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# Difference between revisions of "2015 AMC 10A Problems/Problem 17" ## Problem A line that passes through the origin in tersects both the line $x = 1$ and the line $y=1+ \frac{\sqrt{3}}{3} x$. The three lines create an equilateral triangle. What is the perimeter of the triangle? $\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)} \ 2+2\sqrt{3} \qquad\textbf{(C)} \ 6 \qquad\textbf{(D)} \ 3 + 2\sqrt{3} \qquad\textbf{(E)} \ 6 + \frac{\sqrt{3}}{3}$ ## Solution Since the triangle is equilateral and one of the sides is a vertical line, the other two sides will have opposite slopes. The slope of one of the other lines is $\frac{\sqrt{3}}{3}$ so the other must be $-\frac{\sqrt{3}}{3}$. Since this other line passes through the origin, its equation is simply $y = -\frac{\sqrt{3}}{3}x$. To find two vertices of the triangle, plug in $x=1$ to both the other equations. $y = -\frac{\sqrt{3}}{3}$ $y = 1 + \frac{\sqrt{3}}{3}$ We now have the coordinates of two vertices. $(1, -\frac{\sqrt{3}}{3})$ and $(1, 1 + \frac{\sqrt{3}}{3})$. Apply the distance formula, $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$. $\sqrt{(1-1)^2 + (-\frac{\sqrt{3}}{3} - (1 + \frac{\sqrt{3}}{3}))^2}$ $\sqrt{(-\frac{\sqrt{3}}{3} - 1 - \frac{\sqrt{3}}{3}))^2}$
# Normal Cost Variation Vs. Schedule Variation When a company plans a project, it will estimate the cost of the project before beginning. It may also estimate the amount of time it expects the project to take to complete. However, these estimates are rarely perfect. The variation in the cost estimate is known as the normal cost variance, while the variation in the estimated time to completion is the schedule variance. ## Definitions 1. A project's cost variation is the mathematical difference between the project's earned value at the time of the calculation and the project's actual cost at the time of calculation. A project's schedule variation is the mathematical difference between the project's earned value at the time of calculation and the project's actual value at the time of calculation. A project's earned value is the budgeted cost of the project multiplied by the actual percentage of completion, while its planned value is the budgeted cost of the project multiplied by the expected percentage of completion. ## Calculations 1. Assume your company budgeted the cost of a project at \$600,000 and planned to complete it in eight months. After a month, your company has completed 10 percent of the project with a total cost of \$75,000. Your company's plan shows that 15 percent of the project should have been complete at this point. In this case, the project's actual cost equals \$75,000, its planned value equals \$90,000 (\$600,000 x .15), and its earned value equals \$60,000 (\$600,000 x .10). Then the project's normal cost variance is -\$15,000 (\$60,000 - \$75,000), and its schedule variance is -\$30,000 (\$60,000 - \$90,000). ## Comparison 1. Both the cost variance and the schedule variance incorporate a project's earned value to date. However, while the cost variance finds the difference between earned value and actual cost, the schedule variance finds the difference between the actual earned value and the company's predicted earned value for the project. The cost variance indicates whether the company is within its budget, but the schedule variance indicates whether the company will complete the project on time. ## Considerations 1. When cost variation is negative, it indicates that the project is costing more than the company expected. A schedule variation that is negative indicates that the project is taking longer to complete than the company expected. Positive values for cost variation and schedule variation indicate that the project is under budget and ahead of schedule. Most companies calculate cost variance and schedule variance several times during a project to monitor efficiency and ensure that the project remains within the projected budget.
# Value of Sin 180 The exact value of sin 180 is zero. Sine is one of the primary trigonometric functions which helps in determining the angle or sides of a right-angled triangle. It is also called trigonometric ratio. If theta is an angle, then sine theta is equal to the ratio of perpendicular and hypotenuse of the right triangle. To be noted the value of sin 0 is also 0. In Mathematics, trigonometry is the study of measurements of triangles that deals with the length, height and angles of the triangle. It has an enormous application in diverse fields like Science, Technology, Satellite navigations, and so on to calculate the varying measurements using cosine and sine function. In this article, the value of sin 180 degrees or the value of sin pi is discussed in detail. ## What is the Value of Sine Pi (180°)? Sin 180 is also denoted as sin pi or sin π in radians. Value of Sine 180 Degree (π) is 0 Note: Sin 180° = Sin 0° = 0 ### Sin 180 – Theta One interesting fact related to Sin 180 degrees is sin 180 minus theta is equal to sin theta, where theta is any angle. Sin (180° – Theta) = Sin Theta sin (180° – θ) = sin θ Hence, there are three points that we can conclude here: • 180° – θ will come in IInd quadrant • For any angle subtracted from 180°, sin will not be changed to cos • The sign of sine ratio in the second quadrant will remain positive ## How to derive the value of sin 180 Degrees? The value of sin 180 (sin pi) can be interpreted in terms of different angles like 0°, 90° and 270°. Assume that the unit circle in the Cartesian plane is subdivided into four quadrants. And we know that the value of sin 180 degrees in the Cartesian plane takes place in the second quadrant. Therefore, the value of sin pi in the second quadrant is always positive. From the value of sin 0, we will obtain the value of sin 180. We know that the exact value of sin 0 degree is 0. So, Sin 180 degree is +(sin 0) which is equal to +(0) Therefore, the value of sin 180 degrees = 0. The value of sin pi can be derived from some other trigonometric angles and functions like sine and cosine functions from the trigonometry table. It is known that, 180° – 0° = 180° ———– (1) 270° – 90° = 180°———— (2) ### Sine 180 Degree Derivation: Method 1 Now we can use the above expression (1) in terms of sine functions. From the supplementary angle identity, we know that; Sin (180 – Theta) = Sin Theta Sin A = Sin (180° – A) Therefore, Sin ( 180° – A ) = Sin A Sin ( 180° – 0° ) = Sin 0° Sin 180° = 0 [Since the value Sin 0° is 0] Hence, the value of sin pi is 0 ### Sine Pi Value Derivation: Method 2 From the expression (2), Using complementary angle identity, Sin A = cos (90°- A) we can write the above expression as: Sin 180°= cos (90° – 180°) Sin 180°= cos (-90°) Now, use opposite angle identity cos(-A) = cos A Sin 180°= cos 90° Sin 180°= 0 [Since the value of cos 90 degrees is 0] Therefore, the value of sin 180 is 0. Sin 180° = 0 Similar Articles Value of cos 30 Value of Cos 60 Value of Cos 120 Value of Cos 180 ### Sin 180 and Sin 0 The value of sin 180 and sin 0 is equal to zero. Since 180 degrees angle falls under the second quadrant, therefore, the value of sine theta above 90 degrees changes to cosine function. That means; Sin (90+A) = cos A We can write, sin 180 as; Sin 180 = sin (90+90) = cos 90 Since, we know, cos 90 = 0, therefore, Sin 180 = 0 Also, we have known, sin 0 = 0 Therefore, sin 180 = sin 0. The trigonometric ratios value for different angles and functions are as follows: Trigonometry Ratio Table Angles (In Degrees) 0 30 45 60 90 180 270 360 Angles (In Radians) 0 π/6 π/4 π/3 π/2 π 3π/2 sin 0 1/2 1/√2 √3/2 1 0 −1 0 cos 1 √3/2 1/√2 1/2 0 −1 0 1 tan 0 1/√3 1 √3 Not Defined 0 Not Defined 0 cot Not Defined √3 1 1/√3 0 Not Defined 0 Not Defined cosec Not Defined 2 √2 2/√3 1 Not Defined −1 Not Defined sec 1 2/√3 √2 2 Not Defined −1 Not Defined 1 ## Frequently Asked Questions – FAQs Q1 ### What is the exact value of sin 180 degrees in trigonometry? In trigonometry, the exact value of sin 180 is 0. Q2 ### What is the value of cos 180 degrees? The value of cos 180 degrees is -1. Q3 ### How to find sin 180 value? We can write, sin 180 as; Sin 180 = Sin (90+90) Since, sin (90+a) = cos a So, sin (90+90) = cos 90 As we know cos 90 = 0, therefore, Sin 180 = 0 Q4 ### What is the value of sin pi? In trigonometry, we use pi (π) for 180 degrees to represent the angle in radians. Hence, sin π is equal to sin 180 or sin π = 0. Q5 ### How to calculate the value of sin 120? We can write, sin 120 as; Sin 120 = sin (90+30) Since, sin (90 + A) = cos A, so Sin (90+30) = cos 30 Since, cos 30 = √3/2 Therefore, sin 120 = √3/2 Test your knowledge on Value of sin 180
# At a Glance - Scaling Vectors What's one apple plus another apple? Two apples. What's two oranges less one orange? One orange. What's one griffin plus one minotaur? Any mythology buff could tell you it's a griffotaur, of course. Adding and subtracting vectors is something like adding the griffin and the minotaur. It's possible to add two vectors together or subtract one vector from another. We get another vector in its place. To add vectors, we add the x-components to each other and add the y-components to each other. ### Sample Problem To add the vectors <3, 4> and <5, 6> we add the first components together and add the second components together: < 3, 4> + <5, 6 > = <8, 10>. ### Sample Problem To subtract, we subtract the x-components and then subtract the y-components. <3, 4> – <5, 6> = <-2, -2>. When we add or subtract vectors, remember to deal with the x-components and the y-components separately. Why? Going back to counting fruit, apples are the x-component and oranges are the y-component of vectors. We can't add apples to oranges, but we can add apples to apples and oranges to oranges. Now we're going to work with multiplication of a vector by a real number. To multiply a vector by a number, we multiply each component of the vector by that number. Multiply the minotaur vector by 2. What do we get? A minotaur twice the size. Likewise, if we multiply the griffin by 0.01, we get a pint-sized, cute griffin to take home with us. When we multiply a vector by a real number, we call the number a scalar. We call the real number a scalar because it scales the vector to be a different size. If the scalar is negative it will also change the direction of the vector. ### Sample Problem Take the vector <1, 2>. This vector has magnitude √5 Now multiply the vector by 3 to get a new vector: <1, 2> → <3, 6>. The magnitude of the new vector is The original vector <1, 2> had magnitude √5, The magnitude of the new vector <3, 6> is 3 times greater than the magnitude of the original vector, <1, 2>. In multiplying the vector by 3, we scaled the arrow to be 3 times longer. #### Example 1 Multiply the vector <9, 2> by 5. #### Example 2 What is the magnitude of the vector 5<2, 3>? #### Example 3 If the vector a<4, 3> has magnitude 10, what is a? #### Example 4 By what number must we multiply the vector <3, 5> to get a vector of length 2? #### Example 5 What happens to the magnitude and direction of the vector <2, 2> when it is multiplied by -1 ? #### Exercise 1 Perform the following vector operation. • <2, -10> + <4, 6> #### Exercise 2 What is the sum of these two vectors? • <3, 7> + <-2, -3> #### Exercise 3 What's the difference between these two vectors? • <1, 1> – <-3, -6> #### Exercise 4 Find the sum of the following two vectors. • <2, 2> + <1, 5> #### Exercise 5 Perform the vector operation. • <11, 0.3> – <1, 0.4> #### Exercise 6 What is the resulting vector? • 7<3, 4> #### Exercise 7 What is the resulting vector after performing the multiplication? • 2<2, 8> #### Exercise 8 Perform the scalar multiplication seen here. • -5<-3, -7> #### Exercise 9 Work out the multiplication problem. 43<6, -1> #### Exercise 10 What is the resulting vector after performing the scalar multiplication? • -3 <6, -1> #### Exercise 11 What vector are we left with after doing the following scalar multiplication? • 0<7, 6> #### Exercise 12 • Find \|\|3<7, 10>\|\|. #### Exercise 13 • A vector v has magnitude 10. What is the magnitude of the vector ? #### Exercise 14 • If the vector a<5, 12> has magnitude 39, what is a? #### Exercise 15 • Find the magnitude of the vector (-1)<8, 3>. #### Exercise 16 • Find the magnitude of the vector -2<3, 2>.
# The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 77ft^2, how do you find the dimensions of the rectangle? ##### 1 Answer Mar 26, 2016 Width = $\frac{11}{2} \text{ ft = 5 foot 6 inches}$ Length = $14 \text{ feet}$ #### Explanation: Breaking the question down into its component parts: Let length be $L$ Let width be $w$ Let area be $A$ Length is 3 ft more than: L=" "?+3 twice" "L=2?+3 its width$\text{ } L = 2 w + 3$ Area $= A = 77 = \text{width "xx" Length}$ $A = 77 = w \times \left(2 w + 3\right)$ $2 {w}^{2} + 3 w = 77$ $2 {w}^{2} + 3 w - 77 = 0$ This is a quadratic equation '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Standard form $y = a {x}^{2} + b x + c$ $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ $a = 2 \text{; "b=3"; } c = - 77$ $x = \frac{- \left(3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(2\right) \left(- 77\right)}}{2 \left(2\right)}$ $x = \frac{- \left(3\right) \pm 25}{4} = - 7 \mathmr{and} \frac{11}{2}$ As we can not have a negative area in this context the answer for $x$ is $\frac{11}{2}$ But $\textcolor{b l u e}{x = w \text{ so the width is } \frac{11}{2}}$ $\textcolor{b l u e}{L = 2 w + 3 = 11 + 3 = 14}$
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 9.17: Area of a Circle Difficulty Level: At Grade Created by: CK-12 Estimated14 minsto complete % Progress Practice Area of a Circle MEMORY METER This indicates how strong in your memory this concept is Progress Estimated14 minsto complete % Estimated14 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Do you remember Miguel? He had just finished working on figuring out the circumference of three different on deck pads for the pitchers to use while they warm up. Let’s look at his dilemma again before we look at the area of the pads. Miguel’s latest task is to measure some different “on deck” pads for the pitchers to practice with. An on deck pad is a circular pad that is made up of a sponge and some fake grass. Pitcher’s practice their warm-ups by standing on them. They work on stretching and get ready to “pitch” the ball prior to their turn on the mound. Miguel has three different on deck pads that he is working with. The coach has asked him to measure each one and find the circumference and the area of each. Miguel knows that the circumference is the distance around the edge of the circle. He decides to start with figuring out the circumference of each circle. He measures the distance across each one. The first one measures 4 ft. across. The second measures 5 ft. across. The third one measures 6 ft across. Miguel has completed the first part of this assignment. He knows the circumference of each pad. Now he has to figure out the area of each. Miguel isn’t sure how to do this. He can’t remember how to find the area of a circle. Miguel needs some help. This is where you come in use this Concept to help you learn how to find the area of a circle. When finished, we’ll come back to this problem and you can help Miguel figure out the area of each on deck pad. ### Guidance You have learned how to calculate the circumference of a circle. Let’s take a few minutes to review the terms associated with circles. A circle is a set of connected points equidistant from a center point. The diameter is the distance across the center of the circle and the radius is the distance from the center of the circle to the edge. We also know that the number pi, π\begin{align}\pi\end{align}, is the ratio of the diameter to the circumference. We use 3.14 to represent pi in operations. What does all of this have to do with area? Well, to find the area of a figure, we need to figure out the measurement of the space contained inside a two-dimensional figure. This is the measurement area. This is also the measurement inside a circle. You learned how to find the circumference of a circle, now let’s look at using these parts to find the area of the circle. How do we find the area of a circle? The area of a circle is found by taking the measurement of the radius, squaring it and multiplying it by pi. Here is the formula. A=πr2\begin{align}A= \pi r^2 \end{align} Write this formula down in your notebook. We already know that the symbol π\begin{align}\pi\end{align} represents the number 3.14, so all we need to know to find the area of a circle is its radius. We simply put this number into the formula in place of r\begin{align}r\end{align} and solve for the area, A\begin{align}A\end{align}. Let’s try it out. What is the area of the circle below? We know that the radius of the circle is 12 centimeters. We put this number into the formula and solve for A\begin{align}A\end{align}. AAAA=πr2=π(12)2=144 π=452.16 cm2\begin{align}A & = \pi r^2\\ A & = \pi (12)^2\\ A & = 144 \ \pi\\ A & = 452.16 \ cm^2\end{align} Remember that squaring a number is the same as multiplying it by itself. The area of a circle with a radius of 12 centimeters is 452.16 square centimeters. Some students have formed a circle to play dodge ball. The radius of the circle is 21 feet. What is the area of their dodge ball circle? The dodge ball court forms a circle, so we can use the formula to find its area. We know that the radius of the circle is 21 feet, so let’s put this into the formula and solve for area, A\begin{align}A\end{align}. AAAA=πr2=π(21)2=441 π=1,384.74 ft2\begin{align}A & = \pi r^2\\ A & = \pi (21)^2\\ A & = 441 \ \pi\\ A & = 1,384.74 \ ft^2\end{align} Notice that a circle with a larger radius of 21 feet has a much larger area: 1,384.74 square feet. Sometimes, you will be given a problem with the diameter and not the radius. When this happens, you can divide the measurement of the diameter by two and then use the formula. Find the area of each circle. #### Example A Solution:254.34\begin{align}254.34\end{align} sq. in. #### Example B Solution: 379.94\begin{align}379.94\end{align} sq. in #### Example C Diameter = 8 ft. Solution: 50.24\begin{align}50.24\end{align} sq. feet Here is the original problem once again. Do you remember Miguel? He had just finished working on figuring out the circumference of three different on deck pads for the pitchers to use while they warm up. Let’s look at his dilemma again before we look at the area of the pads. Miguel’s latest task is to measure some different “on deck” pads for the pitchers to practice with. An on deck pad is a circular pad that is made up of a sponge and some fake grass. Pitcher’s practice their warm-ups by standing on them. They work on stretching and get ready to “pitch” the ball prior to their turn on the mound. Miguel has three different on deck pads that he is working with. The coach has asked him to measure each one and find the circumference and the area of each. Miguel knows that the circumference is the distance around the edge of the circle. He decides to start with figuring out the circumference of each circle. He measures the distance across each one. The first one measures 4 ft. across. The second measures 5 ft. across. The third one measures 6 ft across. Miguel has completed the first part of this assignment. He knows the circumference of each pad. Now he has to figure out the area of each. Miguel isn’t sure how to do this. He can’t remember how to find the area of a circle. Miguel needs some help. Now it’s time to help Miguel figure out each area. The first one has a diameter of 4 feet, so it has a radius of 2 ft. Here is the area of the first pad. AAA=πr2=3.14(22)=12.56 sq.feet\begin{align}A&= \pi r^2 \\ A&=3.14(2^2) \\ A&=12.56 \ sq.feet \end{align} The second pad has a diameter of 5 feet, so it has a radius of 2.5 feet. AAA=πr2=3.14(2.52)=19.63 sq.feet\begin{align}A& = \pi r^2 \\ A& =3.14(2.5^2) \\ A&=19.63 \ sq.feet \end{align} The third pad has a diameter of 6 feet, so it has a radius of 3 feet. AAA=πr2=3.14(32)=28.26 sq.feet\begin{align}A & = \pi r^2 \\ A& = 3.14(3^2) \\ A& = 28.26 \ sq.feet \end{align} Miguel is very pleased with his work. He is sure that his coach will be pleased with his efforts as well! ### Vocabulary Here are the vocabulary words in this Concept. Circle a set of connected points that are equidistant from a center point. Diameter the distance across the center of a circle. the distance from the center of the circle to the outer edge. Area the space inside a two-dimensional figure ### Guided Practice Here is one for you to try on your own. Find the area of a circle with a diameter of 10 in. First, we divide the measurement in half to find the radius. 10÷2=5 in\begin{align}10 \div 2 = 5 \ in\end{align} Now we use the formula. AAAA=πr2=3.14(52)=3.14(25)=78.5 square inches\begin{align}A & = \pi r^2 \\ A& = 3.14(5^2) \\ A&=3.14 (25) \\ A&=78.5 \ square \ inches \end{align} ### Video Review Here is a video for review. ### Practice Directions: Find the area of each circle given the radius or diameter. Round to the nearest hundredth when necessary. 1. r=3 in\begin{align}r = 3 \ in\end{align} 2. r=5 in\begin{align}r = 5 \ in\end{align} 3. r=4 ft\begin{align}r = 4 \ ft\end{align} 4. r=7 m\begin{align}r = 7 \ m\end{align} 5. r=6 cm\begin{align}r = 6 \ cm\end{align} 6. r=3.5 in\begin{align}r = 3.5 \ in\end{align} 7. d=16 in\begin{align}d = 16 \ in\end{align} 8. d=14 cm\begin{align}d = 14 \ cm\end{align} 9. d=20 in\begin{align}d = 20 \ in\end{align} 10. d=15 m\begin{align}d = 15 \ m\end{align} 11. d=22 cm\begin{align}d = 22 \ cm\end{align} 12. d=24 mm\begin{align}d = 24 \ mm\end{align} 13. d=48 in\begin{align}d = 48 \ in\end{align} 14. r=16.5 in\begin{align}r = 16.5 \ in\end{align} 15. r=25.75 in\begin{align}r = 25.75 \ in\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
