c = (c 1) p = 10p p 50. 8x x = 6 7x (3 v) = v 9 6v (w 10) = 3w 2 7w 53. n n = -5(n + 2) 54. y + 2 6y = -4(y 3) 55. 3(x + 20) + (x + 20) + x = (w + 9) + 2w = (x 4) + 2x = 7x + 2(-2x 1) 58. 5(x + 1) 3x = (4 3x) Think Outside the Box. Solve each equation. 59. x 2 5x + 4 = x 2 + 3x (4x 3x 2 ) 9 = -3(x + 2x 2 ) The perimeter of the rectangle is the same as the perimeter of the triangle. What is the length of the rectangle? (2x + 2) feet (3x 4) feet (5x 8) feet (3x 1) feet (4x 4) feet Solving Linear Equations with More Than Two Operations page 2.2-8
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# Basic Geometry Concepts (video lessons, diagrams, examples, step-by-step solutions)
Geometric Plane Shapes For Kids – Primary Vocabulary
Geometric Plane Shapes For Kids – Primary Vocabulary
Related Pages
2-D and 3-D Shapes
More Geometry Lessons
These lessons introduces basic geometry terms including: points, lines, line segments, midpoints, rays, planes and space. Building upon these basic ideas, we look into some introductory geometry concepts.
Basic Geometry Geometric Terms Fundamental Concepts of Geometry Angles Geometric Theorems Geometry Worksheets Math Worksheets
The following table gives some geometry concepts, words and notations. Scroll down the page for examples, explanations and solutions.
We may think of a point as a “dot” on a piece of paper or the pinpoint on a board. In geometry, we usually identify this point with a number or letter. A point has no length, width, or height – it just specifies an exact location. It is zero-dimensional.
Every point needs a name. To name a point, we can use a single capital letter. The following is a diagram of points A, B, and M:
We can use a line to connect two points on a sheet of paper. A line is one-dimensional. That is, a line has length, but no width or height. In geometry, a line is perfectly straight and extends forever in both directions. A line is uniquely determined by two points.
Lines need names just like points do, so that we can refer to them easily. To name a line, pick any two points on the line.
The line passing through the points A and B is denoted by
A set of points that lie on the same line are said to be collinear. Pairs of lines can form intersecting lines, parallel lines, perpendicular lines and skew lines.
Because the length of any line is infinite, we sometimes use parts of a line. A
line segment connects two endpoints. A line
segment with two endpoints A and B is denoted by
.
A line segment can also be drawn as part of a line.
The midpoint of a segment divides the segment into two segments
of equal length. The diagram below shows the midpoint M of the line segment . Since M is the midpoint, we know that the lengths AM = MB.
A ray is part of a line that extends without end in one direction. It starts from one endpoint and extends forever in one direction.
A ray starting from point A and passing through B is denoted by
Planes are two-dimensional. A plane has length and width, but no height, and extends infinitely on all sides. Planes are thought of as flat surfaces, like a tabletop. A plane is made up of an infinite amount of lines. Two-dimensional figures are called plane figures.
All the points and lines that lie on the same plane are said to be coplanar.
A plane.
Space is the set of all points in the three dimensions – length, width and height. It is made up of an infinite number of planes. Figures in space are called solids.
Figures in space.
This video explains and demonstrates the fundamental concepts (undefined terms) of geometry: points, lines, ray, collinear, planes, and coplanar. The basic ideas in geometry and how we represent them with symbols.
A point is an exact location in space. They are shown as dots on a plane in 2 dimensions or a dot in space in 3 dimensions. It is labeled with capital letters. It does not take up any space.
A line is a geometric figure that consists of an infinite number of points lined up straight that extend in both directions for ever (indicated by the arrows at the end). A line is identified by a lower case letter or by two points that the line passes through. There is exactly 1 line through two points. All points on the same line are called collinear. Points not on the same line are noncollinear.
Two lines (on the same plane) are either parallel or they will meet at a point of intersection.
A line segment is a part of a line with two endpoints. A line segment
starts and stops at two endpoints.