Front Matter1 Variables2 linear Equations3 Graphs of straight Equations4 Applications of linear Equations5 Exponents and also Roots6 Quadratic Equations7 Polynomials8 Algebraic Fractions9 more About Exponents and Roots Authored in PreTeXt ## Section 5.2 Square Roots and also Cube Roots ### Subsection What is a Square Root? Suppose we would favor to draw a square whose area is 25 square inches. How long need to each next of the square be? Because the formula for the area that a square is $$A=s^2\text,$$ the next $$s$$ should satisfy the equation $$s^2=25\text.$$ We want a number whose square is 25. Together you can probably guess, the size of the square should be 5 inches, due to the fact that $$5^2=25\text.$$ us say the 5 is a square source of 25. You are watching: A rational number whose cube root is a whole number Square Root. The number $$s$$ is called a square source of a number $$b$$ if $$s^2=b\text.$$ Finding a square root of a number is opposing of squaring a number. Example 5.13. Find a square root of 25, and a square source of 144. Solution 5 is a square source of 25 since $$5^2=25\text.$$ 12 is a square source of 144 because $$12^2=144\text.$$ Look Closer. 5 is no the only square source of 25, because $$(-5)^2=25$$ as well. Thus, 25 has actually two square roots, $$5$$ and also $$-5\text.$$ Every positive number has two square roots, one positive and one negative. What is the square source of a number $$n\text?$$ A number whose square is $$n$$ The hopeful square source of a number is called the principal square root. The symbol $$\sqrt\hphantom00$$ denotes the positive or principal square root. Thus, we may write \beginequation\sqrt25 = 5~~~~~~\textand~~~~~~\sqrt144=12\endequation The prize $$\sqrt\hphantom00$$ is referred to as a radical sign, and also the number inside is called the radicand. Square roots are often dubbed radicals. What around the various other square root, the negative one? If we want to suggest the an adverse square root of a number, we place a negative sign exterior the radical sign, like this: \beginequation-\sqrt16=-4~~~~~~\textand~~~~~~ -\sqrt49=-7\endequation If we want to refer to both square roots, we use the prize $$\pm\text,$$ read "plus or minus." for example, \beginequation\pm \sqrt36=\pm 6,~~~~\textwhich means~~~~6~~\textor~~ -6\endequation Note the zero has only one square root: $$\sqrt0=0\text.$$ Example 5.14. Find each square root. $$\displaystyle -\sqrt81 = -9$$$$\displaystyle \pm \sqrt\dfrac64121 = \pm \dfrac811$$Reading concerns analysis Questions2. What is the positive square root of a number called? The major square root 3. The number inside a radical sign Every confident number has actually two square roots, and zero has precisely one square root. What about the square root of a an unfavorable number? for example, can we discover $$\sqrt-4\text?$$ The price is No, because the square of any number is optimistic (or zero). Try this yourself: The just reasonable candidates because that $$\sqrt-4$$ are $$2$$ and $$-2\text,$$ but \beginalign2^2 \amp = \underline\hphantom000000\-2)^2 \amp = \underline\hphantom000000\endalign We cannot find the square root of a negative number. We say that the square root of a an unfavorable number is undefined. Reading inquiries reading Questions4. How execute we uncover the square source of a an unfavorable number? Answer The square root of a an unfavorable number is undefined. ### Subsection Rational and also Irrational Numbers Rational Number. A rational number is one that have the right to be expressed together a quotient (or ratio) of two integers, where the denominator is not zero. The ax "rational" has actually nothing to perform with gift reasonable or logical; it comes from the native ratio. Thus, any portion such as \beginequation\dfrac23,~~~ \dfrac-47,~~~ \textor~~~ \dfrac158\endequation is a rational number. Integers are additionally rational numbers, because any kind of integer can be composed as a portion with a denominator that 1. (For example, \(6= \dfrac61). Every one of the number we have actually encountered prior to this chapter space rational numbers. What is a rational number? One that deserve to be expressed together a quotient of two integers Every portion can be composed in decimal form. Decimal form of a rational Number. The decimal depiction of a rational number has one of two forms. The decimal depiction terminates, or ends.The decimal representation repeats a pattern.Example 5.15. Write the decimal type for each rational number. $$\displaystyle \dfrac34$$$$\displaystyle \dfrac411$$ Solution We have the right to use a calculator or long division to division the molecule by the denominatorto uncover $$\dfrac34=0.75\text.$$We divide 4 by 11 to discover $$\dfrac411=0.363636 \ldots = 0.\overline36\text.$$ The line over the number 36 is dubbed a repeater bar, and also it suggests that those number are recurring forever. How deserve to you identify a reasonable number in its decimal form? It one of two people terminates or repeats a pattern. What about the decimal form for $$\sqrt5\text?$$ If you usage a calculator v an eight-digit display, you will find \beginequation\sqrt5 \approx 2.236~ 068\endequation However, this number is only an approximation, and also not the precise value of $$\sqrt5\text.$$ (Try squaring $$2.236~ 068$$ and you will see that \beginequation2.236~ 068^2 = 5.000~ 000~ 100~ 624\endequation which is not exactly 5, although that is close.) In fact, no matter how countless digits her calculator or computer system can display, you deserve to never find an accurate decimal equivalent for $$\sqrt5\text.$$ $$\sqrt5$$ is an instance of an irrational number. Irrational Number. An irrational number is one that cannot be expressed together a quotient of two integers. Look Closer. There is no end decimal portion that gives the precise value the $$\sqrt5\text.$$ The decimal depiction of an irrational number never ever ends, and also does not repeat any type of pattern! The best we have the right to do is round turn off the decimal type and give an almost right value. Nonetheless, an irrational number still has actually a an accurate location on the number line, just as a reasonable number does. The figure listed below shows the areas of numerous rational and irrational number on a number line. Real Numbers. Each allude on a number line synchronizes either come a reasonable number or an irrational number, and these numbers to fill up the number heat completely. The rational and irrational numbers together make up the real numbers, and also the number heat is sometimes referred to as the real line. What room the rational and irrational numbers with each other called? The real numbers It is vital that you understand the difference between specific value and also an approximation. Example 5.16. We cannot compose down an accurate decimal equivalent for one irrational number. \beginalign\sqrt5~~~~ \amp \textindicates the specific value that the square source of 5\\2.236068~~~~ \amp \text is one approximation to the square source of 5\endalign We regularly use a decimal approximation because that a rational number. \beginalign\dfrac23~~~~ \amp \textindicates the precise value the 2 separated by 3\\0.666667~~~~ \amp \textis one approximation for~\dfrac23\endalign Of course, even though numerous radicals room irrational numbers, some radicals, such together $$\sqrt16=4$$ and $$\sqrt\dfrac925\text,$$ stand for integers or fractions. Integers such together 9 and 25, whose square root are totality numbers, are called perfect squares. ### Subsection order of Operations When we evaluate algebraic expressions the involve radicals, we must follow the order of operations together usual. Square roots occupy the same place as index number in the order of operations: They room computed after parentheses but prior to multiplication. Example 5.17. Find a decimal approximation to three decimal places for $$8-2\sqrt7\text.$$ Solution You may have the ability to enter this expression into your calculator just as the is written. If not, girlfriend must enter the to work in the ideal order. The expression has two terms, 8 and $$-2\sqrt7\text,$$ and the second term is the product that $$\sqrt7$$ v $$-2\text.$$ We have to not begin by subtracting 2 from 8, because multiplication comes before subtraction. First, we uncover an approximation because that $$\sqrt7\text:$$ \beginequation\sqrt7 \approx 2.6457513\endequation Do not round off your approximations at any intermediate action in the difficulty or you will shed accuracy at each step! friend should be able to work straight with the worth on your calculator"s display. Next, us multiply ours approximation by $$-2$$ come find \beginequation-2\sqrt7 \approx -5.2905026\endequation \beginequation8-2\sqrt7 \approx 2.7084974\endequation Rounding to 3 decimal places provides 2.708 Caution 5.18. In Example 5.17, it is not true the $$8-2\sqrt7$$ is same to $$6\sqrt7\text.$$ The stimulate of to work tells united state that we need to perform the multiplication $$-2\sqrt7$$ first, then add the an outcome to $$8\text.$$ You have the right to verify that $$6\sqrt7 \approx 15.874\text,$$ i m sorry is no the very same answer we obtained in Example 5.17. We deserve to now upgrade the order of work by modifying Steps 1 and 2 to incorporate radicals. Order the Operations.Perform any type of operations inside parentheses, or above or listed below a fraction bar.Compute all shown powers and also roots.Perform every multiplications and also divisions in the order in i beg your pardon they take place from left come right.Perform enhancements and subtractions in order native left come right.Reading questions analysis Questions8. When do we evaluate roots in the bespeak of operations? After parentheses but before products and quotients ### Subsection Cube Roots Imagine a cube who volume is 64 cubic inches. What is the length, $$c\text,$$ the one side of this cube? due to the fact that the volume the a cube is provided by the formula $$V=c^3\text,$$ us must find a number the satisfies \beginequationc^3=64\endequation We are looking for a number $$c$$ whose cube is 64. With a little trial and also error we deserve to soon find that $$c=4\text.$$ The number $$c$$ is dubbed the cube source of 64, and also is denoted through $$\sqrt<3>64\text.$$ Cube Root. The number $$c$$ is referred to as a cube source of a number $$b$$ if $$c^3=b\text.$$ Example 5.19.$$\sqrt<3>-64=-4$$ since $$(-4)^3=-64\text.$$$$\sqrt<3>9$$ is one irrational number around equal come $$2.08\text,$$ because $$2.08^3=8.998912\text.$$ Recall that every positive number has actually two square roots, and also that an unfavorable numbers execute not have actually square roots. The case is various with cube roots. Every number has specifically one cube root. The cube source of a optimistic number is positive, and the cube root of a negative number is negative. Just just like square roots, some cube roots room irrational numbers and some room not. Cube roots are treated the exact same as square root in the order of operations. What is the cube source of a number $$n\text?$$ A number whose cube is $$n$$ Subsubsection answers to an abilities Warm-Up$$\displaystyle 4$$$$\displaystyle \dfrac17-k3$$$$\displaystyle 20$$$$\displaystyle 4(-4+m)$$$$\displaystyle -5$$$$\displaystyle \dfrac-c-72$$$$\displaystyle \dfrac59$$$$\displaystyle \dfrac5b+3$$ ### Subsection Lesson Subsubsection activity 1: Roots and also RadicalsExercises Exercises1. Find two square roots for each number. $$\displaystyle 225$$$$\displaystyle \dfrac49$$2. $$\displaystyle \sqrt64$$$$\displaystyle \sqrt-64$$$$\displaystyle -\sqrt64$$$$\displaystyle \pm \sqrt-64$$3. Evaluate every cube root. (Use a calculator if necessary.) $$\displaystyle \sqrt<3>8$$$$\displaystyle \sqrt<3>-125$$$$\displaystyle \sqrt<3>-1$$$$\displaystyle \sqrt<3>50$$4.If $$~p=\sqrtd~\text,$$ then $$~d= \underline\hspace4.545454545454546em$$If $$~v^2=k~\text,$$ then $$~v= \underline\hspace4.545454545454546em$$5.Explain why the square root of a an unfavorable number is undefined.If $$x \gt 0\text,$$ define the difference in between $$\sqrt-x$$ and also $$-\sqrtx\text.$$Explain why you can always simplify $$\sqrtx~\sqrtx\text,$$ as long as $$x$$ is non-negative.Explain why friend can constantly simplify $$\dfracx\sqrtx\text,$$ as lengthy as $$x$$ is positive.6.Make a perform of the squares of every the integers indigenous 1 come 20. These space the first 20 perfect squares. Currently make a list of square roots for these perfect squares.Make a perform of the cubes of every the integers from 1 come 10. These room the first 10 perfect cubes. Now make a list of cube roots because that these perfect cubes.Subsubsection task 2: reasonable NumbersExercises Exercises1. Find the decimal type for every rational number. Does it terminate? $$\displaystyle \dfrac23$$$$\displaystyle \dfrac52$$$$\displaystyle \dfrac1327$$$$\displaystyle \dfrac9622000$$2. Give a decimal equivalent for every radical, and also identify it together rational or irrational. If necessary, round her answers to three decimal places. $$\sqrt1= \underline\hspace6.818181818181818em$$ $$\hphantom0000$$ $$\sqrt6= \underline\hspace6.818181818181818em$$ $$\sqrt2= \underline\hspace6.818181818181818em$$ $$\hphantom0000$$ $$\sqrt7= \underline\hspace6.818181818181818em$$ $$\sqrt3= \underline\hspace6.818181818181818em$$ $$\hphantom0000$$ $$\sqrt8= \underline\hspace6.818181818181818em$$ $$\sqrt4= \underline\hspace6.818181818181818em$$ $$\hphantom0000$$ $$\sqrt9= \underline\hspace6.818181818181818em$$ $$\sqrt5= \underline\hspace6.818181818181818em$$ $$\hphantom0000$$ $$\sqrt10= \underline\hspace6.818181818181818em$$ 3. True or False. If a number is irrational, it cannot be one integer.Every actual number is either rational or irrational.Irrational numbers carry out not have precise location top top the number line.We can not find specific decimal indistinguishable for one irrational number.2.8 is just an approximation for $$\sqrt8\text;$$ the precise value is 2.828427125.$$\sqrt17$$ appears somewhere between 16 and also 18 ~ above the number line.Subsubsection task 3: order of OperationsExercises Exercises1. A radical symbol acts favor parentheses to group operations. Any kind of operations that appear under a radical have to be carry out before analyzing the root. Simplify $$~~\sqrt6^2-4(3)$$Approximate her answer to part (a) to 3 decimal places. Explain why the adhering to is incorrect: \beginalign\sqrt6^2-4(3) \amp = \sqrt6^2-\sqrt4(3)\\\amp = \sqrt36 - \sqrt12\\\amp \approx 6-3.464 = 2.536\endalign 2. Simplify $$~~5-3\sqrt16+2(-3)$$ Evaluate $$~~2x^2-\sqrt9-x~~$$ for $$x=-3\text.$$ Subsubsection Wrap-UpObjectives. In this lesson we exercised the complying with skills: Computing square roots and cube rootsUsing radical notationDistinguishing between rational and irrational numbersDistinguishing in between exact values and approximationsSimplifying expressions involving radicalsQuestions.Explain why you don"t require a calculator to advice $$~\dfrac23\sqrt23\text.$$Explain why you cannot compose down an exact decimal form for $$\sqrt6\text.$$Explain why $$~\sqrt3^2+4^2 \not= 3+4\text.$$ ### Subsection Homework Preview Exercises Exercises1. Simplify. $$\displaystyle 8-3\sqrt25$$$$\displaystyle (3\sqrt16)(-2\sqrt81)$$$$\displaystyle \dfrac3-\sqrt366$$2. Find a decimal approximation rounded to 2 places. $$\displaystyle -3+\sqrt34$$$$\displaystyle \sqrt6^2-4(3)$$$$\displaystyle \sqrt<3>\dfrac185$$3. Simplify. $$\displaystyle 2\sqrtx(8\sqrtx)$$$$\displaystyle \dfrac6m3\sqrtm$$$$\displaystyle \sqrt<3>H(\sqrt<3>H)(\sqrt<3>H)$$4. See more: How To Attach Ski Rope To Tow Ring, Best Way To Attach Tow/Ski Line Evaluate because that $$x=3,~y=5\text.$$ ring to hundredths. $$\displaystyle \sqrtx^2+y^2$$$$\displaystyle (x+y)^2-(x^2+y^2)$$$$\displaystyle \sqrtx+\sqrty-\sqrt(x+y)$$Subsubsection Answers come Homework Preview$$\displaystyle -7$$$$\displaystyle -216$$$$\displaystyle -\dfrac12$$$$\displaystyle 2.83$$$$\displaystyle 4.90$$$$\displaystyle 1.53$$$$\displaystyle 16x$$$$\displaystyle 2\sqrtm$$$$\displaystyle H$$$$\displaystyle 5.83$$$$\displaystyle 30$$$$\displaystyle 1.14$$ ### Exercises Homework 5.2 For difficulties 1–3, simplify. Execute not use a calculator! 1.$$\displaystyle 4-2\sqrt64$$$$\displaystyle \dfrac4-\sqrt642$$2.$$\displaystyle \sqrt9-4(-18)$$$$\displaystyle \sqrt\dfrac4(50)-5616$$3.$$\displaystyle 5\sqrt<3>8-\dfrac\sqrt<3>648$$$$\displaystyle \dfrac3+\sqrt<3>-7296-\sqrt<3>-27$$
# Go Math Grade 5 Chapter 2 Answer Key Pdf Divide Whole Numbers ## Divide Whole Numbers Go Math Grade 5 Chapter 2 Answer Key Pdf Apply the math to real-time examples by learning the tricks using HMH Go Math Grade 5 Answer Key. The quick way of solving math problems will help the students to save time. So, students can practice more questions utilizing the time properly. If you want the best way of learning then you must use Go Math Grade 5 Chapter 2 Divide Whole Numbers Answer Key. Lesson 1: Place the First Digit Lesson 2: Divide by 1-Digit Divisors Lesson 3: Investigate â€¢ Division with 2-Digit Divisors Lesson 4: Partial Quotients Mid-Chapter Checkpoint Lesson 5: Estimate with 2-Digit Divisors Lesson 6: Divide by 2-Digit Divisors Lesson 7: Interpret the Remainder Lesson 9: Problem Solving â€¢ Division Review/Test ### Place the First Digit – Share and Show – Page No. 63 Divide. Question 1. 3)$$\overline { 579 }$$ _____ 193 Explanation: Divide integers 57/3 = 19 Multiply 19 x 3 = 57; Subtract 57 – 57 = 0 Write down 9 and divide integers 9/3 = 3. Multiply 3 x 3 = 9. Subtract 9 – 9 = 0. The remainder is 0. Question 2. 5)$$\overline { 1,035 }$$ _____ 207 Explanation: Divide integers 10/5 = 2 Multiply 2 x 5 = 10; Subtract 10 – 10 = 0 Write down 35 and divide integers 35/5 = 7. Multiply 7 x 5 = 35. Subtract 35 – 35 = 0. The remainder is 0. Go Math Book 5th Grade Place The First Digit Lesson 2.1 Question 3. 8)$$\overline { 1,766 }$$ _____ R _____ 220 R 6 Explanation: Divide integers 17/8 = 2 Multiply 2 x 8 = 16; Subtract 17 – 16 = 1 Write down 16 and divide integers 16/8 = 2. Multiply 2 x 8 = 16. Subtract 16 – 16 = 0. Write down 6; 6 < 8. There are not enough tens So, the remainder is 6 Divide. Question 4. 8)$$\overline { 275 }$$ _____ R _____ 43 R 3 Explanation: Divide integers 27/8 = 3 Multiply 8 x 3 = 24; Subtract 27 – 24= 3 Write down 3 and divide integers 35/8 = 4. Multiply 8 x 4 = 32. Subtract 35 – 32 = 3. The remainder is 3. Question 5. 3)$$\overline { 468 }$$ _____ 155 R 3 Explanation: Divide integers 46/3 = 15 Multiply 3 x 15 = 45; Subtract 46 – 45= 1 Write down 18 and divide integers 18/3 = 5. Multiply 3 x 5 = 15. Subtract 18 – 15 = 3. The remainder is 3. Question 6. 4)$$\overline { 3,220 }$$ _____ 805 Explanation: Divide integers 32/4 = 8 Multiply 4 x 8 = 32; Subtract 32 – 32 = 0 Write down 20 and divide integers 20/4 = 5. Multiply 4 x 5 = 20. Subtract 20 – 20= 0. The remainder is 0. Question 7. 6)$$\overline { 618 }$$ _____ 103 Explanation: Divide integers 61/6 = 10 Multiply 6 x 10 = 60; Subtract 61 – 60 = 1 Write down 18 and divide integers 18/6 = 3. Multiply 6 x 3 = 18. Subtract 18 – 18 = 0. The remainder is 0. Question 8. 4)$$\overline { 716 }$$ _____ 179 Explanation: Divide integers 71/4 = 17 Multiply 4 x 17 = 68; Subtract 71 – 68 = 3 Write down 36 and divide integers 36/4 = 9. Multiply 4 x 9 = 36. Subtract 36 – 36 = 0. The remainder is 0. Question 9. 9)$$\overline { 1,157 }$$ _____ R _____ 128 R 5 Explanation: Divide integers 11/9 = 1 Multiply 9 x 1 = 9; Subtract 11 – 9 = 2 Write down 25 and divide integers 25/9 = 2. Multiply 9 x 2 = 18. Subtract 25 – 18 = 7. Write down 77 and divide integers 77/9 = 8. Multiply 9 x 8 = 72. Subtract 77 – 72= 5. The remainder is 5. Question 10. 6)$$\overline { 6,827 }$$ _____ R _____ 1,137 R 5 Explanation: Divide integers 6/6 = 1 Multiply 6 x 1 = 6; Subtract 6 – 6 = 0 Write down 82 and divide integers 82/6 = 13. Multiply 6 x 13 = 78. Subtract 82 – 78 = 4. Write down 47 and divide integers 47/6 = 7. Multiply 6 x 7 = 42. Subtract 47 – 42= 5. The remainder is 5. 7)$$\overline { 8,523 }$$ _____ R _____ 1,217 R 4 Explanation: Divide integers 8/7 = 1 Multiply 7 x 1 = 7; Subtract 8 – 7 = 1 Write down 15 and divide integers 15/7 = 2. Multiply 7 x 2 = 14. Subtract 15 – 14 = 1. Write down 12 and divide integers 12/7 = 1. Multiply 7 x 1 = 7. Subtract 12 – 7= 5. Write down 53 and divide integers 53/7 = 7. Multiply 7 x 7 = 49. Subtract 53 – 49= 4. The remainder is 4. Practice: Copy and Solve Divide. Question 12. 645 Ã· 8 = _____ R _____ 645 Ã· 8 = 80 R 5 Explanation: Divide integers 64/8 = 8 Multiply 8 x 8 = 64; Subtract 64 – 64 = 0 Write down 05; 5 < 8; There are not enough tens The remainder is 5. Question 13. 942 Ã· 6 = _____ 157 Explanation: Divide integers 9/6 = 1 Multiply 6 x 1 = 6; Subtract 9 – 6 = 3 Write down 34 and divide integers 34/6 = 5. Multiply 6 x 5 = 30. Subtract 34 – 30 = 4. Write down 42 and divide integers 42/6 = 7. Multiply 6 x 7 = 42. Subtract 42 – 42 = 0. The remainder is 0. Question 14. 723 Ã· 7 = _____ R _____ 103 R 2 Explanation: Divide integers 7/7 = 1 Multiply 7 x 1 = 7; Subtract 7 – 7 = 0 Write down 23 and divide integers 23/7 = 3. Multiply 7 x 3 = 21. Subtract 23 – 21 = 2. The remainder is 2. Question 15. 3,478 Ã· 9 = _____ R _____ 386 R 4 Explanation: Divide integers 34/9 = 3 Multiply 9 x 3 = 27; Subtract 34 – 27 = 7 Write down 77 and divide integers 77/9 = 8. Multiply 9 x 8 = 72. Subtract 77 – 72 = 5. Write down 58 and divide integers 58/9 = 6. Multiply 9 x 6 = 54. Subtract 58 – 54= 4. The remainder is 4. Question 16. 3,214 Ã· 5 = _____ R _____ 642 R 4 Explanation: Divide integers 32/5 = 6 Multiply 5 x 6 = 30; Subtract 32 – 30 = 2 Write down 21 and divide integers 21/5 = 4. Multiply 5 x 4 = 20. Subtract 21 – 20 = 1. Write down 14 and divide integers 14/5 = 2. Multiply 5 x 2 = 10. Subtract 14 – 10 = 4. The remainder is 4. Question 17. 492 Ã· 4 = _____ 123 Explanation: Divide integers 4/4 = 1 Multiply 4 x 1 = 4; Subtract 4 – 4 = 0 Write down 9 and divide integers 9/4 = 2. Multiply 4 x 2 = 8. Subtract 9 – 8 = 1. Write down 12 and divide integers 12/4 = 3. Multiply 4 x 3 = 12. Subtract 12 – 12 = 0. The remainder is 0. Question 18. 2,403 Ã· 9 = _____ 267 Explanation: Divide integers 24/9 = 2 Multiply 9 x 2 = 18; Subtract 24 – 18 = 6 Write down 60 and divide integers 60/9 = 6. Multiply 9 x 6 = 54. Subtract 60 – 54 = 6. Write down 63 and divide integers 63/9 = 7. Multiply 9 x 7 = 63. Subtract 63 – 63 = 0. The remainder is 0. Question 19. 