A ray is part of a line with one endpoint and extends in one direction forever.
A plane is a flat 2-dimensional surface. A plane can be identified by 3 points in the plane or by a capital letter. There is exactly 1 plane through three points. The intersection of two planes is a line.
Coplanar points are points in one plane.
An angle consists of two rays with a common endpoint. The two rays are called the sides of the angle and the common endpoint is the vertex of the angle.
Each angle has a measure generated by the rotation about the vertex. The measure is determined by the rotation of the terminal side about the initial side. A counterclockwise rotation generates a positive angle measure. A clockwise rotation generates a negative angle measure. The units used to measure an angle are either in degrees or radians.
Angles can be classified base upon the measure: acute angle, right angle, obtuse angle, and straight angle.
If the sum of measures of two positive angles is 90°, the angles are called complementary.
If the sum of measures of two positive angles is 180°, the angles are called supplementary.
Examples:
The Opposite Angle Theorem (OAT)
When two straight lines cross, opposite angles are equal.
The Angle Sum of a Triangle Theorem
The interior angles of any triangle have a sum of 180°.
The Exterior Angle Theorem (EAT)
Any exterior angle of a triangle is equal to the sum of the opposite interior angles.
Parallel Lines Theorem (PLT)
Whenever a pair of parallel lines is cut by a transversal
a) corresponding angles are equal (PLT-F)
b) alternate angles are equal (PLT-Z)
c) interior angles have a sum of 180° (PLT-C)
Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
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# Which Is the Only Solution to the Equation LOG3(X2 + 6X) = LOG3(2X + 12)?
Which Is the Only Solution to the Equation LOG3(X^2 + 6X) = LOG3(2X + 12)?
In mathematics, equations play a crucial role in solving problems and understanding the relationships between variables. One such equation that requires careful analysis is LOG3(X^2 + 6X) = LOG3(2X + 12). In this article, we will explore the steps to find the only solution to this equation and provide some frequently asked questions related to logarithmic equations.
To solve the equation LOG3(X^2 + 6X) = LOG3(2X + 12), we need to understand the properties of logarithms and apply them appropriately. The equation involves the base 3 logarithm, which means that the logarithm of a number is taken with respect to the base 3. Let’s break down the process step by step:
Step 1: Apply the property of logarithms
The property of logarithms states that if LOGa(b) = LOGa(c), then b = c. In our equation, we have LOG3(X^2 + 6X) = LOG3(2X + 12). Applying the logarithmic property, we can write X^2 + 6X = 2X + 12.
Step 2: Simplify the equation
To solve for X, we need to simplify the equation obtained from step 1. Rearranging the terms, we have X^2 + 4X – 12 = 0.
Step 3: Factor or use the quadratic formula
To find the solutions to a quadratic equation, we can either factor it or use the quadratic formula. Factoring the equation X^2 + 4X – 12 = 0, we get (X – 2)(X + 6) = 0. Therefore, X = 2 or X = -6.
Step 4: Check for extraneous solutions
Now that we have obtained two possible solutions, we need to verify if they satisfy the original equation. Plugging X = 2 into the original equation, we get LOG3(2^2 + 6(2)) = LOG3(2(2) + 12), which simplifies to LOG3(16) = LOG3(16). Both sides of the equation are equal, so X = 2 is a valid solution.
See also Which of the Following Statements Regarding the Different Stages of the Grieving Process Is Correct?
Similarly, plugging X = -6 into the original equation, we get LOG3((-6)^2 + 6(-6)) = LOG3(2(-6) + 12), which simplifies to LOG3(0) = LOG3(0). However, the logarithm of 0 is undefined, so X = -6 is not a valid solution.
Therefore, the only solution to the equation LOG3(X^2 + 6X) = LOG3(2X + 12) is X = 2.
FAQs:
Q1: What are logarithms?
A1: Logarithms are mathematical functions that represent the exponent to which a base number must be raised to obtain a given number. In the equation LOG3(X^2 + 6X) = LOG3(2X + 12), the base is 3.
Q2: What is the property of logarithms used in this equation?