2,205 Ã· 6 = _____ R _____ 367 R 3 Explanation: Divide integers 22/6 = 3 Multiply 6 x 3 = 18; Subtract 22 – 18 = 4 Write down 40 and divide integers 40/6 = 6. Multiply 6 x 6 = 36; Subtract 40 – 36 = 4 Write down 45 and divide integers 45/6 = 7. Multiply 6 x 7 = 42; Subtract 45 – 42 = 3 The remainder is 3. Question 20. 2,426 Ã· 3 = _____ R _____ 808 R 2 Explanation: Divide integers 24/3 = 8 Multiply 3 x 8 = 24; Subtract 24 – 24 = 0 Write down 26 and divide integers 26/3 = 8. Multiply 3 x 8 = 24. Subtract 26 – 24 = 2. The remainder is 2. Question 21. 1,592 Ã· 8 = _____ R _____ 199 Explanation: Divide integers 15/8 = 1 Multiply 8 x 1 = 8; Subtract 15 – 8 = 7 Write down 79 and divide integers 79/8 = 9. Multiply 8 x 9 = 72; Subtract 79 – 72 = 7 Write down 72 and divide integers 72/8 = 9. Multiply 8 x 9 = 72; Subtract 72 – 72 = 0 The remainder is 0. Question 22. 926 Ã· 4 = _____ R _____ 231 R 2 Explanation: Divide integers 9/4 = 2 Multiply 4 x 2 = 8; Subtract 9 – 8 = 1 Write down 12 and divide integers 12/4 = 3. Multiply 4 x 3 = 12; Subtract 12 – 12 = 0 Write down 6 and divide integers 6/4 = 1. Multiply 4 x 1 = 4; Subtract 6 – 4 = 2 The remainder is 2. Question 23. 6,033 Ã· 5 = _____ R _____ 1,206 R 3 Explanation: Divide integers 6/5 = 1 Multiply 5 x 1 = 5; Subtract 6 – 5 = 1 Write down 10 and divide integers 10/5 = 2. Multiply 5 x 2 = 10; Subtract 10 – 10 = 0 Write down 33 and divide integers 33/5 = 6. Multiply 5 x 6 = 30; Subtract 33 – 30 = 3 The remainder is 3. ### Place the First Digit – UNLOCK the Problem – Page No. 64 Question 24. Rosa has a garden divided into sections. She has 125 daisy plants. If she plants an equal number of daisy plants in each section of daisies, will she have any left over? If so, how many daisy plants will be left over? a. What information will you use to solve the problem? Type below: __________ We can use the fact that she has 125 daisy plants and she plants an equal number of the daisy plants in each of 3 sections. Question 24. b. How will you use division to find the number of daisy plants left over? Type below: __________ We have to do 125/3 Divide integers 12/3 = 4 Multiply 3 x 4 = 12; Subtract 12 – 12 = 0 Write down 5 and divide integers 5/3 = 1. Multiply 3 x 1 = 3; Subtract 5 – 3 = 2 The remainder is 2. 41 daisy plants in each section. 2 daisy plants left over Question 24. c. Show the steps you use to solve the problem. Estimate: 120 Ã· 3 = _____ Type below: __________ Divide integers 12/3 = 4 Multiply 3 x 4 = 12; Subtract 12 – 12 = 0 The remainder is 0. Question 24. d. Complete the sentences: Rosa has _____ daisy plants. She puts an equal number in each of _____ sections. Each section has _____ plants. Rosa has _____ daisy plants left over. Type below: __________ Rose has 125 daisy planes. She puts an equal number in each of 3 sections. Each section has 41 plants. Rosa has 2 daisy plants left over. Question 25. One case can hold 3 boxes. Each box can hold 3 binders. How many cases are needed to hold 126 binders? _____ cases 14 cases Explanation: One case can hold 3 boxes. Each box can hold 3 binders. 3 x 3 = 9. For 12 binders, 126/ (3 x 3) = 126/9 = 14 Question 26. Test Prep In which place is the first digit in the quotient 1,497 Ã· 5? Options: a. thousands b. hundreds c. tens d. ones b. hundreds Explanation: 1,497 Ã· 5 = 499. The first digit 4 is in hundreds place. ### Divide by 1-Digit Divisors – Share and Show – Page No. 67 Question 1. 8)$$\overline { 624 }$$ Check _____ 78 Explanation: Divide integers 62/8 = 7 Multiply 8 x 7 = 56; Subtract 62 – 56 = 6 Write down 64 and divide integers 64/8 = 8. Multiply 8 x 8 = 64. Subtract 64 – 64 = 0. The remainder is 0. Check: 78 x 8 = 624; 624 = 624 Question 2. 4)$$\overline { 3,220 }$$ Check _____ 805 Explanation: Divide integers 32/4 = 8 Multiply 4 x 8 = 32; Subtract 32 – 32 = 0 Write down 20 and divide integers 20/4 = 5. Multiply 4 x 5 = 20. Subtract 20 – 20 = 0. The remainder is 0. Check: 805 x 4 = 3,220; 3,220 = 3,220. Question 3. 4)$$\overline { 1,027 }$$ Check _____ R _____ 256 R 3 Explanation: Divide integers 10/4 = 2 Multiply 4 x 2 = 8; Subtract 10 – 8 = 2 Write down 22 and divide integers 22/4 = 5. Multiply 4 x 5 = 20. Subtract 22 – 20= 2. Write down 27 and divide integers 27/4 = 6. Multiply 4 x 6 = 24. Subtract 27 – 24 = 3. The remainder is 3. So, 256 R 3. Check: 256 x 4 = 1,024; 1,024 + 3 = 1,027. 1,027 = 1,027 Divide. Question 4. 6)$$\overline { 938 }$$ _____ R _____ 156 R 2 Explanation: Divide integers 9/6 = 1 Multiply 6 x 1 = 6; Subtract 9 – 6 = 3 Write down 33 and divide integers 33/6 = 5. Multiply 6 x 5 = 30. Subtract 33 – 30 = 3. Write down 38 and divide integers 38/6 = 6. Multiply 6 x 6 = 36. Subtract 38 – 36 = 2. The remainder is 2. So, 156 R 2. Question 5. 4)$$\overline { 762 }$$ _____ R _____ 190 R 2 Explanation: Divide integers 7/4 = 1 Multiply 4 x 1 = 4; Subtract 7 – 4 = 3 Write down 36 and divide integers 36/4 = 9. Multiply 4 x 9 = 36. Subtract 36 – 36 = 0. Write down 2. 2 < 4; There are not enough tens The remainder is 2. So, 190 R 2. Question 6. 3)$$\overline { 5,654 }$$ _____ R _____ 1884 R 2 Explanation: Divide integers 5/3 = 1 Multiply 3 x 1 = 3; Subtract 5 – 3 = 2 Write down 26 and divide integers 26/3 = 8. Multiply 3 x 8 = 24. Subtract 26 – 24 = 2. Write down 25 and divide integers 25/3 = 8. Multiply 3 x 8 = 24. Subtract 25 – 24 = 1. Write down 14 and divide integers 14/3 = 4. Multiply 3 x 4 = 12. Subtract 14 – 12 = 2. The remainder is 2. So, 1884 R 2. Question 7. 8)$$\overline { 475 }$$ _____ R _____ 59 R 3 Explanation: Divide integers 47/8 = 5 Multiply 8 x 5 = 40; Subtract 47 – 40 = 7 Write down 75 and divide integers 75/8 = 9. Multiply 9 x 8 = 72. Subtract 75 – 72 = 3. The remainder is 3. So, 59 R 3. Practice: Copy and Solve Divide. Question 8. 4)$$\overline { 671 }$$ _____ R _____ 167 R 3 Explanation: Divide integers 6/4 = 1 Multiply 4 x 1 = 4; Subtract 6 – 4 = 2 Write down 27 and divide integers 27/4 = 6. Multiply 4 x 6 = 24. Subtract 27 – 24 = 3. Write down 31 and divide integers 31/4 = 7. Multiply 4 x 7 = 28. Subtract 31 – 28 = 3. The remainder is 3. So, 167 R 3. Question 9. 9)$$\overline { 2,023 }$$ _____ R _____ 224 R 7 Explanation: Divide integers 20/9 = 2 Multiply 9 x 2 = 18; Subtract 20 – 18 = 2 Write down 22 and divide integers 22/9 = 2. Multiply 9 x 2 = 18. Subtract 22 – 18 = 4. Write down 43 and divide integers 43/9 = 4. Multiply 9 x 4 = 36. Subtract 43 – 36 = 7. The remainder is 7. So, 224 R 7. Question 10. 3)$$\overline { 4,685 }$$ _____ R _____ 1,561 R 2 Explanation: Divide integers 4/3 = 1 Multiply 3 x 1 = 3; Subtract 4 – 3 = 1 Write down 16 and divide integers 16/3 = 5. Multiply 3 x 5 = 15. Subtract 16 – 15 = 1. Write down 18 and divide integers 18/3 = 6. Multiply 3 x 6 = 18. Subtract 18 – 18 = 0. Write down 5 and divide integers 5/3 = 1. Multiply 3 x 1 = 3. Subtract 5 – 3 = 2. The remainder is 2. So, 1,561 R 2. Question 11. 8)$$\overline { 948 }$$ _____ R _____ 118 R 4 Explanation: Divide integers 9/8 = 1 Multiply 8 x 1 = 8; Subtract 9 – 8 = 1 Write down 14 and divide integers 14/8 = 1. Multiply 8 x 1 = 8. Subtract 14 – 8 = 6. Write down 68 and divide integers 68/8 = 8. Multiply 8 x 8 = 64. Subtract 68 – 64 = 4. The remainder is 4. So, 118 R 4. Question 12. 1,326 Ã· 4 = _____ R _____ 331 R 2 Explanation: Divide integers 13/4 = 3 Multiply 4 x 3 = 12; Subtract 13 – 12 = 1 Write down 12 and divide integers 12/4 = 3. Multiply 4 x 3 = 12. Subtract 12 – 12 = 0. Write down 6 and divide integers 6/4 = 1. Multiply 4 x 1 = 4. Subtract 6 – 4 = 2. The remainder is 2. So, 331 R 2. Question 13. 5,868 Ã· 6 = _____ 978 Explanation: Divide integers 58/6 = 9 Multiply 6 x 9 = 54; Subtract 58 – 54 = 4 Write down 46 and divide integers 46/6 = 7. Multiply 6 x 7 = 42. Subtract 46 – 42 = 4. Write down 48 and divide integers 48/6 = 8. Multiply 6 x 8 = 48. Subtract 48 – 48 = 0. The remainder is 0. So, 978. Question 14. 566 Ã· 3 = _____ R _____ 188 R 2 Explanation: Divide integers 5/3 = 1 Multiply 3 x 1 = 3; Subtract 5 – 3 = 2 Write down 26 and divide integers 26/3 = 8. Multiply 3 x 8 = 24. Subtract 26 – 24 = 2. Write down 26 and divide integers 26/3 = 8. Multiply 3 x 8 = 24. Subtract 26 – 24 = 2. The remainder is 2. So, 188 R 2. Question 15. 3,283 Ã· 9 = _____ R _____ 364 R 7 Explanation: Divide integers 32/9 = 3 Multiply 9 x 3 = 27; Subtract 32 – 27 = 5 Write down 58 and divide integers 58/9 = 6. Multiply 9 x 6 = 54. Subtract 58 – 54 = 4. Write down 43 and divide integers 43/9 = 4. Multiply 9 x 4 = 36. Subtract 43 – 36 = 7. The remainder is 7. So, 364 R 7. Algebra Find the value of n in each equation. Write what n represents in the related division problem. Question 16. n = 4 Ã— 58 Value of n = _______ Represents: _______ Value of n = 232 Represents: dividend Explanation: n = 4 Ã— 58; 232 = 4 x 58; n is the dividend Question 17. 589 = 7 Ã— 84 + n Value of n = _______ Represents: _______ Value of n = 1 Represents: remainder Explanation: 589 = 7 Ã— 84 + n 589 = 588 + n; 589 – 588 = n; 1 = n n is the remainder Go Math Grade 5 Chapter 2 Assessment Question 18. n = 5 Ã— 67 + 3 Value of n = _______ Represents: _______ Value of n = 338 Represents: dividend Explanation: n = 5 Ã— 67 + 3 n = 335 + 3 n = 338 n is the dividend ### Divide by 1-Digit Divisors – Problem Solving – Page No. 68 Use the table to solve 19â€“20. Question 19. If the Welcome gold nugget were turned into 3 equal-sized gold bricks, how many troy ounces would each brick weigh? _____ troy ounces 739 troy ounces Explanation: Welcome gold nugget = 2,217 troy ounces. If it turned into 3 equal-sized gold bricks, 2,217/3 = 739. 739 troy ounces Question 20. Pose a Problem Look back at Problem 19. Write a similar problem by changing the nugget and the number of bricks. Then solve the problem. Type below: __________ 571 troy ounces Explanation: If Welcome Stranger nugget were turned into 4 equal-sized gold bricks, how many troy ounces would each brick weigh? Welcome Stranger nugget = 2,284. If it turned into 4 equal-sized gold bricks, 2,217/3 = 571. 571 troy ounces Question 21. There are 246 students going on a field trip to pan for gold. If they are going in vans that hold 9 students each, how many vans are needed? How many students will ride in the van that isnâ€™t full? The number of vans: _________ _________ students in the van that isnâ€™t full The number of vans: 27 3 students will ride in the van that isnâ€™t full Explanation: There are 246 students going on a field trip to pan for gold. If they are going in vans that hold 9 students each, 246/9 = 27 R 3 The number of vans: 27 3 students will ride in the van that isnâ€™t full Question 22. One crate can hold 8 cases of trading cards. How many crates are needed to hold 128 cases of trading cards? _____ crates 16 crates Explanation: One crate can hold 8 cases of trading cards. To hold 128 cases of trading cards, 128/8 = 16 crates needed. Question 23. Test Prep At a bake sale, a fifth-grade class sold 324 cupcakes in packages of 6. How many packages of cupcakes did the class sell? Options: a. 1,944 b. 108 c. 64 d. 54 d. 54 Explanation: At a bake sale, a fifth-grade class sold 324 cupcakes in packages of 6. 324/6 = 54 ### Division with 2-Digit Divisors – Share and Show – Page No. 71 Use the quick picture to divide. Question 1. 143 Ã· 13 = _____ 143 Ã· 13 = 11 Explanation: 143 = 100 + 40 + 3 Model the first partial quotient by making a rectangle with the hundred and 3 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10) = 100 + 30 = 130. The rectangle shows 10 groups of 13. Model the second partial quotient by making a line with the ten and 3 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1) = 10 + 3 = 13 130 + 13 = 143; So, the answer is 10 + 1 = 11 Divide. Use base-ten blocks. Question 2. 168 Ã· 12 = _____ 168 Ã· 12 = 14 Explanation: 168 Ã· 12 Model the first partial quotient by making a rectangle with the hundred and 2 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10) = 100 + 20 = 120. The rectangle shows 10 groups of 12. Model the second partial quotient by making a line with the ten and 2 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1) = 10 + 2 = 12. Repeat the above step more three times to get 120 + 12 + 12 + 12 + 12 = 168; So, the answer is 10 + 1 + 1 + 1 + 1 = 14 Question 3. 154 Ã· 14 = _____ 154 Ã· 14 = 11 Explanation: 154 Ã· 14 Model the first partial quotient by making a rectangle with the hundred and 4 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10 + 10) = 100 + 40 = 140. The rectangle shows 10 groups of 14. Model the second partial quotient by making a line with the ten and 4 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1 + 1) = 10 + 4 = 14. Repeat the above step more three times to get 140 + 14 = 154; So, the answer is 10 + 1 = 11 Question 4. 187 Ã· 11 = _____ 187 Ã· 11 = 17 Explanation: 187 Ã· 11 = Model the first partial quotient by making a rectangle with the hundred and 1 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10) = 100 + 10 = 110. The rectangle shows 10 groups of 11. Model the second partial quotient by making a line with the ten and 1 ones. In the Record section, cross out the ten and ones you use. 10 + (1) = 10 + 1 = 11. Repeat the above step more six times to get 110 + 11 + 11 + 11 + 11 + 11 + 11 + 11 = 187; So, the answer is 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 17 Divide. Draw a quick picture. Question 5. 165 Ã· 11 = _____ 165 Ã· 11 = 15 Explanation: 165 Ã· 11 Model the first partial quotient by making a rectangle with the hundred and 1 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10) = 100 + 10 = 110. The rectangle shows 10 groups of 11. Model the second partial quotient by making a line with the ten and 1 ones. In the Record section, cross out the ten and ones you use. 10 + (1) = 10 + 1 = 11. Repeat the above step more four times to get 110 + 11 + 11 + 11 + 11 + 11 = 165; So, the answer is 10 + 1 + 1 + 1 + 1 + 1 = 15 Question 6. 216 Ã· 18 = _____ 216 Ã· 18 = 12 Explanation: 216 Ã· 18 Model the first partial quotient by making a rectangle with the hundred and 8 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10 + 10 + 10 + 10 + 10 + 10) = 100 + 80 = 180. The rectangle shows 10 groups of 18. Model the second partial quotient by making a line with the ten and 8 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) = 10 + 8 = 18. Repeat the above step to get 180 + 18 + 18Â = 216; So, the answer is 10 + 1 + 1 = 12 196 Ã· 14 = _____ 196 Ã· 14 = 14 Explanation: 196 Ã· 14 Model the first partial quotient by making a rectangle with the hundred and 4 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10 + 10) = 100 + 40 = 140. The rectangle shows 10 groups of 14. Model the second partial quotient by making a line with the ten and 4 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1 + 1) = 10 + 4 = 14. Repeat the above step more three times to get 140 + 14 + 14 + 14 + 14Â = 196; So, the answer is 10 + 1 + 1 + 1 + 1 = 14 Question 8. 195 Ã· 15 = _____ 195 Ã· 15 = 13 Explanation: 195 Ã· 15 Model the first partial quotient by making a rectangle with the hundred and 5 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10 + 10 + 10) = 100 + 50 = 150. The rectangle shows 10 groups of 15. Model the second partial quotient by making a line with the ten and 5 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1 + 1 + 1) = 10 + 5 = 15. Repeat the above step more three times to get 150 + 15 + 15 + 15Â = 195; So, the answer is 10 + 1 + 1 + 1 = 13 Question 9. 182 Ã· 13 = _____ 182 Ã· 13 = 14 Explanation: 182 Ã· 13 Model the first partial quotient by making a rectangle with the hundred and 3 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10 + 10 ) = 100 + 30 = 130. The rectangle shows 10 groups of 13. Model the second partial quotient by making a line with the ten and 3 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1 + 1) = 10 + 3 = 13. Repeat the above step more four times to get 130 + 13 + 13 + 13 + 13 = 182; So, the answer is 10 + 1 + 1 + 1 + 1 = 14 Question 10. 228 Ã· 12 = _____ 228 Ã· 12 = 19 Explanation: 228 Ã· 12 Model the first partial quotient by making a rectangle with the hundred and 2 tens. In the Record section, cross out the hundred and tens you use. (10 x 10) + (10 + 10) = 100 + 20 = 120. The rectangle shows 10 groups of 12. Model the second partial quotient by making a line with the ten and 2 ones. In the Record section, cross out the ten and ones you use. 10 + (1 + 1) = 10 + 2 = 12. Repeat the above step more eight times to get 120 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 = 228; So, the answer is 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 19 ### Division with 2-Digit Divisors – Connect to Social Studies – Page No. 72 Pony Express The Pony Express used men riding horses to deliver mail between St. Joseph, Missouri, and Sacramento, California, from April, 1860 to October, 1861. The trail between the cities was approximately 2,000 miles long. The first trip from St. Joseph to Sacramento took 9 days 23 hours. The first trip from Sacramento to St. Joseph took 11 days 12 hours. Solve. Question 11. Suppose two Pony Express riders rode a total of 165 miles. If they replaced each horse with a fresh horse every 11 miles, how many horses would they have used? _____ horses 16 horses Explanation: Suppose two Pony Express riders rode a total of 165 miles. If they replaced each horse with a fresh horse every 11 miles. Then, 16 horses used. Question 12. Suppose a Pony Express rider was paid $192 for 12 weeks of work. If he was paid the same amount each week, how much was he paid for each week of work?$ _____ $16 Explanation: Suppose a Pony Express rider was paid$192 for 12 weeks of work. For each week. $192/12 =$16. Question 13. Suppose three riders rode a total of 240 miles. If they used a total of 16 horses, and rode each horse the same number of miles, how many miles did they ride before replacing each horse? _____ miles 15 miles Explanation: Assuming each horse was only ridden once then a total of 16 horses were ridden for a total of 240 miles 240 miles/16 horses = 15 miles/horse if each horse was ridden more than once before being replaced the distance between replacements could be reduced. The fact that there were 3 riders is irrelevant. Question 14. Suppose it took 19 riders a total of 11 days 21 hours to ride from St. Joseph to Sacramento. If they all rode the same number of hours, how many hours did each rider ride? _____ hours 15 hours Explanation: Suppose it took 19 riders a total of 11 days 21 hours to ride from St. Joseph to Sacramento. (11 x 24 + 21)/19 = (264 + 21)/19 = 285/19 = 15 hours. ### Partial Quotients – Share and Show – Page No. 75 Divide. Use partial quotients. Question 1. 18)$$\overline { 648 }$$ _____ 36 Explanation: Multiply 18 x 10 = 180; Subtract: 648 – 180 = 468. partial quotient = 10 Multiply 18 x 10 = 180; Subtract: 468 – 180= 288. partial quotient = 10 Multiply 18 x 10 = 180; Subtract: 288- 180= 108. partial quotient = 10 Multiply 18 x 6 = 108; Subtract: 108 – 108 = 0. partial quotient = 6; The remainder is 0; Add the partial quotient to find the whole number quotient; 10 + 10 + 10 + 6 = 36 R 0 Question 2. 62)$$\overline { 3,186 }$$ _____ R _____ Explanation: Multiply 62 x 10 = 620; Subtract: 3,186 – 620 = 2,566. partial quotient = 10 Multiply 62 x 10 = 620; Subtract: 2,566 – 620 = 1,946. partial quotient = 10 Multiply 62 x 10 = 620; Subtract: 1,946 – 620 = 1,326. partial quotient = 10 Multiply 62 x 10 = 620; Subtract: 1,326 – 620 = 706. partial quotient = 10 Multiply 62 x 10 = 620; Subtract: 706 – 620 = 86. partial quotient = 10 Multiply 62 x 1 = 62; Subtract: 86 – 62 = 24. partial quotient = 1 The remainder is 24; Add the partial quotient to find the whole number quotient; 10 + 10 + 10Â + 10 + 1 = 51 R 24 Question 3. 858 Ã· 57 _____ R _____ Explanation: Multiply 57 x 10 = 570; Subtract: 858 – 570 = 288. partial quotient = 10 Multiply 57 x 5 = 285; Subtract: 288 – 285 = 3. partial quotient = 5 The remainder is 3; Add partial quotient to find the wholenumber quotient; 10 + 5 = 15 R 3 Divide. Use partial quotients. Question 4. 73)$$\overline { 584 }$$ _____ 8 Explanation: Multiply 73 x 8 = 584; Subtract: 584 – 584 = 0. partial quotient = 8 The remainder is 0; Question 5. 51)$$\overline { 1,831 }$$ _____ R _____ 35 R 46 Explanation: Multiply 51 x 10 = 510; Subtract: 1,831 – 510 = 1,321. partial quotient = 10 Multiply 51 x 10 = 510; Subtract: 1,321 – 510 = 811. partial quotient = 10 Multiply 51 x 10 = 510; Subtract: 811 – 510 = 301. partial quotient = 10 Multiply 51 x 5 = 255; Subtract: 301 – 255 = 46. partial quotient = 5 The remainder is 46; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 5 = 35 R 46 Question 6. 82)$$\overline { 2,964 }$$ _____ R _____ 36 R 12 Explanation: Multiply 82 x 10 = 820; Subtract: 2,964 – 820 = 2,144. partial quotient = 10 Multiply 82 x 10 = 820; Subtract: 2,144 – 820 = 1,324. partial quotient = 10 Multiply 82 x 10 = 820; Subtract: 1,324 – 820 = 504. partial quotient = 10 Multiply 82 x 6 = 492; Subtract: 504 – 492= 12. partial quotient = 6 The remainder is 12; Add partial quotient to find the whole number quotient; 10 + 10 + 10 + 6 = 36 R 12 Question 7. 892 Ã· 26 _____ R _____ 34 R 8 Explanation: Multiply 26 x 10 = 260; Subtract: 892 – 260 = 632. partial quotient = 10 Multiply 26 x 10 = 260; Subtract: 632 – 260 = 372. partial quotient = 10 Multiply 26 x 10 = 260; Subtract: 372 – 260 = 112. partial quotient = 10 Multiply 26 x 4 = 104; Subtract: 112 – 104 = 8. partial quotient = 4 The remainder is 8; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 4 = 34 R 8 Question 8. 1,056 Ã· 48 _____ 22 Explanation: Multiply 48 x 10 = 480; Subtract: 1,056 – 480 = 576. partial quotient = 10 Multiply 48 x 10 = 480; Subtract: 576 – 480 = 96. partial quotient = 10 Multiply 48 x 2 = 96; Subtract: 96 – 96 = 0. partial quotient = 2 The remainder is 0; Add partial quotient to find the wholenumber quotient; 10 + 10 + 2 = 22 Question 9. 2,950 Ã· 67 _____ R _____ 44 R 2 Explanation: Multiply 67 x 10 = 670; Subtract: 2,950 – 670 = 2,280. partial quotient = 10 Multiply 67 x 10 = 670; Subtract: 2,280 – 670 = 1,610. partial quotient = 10 Multiply 67 x 10 = 670; Subtract: 1,610 – 670 = 940. partial quotient = 10 Multiply 67 x 10 = 670; Subtract: 940 – 670 = 270. partial quotient = 10 Multiply 67 x 4= 268; Subtract: 270 – 268 = 2. partial quotient = 4 The remainder is 2; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 10 + 4 = 44 R 2 Practice: Copy and Solve Divide. Use partial quotients. Question 10. 653 Ã· 42 _____ R _____ 15 R 23 Explanation: Multiply 42 x 10 = 420; Subtract: 653 – 420 = 233. partial quotient = 10 Multiply 42 x 5 = 210; Subtract: 233 – 210 = 23. partial quotient = 5 The remainder is 23; Add partial quotient to find the wholenumber quotient; 10 + 5 = 15 R 23 Question 11. 946 Ã· 78 _____ R _____ 12 R 10 Explanation: Multiply 78 x 10 = 780; Subtract: 946 – 780 = 166. partial quotient = 10 Multiply 78 x 2 = 156; Subtract: 166 – 156 = 10. partial quotient = 2 The remainder is 10; Add partial quotient to find the wholenumber quotient; 10 + 2 = 12 R 10 412 Ã· 18 _____ R _____ 22 R 16 Explanation: Multiply 18 x 10 = 180; Subtract: 412 – 180 = 232. partial quotient = 10 Multiply 18 x 10 = 180; Subtract: 232 – 180 = 52. partial quotient = 10 Multiply 18 x 2 = 36; Subtract: 52 – 36 = 16. partial quotient = 2 The remainder is 16; Add partial quotient to find the wholenumber quotient; 10 + 10 + 2 = 22 R 16 Question 13. 871 Ã· 87 _____ R _____ 10 R 1 Explanation: Multiply 87 x 10 = 870; Subtract: 871 – 870 = 1. partial quotient = 10 The remainder is 1; 10 R 1 Question 14. 1,544 Ã· 34 _____ R _____ 45 R 14 Explanation: Multiply 34 x 10 = 340; Subtract: 1,544 – 340 = 1,204. partial quotient = 10 Multiply 34 x 10 = 340; Subtract: 1,204 – 340 = 864. partial quotient = 10 Multiply 34 x 10 = 340; Subtract: 864 – 340 = 524. partial quotient = 10 Multiply 34 x 10 = 340; Subtract: 524 – 340 = 184. partial quotient = 10 Multiply 34 x 5 = 170; Subtract: 184 – 170 = 14. partial quotient = 5 The remainder is 14; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 10 + 5 = 45 R 14 Question 15. 2,548 Ã· 52 _____ R _____ 47 R 14 Explanation: Multiply 52 x 10 = 520; Subtract: 2,548 – 520 = 2028. partial quotient = 10 Multiply 52 x 10 = 520; Subtract: 2028- 520 = 1508. partial quotient = 10 Multiply 52 x 10 = 520; Subtract: 1508- 520 = 988. partial quotient = 10 Multiply 52 x 10 = 520; Subtract: 988 – 520 = 468. partial quotient = 10 Multiply 52 x 9 = 468; Subtract: 468 – 468= 0. partial quotient = 9 The remainder is 0; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 10 + 9 = 49 R 0 Question 16. 2,740 Ã· 83 _____ R _____ 33 R 1 Explanation: Multiply 83 x 10 = 830; Subtract: 2,740 – 830= 1910. partial quotient = 10 Multiply 83 x 10 = 830; Subtract: 1910 – 830= 1080. partial quotient = 10 Multiply 83 x 10 = 830; Subtract: 1080 – 830= 250. partial quotient = 10 Multiply 83 x 3 = 249; Subtract: 250 – 249 = 1. partial quotient = 3 The remainder is 1; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 3 = 33 R 1 Question 17. 4,135 Ã· 66 _____ R _____ 62 R 43 Explanation: Multiply 66 x 10 = 660; Subtract: 4,135 – 660 = 3475. partial quotient = 10 Multiply 66 x 10 = 660; Subtract: 3475 – 660 = 2815. partial quotient = 10 Multiply 66 x 10 = 660; Subtract: 2815 – 660 = 2155. partial quotient = 10 Multiply 66 x 10 = 660; Subtract: 2155 – 660 = 1495. partial quotient = 10 Multiply 66 x 10 = 660; Subtract: 1495 – 660 = 835. partial quotient = 10 Multiply 66 x 10 = 660; Subtract: 835 – 660 = 175. partial quotient = 10 Multiply 66 x 2 = 132; Subtract: 175 – 132 = 43. partial quotient = 2 The remainder is 43; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 10 + 10 + 10 + 2 = 62 R 43 ### Partial Quotients – Problem Solving – Page No. 76 Use the table to solve 18â€“20 and 22. Question 18. How many years would it take for a person in the United States to eat 855 pounds of apples? _____ years 45 years Explanation: Each year a person eats 19 pounds of apples. So, to eat 855 pounds of apples, it takes 855/19 = 45 years. Question 19. How many years would it take for a person in the United States to eat 1,120 pounds of turkey? _____ years 80 years Explanation: Each year a person eats 14 pounds of turkey. So, to eat 1,120 pounds of turkey, it takes 1,120/14 = 80 years. Question 20. If 6 people in the United States each eat the average amount of popcorn for 5 years, how many quarts of popcorn will they eat? _____ quarts 2,040 quarts Explanation: 1 person eats 68 quarts of popcorn each year. 6 people = 6 x 68 quarts of popcorn = 408 quarts of popcorn for each year. For 5 years, they will eat popcorn = 5 x 408 = 2,040 quarts Question 21. In a study, 9 people ate a total of 1,566 pounds of potatoes in 2 years. If each person ate the same amount each year, how many pounds of potatoes did each person eat in 1 year? _____ pounds 87 pounds Explanation: 9 people ate a total of 1,566 pounds of potatoes in 2 years. If each person ate the same amount each year, 1,566/2 = 783. To calculate how many pounds of potatoes did each person eat in 1 year, 783/9 = 87 pounds. Question 22. Sense or Nonsense? In the United States, a person eats more than 40,000 pounds of bread in a lifetime if he or she lives to be 80 years old. Does this statement make sense, or is it nonsense? Explain. __________ nonsense; 40,000 pounds / 80 years = 4,000 pounds / 8 years = 2,000 pounds (1 ton) / 4 years = 1,000 pounds / 2 years = 1,000 pounds / 2 years = 500 pounds per year = almost 1 and 1/2 pounds of bread every day of your life. Question 23. Test Prep The school auditorium has 448 seats arranged in 32 equal rows. How many seats are in each row? Options: a. 14,336 b. 480 c. 416 d. 14 d. 14 Explanation: The school auditorium has 448 seats arranged in 32 equal rows. Each row = 448/32 = 14 ### Mid-Chapter Checkpoint – Page No. 77 Concepts and Skills Question 1. Explain how estimating the quotient helps you place the first digit in the quotient of a division problem. Type below: __________ Let’s do 5980 divided by 347 Estimate: 6000/300 = 20 So, I now know my first digit will go into the 10’s place or 57890 divided by 34 that is 60,000 divided by 30 = 2000 my first digit goes into the thousands place. Question 2. Explain how to use multiplication to check the answer to a division problem. Type below: __________ Take 739/9 = 82 R 1. Check: 9 x 82 + 1 = 739. divisor x quotient + remainder = dividend. Divide. Question 3. 633 Ã· 3 = _____ 211 Explanation: Divide integers 6/3 = 2 Multiply 3 x 2 = 6; Subtract 6 â€“ 6 = 0 Write down 3 and divide integers 3/3 = 1. Multiply 3 x 1 = 3. Subtract 3 – 3 = 0. Write down 3 and divide integers 3/3 = 1. Multiply 3 x 1 = 3. Subtract 3 – 3 = 0. The remainder is 0. Question 4. 487 Ã· 8 = _____ R _____ 60 R 7 Explanation: Divide integers 48/8 = 6 Multiply 8 x 6 = 48; Subtract 48 â€“ 48 = 0 Write down 7;7 < 8. The remainder is 7. So, 60 R 7. Question 5. 1,641 Ã· 4 = _____ R _____ 410 R 1 Explanation: Divide integers 16/4 = 4 Multiply 4 x 4 = 16; Subtract 16 â€“ 16 = 0 Write down 4 and divide integers 4/4 = 1. Multiply 4 x 1 = 4; Subtract 4 â€“ 4 = 0 Write down 1; 1<4 The remainder is 1. So, 410 R 1. Question 6. 2,765 Ã· 9 = _____ R _____ 307 R 2 Explanation: Divide integers 27/9 = 3 Multiply 9 x 3 = 27; Subtract 27 â€“ 27 = 0 Write down 65 and divide integers 65/9 = 7. Multiply 9 x 7 = 63. Subtract 65 â€“ 63 = 2. The remainder is 2. So, 307 R 2. Divide. Use partial quotients. Question 7. 156 Ã· 13 = _____ 12 Explanation: Multiply 13 x 10 = 130; Subtract: 156 â€“ 130 = 26. partial quotient = 10 Multiply 13 x 2 = 26; Subtract: 26 â€“ 26 = 0. partial quotient = 2 The remainder is 0; Add the partial quotient to find the whole number quotient; 10 +2 = 12 R 0 Question 8. 318 Ã· 53 = _____ 6 Explanation: Multiply 53 x 6 = 318; Subtract: 318 â€“ 318= 0. partial quotient = 6 The remainder is 0; quotient = 6 Question 9. 1,562 Ã· 34 = _____ r _____ 45 R 32 Explanation: Multiply 34Â x 10 = 340; Subtract: 1,562 â€“ 340 = 1,222. partial quotient = 10 Multiply 34Â x 10 = 340; Subtract: 1,222 â€“ 340 = 882. partial quotient = 10 Multiply 34Â x 10 = 340; Subtract: 882 â€“ 340 = 542. partial quotient = 10 Multiply 34Â x 10 = 340; Subtract: 542 â€“ 340 = 202. partial quotient = 10 Multiply 34Â x 5 = 170; Subtract: 202 â€“ 170 = 32. partial quotient = 5 The remainder is 32; Add partial quotient to find the whole number quotient; 10 + 10 + 10 + 10 + 5 = 45 R 32 Question 10. 4,024 Ã· 68 = _____ r _____ 59 R 12 Explanation: Multiply 68 x 10 = 680; Subtract: 4,024 â€“ 680 = 3,344. partial quotient = 10 Multiply 68 x 10 = 680; Subtract: 3,344 â€“ 680= 2664. partial quotient = 10 Multiply 68 x 10 = 680; Subtract: 2664 â€“ 680 = 1984. partial quotient = 10 Multiply 68 x 10 = 680; Subtract: 1984 â€“ 680= 1304. partial quotient = 10 Multiply 68 x 10 = 680; Subtract: 1304 â€“ 680 = 624. partial quotient = 10 Multiply 68 x 9 = 612; Subtract: 624 â€“ 612 = 12. partial quotient = 9 The remainder is 12; Add partial quotient to find the wholenumber quotient; 10 + 10 + 10 + 10 + 10 + 9 = 59 R 12 ### Mid-Chapter Checkpoint – Page No. 78 Question 11. Emma is planning a party for 128 guests. If 8 guests can be seated at each table, how many tables will be needed for seating at the party? _____ tables 16 tables Explanation: Emma is planning a party for 128 guests. If 8 guests can be seated at each table 128/8 = 16. Question 12. Tickets for the basketball game cost $14 each. If the sale of the tickets brought in$2,212, how many tickets were sold? _____ tickets 158 tickets Explanation: Tickets for the basketball game cost $14 each. If the sale of the tickets brought in$2,212, $2,212/$14 = 158 Question 13. Margo used 864 beads to make necklaces for the art club. She made 24 necklaces with the beads. If each necklace has the same number of beads, how many beads did Margo use for each necklace? _____ Explanation: Margo used 864 beads to make necklaces for the art club. She made 24 necklaces with the beads. If each necklace has the same number of beads, 864/24 = 36 beads Go Math Grade 5 Chapter 2 Test Pdf Question 14. Angie needs to buy 156 candles for a party. Each package has 8 candles. How many packages should Angie buy? _____ packages 20 packages Explanation: Angie needs to buy 156 candles for a party. Each package has 8 candles. 156/8 = 20 Question 15. Max delivers 8,520 pieces of mail in one year. About how many pieces of mail does he deliver in 2 months? Explain your steps. _____ pieces 1,420 pieces Explanation: Max delivers 8,520 pieces of mail in one year. So, for 12 months, 8,520/12 = 710. To deliver in 2 months, 710 x 2 = 1,420 ### Share and Show – Page No. 81 Use compatible numbers to find two estimates. Question 1. 22)$$\overline { 154 }$$ 140 Ã· 20 = _____ 160 Ã· 20 = _____ Estimate: _____ ; _____ 140 Ã· 20 = 7 160 Ã· 20 = 8 Estimate: 7 ; 8 Explanation: Two sets of compatible numbers to find two different estimates are 140 Ã· 20 14 Ã· 2 = 7 140 Ã· 20 = 7 160 Ã· 20 16 Ã· 2 = 8 160 Ã· 20Â = 8 Question 2. 68)$$\overline { 503 }$$ Estimate: _____ ; _____ 476 Ã· 68= 7 544 Ã· 68 = 8 Estimate: 7 ; 8 Explanation: Multiples of 68: 68 136 204 272 340 408 476 544 Find multiples that are close to the dividend. Use either or both numbers to estimate the quotient. 476/68 = 7 544/68 = 8 The quotient is between 7 and 8. Question 3. 81)$$\overline { 7,052 }$$ Estimate: _____ ; _____ 6400 Ã· 80 = 80 7200 Ã· 80 = 90 Estimate: 80 ; 90 Explanation: 6400/80 64/8 = 8 640 / 80 = 8 6400/80 = 800 7200/80 72/8 = 9 720/80 =9 7200/80 = 90 Estimate: 80, 90 Question 4. 33)$$\overline { 291 }$$ Estimate: _____ ; _____ 240 Ã· 30= 8 270 Ã· 30 = 9 Estimate: 8 ; 9 Explanation: 240/30 24/3 = 8 240/30 = 8 270/30 27/3 = 9 270/30 = 9 Estimate: 8, 9 Question 5. 58)$$\overline { 2,365 }$$ Estimate: _____ ; 2400 Ã· 60= 40 3000 Ã· 60 = 50 Estimate: 40 ; 50 Explanation: 2400/60 24/6 = 4 240/60 = 4 2400/60 = 40 3000/60 30/6 = 5 300/60 = 5 3000/60 = 50 Estimate: 40, 50 Question 6. 19)$$\overline { 5,312 }$$ Estimate: _____ ; _____ 5300 Ã· 20= 7 5320 Ã· 20 = 8 Estimate: 265 ; 266 Explanation: 5300/20 5300/20 = 265 5320/20 5320/20 = 266 Use compatible numbers to find two estimates. Question 7. 42)$$\overline { 396 }$$ Estimate: _____ ; _____ 360 Ã· 40 = 9 400 Ã· 40 = 10 Estimate: 9 ; 10 Explanation: 360/40 = 9 400/40 = 10 Estimate: 9,10 Question 8. 59)$$\overline { 413 }$$ Estimate: _____ ; _____ 420 Ã· 60= 7 480 Ã· 60 = 8 Estimate: 7 ; 8 Explanation: 420/60 = 7 480/60 = 8 Question 9. 28)$$\overline { 232 }$$ Estimate: _____ ; _____ 240 Ã· 30 = 8 270Ã· 30 = 9 Estimate: 8 ; 9 Explanation: 240/30 = 8 270/30 = 9 Estimate: 8 ; 9 How To Divide 5th Grade Lesson 2.3 Question 10. 22)$$\overline { 368 }$$ Estimate: _____ ; _____ 320 Ã· 20= 16 340 Ã· 20 = 17 Estimate: 16 ; 17 Explanation: 320/20 = 16 340/20 = 17 Estimate: 16 ; 17 Question 11. 78)$$\overline { 375 }$$ Estimate: _____ ; _____ 320 Ã· 80 = 4 400 Ã· 80 = 5 Estimate: 4 ; 5 Explanation: 320/80 = 4 400/80 = 5 Estimate: 16 ; 17 Question 12. 88)$$\overline { 6,080 }$$ Estimate: _____ ; _____ 6210Ã· 90= 69 6300 Ã· 90 = 70 Estimate: 69 ; 70 Explanation: 6210/90 = 69 6300/90 = 70 Question 13. 5,821 Ã· 71 Estimate: _____ ; _____ 5180 Ã· 70 = 74 5250Ã· 70 = 75 Estimate: 74 ; 75 Explanation: 5180/70 = 74 5250/70 = 75 Estimate: 74 ; 75 Question 14. 1,565 Ã· 67 Estimate: _____ ; _____ 1610 Ã· 70 = 23 1680 Ã· 70 = 24 Estimate: 23 ; 24 Explanation: 1610/70 = 23 1680/70 = 24 Estimate: 23 ; 24 Question 15. 7,973 Ã· 91 Estimate: _____ ; _____ 476 Ã· 90 = 87 544 Ã· 90 = 88 Estimate: 87 ; 88 Explanation: 6960/90 = 87 7920/90 = 88 Estimate: 87 ; 88 Use compatible numbers to estimate the quotient. Question 16. 19)$$\overline { 228 }$$ Estimate: _____ 240 Ã· 20 = 12 260 Ã· 20 = 13 Estimate: 12 ; 13 Explanation: 240/20 = 12 260/20 = 13 Estimate: 12 ; 13 Question 17. 25)$$\overline { 595 }$$ Estimate: $_____ Answer: 575 Ã· 25 = 23 600 Ã· 25 = 24 Estimate: 23 ; 24 Explanation: 575/25 = 23 600/25 = 24 Estimate: 23 ; 24 Question 18. 86)$$\overline { 7,130 }$$ Estimate: _____ Answer: 7380 Ã· 90 = 82 7470 Ã· 90 = 83 Estimate: 82 ; 83 Explanation: 7380/90 = 82 7470/90 = 83 Estimate: 82 ; 83 Question 19. 83)$$\overline { 462 }$$ Estimate: _____ Answer: 400 Ã· 80 = 5 480 Ã· 80 = 6 Estimate: 5 ; 6 Explanation: 400/80 = 5 480/80 = 6 Estimate: 5 ; 6 Question 20. 27)$$\overline { 9,144 }$$ Estimate: _____ Answer: 10,140 Ã· 30 = 338 10,170 Ã· 30 = 339 Estimate: 338 ; 339 Explanation: 10,140/30 = 338 10,170/30 = 339 Estimate: 338 ; 339 Question 21. 68)$$\overline { 710 }$$ Estimate: _____ Answer: 700 Ã· 70 = 10 770 Ã· 70 = 11 Estimate: 10 ; 11 Explanation: 700/70 = 10 770/70 = 11 Estimate: 10 ; 11 Question 22. 707 Ã· 36 Estimate: _____ Answer: 760 Ã· 40 = 19 800 Ã· 40 = 20 Estimate: 19 ; 20 Explanation: 760/40 = 19 800/40 = 20 Estimate: 19 ; 20 Question 23. 1,198 Ã· 41 Estimate: _____ Answer: 1160 Ã· 40 = 29 1200 Ã· 40 = 30 Estimate: 29 ; 30 Explanation: 1160/40 = 29 1200/40 = 30 Estimate: 29 ; 30 Question 24. 5,581 Ã· 72 Estimate: _____ Answer: 5390 Ã· 70 = 77 5460 Ã· 70 = 78 Estimate: 77 ; 78 Explanation: 5390/70 = 77 5460/70 = 78 Estimate: 77 ; 78 ### Problem Solving – Page No. 82 Use the picture to solve 25â€“26. Question 25. About how many meters tall is each floor of the Williams Tower? _____ m Answer: 4.29 meters Explanation: Williams Tower has 275 meters and 64 floors. 275/64 = 4.29 meters Question 26. About how many meters tall is each floor of the Chrysler Building? _____ m Answer: 4.142 m Explanation: Chrysler Building has 319 meters and 77 floors 319/77 = 4.142 Question 27. Eli needs to save$235. To earn money, he plans to mow lawns and charge $21 for each. Write two estimates Eli could use to determine the number of lawns he needs to mow. Decide which estimate you think is the better one for Eli to use. Explain your reasoning. Type below: __________ Answer: 220/20 = 11 Explanation: Calculate$235/$21 210/21 = 10 220/20 = 11 number 220 is closer to 235. So, the better estimate is 220/20 = 11. Question 28. Explain how you know whether the quotient of 298 Ã· 31 is closer to 9 or to 10. Type below: __________ Answer: 270/30 = 9 310/31 = 10 298 is closer to 270. So, the quotient is closer to 9 than 10. Question 29. Test Prep Anik built a tower of cubes. It was 594 millimeters tall. The height of each cube was 17 millimeters. About how many cubes did Anik use? Options: a. 10 b. 16 c. 30 d. 300 Answer: c. 30 Explanation: 594/17 540/18 = 30 600/15 = 40 So, Anik use 30 cubes ### Share and Show – Page No. 85 Divide. Check your answer. Question 1. 28)$$\overline { 620 }$$ _____ R _____ Answer: 22 R 4 Explanation: Divide integers 62/28 = 2 Multiply 28 x 2 = 56; Subtract 62 â€“ 56 = 6 Write down 60 and divide integers 60/28 = 2. Multiply 28 x 2 = 56. Subtract 60 â€“ 56 = 4. The remainder is 4. So, 22 R 4. Check: 22 x 28 = 616; 616 + 4 = 620 620 = 620 Question 2. 64)$$\overline { 842 }$$ _____ R _____ Answer: 13 R 10 Explanation: Divide integers 84/64 = 1 Multiply 64 x 1 = 64; Subtract 84 â€“ 64 = 20 Write down 202 and divide integers 202/64 = 3. Multiply 64 x 3 = 192. Subtract 202 â€“ 192Â = 10. The remainder is 10. So, 13 R 10. Check: 64 x 13 = 832; 832 + 10 = 842 842 = 842 Go Math Grade 5 Lesson 2.4 Answer Key Question 3. 53)$$\overline { 2,340 }$$ _____ R _____ Answer: 44 R 8 Explanation: Divide integers 234/53 = 4 Multiply 53 x 4 = 212; Subtract 234 â€“ 212 = 22 Write down 220 and divide integers 220/53 = 4. Multiply 53 x 4 = 212. Subtract 220 â€“ 212 = 8. The remainder is 8. So, 44 R 8. Check: 53 x 44 = 2332; 2332 + 8 = 2340 2340 = 2340 Question 4. 723 Ã· 31 _____ R _____ Answer: 23 R 10 Explanation: Divide integers 72/31 = 2 Multiply 31 x 2 = 62; Subtract 72 â€“ 62 = 10 Write down 103 and divide integers 103/31 = 3. Multiply 31 x 3 = 93. Subtract 103 â€“ 93 = 10. The remainder is 10. So, 23 R 10. Check: 31 x 23 = 713; 713 + 10 = 723 723 = 723 Question 5. 1,359 Ã· 45 _____ R _____ Answer: 30 R 9 Explanation: Divide integers 135/45 = 3 Multiply 45 x 3 = 62; Subtract 135 â€“ 135 = 0 Write down 9; 9<45 The remainder is 9. So, 30 R 9. Check: 45 x 30 = 1350; 1350 + 9 = 1359 1359 = 1359 Question 6. 7,925 Ã· 72 _____ R _____ Answer: 110 R 5 Explanation: Divide integers 79/72 = 1 Multiply 72 x 1 = 72; Subtract 79 â€“ 72 = 7 Write down 72 and divide integers 72/72= 1. Multiply 72 x 1 = 72; Subtract 72 â€“ 72 = 0 Write down 5; 5<72 The remainder is 5. So, 110 R 5. Check: 72 x 110 = 7920; 7920 + 5 = 7925 7925 = 7925 On Your Own Divide. Check your answer. Question 7. 16)$$\overline { 346 }$$ _____ R _____ Answer: Explanation: Divide integers 34/16 = 2 Multiply 16 x 2 = 32; Subtract 34 â€“ 32 = 2 Write down 26 and divide integers 26/16= 1. Multiply 16 x 1 = 16; Subtract 26 â€“ 16 = 10 The remainder is 10. So, 21 R 10. Check: 16 x 21 = 336; 336 + 10 = 346 346 = 346 Question 8. 34)$$\overline { 241 }$$ _____ R _____ Answer: 7 R 3 Explanation: Divide integers 241/34 = 7 Multiply 34 x 7 = 238; Subtract 241 â€“ 238= 3 The remainder is 3. So, 7 R 3 Check: 34Â x 7 = 238; 238 + 3 = 241 241 = 241 Question 9. 77)$$\overline { 851 }$$ _____ R _____ Answer: 11 R 4 Explanation: Divide integers 85/77 = 1 Multiply 77 x 1 = 77; Subtract 85Â â€“ 77 = 8 Write down 81 and divide integers 81/77= 1. Multiply 77 x 1 = 77; Subtract 81 â€“ 77 = 4 The remainder is 4. So, 11 R 4. Check: 77 x 11 = 847; 847 + 4 = 851 851 = 851 Question 10. 21)$$\overline { 1,098 }$$ _____ R _____ Answer: 52 R 6 Explanation: Divide integers 109/21 = 5 Multiply 21 x 5 = 105; Subtract 109 â€“ 105= 4 Write down 48 and divide integers 48/21 = 2. Multiply 21 x 2 = 42; Subtract 48 â€“ 42 = 6 The remainder is 6. So, 52 R 6. Check: 21 x 52 = 1092; 1092 + 6 = 1098 1098 = 1098 Question 11. 32)$$\overline { 6,466 }$$ _____ R _____ Answer: 202 R 2 Explanation: Divide integers 64/32= 2 Multiply 32 x 2 = 64; Subtract 64 â€“ 64 = 0 Write down 66 and divide integers 66/32 = 2. Multiply 32 x 2 = 64; Subtract 66 â€“ 64 = 2 The remainder is 2. So, 202 R 2. Check: 32 x 202 = 6464; 6464 + 2 = 6466 6466 = 6466 Question 12. 45)$$\overline { 9,500 }$$ _____ R _____ Answer: 211 R 5 Explanation: Divide integers 95/45 = 2 Multiply 45 x 2 = 90; Subtract 95 â€“ 90 = 5 Write down 50 and divide integers 50/45 = 1. Multiply 45 x 1 = 45; Subtract 50 â€“ 45 = 5 Write down 50 and divide integers 50/45 = 1. Multiply 45 x 1 = 45; Subtract 50 â€“ 45 = 5 The remainder is 5. So, 211 R 5. Check: 45 x 211 = 9495; 9495 + 5 = 9500 9500 = 9500 Question 13. 483 Ã· 21 _____ Answer: 23 Explanation: Divide integers 48/21 = 2 Multiply 21 x 2 = 42; Subtract 48 â€“ 42 = 6 Write down 63 and divide integers 63/21 = 3. Multiply 21 x 3 = 63; Subtract 63 â€“ 63 = 0 The remainder is 0. So, 23 R 0. Check: 23 x 21 = 483; 483 = 483 Question 14. 2,292 Ã· 19 _____ R _____ Answer: 120 R 12 Explanation: Divide integers 22/19 = 1 Multiply 19 x 1 = 19; Subtract 22Â â€“ 19 = 3 Write down 39 and divide integers 39/19 = 2. Multiply 19 x 2 = 38; Subtract 39 â€“ 38 = 1 Write down 12; 12<19 The remainder is 12. So, 120 R 12. Check: 19 x 120 = 2280; 2280 + 12 = 2,292 2,292 = 2,292 Question 15. 4,255 Ã· 30 _____ R _____ Answer: 141 R 25 Explanation: Divide integers 42/30 = 1 Multiply 30 x 1 = 30; Subtract 42 â€“ 30 = 12 Write down 125 and divide integers 125/30 = 4. Multiply 30 x 4 = 120; Subtract 125 â€“ 120 = 5 Write down 55 and divide integers 55/30 = 1. Multiply 30 x 1 = 30; Subtract 55 â€“ 30 = 25 The remainder is 25. So, 141 R 25. Check: 30 x 141 = 4230; 4230 + 25 = 4,255 4,255 = 4,255 Practice: Copy and Solve Divide. Check your answer. Question 16. 775 Ã· 35 _____ R _____ Answer: 22 R 5 Explanation: Divide integers 77/35 = 2 Multiply 35 x 2 = 70; Subtract 77 â€“ 70 = 7 Write down 75 and divide integers 75/35 = 2. Multiply 35 x 2 = 70; Subtract 75 â€“ 70 = 5 The remainder is 5. So, 22 R 5. Check: 22 x 35 = 770; 770 + 5 = 775 775 = 775 Go Math Grade 5 Student Edition Question 17. 820 Ã· 41 _____ Answer: 20 Explanation: Divide integers 82/41 = 2 Multiply 41 x 2 = 82; Subtract 82 â€“ 82= 0 The remainder is 0. So, 20 R 0. Check: 41 x 20 = 820; 820 = 820 Question 18. 805 Ã· 24 _____ R _____ Answer: 33 R 13 Explanation: Divide integers 80/24 = 3 Multiply 24 x 3 = 72; Subtract 80 â€“ 72 = 8 Write down 85 and divide integers 85/24 = 3. Multiply 24 x 3 = 72; Subtract 85 â€“ 72 = 13 The remainder is 13. So, 33 R 13. Check: 24 x 33 = 792; 792 + 13 = 805 805 = 805 Question 19. 1,166 Ã· 53 _____ R _____ Answer: 22 R 0 Explanation: Divide integers 116/53 = 2 Multiply 53 x 2 = 106; Subtract 116 â€“ 106= 10 Write down 106 and divide integers 106/53 = 2. Multiply 53 x 2 = 106; Subtract 106 â€“ 106= 0 The remainder is 0. So, 22 R 0. Check: 53 x 22 = 1166; 1166 = 1166 Question 20. 