A2: The property of logarithms used in this equation is the equality property, which states that if LOGa(b) = LOGa(c), then b = c.
Q3: Why is X = -6 not a valid solution?
A3: X = -6 is not a valid solution because plugging it into the original equation leads to LOG3(0) = LOG3(0), which is undefined. Logarithms of 0 are not defined in mathematics.
Q4: Are there any other methods to solve this equation?
A4: Yes, instead of factoring, you can also use the quadratic formula to solve for X. The quadratic formula is X = (-b ± √(b^2 – 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation.
Q5: Can logarithmic equations have multiple solutions?
A5: Yes, logarithmic equations can have multiple solutions. However, in this particular equation, there is only one valid solution, which is X = 2.
In conclusion, the only solution to the equation LOG3(X^2 + 6X) = LOG3(2X + 12) is X = 2. By understanding the properties of logarithms and following the steps outlined above, we can successfully solve this equation.
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# Divisibility and Factors - PowerPoint PPT Presentation
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Divisibility and Factors. 4.1. Divisibility Rules of 2 and 4. A number is divisible by 2 if the last digit if divisible by 2 A number is divisible by 4 if the last two digits are divisible by 4. Divisibility Rules of 3 and 9.
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Divisibility and Factors
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## Divisibility and Factors
4.1
### Divisibility Rules of 2 and 4
• A number is divisible by 2 if the last digit if divisible by 2
• A number is divisible by 4 if the last two digits are divisible by 4
### Divisibility Rules of 3 and 9
• A number is divisible by 3 is the sum of the digits are divisible by 3
• A number is divisible by 9 if the sum of a the digits are divisible by 9
### Divisibility of 5 and 10
• If a number ends in a 5 or 0 its divisible by 5
• If a number ends in a 0 its divisible by 10
### Divisibility of 6
• If it ends in an even number and if the sum of the digits is divisible by 3.
### Examples:
1.) Is 160 divisible by 5
2.) is 56 divisible by 10
3.) is 53 divisible by 2
4.) is 1118 divisible by 2
5.) is 64 divisible by 9
6.) is 472 divisible by 3
7.) Is 174 divisible by 3
8.)Is 43, 542 divisible by 9
### Factor
• A number that will divide with out a remainder into another
### List the positive factors of each number
• 10
• 21
• 24
• 31
What are possible arrangements if there are 36 students singing at a concert. There must be at least 5 students in each row and the same number of students in each row.
### Homework
TB Pg 174 (2-52) even
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# How do you solve 2/5=(y-26)/(y+11)?
Apr 8, 2018
$\frac{152}{3}$
#### Explanation:
First, multiply both sides by $5 \left(y + 11\right)$ to get rid of all denominators.
$\frac{2}{\cancel{5}} \times \cancel{5} \left(y + 11\right) = \frac{y - 26}{\cancel{\left(y + 11\right)}} \times 5 \cancel{\left(y + 11\right)}$
Now simplify to get: $2 \left(y + 11\right) = 5 \left(y - 26\right)$
Now multiply both sides out. $2 y + 22 = 5 y - 130$
Now place constants on the same side and the variables on the same side. $3 y = 152$
Now, just divide both side by 3 and you get your answer.
$y = \frac{152}{3}$
Now, 152 does not evenly divide by three, so we can present it as an improper fraction ($\frac{152}{3}$) or a mixed fraction ($50 \frac{2}{3}$).
Apr 8, 2018
$y = 50 \frac{2}{3} \approx 50.7$
#### Explanation:
$\frac{2}{5} = \frac{y - 26}{y + 11}$
First, cross multiply;
$2 \left(y + 11\right) = 5 \left(y - 26\right)$
Secondly, eliminate the brackets;
$2 y + 22 = 5 y - 130$
Thirdly, collect like terms;
$2 y - 5 y = - 130 - 22$
$- 3 y = - 152$
Divide both sides by $- 3$
$\frac{- 3 y}{- 3} = \frac{- 152}{- 3}$
(cancel(-3)y)/cancel(-3) = (cancel-152)/(cancel-3)
$y = \frac{152}{3}$
$y = 50 \frac{2}{3} \approx 50.7$
Apr 8, 2018
$y = \frac{152}{3}$
#### Explanation:
=$\frac{2}{5} = \frac{y - 26}{y + 11}$
=$2 y + 22 = 5 y - 130$
=$5 y - 2 y = 130 + 22$
=$3 y = 152$
=$y = \frac{152}{3}$
Hope it helps!