1,989 Ã· 15 _____ R _____ Answer: 132 R 9 Explanation: Divide integers 19/15 = 1 Multiply 15 x 1 = 15; Subtract 19 â€“ 15 = 4 Write down 48 and divide integers 48/15 = 3. Multiply 15 x 3 = 45; Subtract 48 â€“ 45 = 3 Write down 39 and divide integers 39/15 = 2. Multiply 15 x 2 = 30; Subtract 39 â€“ 30= 9 The remainder is 9. So, 132 R 9. Check: 15 x 132 = 1980; 1980 + 9 = 1989 1989 = 1989 Question 21. 3,927 Ã· 35 _____ R _____ Answer: 112 R 7 Explanation: Divide integers 39/35 = 1 Multiply 35 x 1 = 35; Subtract 39 â€“ 35 = 4 Write down 42 and divide integers 42/35 = 1. Multiply 35 x 1 = 35; Subtract 42 â€“ 35 = 7 Write down 77 and divide integers 77/35 = 2. Multiply 35 x 2 = 70; Subtract 77 â€“ 70 = 7 The remainder is 7. So, 112 R 7. Check: 35 x 112 = 3920; 3920 + 7 = 3927 3927 = 3927 ### Problem Solving – Page No. 86 Use the list at the right to solve 22â€“24. Question 22. A smoothie shop receives a delivery of 980 ounces of grape juice. How many Royal Purple smoothies can be made with the grape juice? _____ smoothies Answer: 45 smoothies Explanation: A smoothie shop receives a delivery of 980 ounces of grape juice. 980 ounces of grape juice/22 ounces of grape juice = 45 Question 23. The shop has 1,260 ounces of cranberry juice and 650 ounces of passion fruit juice. If the juices are used to make Crazy Cranberry smoothies, which juice will run out first? How much of the other juice will be left over? Type below: _________ Answer: The shop has 1,260 ounces of cranberry juice and 650 ounces of passion fruit juice. If the juices are used to make Crazy Cranberry smoothies, passion fruit juice will run out first. Because 650<1,260. So, passion fruit juice will run out first. 1,260 – 650 =610 Crazy Cranberry juice will be left over. Question 24. In the refrigerator, there are 680 ounces of orange juice and 410 ounces of mango juice. How many Orange Tango smoothies can be made? Explain your reasoning. _____ smoothies Answer: In the refrigerator, there are 680 ounces of orange juice and 410 ounces of mango juice. So, 410 Orange Tango smoothies can be made. Because there are 410 ounces of mango juices available. Question 25. Test Prep James has 870 action figures. He decides to divide them equally among 23 boxes. How many action figures will James have left over? Options: a. 19 b. 23 c. 31 d. 37 Answer: d. 37 Explanation: James has 870 action figures. He decides to divide them equally among 23 boxes. 870/23 = 37 ### Share and Show – Page No. 89 Interpret the remainder to solve. Question 1. Erika and Bradley want to hike the Big Cypress Trail. They will hike a total of 75 miles. If Erika and Bradley plan to hike for 12 days, how many miles will they hike each day? a. Divide to find the quotient and remainder. _____ R _____ Answer: 6 R 3 Explanation: 75/12 = 6 The remainder is 3 6 R 3 Question 1. b. Decide how to use the quotient and remainder to answer the question. Type below: _________ Answer: 75/12 = 6Â 1/4 So, Each day they will hike 6$$\frac{1}{4}$$ miles. Question 2. What if Erika and Bradley want to hike 14 miles each day? How many days will they hike exactly 14 miles? _____ days Answer: 196 days Explanation: If Erika and Bradley want to hike 14 miles each day, 14 x 14 = 196 days Question 3. Dylanâ€™s hiking club is planning to stay overnight at a camping lodge. Each large room can hold 15 hikers. There are 154 hikers. How many rooms will they need? _____ rooms Answer: 11 rooms Explanation: Dylanâ€™s hiking club is planning to stay overnight at a camping lodge. Each large room can hold 15 hikers. There are 154 hikers. So, 154/15 = 10 and the remainder is 4. Dylanâ€™s hiking club require 10 rooms for 150 hikers and other room for 4 hikers. So, in total they need 10 + 1 = 11 rooms. On Your Own Interpret the remainder to solve. Question 4. The students in a class of 24 share 84 cookies equally among them. How many cookies did each student eat? _____ $$\frac{â–¡}{â–¡}$$ cookies Answer: 3$$\frac{1}{2}$$ cookies Explanation: The students in a class of 24 share 84 cookies equally among them. So, 84/24 = 3$$\frac{12}{24}$$ = 3$$\frac{1}{2}$$ Question 5. A campground has cabins that can each hold 28 campers. There are 148 campers visiting the campground. How many cabins are full if 28 campers are in each cabin? _____ cabins Answer: 5$$\frac{1}{7}$$ cabins Explanation: A campground has cabins that can each hold 28 campers. There are 148 campers visiting the campground. 184/28 = 5$$\frac{1}{7}$$ Question 6. A total of 123 fifth-grade students are going to Fort Verde State Historic Park. Each bus holds 38 students. All of the buses are full except one. How many students will be in the bus that is not full? _____ students Answer: 9 students Explanation: A total of 123 fifth-grade students are going to Fort Verde State Historic Park. Each bus holds 38 students. 123/38 = 3 and the remainder is 9. 3 x 38 = 114 students. 1 bus is not full. So, 9 students will be in the bus that is not full Question 7. Whatâ€™s the Error? Sheila is going to divide a 36-inch piece of ribbon into 5 equal pieces. She says each piece will be 7 inches long. Type below: _________ Answer: Sheila is going to divide a 36-inch piece of ribbon into 5 equal pieces. 36/5 = 7$$\frac{1}{5}$$. She said each piece will be 7 inches long and forgot about $$\frac{1}{5}$$ part. ### UNLOCK the Problem – Page No. 90 Question 8. Maureen has 243 ounces of trail mix. She puts an equal number of ounces in each of 15 bags. How many ounces of trail mix does Maureen have left over? a. What do you need to find? Answer: We need to find how many ounces of trail mix does Maureen have left over? Question 8. b. How will you use division to find how many ounces of trail mix are left over? Type below: _________ Answer: The division is 243/15 Question 8. c. Show the steps you use to solve the problem. Type below: _________ Answer: 243/15 Divide integers 24/15 = 1 Multiply 15 x 1 = 15; Subtract 24 â€“ 15 = 9 Write down 93 and divide integers 93/3 = 6. Multiply 15 x 6 = 90. Subtract 93 â€“ 90 = 3. The remainder is 3. So, 16 R 3. Question 8. d. Complete the sentences. Maureen has _______ ounces of trail mix. She puts an equal number in each of _______ bags. Each bag has _______ ounces. Maureen has _______ ounces of trail mix left over. Type below: _________ Answer: Maureen has 243 ounces of trail mix. She puts an equal number in each of 15 bags. Each bag has 16 ounces. Maureen has 3 ounces of trail mix left over. Question 8. e. Fill in the bubble completely to show your answer. Options: a. 3 ounces b. 15 ounces c. 16 ounces d. 17 ounces Answer: c. 16 ounces Question 9. Mr. Field wants to give each of his 72 campers a certificate for completing an obstacle course. If there are 16 certificates in one package, how many packages will Mr. Field need? Options: a. 4 b. 5 c. 16 d. 17 Answer: b. 5 Explanation: Mr. Field wants to give each of his 72 campers a certificate for completing an obstacle course. If there are 16 certificates in one package, 72/16 = 4.5 Question 10. James has 884 feet of rope. There are 12 teams of hikers. If James gives an equal amount of rope to each team, how much rope will each team receive? Options: a. 12 b. 73 c. 73 $$\frac{2}{3}$$ d. 74 Answer: b. 73 Explanation: James has 884 feet of rope. There are 12 teams of hikers. If James gives an equal amount of rope to each team, 884/12 = 73 ### Share and Show – Page No. 92 Adjust the estimated digit in the quotient, if needed. Then divide. Question 1. 4 41)$$\overline { 1,546 }$$ _____ R _____ Answer: 37 R 29 Explanation: 41 x 4 = 164; Subtract: 154 – 164 the estimate too high. Change the quotient to 3 41 x 3 = 123; Subtract: 154 – 123 = 31 Write down 316 and divide integers 316/41 41 x 7 = 287; Subtract: 316 – 287 = 29 37 R 29 Question 2. 2 16)$$\overline { 416 }$$ _____ Answer: 26 Explanation: 16 x 2 = 32; Subtract: 41 – 32 = 9 Write down 96 and divide integers 96/16 16 x 6 = 96; Subtract: 96 – 96 = 0 26 Question 3. 9 34)$$\overline { 2,831 }$$ _____ R _____ Answer: 83 R 9 Explanation: 34 x 9 = 306; Subtract: 283 – 306 the estimate too high. Change the quotient to 8 34 x 8 = 272; Subtract: 283 – 272 = 11 Write down 111 and divide integers 111/34 34 x 3 = 102; Subtract: 111 – 102 = 9 83 R 9 Divide. Question 4. 19)$$\overline { 915 }$$ _____ R _____ Answer: 48 R 3 Explanation: 900/18 = 50 19 xÂ 5 = 95; Subtract: 91 – 95 the estimate too high. Change the quotient to 4 19 xÂ 4 = 76; Subtract: 91 – 76 = 15 Write down 155 and divide integers 155/19 19 x 7 = 133; Subtract: 155 – 133 = 22 22 > 19; So Change the quotient to 8 19 x 8 = 152; Subtract: 155 – 152 = 3 48 R 3 Question 5. 28)$$\overline { 1,825 }$$ _____ R _____ Answer: Explanation: 1800/30 = 60 28 x 6 = 168; Subtract: 182 – 168 = 14 Write down 145 and divide integers 145/28 28 x 5 = 140; Subtract: 145 – 140 = 5 65 R 5 Question 6. 45)$$\overline { 3,518 }$$ _____ R _____ Answer: Explanation: 3600/40 = 90 45 x 9 = 405; Subtract: 351 – 405 the estimate too high. Change the quotient to 7 45 x 7 = 315; Subtract: 351 – 315 = 36 Write down 368 and divide integers 368/45 45 x 8 = 360; Subtract: 368 – 315 = 8 78 R 8 ### On Your Own – Page No. 93 Adjust the estimated digit in the quotient, if needed. Then divide. Question 7. 2 26)$$\overline { 541 }$$ _____ R _____ Answer: 20 R 21 Explanation: 500/25 = 2 26 x 2 = 52; Subtract: 54 – 52 = 2 Write down 21 and divide integers 21/26 20 R 21 Question 8. 1 43)$$\overline { 688 }$$ _____ Answer: 16 Explanation: 800/40 = 20 43 x 2 = 86; Subtract: 68 – 86 the estimate is too high. Change the quotient to 1 43 x 1 = 43; Subtract: 68 – 43 = 25 Write down 258 and divide integers 258/43 43 x 6 = 258; Subtract: 258 – 258 = 0 So, 16 Question 9. 6 67)$$\overline { 4,873 }$$ _____ R _____ Answer: 72 R 49 Explanation: 4800/70 = 60 67 x 6 = 402; Subtract: 487 – 402 = 85 the estimate is too low. Change the quotient to 7 67 x 7 = 469; Subtract: 487 – 469 = 18 Write down 183 and divide integers 183/67 67 x 2 = 134; Subtract: 183 – 134 = 49 72 R 49 Question 10. 15)$$\overline { 975 }$$ _____ Answer: 65 Explanation: 15 x 6 = 90; Subtract 97 – 90 = 7 Write down 75 and divide integers 75/15 15 x 5 = 75; Subtract: 75 – 75 = 0 So, 65 Question 11. 37)$$\overline { 264 }$$ _____ R _____ Answer: 7 R 5 Explanation: 240/40 = 6 37 x 6 = 222; Subtract: 264 – 222 = 42 The estimate is too low. Change the quotient to 7 37 x 7 = 259; Subtract: 264 – 259 = 5 7 R 5 Question 12. 22)$$\overline { 6,837 }$$ _____ R _____ Answer: 310 R 17 Explanation: 6300/20 = 325 22 x 3 = 66; Subtract: 68 – 66 = 2 Write down 23 and divide integers 23/22 22 x 1 = 22; Subtract: 23 – 22 = 1 Write down 17; 17 < 22 310 R 17 Practice: Copy and Solve Divide. Question 13. 452 Ã· 31 _____ $$\frac{â–¡}{â–¡}$$ Answer: 14$$\frac{18}{31}$$ Explanation: Divide integers 45/31 = 1 Multiply 31 x 1 = 31; Subtract 45 â€“ 31 = 14 Write down 142 and divide integers 142/31 = 4. Multiply 31 x 4 = 124. Subtract 142 â€“ 124 = 18. The remainder is 18. So, 14 R 18. 14$$\frac{18}{31}$$ Question 14. 592 Ã· 74 _____ Answer: 8 Explanation: Divide integers 592/74 = 8 So, 8. Question 15. 785 Ã· 14 _____ R _____ Answer: 56$$\frac{1}{14}$$ Explanation: Divide integers 78/14 = 5 Multiply 14 x 5 = 70; Subtract 78 â€“ 70 = 8 Write down 85 and divide integers 85/14 = 6. Multiply 14 x 6 = 84. Subtract 85 â€“ 84 = 1. The remainder is 1. So, 56 R 1. 56$$\frac{1}{14}$$ Question 16. 601 Ã· 66 _____ R _____ Answer: 9 R 7 Explanation: Divide integers 601/66 = 9 Multiply 66 x 9 = 594 ; Subtract 601 â€“ 594= 7 The remainder is 7. So, 9 R 7. 9$$\frac{7}{66}$$ Question 17. 1,067 Ã· 97 _____ Answer: 11 Explanation: Divide integers 106/97 = 1 Multiply 97 x 1 = 97; Subtract 106 – 97 = 9 Write down 97 and divide integers 97/97 = 1 Multiply 97 x 1 = 97; Subtract 97 – 97 = 0 The remainder is 0. So, 11 is the answer. Question 18. 2,693 Ã· 56 _____ R _____ Answer: 48 R 5 Explanation: Divide integers 269/56 = 4 Multiply 56 x 4 = 224; Subtract 269 â€“ 224 = 45 Write down 453 and divide integers 453/56 = 8 Multiply 56 x 8 = 448. Subtract 453 â€“ 448 = 5. The remainder is 5. So, 48 R 5. Question 19. 1,488 Ã· 78 _____ R _____ Answer: 19 R 6 Explanation: Divide integers 148/78 = 1 Multiply 78 x 1 = 78; Subtract 148 â€“ 78 = 70 Write down 708 and divide integers 708/78 = 9. Multiply 78 x 9 = 702. Subtract 708 â€“ 702 = 6. The remainder is 6. So, 19 R 6. Question 20. 2,230 Ã· 42 _____ R _____ Answer: 53 R 4 Explanation: Divide integers 223/42 = 5 Multiply 42 x 5 = 210; Subtract 223 â€“ 210 = 13 Write down 130 and divide integers 130/42 = 3. Multiply 42 x 3 = 126. Subtract 130 â€“ 126 = 4. The remainder is 4. So, 53 R 4. Question 21. 4,295 Ã· 66 _____ R _____ Answer: 65 R 5 Explanation: Divide integers 429/66 = 6 Multiply 66 x 6 = 396; Subtract 429 â€“ 396 = 33 Write down 335 and divide integers 335/66 = 5. Multiply 66 x 5 = 330. Subtract 335 – 330 = 5. The remainder is 5 So, 65 R 5 Algebra Write the unknown number for each â– . Question 22. â– Ã· 33 = 11 â– = _____ Answer: 363 Explanation: n Ã· 33 = 11 n = 11 x 33 = 363 Question 23. 1,092 Ã· 52 = â– â– = _____ Answer: 21 Explanation: 1,092 Ã· 52 = 21 Question 24. 429 Ã· â– = 33 â– = _____ Answer: 13 Explanation: 429 Ã· n = 33 n = 429 Ã· 33 n = 13 ### UNLOCK the Problem – Page No. 94 Question 25. A banquet hall serves 2,394 pounds of turkey during a 3-week period. If the same amount is served each day, how many pounds of turkey does the banquet hall serve each day? a. What do you need to find? Type below: _________ Answer: How many Lbs at turkey do they serve each day? Question 25. b. What information are you given? Type below: _________ Answer: Every 3 weeks, serves 2,394 lbs. Question 25. c. What other information will you use? Type below: _________ Answer: Same each day, 3 weeks = 21 days Question 25. d. Find how many days there are in 3 weeks. There are ______ days in 3 weeks. Type below: _________ Answer: There are ______ days in 3 weeks Explanation: 1 week = 7 days. 3 x 7 = 21 days Question 25. e. Divide to solve the problem. Type below: _________ Answer: 2394/3 = t t = 798 798/7 = 114 pounds Question 25. f. Fill in the bubble for the correct answer choice. Options: a. 50,274 pounds b. 798 pounds c. 342 pounds d. 114 pounds Answer: d. 114 pounds Adjust Quotients Lesson 2.8 Go Math Grade 5 Partial Quotients Question 26. Marcos mixes 624 ounces of lemonade. He wants to fill the 52 cups he has with equal amounts of lemonade. How much lemonade should he put in each cup? Options: a. 8 ounces b. 12 ounces c. 18 ounces d. 20 ounces Answer: b. 12 ounces Explanation: Marcos mixes 624 ounces of lemonade. He wants to fill the 52 cups he has with equal amounts of lemonade. 624/52 = 12 ounces. 12 ounces should he put in each cup Question 27. The Box of Sox company packs 18 pairs of socks in a box. How many boxes will the company need to pack 810 pairs of socks? Options: a. 40 b. 45 c. 55 d. 56 Answer: b. 45 Explanation: The Box of Sox company packs 18 pairs of socks in a box. So, for 810 pairs of socks, 810/18 = 45 ### Share and Show – Page No. 97 Question 1. Paula caught a tarpon with a weight that was 10 times as great as the weight of a permit fish she caught. The total weight of the two fish was 132 pounds. How much did each fish weigh? First, draw one box to represent the weight of the permit fish and ten boxes to represent the weight of the tarpon. Next, divide the total weight of the two fish by the total number of boxes you drew. Place the quotient in each box. Last, find the weight of each fish. The permit fish weighed _____ pounds. The tarpon weighed _____ pounds. Type below: _________ Answer: The permit fish weighed 12 pounds. The tarpon weighed 120 pounds. Explanation: Let S be the weight of a permit fish Paula caught. The weight of the tarpon is 10 times as great as the weight of a permit fish she caught = 10 S The total weight of the two fish was 132 pounds. S + 10S = 132 11S = 132 S = 132/11 = 12 So, Paula caught a fish with the weight of 12 pounds. The tarpon weighted 120 pounds. Question 2. What if the weight of the tarpon was 11 times the weight of the permit fish, and the total weight of the two fish was 132 pounds? How much would each fish weigh? permit fish: _________ pounds tarpon: _________ pounds Answer: permit fish: 11 pounds tarpon: 11 x 11 = 121 pounds Explanation: Let S be the weight of a permit fish Paula caught. The weight of the tarpon is 11 times as great as the weight of a permit fish she caught = 11S Total weight is 132 11S + S = 132 12S = 132 S = 132/12 = 11. permit fish: 11 pounds tarpon: 11 x 11 = 121 pounds Question 3. Jon caught four fish that weighed a total of 252 pounds. The kingfish weighed twice as much as the amberjack and the white marlin weighed twice as much as the kingfish. The weight of the tarpon was 5 times the weight of the amberjack. How much did each fish weigh? amberjack: _________ pounds kingfish: _________ pounds marlin: _________ pounds tarpon: _________ pounds Answer: amberjack: 21 pounds kingfish: 42 pounds marlin: 84 pounds tarpon: 105 pounds Explanation: Let S be the weight of the amberjack. The kingfish weighed twice as much as the amberjack = 2S The white marlin weighed twice as much as the kingfish = 2 X 2S = 4S The weight of the tarpon was 5 times the weight of the amberjack = 5S Total weight = 252 pounds. 2S + 4S + 5S + S = 252 12S = 252 S = 252/12 S = 21. The kingfish weighed twice as much as the amberjack = 2S = 2 x 21 = 42 pounds. The white marlin weighed twice as much as the kingfish = 2 X 2S = 4S = 4 x 21 = 84 pounds. The weight of the tarpon was 5 times the weight of the amberjack = 5S = 5 x 21 = 105 pounds. ### On Your Own – Page No. 98 Use the table to solve 4â€“7. Question 4. Kevin is starting a saltwater aquarium with 36 fish. He wants to start with 11 times as many damselfish as clown fish. How many of each fish will Kevin buy? How much will he pay for the fish? Type below: _________ Answer: Kevin is starting a saltwater aquarium with 36 fish. He uses 1 damselfish and 11 clown fish. So, three groups form for 36 fishes. 1 damselfish and 11 clown fish =$7 + (11 x $20) = 7 + 220 = 227 He buys 3 damselfish and 33 clown fish. 3 x 7 = 21$ for damselfish and 33 x 20 = 660 for clown fish. Each fish = 681/36 = 19 Kevin used a store coupon to buy a 40-gallon tank, an aquarium light, and a filtration system. He paid a total of $240. How much money did Kevin save by using the coupon?$ _____ $25 Explanation: 40-gallon tank =$170 aquarium light = $30 filtration system =$65 170 + 30 + 65 = 265 He paid a total of $240; 265 -240 =$25 Kevin save by using the coupon $25. Question 6. Kevin bought 3 bags of gravel to cover the bottom of his fish tank. He has 8 pounds of gravel left over. How much gravel did Kevin use to cover the bottom of the tank? _____ pounds Answer: 37 pounds. Explanation: 15lb bag of gravel =$13. 3 bags of gravel = 45lb. He has 8 pounds of gravel left over = 45 – 8 = 37 Kevin use 37 pounds of gravel to cover the bottom of the tank Question 7. Pose a Problem Look back at Problem 6. Write a similar problem by changing the number of bags of gravel and the amount of gravel left. Type below: _________ If he bought 5 bags of gravel to cover the bottom of his fish tank. He has 10 pounds of gravel left over. 5 bags of gravel = 15 x 5 = 75lbs He has 10 pounds of gravel left over = 75 – 10 lbs = 65 lbs 65lbs Explanation: Question 8. Test Prep Captain James offers a deep-sea fishing tour. He charges $2,940 for a 14-hour trip. How much does each hour of the tour cost? Options: a.$138 b. $201 c.$210 d. $294 Answer: c.$210 Explanation: Captain James offers a deep-sea fishing tour. He charges $2,940 for a 14-hour trip. Each hour =$2940/14 = 210 ### Chapter Review/Test – Vocabulary – Page No. 99 Choose the best term from the box. Question 1. You can to estimate quotients because they are easy use _________ to compute with mentally Compatible Numbers Question 2. To decide where to place the first digit in the quotient, you can estimate or use _________ Place Value Concepts and Skills Use compatible numbers to estimate the quotient. Question 3. 522 Ã· 6 = _____ 90 Explanation: 522 is close to 540. 540 Ã· 6 = 90 Question 4. 1,285 Ã· 32 = _____ 40 Explanation: 1,280 Ã· 32 = 40 Question 5. 6,285 Ã· 89 = _____ 70 Explanation: 6,300 Ã· 90 = 70 Question 6. 2)$$\overline { 554 }$$ = _____ 277 Explanation: Divide integers 5/2 = 2 Multiply 2 x 2 = 4; Subtract 5 – 4 = 1 Write down 15 and divide integers 15/2 = 7. Multiply 2 x 7 = 14. Subtract 15 – 14 = 1 Write down 14 and divide integers 14/2 = 7. Multiply 2 x 7 = 14. Subtract 14 – 14 = 0 The remainder is 0. Question 7. 8)$$\overline { 680 }$$ = _____ 85 Explanation: Divide integers 68/8 = 8 Multiply 8 x 8 = 64; Subtract 68 – 64 = 4 Write down 40 and divide integers 40/8 = 5. Multiply 8 x 5 = 40. Subtract 40 – 40 = 0 The remainder is 0. Question 8. 5)$$\overline { 462 }$$ = _____ R _____ 92 R 2 Explanation: Divide integers 46/5 = 9 Multiply 5 x 9 = 45; Subtract 46 – 45 = 1 Write down 12 and divide integers 12/5 = 2. Multiply 5 x 2 = 10. Subtract 12 – 10 = 2 The remainder is 2. So, 92 R 2 Check: (92 x 5) + 2 = 460 + 2 = 462 Question 9. 522 Ã· 18 = _____ 29 Explanation: Divide integers 52/18 = 2 Multiply 18 x 2 = 36; Subtract 52 – 36 = 16 Write down 162 and divide integers 162/8 = 9. Multiply 8 x 9 = 162. Subtract 162 – 162 = 0 The remainder is 0 Question 10. 529 Ã· 37 = _____ R _____ 14 R 11 Explanation: Divide integers 52/37 = 1 Multiply 37 x 1 = 37; Subtract 52 – 37 = 15 Write down 159 and divide integers 159/37 = 4. Multiply 37 x 4 = 148. Subtract 159 – 148 = 11 The remainder is 11. So, 14 R 11 Check: (14 x 37) + 11 = 518 + 11 = 529 Question 11. 987 Ã· 15 = _____ R _____ 65 R 12 Explanation: Divide integers 98/15 = 6 Multiply 15 x 6 = 90; Subtract 98 – 90 = 8 Write down 87 and divide integers 87/15 = 5. Multiply 15 x 5 = 75. Subtract 87 – 75 = 12 The remainder is 12. So, 65 R 12 Check: (15 x 65) + 12 = 975 + 12 = 987 Question 12. 1,248 Ã· 24 = _____ 52 Explanation: Divide integers 124/24 = 5 Multiply 24 x 5 = 120; Subtract 124 – 120 = 4 Write down 48 and divide integers 48/24 = 2. Multiply 24 x 2 = 48. Subtract 48 – 48 = 0 The remainder is 0 Question 13. 5,210 Ã· 17 = _____ R _____ 306 R 8 Explanation: Divide integers 52/17 = 3 Multiply 17 x 3 = 51; Subtract 52 – 51 = 1 Write down 110 and divide integers 110/17 = 6. Multiply 17 x 6 = 102. Subtract 110 – 102 = 8 The remainder is 8. So, 306 R 8 Check: (306 x 17) + 8 = 5202 + 8 = 5210 Question 14. 8,808 Ã· 42 = _____ R _____ 209 R 30 Explanation: Divide integers 88/42 = 2 Multiply 42 x 2 = 84; Subtract 88 – 84 = 4 Write down 408 and divide integers 408/42 = 9. Multiply 42 x 9 = 378. Subtract 408 – 378 = 30 The remainder is 30. So, 209 R 30 Check: (209 x 42) + 30 = 8778 + 30 = 8808 ### Chapter Review/Test – Page No. 100 Question 15. Samira bought 156 ounces of trail mix. She wants to divide the amount equally into 24 portions. How many ounces of trail mix will be in each portion? Options: A. 6 ounces B. 6 $$\frac{1}{2}$$ ounces C. 7 ounces D. 12 ounces B. 6 $$\frac{1}{2}$$ ounces Explanation: Samira bought 156 ounces of trail mix. She wants to divide the amount equally into 24 portions.156/24 = 6.5 = 6 $$\frac{1}{2}$$ ounces Question 16. A school band performed 6 concerts. Every seat for each performance was sold. If a total of 1,248 seats were sold for all 6 concerts, how many seats were sold for each performance? Options: A. 28 B. 200 C. 206 D. 208 D. 208 Explanation: A school band performed 6 concerts. Every seat for each performance was sold. If a total of 1,248 seats were sold for all 6 concerts, then 1,248/6 = 208. Question 17. Dylanâ€™s dog weighs 12 times as much as his pet rabbit. The dog and rabbit weigh 104 pounds altogether. How much does Dylanâ€™s dog weigh? Options: A. 104 pounds B. 96 pounds C. 88 pounds D. 8 pounds D. 8 pounds Explanation: Dylanâ€™s dog weighs 12 times as much as his pet rabbit. The dog and rabbit weigh 104 pounds altogether. rabbit weight = S dog weighs = 12S S + 12S = 104; 13S = 104; S = 104/13 =8. Question 18. Jamie is sewing 14 identical costumes for the school play. She needs 210 buttons to complete all of the costumes. How many buttons will she sew onto each costume? Options: A. 15 B. 14 C. 11 D. 9 A. 15 Explanation: Jamie is sewing 14 identical costumes for the school play. She needs 210 buttons to complete all of the costumes. 210/14 = 15 ### Chapter Review/Test – Page No. 101 Question 19. A book publishing company is shipping an order of 300 books. The books are packaged in boxes that each can hold 24 books. How many boxes are needed to ship the order of books? Options: A. 10 B. 11 C. 12 D. 13 D. 13 Explanation: A book publishing company is shipping an order of 300 books. The books are packaged in boxes that each can hold 24 books. 300/24 = 12.5. That is 12 and above boxes. So, the answer is 13 Question 20. Richard is planning a trip to Italy. He thinks he will need $2,750 for the trip. If the trip is 40 weeks away, which is the best estimate of how much money Richard needs to save each week? Options: A.$60 B. $70 C.$600 D. $700 Answer: B.$70 Explanation: Richard is planning a trip to Italy. He thinks he will need $2,750 for the trip. If the trip is 40 weeks away,$2,750/40 = $68.75. That is equal to$70 Question 21. A school club raises $506 to spend on a field trip. There are 23 people going on the trip. How much money did the club raise for each person going? Options: A.$27 B. $22 C.$18 D. $12 Answer: B.$22 Explanation: A school club raises $506 to spend on a field trip. There are 23 people going on the trip.$506/23 = \$22. Question 22. A local orange grower processes 2,330 oranges from his grove this year. The oranges are packaged in crates that each hold 96 oranges. All but one crate is full. How many oranges are in this last crate? Options: A. 24 B. 25 C. 26 D. 27 C. 26 Explanation: 2330 oranges / 96 orange/crate = 24.2708333 crates the decimal portion is the fraction of 96 in the last crate= 96 x .2708333 = 26 oranges in the last crate. ### Chapter Review/Test – Page No. 102 Constructed Response Question 23. On Monday, 1,900 bottles of perfume are delivered to a warehouse. The bottles are packed in boxes. Each box can hold 32 bottles. How many boxes were delivered? Explain how you found your answer. _____ boxes I need to divide 1,900 by 32, which is 59 r12. That means the bottles will completely fill 59 boxes. But there will be 12 bottles left over. These would be packed in another box, which makes a total of 60 boxes. Question 24. Quincy needs 322 yards of ribbon to decorate quilts for a craft fair.The ribbon comes in rolls of 15 yards. _____ rolls I need to divide 322 by 15. The answer is 21Â R 7. Since he can’t buy a part of a roll, I need to add 1 to the quotient. So, the final answer is 22. Question 24. B. Alice needs twice as many yards of ribbon as Quincy. How many rolls of ribbon does Alice need? Explain your answer. _____ rolls Twice the length of 322 yards is 644 yards. If I divide 644 by 15, the answer is 42 R 14. Alice needs to buy 43 rolls of ribbon. The remainder doubled is still less than the amount In 1 roll. Question 24. C Elena needs yellow, red, and blue ribbon. She needs 285 yards of the three colors combined. Suggest numbers of rolls of each color that would give her enough ribbon. (HINT: Break apart the 285 yards into any combination of 3 groups that total this amount.) Type below: _________
If you didn’t have contact with this subject, this Quotient Calculator is the place to make contact and learn about it. Through this post, you will find the answer to the quotient definition, its rights, quotient in mathematics, how to use the calculator step by step, all formula’s you need, what does quotient mean and more. Besides this, there are other free tools and info on our site, such as Associative Property, or Fundamental Counting Principle. These are some useful posts where you can learn interesting facts. Also, head to our free Angular Resolution, Golden ratio and Population Density Calculator to solve all your related problems and answer your questions. ## What is a Quotient? Quotient definition and what quotient mean In following section we will define quotient. We get this value when we divide one integer by another. And this is the easiest way to explain the definition. For instance, when we divide 6 by 3, we obtain 2. We can use a decimal value or an integer as the quotient. For exact divisions, such as 10÷ 5 = 2, the quotient is an integer, but for divisions like 12 ÷ 5 = 2.4, it is a decimal. It can be more than the divisor, but it can never be greater than the dividend. Let’s learn more about the quotient and how to calculate it step by step. Quotient in math The result of dividing an integer by any divisor is quotient in math. It refers to how many times the divisor appears in the dividend. ## Quotient Rule The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator, is the derivative of a quotient, according to the Quotient Rule. ## Quotient and Remainder When faced with a division dilemma, we must determine how many copies of anything we can have if we distribute a number evenly. For instance, let’s imagine you bought a big pizza with 8 pieces for your family of four. Then, we divide to determine how many slices each participant will receive: 8 \div 4=2 What if there were three persons instead of two? Everyone could, of course, have two slices, but that would leave two slices in the box. So now we have to decide, to save them for later or chop them up into smaller pieces. That’s when the remainder and quotient come in help. As stated in the preceding section, the quotient is the outcome of division. However, not all integers are divisible by one another (like 4 divided by 3 or 25 divided by 2). In such instances, we can disperse as much as we can until we’re left with the remaining few, non-divisible parts: the residue. In the pizza example above, an 8-slice pizza divided among three individuals yields two slices per person, with two slices remaining. Other instances are 4 divided by 3, which yields 1 with a remnant of 1, and 25 divided by 2, which yields 12 with a remaining of 1. In arithmetic, we write the result of such division by using a capital R to separate the quotient and remainder. However, when working with negative numbers, we must be cautious with remainders. In mathematics, the remnant is usually defined as a positive number less than the divisor’s (absolute value). Nevertheless, some applications are permitted negative remainders (particularly in the computer sciences). ## Quotient Formula The given formula is: Quotient=\frac {Dividend}{Divisor} ## Quotient Calculator – How to Use? The division formula with the names of its subsequent components may be found at the top of our tool, and we apply it in the calculator below. If you want to solve your math problem try our calculator for free and follow the given steps: • Input first number in section dividend; • Next, input the second number in the section divisor; • Read off the results, the calculator will calculate it automatically. Remember, as indicated in point 3 and the preceding section, that the quotient calculator might give you an answer in a fractional mixed (integer or decimal), or a quotient and remainder. Also, you may anticipate two variations of the remainder for general integers (i.e., negative numbers), but the quotient calculator just delivers the division result without any additional output for decimals (there are no remainders in this case). Regardless of the format you pick, we hope our math calculator provides you with the information you want and answers all your questions. ## Quotient Example We will now do an equation to show you how our calculator works, follow these steps: • as we said in the first line we input dividend (for us the example is number 14); • we are now inputting the divisor (number 5); • and now we get the result of our quotient and reminder. ## FAQ ### How to find the quotient? After the division operation is done, the quotient is obtained. This indicates that the result is the quotient when a dividend is divided by a divisor. ### What is the quotient of 32 and 6? To find the quotient of 36 and 6, it is easiest for you to try our math calculator. You will get the number 6 if you enter the digits correctly. ### What is the fractional result of 3 and 4? The fraction result of 3 and 4 is 3/4. ### How can you find the quotient and remainder of numbers 7 and 2? To solve the quotient and remainder, follow the following formula: remainder = dividend % divisor; quotient = dividend/divisor ### What is a partial quotient? A partial fraction is a method of solving big division problems by using partial quotients.
# How do you factor 10x^2 - 7x - 12? May 31, 2015 Use a variant of the AC Method. $A = 10$, $B = 7$, $C = 12$ Look for a pair of factors of $A C = 10 \times 12 = 120$ whose difference is $7$. We look for the difference rather than the sum because the sign of the constant term is negative. The pair $15 , 8$ works: $15 \times 8 = 120$ $15 - 8 = 7$ Now use that pair to split the middle term, then factor by grouping... $10 {x}^{2} - 7 x - 12 = 10 {x}^{2} - 15 x + 8 x - 12$ $= \left(10 {x}^{2} - 15 x\right) + \left(8 x - 12\right)$ $= 5 x \left(2 x - 3\right) + 4 \left(2 x - 3\right)$ $= \left(5 x + 4\right) \left(2 x - 3\right)$ May 31, 2015 The method George used is the popular factoring AC Method (YouTube). To avoid the lengthy factoring by grouping, you may use the new AC Method. $f \left(x\right) = 10 {x}^{2} - 7 x - 12 = 10 \left(x + p\right) \left(x + q\right) .$ Convert y to $y ' = {x}^{2} - 7 x - 120$ = (x + p')(x + q') Find p' and q' by composing factor pairs of (a.c = -120). a and c have different signs. Proceed: ...(-4, 30)(-5, 24)(-8, 15). This sum is 7 = -b. Then p' = 8 and q' = -15. Then, $p = \frac{p '}{a} = \frac{8}{10} = \frac{4}{5}$ and $q = - \frac{15}{10} = - \frac{3}{2.}$ Factored form: $f \left(x\right) = 10 \left(x + \frac{4}{5}\right) \left(x - \frac{3}{2}\right) = \left(5 x + 4\right) \left(2 x - 3\right) .$ Check by developing: f(x) = 10x^2 - 15x + 8x - 12. OK. This new AC Method is fast, systematic, no guessing, no factoring by grouping.
Вы находитесь на странице: 1из 3 # Introduction: Standard Deviation Standard deviation is a measure of central tendency. Central tendency refers to the quantity that tells us as to by how much are the data entries away from the mean of the data set. Standard deviation is usually denoted by the Greek letter ‘sigma’ s or simply by "s.d.". It is the positive square root of the arithmetic mean of the squared deviations from the mean of the distribution. It is considered as the most reliable measure of variability. It is the square root of the Variance. SD is a Bell Shaped Curve in a graph of normal distribution illustrates how much deviation or spread occurred or can be occurred from the expected value or mean of data set. The main purpose of estimating the sample standard deviation is to measure how widely the individual sample values are being dispersed from the mean. Formula: ## where ∑-“sum of” x- is a value in the data set μ- mean of the data set N- number of data points in the population Example 1 6,2,3,1 ## Step 1: Find the mean. μ= 6+2+3+1=12=3 4 4 Step 2: For each data point, find the square of its distance to the mean. 6 ∣6−3∣2=32=9 2 2−3∣2=12=1 3 ∣3−3∣2=02=0 1 1−3∣2=-22=4 ## Step 3: Sum the values from Step 2. Add up all of the squared distances from the data points to the mean from Step 2 ∑∣x−μ∣2=9+1+0+4=14 ## Step 4: Divide by the number of data points. Divide the sum from Step 3 by the number of data points (N=4) ∑∣x−μ∣2=14=3.5 N 4 ## Take the square root of the number we found in Step 4: √∑∣x−μ∣2=14=√3.5≈1.87 N 4 The standard deviation is 1.87 Example: ## Using our set of data based on the grades of student. 86 75 76 82 91 X (X – μ) (X – μ) 86 4 16 75 -7 49 76 -6 36 82 0 0 91 9 81 X=410 (X – μ) = 182 ## Solving using this formula: = 182 5 σ 2 = 36.4 Standard Deviation σ = √36.4 = 6.03 Grouped Data: ## where ∑-“sum of” F- Frequency M- Mid point n- number of samples Example In a class of students, 9 students scored 50 to 60, 7 students scored 61 to 70, 9 students scored 71 to 85, 12 students scored 86 to 95 and 8 students scored 96 to 100 in the subject of mathematics. Estimate the standard deviation? Range Frequency 50-60 9 61-70 7 71-85 9 86-95 12 96-100 8 ## Step 1: find the mid-point M for each group (50 + 60)/2 = 55 (61 + 70)/2 = 65.5 (71 + 85)/2 = 78 (86 + 95)/2 = 90.5 (96 + 100)/2 = 98 The mid points are 55, 65.5, 78, 90.5 & 98 for the group of students 50 to 60, 61 to 70, 71 to 85, 86 to 95 & 96 to 100 respectively Step 2: calculate the number of samples n n = 9 + 7 + 9 + 12 + 8 n = 45 ## Step 3: find the grouped data mean μ μ = (55 x 9 + 65.5 x 7 + 78 x 9 + 90.5 x 12 + 98 x 8 )/n = 3525.5/45 μ = 78.3444 ## = (((9 x 552) + (7 x 65.52) + (9 x 782) + (12 x 90.52) + (8 x 982)) - (45 x 78.34442))/(45 - 1) = (287127.75 - 276203.025)/ 44 σ2 = 248.2820 ## Step 5: find standard deviation for the grouped data Take square root of variance σ = 15.7569
# Measuring Angles Inside Shapes 6 teachers like this lesson Print Lesson ## Objective SWBAT find the sum of angles. #### Big Idea Students will take turns measuring and finding the sum of angles within a variety of shapes. ## Opening 20 minutes Today's Number Talk For a detailed description of the Number Talk procedure, please refer to the Number Talk Explanation. For this Number Talk, I am encouraging students to represent their thinking using a number line model. For each task today, students shared their strategies with peers (sometimes within their group, sometimes with someone across the room). It was great to see students inspiring others to try new methods and it was equally as great to see students examining each other work for possible mistakes! Getting Started Prior to the lesson, I placed magnetic money and fractions on the board to help students conceptualize our number talk today. I invited students to get a Student Number Line and Hundred Grids. I then drew a Number Line on the Board and marked 0, 1, and 2 on the line. I asked students to do the same on their own number lines. To begin, I asked students to add 1/4 + 0.6 on their number lines and hundreds grids. During this time, some students chose to work alone while others worked with a partner in their math groups. I took this time to conference with students. Next, some students volunteered to explain their reasoning out loud while I modeled their thinking on the board: 1:4 + 0.6 Teacher Demonstration Number Line Others watched carefully, checking their own number lines and hundreds grids to make sure they agreed with the thinking of other students. Here are a few examples of student work during this time: Next, we moved on to adding 1/10 + 1.5. Most students converted 1.5 to 1 5/10 and then used their number lines to take a jump of 1 5/10 and then a jump of 1/10. Here is an example of student number line and hundreds grid: Again, a few students explained how they solved this problem while I modeled their thinking on the board. Here's the end result of the last number talk task: 1:10 +1.5 Teacher Demonstration Number Line. You'll notice a list of patterns off to the right side of the picture listed above. One student had pointed out 1 6/10 = 1 3/5 which is equal to 1.6 because 1.6 = 1 6/10. Then, we discussed how 3/5 is equivalent to 0.6. I asked a few students to grab a calculator to divide 3 by 5 (3/5). They got 0.6. I asked: Can anyone else think of an equivalent fraction to 6/10? I wonder if all fractions equivalent to 6/10 are also equal to the decimal number, 0.6? Students took turns providing equivalent fractions. The students with the calculators would then check the decimal equivalency by dividing the numerator by the denominator. Each time they got 0.6! This was a fun and exciting moment for students! ## Teacher Demonstration 20 minutes Reasoning For today's lesson, I wanted to provide students with more practice using the protractor and with an opportunity to discover the pattern that all angles in a circle add up to 360 degrees. So I created shape puzzles for students to investigate. For each shape puzzle, I drew two lines intersecting in the middle of the shape. This resulted in four angles that would always have a sum of 360 degrees. After measuring and adding all four angles of several shapes, students began to realize that they always equal 360 degrees! Review To connect today's lesson with previous lessons, we began by singing our fun Angles Song Next, we reflected upon the Complementary & Supplementary Angle Poster from yesterday. Students added on the following observations on the Supplementary Angles side, "One angle is acute and the other is obtuse," and, "They can also be two right angles." Then, I reminded students of our current goal: I can find the sum of angles. Activity Explanation I pointed out the following shapes on the counter: Shape Puzzles. I wanted groups of 2-3 students to be able to choose one shape at a time from the counter to investigate, so I printed three copies of each shape to make sure students didn't have to wait on other groups to finish in order to continue their investigation. Using the Group Chart, I modeled how to record the figure in the first column. Using the pentagon, I then modeled how to measure each angle on the inside and showed students how to write an addition equation using the measurement of all the angles. I continued: Once you're done investigating all the figures on the back counter, write down your observations, and then it's your turn! Pointing to the rectangle at the bottom of the page, I explained: You get to split this rectangle up in a similar manner using a ruler or protractor. Students were ready to investigate! Attending to Precision For today's lesson in particular, students will be engaged in Math Practice 6: Attend to precision. All the angles of inside each shape is supposed to add up to 360 degrees. When students are off by just a degree or two, they'll end up with sums close to 360, such as 362 or 359. With time, students will see the pattern and will realize how important it is to attend to precision in order to get the sum of 360 degrees exactly! ## Student Practice 50 minutes Choosing Partners Picking math partners is always easy as I already have students placed in desk groups based upon behavior, abilities, and communication skills. Monitoring Student Understanding While students were working, I conferenced with every group. My goal was to support students by providing them with the opportunity to explain their thinking and by asking guiding questions. I also wanted to encourage students to construct viable arguments by using evidence to support their thinking (Math Practice 3). 1. What are you lining up? (encouraging vocabulary use) 2. Can you explain what you are noticing? 3. What do you think? Why do you think that? 4. How do you know it's not ______? (non-example) 5. Is this acute or obtuse? How do you know? 6. What are you finding out about the sum of angles inside each figure? 7. Can you explain your thinking to your partner to make sure he agrees? 8. What is the sum of all the angles? 9. What did you notice about the two angles across from each other? Conferences Here, Examining Angles Closer, a student explains how the length of angle arms does not effect the angle measurement. I also like watching them use the protractor to measure two angles at once. Here, Connecting the Sum of Angles with 360 Degrees, two students explain why all the angles add up to 360 degrees. Student Work Here's an example of student work during this time: Example of Student Work. ## Closing 10 minutes After today's investigation, I invited students to join me on the front carpet to discuss observations. One student pointed out that the arms of an angle can keep going on forever and it doesn't change the angle measurement. I drew a picture to help other students understand this student's thinking: Comparing Angle Arm Sizes. I drew a picture of two intersecting lines and labeled the angles A, B, C, and D. I then asked: What did you notice when measuring angles the result from two lines intersecting? I labeled the drawing, Sharing Observations, as students shared the following observations: 1. Angles A and C are congruent. 2. Angles D and B are congruent too! 3. Angles A and B have a sum of 180 degrees. 4. Angles D and C have a sum of 180 degrees too! 5. Altogether, A + B + C + D = 360 degrees. I then asked: Based on your investigation today, do you think this is true with all intersecting lines?Most students said, "Yes!"