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BASIC DEFINITIONS - Differential Equations - Calculus AB and Calculus BC
## CHAPTER 9 Differential Equations
Concepts and Skills
In this chapter, we review how to write and solve differential equations, specifically,
• writing differential equations to model dynamic situations;
• understanding a slope field as a graphical representation of a differential equation and its solutions;
• finding general and particular solutions of separable differential equations;
• and using differential equations to analyze growth and decay.
We also review two additional BC Calculus topics:
• Euler’s method to estimate numerical solutions
• and using differential equations to analyze logistic growth and decay.
### A. BASIC DEFINITIONS
Differential equation
A differential equation (d.e.) is any equation involving a derivative. In §E of Chapter 5 we solved some simple differential equations. In Example 50, we were given the velocity at time t of a particle moving along the x-axis:
From this we found the antiderivative:
If the initial position (at time t = 0) of the particle is x = 3, then
x(0) = 0 − 0 + C = 3,
and C = 3. So the solution to the initial-value problem is
A solution of a d.e. is any function that satisfies it. We see from (2) above that the d.e. (1) has an infinite number of solutions—one for each real value of C. We call the family of functions (2) the general solution of the d.e. (1). With the given initial condition x(0) = 3, we determined C, thus finding the unique solution (3). This is called the particular solution.
Note that the particular solution must not only satisfy the differential equation and the initial condition, but the function must also be differentiable on an interval that contains the initial point. Features such as vertical tangents or asymptotes restrict the domain of the solution. Therefore, even when they are defined by the same algebraic representation, particular solutions with different initial points may have different domains. Determining the proper domain is an important part of finding the particular solution.
In §A of Chapter 8 we solved more differential equations involving motion along a straight line. In §B we found parametric equations for the motion of a particle along a plane curve, given d.e.’s for the components of its acceleration and velocity.
Rate of Change
A differential equation contains derivatives. A derivative gives information about the rate of change of a function. For example:
(1) If P is the size of a population at time t, then we can interpret the d.e.
as saying that at any time t the rate at which the population is growing is proportional (3.25%) to its size at that time.
(2) The d.e. tells us that at any time t the rate at which the quantity Q is decreasing is proportional (0.0275%) to the quantity existing at that time.
(3) In psychology, one typical stimulus-response situation, known as logarithmic response, is that in which the response y changes at a rate inversely proportional to the strength of the stimulus x. This is expressed neatly by the differential equation
If we suppose, further, that there is no response when x = x0, then we have the condition y = 0 when x = x0.
(4) We are familiar with the d.e.
for the acceleration due to gravity acting on an object at a height s above ground level at time t. The acceleration is the rate of change of the object’s velocity.
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# 7.
3 Proving Trigonometric Identities
When listening to Mr. Burger on how to prove trig identities he stated that you might want to work with both sides and come to common end statement. I think of proving trig identities the same way you did proofs in geometry. You typically want to work with one side, massage it, and hopefully you will create the expression on the other side of the equal sign. These types of problems should be viewed as ‘given a problem and its answer, how do you get to the answer?’. You want to keep in mind all of the trig identities you have been exposed to thus far to assist you in proving trig identities. Example Prove:
1 + tan θ = tan θ 1 + cot θ
By examining both sides of the equal sides, it appears that you want to begin with the left side in order to create the right side. sin θ 1+ 1 + tan θ cos θ = 1 + cot θ 1 + cos θ sin θ cos θ sin θ + = cosθ cosθ sin θ cos θ + sin θ sin θ cos θ + sin θ cos θ = sin θ + cos θ sin θ =
cosθ + sin θ sin θ + cosθ ÷ cos θ sin θ cos θ + sin θ sin θ · cos θ sin θ + cos θ sin θ = tan θ cos θ
=
=
The reason why it was best to convert in terms of sine and cosine is because the resultant tan θ is a trig function that can be expressed that way. There will be times when you will have to begin with the right side of the equal sign and work your way to create the left side of the equal sign.