# Lesson 4: Chapter 4 Sections 1-2 Size: px Start display at page: Transcription 1 Lesson 4: Chapter 4 Sections 1-2 Caleb Moxley BSC Mathematics 14 September 15 2 4.1 Randomness What s randomness? 3 4.1 Randomness What s randomness? Definition (random) A phenomenon is random if individual outcomes are uncertain (in some sense) but nonetheless display a regular distribution of outcomes when trails are repeated a large number of times. 4 4.1 Randomness What s randomness? Definition (random) A phenomenon is random if individual outcomes are uncertain (in some sense) but nonetheless display a regular distribution of outcomes when trails are repeated a large number of times. Example (deck of cards) Imagine drawing a card from a 52-card deck. 5 4.1 Randomness What s randomness? Definition (random) A phenomenon is random if individual outcomes are uncertain (in some sense) but nonetheless display a regular distribution of outcomes when trails are repeated a large number of times. Example (deck of cards) Imagine drawing a card from a 52-card deck. What card would you draw? 6 4.1 Randomness What s randomness? Definition (random) A phenomenon is random if individual outcomes are uncertain (in some sense) but nonetheless display a regular distribution of outcomes when trails are repeated a large number of times. Example (deck of cards) Imagine drawing a card from a 52-card deck. What card would you draw? If you repeated this process 52 times, how often would you expect to have drawn an ace? 7 4.1 Randomness When the same random phenomenon is repeated a large number of times, we start to see the pattern/distribution that arises among the possible outcomes. 8 4.1 Randomness When the same random phenomenon is repeated a large number of times, we start to see the pattern/distribution that arises among the possible outcomes. Definition (probability) The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions. 9 4.1 Randomness When the same random phenomenon is repeated a large number of times, we start to see the pattern/distribution that arises among the possible outcomes. Definition (probability) The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions. Example (deck of cards) What s the probability of drawing an ace from a 52-card deck? 10 4.1 Randomness When the same random phenomenon is repeated a large number of times, we start to see the pattern/distribution that arises among the possible outcomes. Definition (probability) The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions. Example (deck of cards) What s the probability of drawing an ace from a 52-card deck? You often can calculate the probability of an outcome or event if you make assumptions about the phenomenon rather than having to perform a long series of repetitions. 11 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. 12 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. People often underestimate how many times you d need to repeat the phenomenon before the probabilities converge to their true probabilities. 13 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. People often underestimate how many times you d need to repeat the phenomenon before the probabilities converge to their true probabilities. This notion is completely analogous to the need to take large samples to get a reliable estimate of a population parameter based on a sample statistic. 14 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. People often underestimate how many times you d need to repeat the phenomenon before the probabilities converge to their true probabilities. This notion is completely analogous to the need to take large samples to get a reliable estimate of a population parameter based on a sample statistic. 1 Performing many trials can be difficult, so estimating probabilities or calculating them using mathematical assumptions is often necessary. 15 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. People often underestimate how many times you d need to repeat the phenomenon before the probabilities converge to their true probabilities. This notion is completely analogous to the need to take large samples to get a reliable estimate of a population parameter based on a sample statistic. 1 Performing many trials can be difficult, so estimating probabilities or calculating them using mathematical assumptions is often necessary. 2 Trials must always be independent, i.e the outcomes of one trail must not depend on the outcome of any others. 16 4.1 Randomness In determining probabilities based on a long series of repetitions, it s important that the series be very long. People often underestimate how many times you d need to repeat the phenomenon before the probabilities converge to their true probabilities. This notion is completely analogous to the need to take large samples to get a reliable estimate of a population parameter based on a sample statistic. 1 Performing many trials can be difficult, so estimating probabilities or calculating them using mathematical assumptions is often necessary. 2 Trials must always be independent, i.e the outcomes of one trail must not depend on the outcome of any others. 3 Often we can imitate random behavior rather than actually performing the trials. We call this simulation. 17 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. 18 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like 19 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, 20 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, natural/environmental random phenomena, 21 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, natural/environmental random phenomena, mortality, 22 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, natural/environmental random phenomena, mortality, traffic/queuing theory, 23 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, natural/environmental random phenomena, mortality, traffic/queuing theory, ergodicity/particle movement/brownian motion, 24 4.1 Randomness A rigorous study of games of chance wasn t realized until the 16th and 17th Centuries. Since then, the study of probability has grown to a discipline that helps us understand things like gambling, natural/environmental random phenomena, mortality, traffic/queuing theory, ergodicity/particle movement/brownian motion, finance, and much more. 25 4.1 Randomness Mastery Question: Example (independence) Are the following independent trails? 26 4.1 Randomness Mastery Question: Example (independence) Are the following independent trails? The number of holiday cards you receive over a period of ten years. 27 4.1 Randomness Mastery Question: Example (independence) Are the following independent trails? The number of holiday cards you receive over a period of ten years. The temperature at Vulcan on January 1st each year for three years. 28 4.1 Randomness Mastery Question: Example (independence) Are the following independent trails? The number of holiday cards you receive over a period of ten years. The temperature at Vulcan on January 1st each year for three years. The results of five coin tosses where the coin is 40% likely to land on heads and 60% likely to land on tails. 29 4.1 Randomness Mastery Question: Example (random) Are the following random phenomena? If so, explain how a probability applies to them. The number of cars a particular family owns. 30 4.1 Randomness Mastery Question: Example (random) Are the following random phenomena? If so, explain how a probability applies to them. The number of cars a particular family owns. Whether or not a medical student passes her board exams. 31 4.1 Randomness Mastery Question: Example (random) Are the following random phenomena? If so, explain how a probability applies to them. The number of cars a particular family owns. Whether or not a medical student passes her board exams. The numbers selected in a lottery drawing. 32 A probability model consists of a list of possible outcomes and the corresponding probability of each outcome. 33 A probability model consists of a list of possible outcomes and the corresponding probability of each outcome. This two-part description fully characterizes a phenomenon s randomness. 34 A probability model consists of a list of possible outcomes and the corresponding probability of each outcome. This two-part description fully characterizes a phenomenon s randomness. Definition (outcome) An outcome of a random phenomenon is a possible result from a trail which cannot be broken down into simpler results. 35 A probability model consists of a list of possible outcomes and the corresponding probability of each outcome. This two-part description fully characterizes a phenomenon s randomness. Definition (outcome) An outcome of a random phenomenon is a possible result from a trail which cannot be broken down into simpler results. Are the following outcomes of rolling a six-sided die or not? 36 A probability model consists of a list of possible outcomes and the corresponding probability of each outcome. This two-part description fully characterizes a phenomenon s randomness. Definition (outcome) An outcome of a random phenomenon is a possible result from a trail which cannot be broken down into simpler results. Are the following outcomes of rolling a six-sided die or not? rolling a three rolling an odd number rolling an even number less than four 37 Definition (sample space) The sample space of a random phenomenon is the set of all possible outcomes. 38 Definition (sample space) The sample space of a random phenomenon is the set of all possible outcomes. Definition (event) An event is an outcome, a set of outcomes, or the entire collection of all possible outcomes of a random phenomenon. That is, it s any subset of the sample space - possibly the whole sample space. 39 Definition (sample space) The sample space of a random phenomenon is the set of all possible outcomes. Definition (event) An event is an outcome, a set of outcomes, or the entire collection of all possible outcomes of a random phenomenon. That is, it s any subset of the sample space - possibly the whole sample space. Example (number of children) A population of four families have 0, 2, 2, 1 children in each family. If we selected one family from this population at random (with equal likelihood), what is the sample space for this random trail? 40 Definition (sample space) The sample space of a random phenomenon is the set of all possible outcomes. Definition (event) An event is an outcome, a set of outcomes, or the entire collection of all possible outcomes of a random phenomenon. That is, it s any subset of the sample space - possibly the whole sample space. Example (number of children) A population of four families have 0, 2, 2, 1 children in each family. If we selected one family from this population at random (with equal likelihood), what is the sample space for this random trail? What are the outcomes? 41 Definition (sample space) The sample space of a random phenomenon is the set of all possible outcomes. Definition (event) An event is an outcome, a set of outcomes, or the entire collection of all possible outcomes of a random phenomenon. That is, it s any subset of the sample space - possibly the whole sample space. Example (number of children) A population of four families have 0, 2, 2, 1 children in each family. If we selected one family from this population at random (with equal likelihood), what is the sample space for this random trail? What are the outcomes? What is the set of all possible events? 42 The following facts are true for all probabilities and all probability models. 43 The following facts are true for all probabilities and all probability models. Any probability is a number between 0 and 1. 44 The following facts are true for all probabilities and all probability models. Any probability is a number between 0 and 1. The event containing all possible outcomes (i.e. the entire sample space) must have a probability equal to 1. 45 The following facts are true for all probabilities and all probability models. Any probability is a number between 0 and 1. The event containing all possible outcomes (i.e. the entire sample space) must have a probability equal to 1. If two events are disjoint (i.e. if they have no outcomes in common), then the probability that one or the other occurs is the sum of their individual probabilities. 46 The following facts are true for all probabilities and all probability models. Any probability is a number between 0 and 1. The event containing all possible outcomes (i.e. the entire sample space) must have a probability equal to 1. If two events are disjoint (i.e. if they have no outcomes in common), then the probability that one or the other occurs is the sum of their individual probabilities. The probability that an event does not occur is 1 minus the probability that the event does occur. 47 The following facts are true for all probabilities and all probability models. Any probability is a number between 0 and 1. The event containing all possible outcomes (i.e. the entire sample space) must have a probability equal to 1. If two events are disjoint (i.e. if they have no outcomes in common), then the probability that one or the other occurs is the sum of their individual probabilities. The probability that an event does not occur is 1 minus the probability that the event does occur. We summarize these facts in a table on the next slide. 48 Rule 1 0 P(A) 1 Rule 2 P(S) = 1 Rule 3 P(A) + P(B) = P(A or B) if A and B are disjoint Rule 4 P(A c ) = 1 P(A) 49 Rule 1 0 P(A) 1 Rule 2 P(S) = 1 Rule 3 P(A) + P(B) = P(A or B) if A and B are disjoint Rule 4 P(A c ) = 1 P(A) It s important to understand how these identities arise, and looking at Venn diagrams can be useful for these and other identities. 50 Use probability identities and Venn diagrams to help answer the following questions. 51 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other 52 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? 53 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? 0.82. 54 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? What s the probability that a student has either red or brown hair? 55 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? What s the probability that a student has either red or brown hair? 0.36. 56 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? What s the probability that a student has either red or brown hair? What s the probability that a student has neither black nor blond hair? 57 Use probability identities and Venn diagrams to help answer the following questions. Example (hair color) The table gives the proportion of hair colors in a class. Black Blond Brown Red White/Grey Other What s the probability that a student has a known hair color? What s the probability that a student has either red or brown hair? What s the probability that a student has neither black nor blond hair? 0.58. 58 When a sample space has a finite possible number of outcomes: 59 When a sample space has a finite possible number of outcomes: Assign a probability to each outcome using some rule/description of the phenomenon. These probabilities must sum to 1 and be between 0 and 1. 60 When a sample space has a finite possible number of outcomes: Assign a probability to each outcome using some rule/description of the phenomenon. These probabilities must sum to 1 and be between 0 and 1. The probability of an event can be calculated as the sum of the probabilities of the outcomes which make it up. 61 When a sample space has a finite possible number of outcomes: Assign a probability to each outcome using some rule/description of the phenomenon. These probabilities must sum to 1 and be between 0 and 1. The probability of an event can be calculated as the sum of the probabilities of the outcomes which make it up. Example (rolling dice) Create the probability model for rolling two dice. 62 When a sample space has a finite possible number of outcomes: Assign a probability to each outcome using some rule/description of the phenomenon. These probabilities must sum to 1 and be between 0 and 1. The probability of an event can be calculated as the sum of the probabilities of the outcomes which make it up. Example (rolling dice) Create the probability model for rolling two dice 63 To construct the previous probability model for two dice, we used the following rule. 64 To construct the previous probability model for two dice, we used the following rule. Theorem (equally likely outcomes) If a random phenomenon has k possible outcomes, all equally likely, then each individual outcome has probability 1 k. The probability of any event A is P(A) = count of outcomes in A k 65 To construct the previous probability model for two dice, we used the following rule. Theorem (equally likely outcomes) If a random phenomenon has k possible outcomes, all equally likely, then each individual outcome has probability 1 k. The probability of any event A is P(A) = count of outcomes in A k We didn t use this rule exactly because we pinched together certain outcomes since their result was the same in some sense, but the idea is the same. 66 When a phenomenon includes two procedures like our rolling of two dice, it can be difficult to assign probabilities to the sample space. However, this process simplifies when you can assume that the events are independent. 67 When a phenomenon includes two procedures like our rolling of two dice, it can be difficult to assign probabilities to the sample space. However, this process simplifies when you can assume that the events are independent. Definition (independence) Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, then P(A and B) = P(A) P(B). We call this the multiplication rule for independent events. 68 Example (independent events) Are the following independent events or not? 69 Example (independent events) Are the following independent events or not? 1 Drawing an ace from one deck of cards and then drawing a king from another deck. 70 Example (independent events) Are the following independent events or not? 1 Drawing an ace from one deck of cards and then drawing a king from another deck. 2 Drawing an ace from one deck of cards and then drawing a king from the same deck without replacing the ace first. 71 Example (independent events) Are the following independent events or not? 1 Drawing an ace from one deck of cards and then drawing a king from another deck. 2 Drawing an ace from one deck of cards and then drawing a king from the same deck without replacing the ace first. 3 Event A is disjoint from Event B. Are they independent? 72 With all these rules at our disposal, we can now calculate probabilities of somewhat complicated events. 73 With all these rules at our disposal, we can now calculate probabilities of somewhat complicated events. What s the probability that in a room of 25 people at least two people share a birthday? 74 With all these rules at our disposal, we can now calculate probabilities of somewhat complicated events. What s the probability that in a room of 25 people at least two people share a birthday? 56.87%. ### Probability. Ms. Weinstein Probability & Statistics Probability Ms. Weinstein Probability & Statistics Definitions Sample Space The sample space, S, of a random phenomenon is the set of all possible outcomes. Event An event is a set of outcomes of a random ### Probability Models. Section 6.2 Probability Models Section 6.2 The Language of Probability What is random? Empirical means that it is based on observation rather than theorizing. 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Section 7.1 - Experiments, Sample Spaces, ### CHAPTER 2 PROBABILITY. 2.1 Sample Space. 2.2 Events CHAPTER 2 PROBABILITY 2.1 Sample Space A probability model consists of the sample space and the way to assign probabilities. Sample space & sample point The sample space S, is the set of all possible outcomes ### Math Exam 2 Review. NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5. Math 166 Spring 2007 c Heather Ramsey Page 1 Math 166 - Exam 2 Review NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5. Section 7.1 - Experiments, Sample Spaces, ### Math 147 Lecture Notes: Lecture 21 Math 147 Lecture Notes: Lecture 21 Walter Carlip March, 2018 The Probability of an Event is greater or less, according to the number of Chances by which it may happen, compared with the whole number of ### "Well, statistically speaking, you are for more likely to have an accident at an intersection, so I just make sure that I spend less time there. 6.2 Probability Models There was a statistician who, when driving his car, would always accelerate hard before coming to an intersection, whiz straight through it, and slow down again once he was beyond ### STAT 3743: Probability and Statistics STAT 3743: Probability and Statistics G. Jay Kerns, Youngstown State University Fall 2010 Probability Random experiment: outcome not known in advance Sample space: set of all possible outcomes (S) Probability ### If a regular six-sided die is rolled, the possible outcomes can be listed as {1, 2, 3, 4, 5, 6} there are 6 outcomes. Section 11.1: The Counting Principle 1. Combinatorics is the study of counting the different outcomes of some task. For example If a coin is flipped, the side facing upward will be a head or a tail the ### Module 4 Project Maths Development Team Draft (Version 2) 5 Week Modular Course in Statistics & Probability Strand 1 Module 4 Set Theory and Probability It is often said that the three basic rules of probability are: 1. Draw a picture 2. Draw a picture 3. Draw ### Chapter 4: Probability and Counting Rules Chapter 4: Probability and Counting Rules Before we can move from descriptive statistics to inferential statistics, we need to have some understanding of probability: Ch4: Probability and Counting Rules ### Exam III Review Problems c Kathryn Bollinger and Benjamin Aurispa, November 10, 2011 1 Exam III Review Problems Fall 2011 Note: Not every topic is covered in this review. Please also take a look at the previous Week-in-Reviews ### Simple Probability. Arthur White. 28th September 2016 Simple Probability Arthur White 28th September 2016 Probabilities are a mathematical way to describe an uncertain outcome. For eample, suppose a physicist disintegrates 10,000 atoms of an element A, and ### Probability with Set Operations. MATH 107: Finite Mathematics University of Louisville. March 17, Complicated Probability, 17th century style Probability with Set Operations MATH 107: Finite Mathematics University of Louisville March 17, 2014 Complicated Probability, 17th century style 2 / 14 Antoine Gombaud, Chevalier de Méré, was fond of gambling ### Probability - Grade 10 * OpenStax-CNX module: m32623 1 Probability - Grade 10 * Rory Adams Free High School Science Texts Project Sarah Blyth Heather Williams This work is produced by OpenStax-CNX and licensed under the Creative ### Section 6.5 Conditional Probability Section 6.5 Conditional Probability Example 1: An urn contains 5 green marbles and 7 black marbles. Two marbles are drawn in succession and without replacement from the urn. a) What is the probability ### Objective: Determine empirical probability based on specific sample data. (AA21) Do Now: What is an experiment? List some experiments. What types of things does one take a "chance" on? Mar 1 3:33 PM Date: Probability - Empirical - By Experiment Objective: Determine empirical probability ### 1. The chance of getting a flush in a 5-card poker hand is about 2 in 1000. CS 70 Discrete Mathematics for CS Spring 2008 David Wagner Note 15 Introduction to Discrete Probability Probability theory has its origins in gambling analyzing card games, dice, roulette wheels. Today ### Probability: Terminology and Examples Spring January 1, / 22 Probability: Terminology and Examples 18.05 Spring 2014 January 1, 2017 1 / 22 Board Question Deck of 52 cards 13 ranks: 2, 3,..., 9, 10, J, Q, K, A 4 suits:,,,, Poker hands Consists of 5 cards A one-pair ### What Do You Expect? Concepts Important Concepts What Do You Expect? Concepts Examples Probability A number from 0 to 1 that describes the likelihood that an event will occur. Theoretical Probability A probability obtained by analyzing ### Contents 2.1 Basic Concepts of Probability Methods of Assigning Probabilities Principle of Counting - Permutation and Combination 39 CHAPTER 2 PROBABILITY Contents 2.1 Basic Concepts of Probability 38 2.2 Probability of an Event 39 2.3 Methods of Assigning Probabilities 39 2.4 Principle of Counting - Permutation and Combination 39 2.5 ### [Independent Probability, Conditional Probability, Tree Diagrams] Name: Year 1 Review 11-9 Topic: Probability Day 2 Use your formula booklet! Page 5 Lesson 11-8: Probability Day 1 [Independent Probability, Conditional Probability, Tree Diagrams] Read and Highlight Station ### PROBABILITY Case of cards WORKSHEET NO--1 PROBABILITY Case of cards WORKSHEET NO--2 Case of two die Case of coins WORKSHEET NO--3 1) Fill in the blanks: A. The probability of an impossible event is B. The probability of a sure ### Question of the Day. Key Concepts. Vocabulary. Mathematical Ideas. QuestionofDay QuestionofDay Question of the Day There are 31 educators from the state of Nebraska currently enrolled in Experimentation, Conjecture, and Reasoning. What is the probability that two participants in our ### Discrete Mathematics and Probability Theory Spring 2016 Rao and Walrand Note 13 CS 70 Discrete Mathematics and Probability Theory Spring 2016 Rao and Walrand Note 13 Introduction to Discrete Probability In the last note we considered the probabilistic experiment where we flipped a ### The topic for the third and final major portion of the course is Probability. We will aim to make sense of statements such as the following: CS 70 Discrete Mathematics for CS Spring 2006 Vazirani Lecture 17 Introduction to Probability The topic for the third and final major portion of the course is Probability. We will aim to make sense of ### 1. How many subsets are there for the set of cards in a standard playing card deck? How many subsets are there of size 8? Math 1711-A Summer 2016 Final Review 1 August 2016 Time Limit: 170 Minutes Name: 1. How many subsets are there for the set of cards in a standard playing card deck? How many subsets are there of size 8? ### heads 1/2 1/6 roll a die sum on 2 dice 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 1, 2, 3, 4, 5, 6 heads tails 3/36 = 1/12 toss a coin trial: an occurrence trial: an occurrence roll a die toss a coin sum on 2 dice sample space: all the things that could happen in each trial 1, 2, 3, 4, 5, 6 heads tails 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 example of an outcome: ### Textbook: pp Chapter 2: Probability Concepts and Applications 1 Textbook: pp. 39-80 Chapter 2: Probability Concepts and Applications 2 Learning Objectives After completing this chapter, students will be able to: Understand the basic foundations of probability analysis. ### Midterm 2 Practice Problems Midterm 2 Practice Problems May 13, 2012 Note that these questions are not intended to form a practice exam. They don t necessarily cover all of the material, or weight the material as I would. They are ### I. WHAT IS PROBABILITY? C HAPTER 3 PROAILITY Random Experiments I. WHAT IS PROAILITY? The weatherman on 10 o clock news program states that there is a 20% chance that it will snow tomorrow, a 65% chance that it will rain and ### Probability Test Review Math 2. a. What is? b. What is? c. ( ) d. ( ) Probability Test Review Math 2 Name 1. Use the following venn diagram to answer the question: Event A: Odd Numbers Event B: Numbers greater than 10 a. What is? b. What is? c. ( ) d. ( ) 2. In Jason's homeroom ### The probability set-up CHAPTER 2 The probability set-up 2.1. Introduction and basic theory We will have a sample space, denoted S (sometimes Ω) that consists of all possible outcomes. For example, if we roll two dice, the sample ### Name: Class: Date: 6. An event occurs, on average, every 6 out of 17 times during a simulation. The experimental probability of this event is 11 Class: Date: Sample Mastery # Multiple Choice Identify the choice that best completes the statement or answers the question.. One repetition of an experiment is known as a(n) random variable expected value ### Probability - Chapter 4 Probability - Chapter 4 In this chapter, you will learn about probability its meaning, how it is computed, and how to evaluate it in terms of the likelihood of an event actually happening. A cynical person ### Chapter 3: Elements of Chance: Probability Methods Chapter 3: Elements of Chance: Methods Department of Mathematics Izmir University of Economics Week 3-4 2014-2015 Introduction In this chapter we will focus on the definitions of random experiment, outcome, ### The Teachers Circle Mar. 20, 2012 HOW TO GAMBLE IF YOU MUST (I ll bet you \$5 that if you give me \$10, I ll give you \$20.) The Teachers Circle Mar. 2, 22 HOW TO GAMBLE IF YOU MUST (I ll bet you \$ that if you give me \$, I ll give you \$2.) Instructor: Paul Zeitz (zeitzp@usfca.edu) Basic Laws and Definitions of Probability If ### Section : Combinations and Permutations Section 11.1-11.2: Combinations and Permutations Diana Pell A construction crew has three members. A team of two must be chosen for a particular job. In how many ways can the team be chosen? How many words ### EECS 203 Spring 2016 Lecture 15 Page 1 of 6 EECS 203 Spring 2016 Lecture 15 Page 1 of 6 Counting We ve been working on counting for the last two lectures. We re going to continue on counting and probability for about 1.5 more lectures (including ### Raise your hand if you rode a bus within the past month. Record the number of raised hands. 166 CHAPTER 3 PROBABILITY TOPICS Raise your hand if you rode a bus within the past month. Record the number of raised hands. Raise your hand if you answered "yes" to BOTH of the first two questions. Record ### Introduction to Probability and Statistics I Lecture 7 and 8 Introduction to Probability and Statistics I Lecture 7 and 8 Basic Probability and Counting Methods Computing theoretical probabilities:counting methods Great for gambling! Fun to compute! If outcomes ### CS 361: Probability & Statistics January 31, 2018 CS 361: Probability & Statistics Probability Probability theory Probability Reasoning about uncertain situations with formal models Allows us to compute probabilities Experiments will ### Business Statistics. Chapter 4 Using Probability and Probability Distributions QMIS 120. Dr. Mohammad Zainal Department of Quantitative Methods & Information Systems Business Statistics Chapter 4 Using Probability and Probability Distributions QMIS 120 Dr. Mohammad Zainal Chapter Goals After completing this chapter, ### The next several lectures will be concerned with probability theory. We will aim to make sense of statements such as the following: CS 70 Discrete Mathematics for CS Fall 2004 Rao Lecture 14 Introduction to Probability The next several lectures will be concerned with probability theory. We will aim to make sense of statements such ### Probability and Counting Rules. Chapter 3 Probability and Counting Rules Chapter 3 Probability as a general concept can be defined as the chance of an event occurring. Many people are familiar with probability from observing or playing games of ### Before giving a formal definition of probability, we explain some terms related to probability. probability 22 INTRODUCTION In our day-to-day life, we come across statements such as: (i) It may rain today. (ii) Probably Rajesh will top his class. (iii) I doubt she will pass the test. (iv) It is unlikely
# How do you simplify 6^-6/6^-5? Dec 26, 2016 $= \frac{1}{6}$ #### Explanation: There are 2 laws of indices going on here - you can apply them in any order. Use whichever method you prefer. Answers are usually given with positive indices. (An exception in with scientific notation where a negative index means the number is a decimal fraction.) When dividing, subtract the indices of like bases: $\rightarrow \text{ } {x}^{m} / {x}^{n} = {x}^{m - n}$ Change a negative to a positive index by using the reciprocal $\rightarrow \text{ "x^-m = 1/x^m" and } \frac{1}{x} ^ - n = {x}^{n}$ $= \frac{\textcolor{b l u e}{{6}^{-} 6}}{6} ^ - 5 = \frac{1}{\textcolor{b l u e}{{6}^{6}} \times {6}^{-} 5}$ $= \frac{1}{6}$ OR: (6^-6)/6^-5 = 6^(-6-(-5) $= {6}^{- 6 + 5}$ $= {6}^{-} 1$ $\frac{1}{6}$ OR: $\frac{{6}^{-} 6}{6} ^ - 5$ $= \frac{{6}^{5}}{6} ^ 6$ $= \frac{1}{6}$
# “The Mystery of the Square Root of 6 Unveiled” The square root of 6 has puzzled mathematicians for centuries. It is an irrational number – meaning that it cannot be expressed as a simple fraction – and its decimal representation goes on indefinitely without repeating. But what exactly is the square root of 6? And how can we make sense of this mysterious number? To answer these questions, let’s start by considering what it means to take a square root. The square root of a number is the value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, since 3 x 3 = 9. In the case of 6, however, there is no whole number that satisfies this property. If we try to find the square root of 6 using a calculator or computer program, we’ll get a long, non-repeating decimal: √6 ≈ 2.449489742783178 This number may seem arbitrary, but it has some interesting properties. For one thing, it is an example of a quadratic irrational – a type of irrational number that can be expressed as the solution to a quadratic equation with rational coefficients. Specifically, the square root of 6 is the positive root of the equation x^2 = 6. But what does this mean in practical terms? One way to interpret the square root of 6 is as the length of the diagonal of a square with area 6. To see why this is true, imagine that we have a square with sides of length 6. By the Pythagorean theorem (which states that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides), the length of the diagonal of this square is: √(6^2 + 6^2) = √72 We can simplify this expression by factoring out 6 from both terms: √(6^2 + 6^2) = 6√2 Thus, the length of the diagonal of this square is 6√2. Since the area of a square is equal to the length of one side squared, we can solve for the side length of a square with area 6 by taking the square root of 6: side length = √6 And since the diagonal of this square is 6√2, we can conclude that: diagonal length = √(6 x 2) = √12 = 2√3 This means that the square root of 6 is also related to the square roots of other numbers – specifically, the square root of 2 and the square root of 3. In fact, we can express the square root of 6 in terms of these two numbers using the Pythagorean theorem: √6 = √(2 x 3) = √2 x √3 This equation shows that the square root of 6 can be decomposed into two smaller square roots. This property is related to the concept of prime factorization – the idea that every positive integer can be expressed as a unique product of prime numbers. In this case, the prime factors are 2 and 3, and the square root of 6 is their product. But why stop at just two factors? We can continue this process and express the square root of 6 as a product of many more square roots. For example: √6 = √(2 x 3) = √2 x √3 = √(2 x 2 x 3) = √2 x √2 x √3 = √(2 x 2 x 3 x 3) = √2 x √2 x √3 x √3 = √(2^2 x 3^2) = 2√3 By decomposing the square root of 6 in this way, we can see that it is related not just to two numbers, but to an entire network of factors and products. This illustrates the interconnectedness of mathematical ideas – how one concept can lead to another, and how seemingly disparate pieces of knowledge can be combined to form a larger whole. In conclusion, the mystery of the square root of 6 has been unveiled – or at least, its properties have been explained. We now know that it is an irrational number that represents the length of the diagonal of a square with area 6, and that it can be decomposed into smaller factors such as the square root of 2 and the square root of 3. But beyond its practical uses, the square root of 6 also serves as a reminder of the elegance and complexity of mathematics – a subject that continues to surprise and amaze us with its beauty.