Try the following: Prove. 1 − cos x = sin x tan x 1. cos x 2. cot α + tan α = sec α csc α sec 2 θ − tan 2 θ =1 2 2 2 2sin θ + 2 cos θ
3.
Answers: 1 1 cos 2 x − cos x = − cos x cos x cos x 1 − cos 2 x = cos x sin 2 x sin x sin x sin x = = = sin x = tan x sin x = sin x tan x cos x cos x cos x
cot α + tan α = cos α sin α + sin α cos α cos 2 α sin 2 α = + sin α cos α sin α cos α cos 2 α + sin 2 α 1 1 1 = = ⋅ = csc α sec α = secα cscα = sin α cos α sin α cos α sin α cos α
sec 2 θ − tan 2 θ 1 = 2 2 2 2sin θ + 2 cos θ 2 ( sin θ + cos 2 θ ) = 1 1 = 2(1) 2
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# Calculus 3 : Integration
## Example Questions
← Previous 1 3 4 5 6
### Example Question #9 : How To Find Position
Function gives the velocity of a particle as a function of time.
Find the equation that models the particle's postion as a function of time.
Explanation:
Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.
We are starting with the following
We need to perform the following:
Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.
Which makes our position function, h(t), the following:
### Example Question #10 : How To Find Position
Consider the velocity function modeled in meters per second by v(t).
Find the position of a particle whose velocity is modeled by after seconds.
Explanation:
Recall that velocity is the first derivative of position, so to find the position function we need to integrate .
Becomes,
Then, we need to find
### Example Question #1 : Integration
Find the position function given the velocity function:
Explanation:
To find the position from the velocity function, integrate
by increasing the exponent of each t term and then dividing that term by the new exponent value.
### Example Question #2 : Integration
Consider the velocity function given by
Find the position of a particle after seconds if its velocity can be modeled by and the graph of its position function passes through the point .
Explanation:
Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)
So we get:
What we ultimately need is p(5), but first we need to find c: Use the point (2,2)
So our position function is:
### Example Question #3 : Integration
The velocity equation of an object is given by the equation . What is the position of the object at time if the initial position of the object is ?
Explanation:
The position of the object can be found by integrating the velocity equation and solving for . To integrate the velocity equation we first rewrite the equation.
To integrate this equation we must use the power rule where,
.
Applying this to the velocity equation gives us,
.
We must solve for the value of by using the initial position of the object.
Therefore, and .
### Example Question #4 : Integration
The acceleration of an object is given by the equation . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?
Explanation:
To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that
Therefore integrating the acceleration equation gives us
We can solve for the value of by using the initial velocity of the object.
Therefore and
To find the position of the object we integrate the velocity equation.
We can solve for this new value of by using the object's initial position
Therefore and
### Example Question #5 : Integration
The velocity of an object is given by the equation . What is the position of the object at time if the object has a position of and time ?
Explanation:
To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if
Using this rule we find that
Using the position of the object at time we can solve for
Therefore and
We can now find the position at time .
### Example Question #31 : How To Find Position
The velocity of an object is . What is the position of the object if its initial position is ?
Explanation:
The position is the integral of the velocity. By integrating with the power rule we can find the object's position.
The power rule is where
.
Therefore the position of the object is
.
We can solve for the constant using the object's initial position.
Therefore and .
### Example Question #39 : How To Find Position
If the velocity function of a car is , what is the position when ?
Explanation:
To find the position from the velocity function, take the integral of the velocity function.
Substitute .
### Example Question #6 : Integration
If the velocity of a particle is , and its position at is , what is its position at ?
Explanation:
This problem can be done using the fundamental theorem of calculus. We hvae
where is the position at time . So we have
so we need to solve for :
So then the position at is
← Previous 1 3 4 5 6
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