Tag Archives: rule of four Graphing Ratios and Proportions Last week, some colleagues and I were pondering the difficulties many middle school students have solving ratio and proportion problems. Here are a few thoughts we developed to address this and what we think might be an uncommon graphical extension (for most) as a different way to solve. For context, consider the equation $\displaystyle \frac{x}{6} = \frac{3}{4}$. (UNFORTUNATE) STANDARD METHOD: The default procedure most textbooks and students employ is cross-multiplication. Using this, a student would get $\displaystyle 4x=18 \longrightarrow x = \frac{18}{4} = \frac{9}{2}$ While this delivers a quick solution, we sadly noted that far too many students don’t really seem to know why the procedure works. From my purist mathematical perspective, the cross-multiplication procedure may be an efficient algorithm, but cross-multiplication isn’t actually a mathematical function. Cross-multiplication may be the result, but it isn’t what happens. METHOD 2: In every math class I teach at every grade level, my mantra is to memorize as little as possible and to use what you know as broadly as possible. To avoid learning unnecessary, isolated procedures (like cross-multiplication), I propose “fraction-clearing”–multiplying both sides of an equation by common denominatoras a universal technique in any equation involving fractions. As students’ mathematical and symbolic sophistication grows, fraction-clearing may occasionally yield to other techniques, but it is a solid, widely-applicable approach for developing algebraic thinking. From the original equation, multiply both sides by common denominator, handle all of the divisions first, and clean up. For our example, the common denominator 24 will do the trick. $\displaystyle 24 \cdot \frac{x}{6} = 24 \cdot \frac{3}{4}$ $4 \cdot x = 6 \cdot 3$ $\displaystyle x = \frac{9}{2}$ Notice that the middle line is precisely the result of cross-multiplication. Fraction-clearing is the procedure behind cross-multiplication and explains exactly why it works: You have an equation and apply the same operation (in our case, multiplying by 24) to both sides. As an aside, I’d help students see that multiplying by any common denominator would do the trick (for our example, 12, 24, 36, 48, … all work), but the least common denominator (12) produces the smallest products in line 2, potentially simplifying any remaining algebra. Since many approaches work, I believe students should be free to use ANY common denominator they want. Eventually, they’ll convince themselves that the LCD is just more efficient, but there’s absolutely no need to demand that of students from the outset. METHOD 3: Remember that every equation compares two expressions that have the same measure, size, value, whatever. But fractions with differing denominators (like our given equation) are difficult to compare. Rewrite the expressions with the same “units” (denominators) to simplify comparisons. Fourths and sixths can both be rewritten in twelfths. Then, since the two different expressions of twelfths are equivalent, their numerators must be equivalent, leading to our results from above. $\displaystyle \frac{2}{2} \cdot \frac{x}{6} = \frac{3}{3} \cdot \frac{3}{4}$ $\displaystyle \frac{2x}{12} = \frac {9}{12}$ $2x=9$ $\displaystyle x = \frac{9}{2}$ I find this approach more appealing as the two fractions never actually interact. Fewer moving pieces makes this approach feel much cleaner. UNCOMMON(?) METHOD 4: Graphing A fundamental mathematics concept (for me) is the Rule of 4 from the calculus reform movement of the 1990s. That is, mathematical ideas can be represented numerically, algebraically, graphically, and verbally. [I’d extend this to a Rule of 5 to include computer/CAS representations, but that’s another post.] If you have difficulty understanding an idea in one representation, try translating it into a different representation and you might gain additional insights, or even a solution. At a minimum, the act of translating the idea deepens your understanding. One problem many students have with ratios is that teachers almost exclusively teach them as an algebraic technique–just as I have done in the first three methods above. In my conversation this week, I finally recognized this weakness and wondered how I could solve ratios using one of the missing Rules: graphically. Since equivalent fractions could be seen as different representations of the slope of a line through the origin, I had my answer. Students learning ratios and proportions may not seen slope yet and may or may not have seen an xy-coordinate grid, so I’d avoid initial use of any formal terminology. I labeled my vertical axis “Top,” and the horizontal “Bottom”. More formal names are fine, but unnecessary. While I suspect most students might think “top” makes more sense for a vertical axis and “bottom” for the horizontal, it really doesn’t matter which axis receives which label. In the purely numeric fraction in our given problem, $\displaystyle \frac{x}{6} = \frac{3}{4}$, “3” is on top, and “4” is on the bottom. Put a point at the place where these two values meet. Finally draw a line connecting your point and the origin. The other fraction has a “6” in the denominator. Locate 6 on the “bottom axis”, trace to the line, and from there over to the “top axis” to find the top value of 4.5. Admittedly, the 4.5 solution would have been a rough guess without the earlier solutions, but the graphical method would have given me a spectacular estimate. If the graph grid was scaled by 0.5s instead of by 1s and the line was drawn very carefully, this graph could have given an exact answer. In general, solutions with integer-valued unknowns should solve exactly, but very solid approximations would always result. CONCLUSION: Even before algebraic representations of lines are introduced, students can leverage the essence of that concept to answer proportion problems. Serendipitously, the graphical approach also sets the stage for later discussions of the coordinate plane, slope, and linear functions. I could also see using this approach as the cornerstone of future class conversations and discoveries leading to those generalizations. I suspect that students who struggle with mathematical notation might find greater understanding with the graphical/visual approach. Eventually, symbolic manipulation skills will be required, but there is no need for any teacher to expect early algebra learners to be instant masters of abstract notation. Exploring Sequences and Lines Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations. There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence. What do all members of this family have in common? As with most great math problems, the problem is simply stated and can be approached from many different perspectives. In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further. WARNING: Problem Solution Follows MOST COMMON STUDENT APPROACH: Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences. After a few tries, most get a little frustrated as no obvious pattern emerges from the equations. It is this point that is typically the most difficult for me as a teacher. I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own. Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?” Eventually, most decide to graph their equations to see if anything pops out. The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator. Two lines intersecting in a point is common. Three or more in a single point almost always indicates something interesting. Seven lines through a single point is screaming for attention! From this graph, all lines in this family apparently contain the point (1,-2). That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician. Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family. NOTE: All graphs my students have produced over the years have always contained specific equations. I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d). UNIQUE SOLUTION METHODS FROM STUDENTS: All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A. Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get $Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$. Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family. A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2. Method II: A few students every year work algebraically from properties of arithmetic sequences. For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$. This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences. Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family. Method III: This year, I had a student take an approach I’d never seen before. She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n. This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ . If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha. Still, this system is pretty easy to manipulate. If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving $m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$. Solving for y shows that all of the coefficients simplify surprisingly easily. $((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$ $(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$ From here, determining $x=1$ is easy, proving the relationship. SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS: Approach A: High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems. One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$. Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$. Obviously, the only thing these two lines have in common is the point (1,-2). A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property. Approach B: A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ Collecting like terms gives $(x+y+1)\cdot a+(y+2)\cdot d=0$. The values of a and d must remain as parameters to include all possible arithmetic sequences. Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$. EXTENSION: We once had a test question at the end of the unit containing this exercise. Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2). It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2). The two most common responses we’ve seen involve a reflection or a vertical translation. (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick: $Ax-By+C=0$. Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.
# Divisibility - Math Olympiad Show that for any positive integer $m$, there is an infinite number of pairs of integers $(x,y)$ satisfying the conditions: i) $\gcd(x,y)=1$; ii) $y \mid x^2+m$; iii) $x \mid y^2+m$. - ## 2 Answers I lead off by settling the case $m=1$ by applying a technique from an old answer by Bill Dubuque. Only posting this, because I think this has useful, generalizable ideas (but won't take us the distance). We observe that $(x,y)=(5,2)$ is a solution. In our case $m=1$ we can painlessly conclude that $\gcd(x,m)=\gcd(y,m)=1$. Therefore also $\gcd(x,x^2+m)=\gcd(y,y^2+m)=1$. Therefore the conditions ii)- iii) can be combined to read (keeping condition i) $$xy\mid(x^2+1)(y^2+1),$$ which is equivalent to the requirement $$xy\mid x^2+y^2+1.$$ Given that the single solution $(x,y)=(5,2)$ satisfies the equation $$3xy=x^2+y^2+1 \qquad()$$ we are lead to the conclusion that if equation $()$ has an infinite number of integer solutions (necessarily coprime), then that settles the claim in the case $m=1$. Here's the key idea. If $(x,y)$ is a solution of $(*)$, then $(3x-y,x)$ is another. This follows immediately from the observation that for a fixed $x$ the two solutions for $y$ of the quadratic equation sum up to $3x$ (=the negative of the coefficient of the linear term). Using this recursively allows us to generate an infinite number of solutions: $$(5,2)\mapsto (13,5)\mapsto (34,13)\mapsto (89,34)\mapsto\cdots$$ I leave it as an exercise for the reader to show that the numbers in this sequence also appear in the Fibonacci sequence. To settle the general case we need a mechanism for finding a single solution $(x,y)$ such that both $x$ and $y$ are coprime to $m$. The above machinery most likely then spews out an infinite family of solutions. Edit: $(x,y)=(1,1)$ should work as the initial solution as per suggestion by Ivan Loh! LOL :-) :D - So, just to prove that $x^2+y^2+m=dxy$ has infinite comprime solutions, for any $d$... How to do this? – leticia Mar 15 '13 at 22:59 @Jyrki Lahtonen $(x,y)=(1,1)$ works... – Ivan Loh Mar 16 '13 at 0:13 This method is also known as Vieta Jumping In this case, we are not looking for a contradiction. Rather, we descend to the base case, then reverse the descent to get all solutions. So in fact you have found all positive integer solutions, assuming you start with $(1,1)$. – Ivan Loh Mar 16 '13 at 0:20 @leticiamat If you read the history of Vieta jumping at the link I provided above, this method was introduced to Olympiad mathematics via IMO 1988 Q6. It has since become a standard technique for solving such Olympiad number theory questions, so this is probably how you are meant to solve it. – Ivan Loh Mar 16 '13 at 0:27 @Ivan: Thanks for the link. I would suggest that the techniques was useful already at IMO1981 (Georgetown), when the equation $(m^2-mn-n^2)^2=1$ was studied. Then the sequence of solutions was formed by pairs of consecutive Fibonacci numbers. Don't remember the number of the question, but "I was there!" – Jyrki Lahtonen Mar 16 '13 at 5:37 We will also add the condition (4) that $\gcd(xy , m) = 1$ (as pointed out by ThomasAndrews) Observe that $(x, y) = (m+1, (m+1)^2 + m)$ is a solution to the conditions. Condition 1 is easy to verify. Condition 2 is trivial. Condition 3 follows easily. [You could also show that the polynomial $(m+1)$ is a factor of the polynomial $[ (m+1)^2 + m ] ^2 + m$ by substituting $m=-1$ to obtain $[ (-1 + 1)^2 + 1]^2 -1 = 0$ and applying the Remainder Factor Theorem.] Condition 4 is trivial. Note: For $m=1$, we get $(x, y) = (2, 5)$. We now proceed by Vieta's Root Jumping (as Jyrki suggested). Given any solution $(x, y)$, we know that $$xy \mid (x^2 + m)(y^2 + m) \Leftrightarrow xy \mid mx^2 + my^2 + m^2 \Leftrightarrow xy \mid x^2 + y^2 + m$$ Hence, there exists a $D$ such that $Dxy = x^2 + y^2 + m$. Claim: $(y, Dx-y)$ is a another solution. You can either check this through algebraic manipulation, or realize that $Dx-y$ is the other integer root to the above equation (when viewed as a quadratic in $y$). Remember to check all the conditions. As Ivan suggested, to show that we have distinct solutions, we will show that $x < y \Rightarrow y < Dy-x$, and thus the Veita jumping gives us a strictly increasing sequence. From the product of roots, we get that $(Dx-y) \times x = y^2 + m \geq y^2$. Hence, $x< y \Rightarrow Dy-x \geq \frac {y^2}{x} > \frac {y^2}{y} = y$. - (+1) but can't you just start with $(1,1)$? Take $x \leq y$ so since the product of roots is $y^2+m \geq x^2+1$ the other root is bigger. – Ivan Loh Mar 16 '13 at 0:51 Of course you can, and you can get to $(1, 1)$ by 'backwards' Vieta jumping. I actually started off with $(1, 1)$ and got $(1, m+1)$ and then $(m+1, m^2+3m+1)$. – Calvin Lin Mar 16 '13 at 0:59 there's one step you assume $(xy,m)=1$. It's true, but don't you have to prove that it is true that $(Dx-y,m)=1$? Specifically, when you have $xy\|(x^2+m)(y^2+m)$ and deduce that $xy\|(x^2+y^2+m)$ you are using that $(xy,m)=1$. – Thomas Andrews Mar 26 '13 at 14:28 @ThomasAndrews Agreed. Will edit. – Calvin Lin Mar 26 '13 at 15:24 Also, isn't $Dx-y = \frac{x^2+m}{y}$? If so, how is $(Dx-y)\times x = y^2 +m$? Should that be $(Dx-y)\times y = x^2 + m$? – Thomas Andrews Mar 26 '13 at 15:28
# Division '÷' \| Basics of Arithmetic See our other arithmetic pages, for discussion and examples of: Addition (+), Subtraction (-) and Multiplication (×). ## Division In handwriting the usual sign for division is (÷) on a spreadsheet and some other computer applications the ‘/’ symbol is used to denote division. Division is the opposite of multiplication in mathematics. Division allows us to divide or 'share' numbers to find an answer. Multiplication gives us a quick way of doing multiple additions and division gives us a quick way of doing multiple subtractions. Division is often considered the hardest of the four main arithmetic functions, remember all division calculations can easily be carried out on a calculator. When you think about division think about sharing a number equally by the number of times the sum says. For example 10 sweets divided by 2 children = 5 sweets each. ## Some Quick Rules about Division: • When you divide 0 by another number the answer is always 0. For example: 0 ÷ 2 = 0. That is 0 sweets shared equally among 2 children - each child gets 0 sweets. • When you divide a number by 0 you are not dividing at all (this is quite a problem in mathematics). 2 ÷ 0 is not possible. You have 2 sweets but no children to divide them to. You cannot divide by 0. • When you divide by 1 the answer is the same as the number you were dividing. 2 ÷ 1 = 2. Two sweets divided by one child. • When you divide by 2 you are halving the number. 2 ÷ 2 = 1. • Any number divided by the same number is 1. 20 ÷ 20 = 1. Twenty sweets divided by twenty children - each child gets one sweet. • Numbers must be divided in the correct order. 10 ÷ 2 = 5 whereas 2 ÷ 10 = 0.2. Ten sweets divided by two children is very different to 2 sweets divided by 10 children. • All fractions such as ½, ¼ and ¾ are division sums. ½ is 1 ÷ 2. One sweet divided by two children. ### Multiple Subtractions Division sums can be a useful and quicker way of performing multiple subtraction sums. For example: If John has 10 gallons of fuel in his car and uses 2 gallons a day how many days before he runs out? We can work this problem out by doing a series of subtractions, or by counting backwards in steps of 2. • On day 1 John starts with 10 gallons and ends with 8 gallons. 10 - 2 = 8 • On day 2 John starts with 8 gallons and ends with 6 gallons. 8 - 2 = 6 • On day 3 John starts with 6 gallons and ends with 4 gallons. 6 - 2 = 4 • On day 4 John starts with 4 gallons and ends with 2 gallons. 4 - 2 = 2 • On day 5 John starts with 2 gallons and ends with 0 gallons. 2 - 2 = 0 John runs out of fuel on day 5. A quicker way of preforming this sum would be to divide 10 by 2. 10 ÷ 2 = 5. That is, how many times does 2 go into 10. The multiplication table (see multiplication) can be used to help us find the answer to simple division sums. In the example above we needed to calculate 10 ÷ 2. To do this, using the multiplication table locate the column for 2 (the red shaded heading). Work down the column until you find the number you are looking for, 10. Move across the row to the left to see the answer (the red shaded heading) 5. ### Multiplication Table × 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100 We can work out other simple division sums using the same method. 56 ÷ 8 = 7 for example. Find 7 on the top row, look down the column until you find 56 find the corresponding row number, 8 If possible you should try to memorise the multiplication table above, it makes solving simple multiplication and division sums much quicker. ## Dividing Larger Numbers Although you can quickly preform division sums on a calculator by learning how to calculate division manually can help in situations when you don't have a calculator to hand or don't want to use one. Division sums can look daunting but in fact, as with most mathematics, they are logical. Dave’s car needs new tyres, all four driving tyres and the spare need to be replaced. Dave has had a quote from a local garage for £480 to include the tyres, fitting and disposal of the old tyres. How much does each tyre cost? The problem we need to calculate here is 480 ÷ 5. This is the same as saying how many times will 5 go into 480? Traditionally division sums are written like: 5 4 8 0 We start by dividing 5 by 4 and immediately hit a problem. 5 does not divide by 4 to leave a whole number, as 5 is greater than 4. We are only interested in whole numbers and as 5 does not go into 4 we put a 0 in the first (hundreds) column. For help with the hundreds, tens and units columns see our page on numbers. Hundreds Tens Units 0 5 4 8 0 The next step is to use the next number to see how many times 5 goes into 48. 5 does go into 48 as 48 is greater than 5, however we need to know how many times it goes. If we refer to our multiplication table we can see that 9 × 5 = 45 and 10 × 5 = 50. 48; the number we’re looking for falls between these two values. We take the lower value as it is smaller than 48 and therefore will go. 5 goes into 45 9 times, but not exactly as it leaves a remainder of 3. The remainder what is left when we subtract the number we have found from the number we are dividing into: 48 - 45 = 3. So 5 × 9 = 45, + 3 to get 48. We can enter 9 as our answer for the second part of the sum and bring our remainder in front of our last number. Our last number becomes 30. Hundreds Tens Units 0 9 5 4 8 30 We now divide 30 by 5 or find out how many times 5 goes into 30. Using our multiplication table we can see the answer is exactly 6, with no remainder. 5 × 6 = 30. Hundreds Tens Units 0 9 6 5 4 8 30 As there are no remainders left we have finished the sum and have the answer 96. Dave’s new tyres are going to cost £96 each. 480 ÷ 5 = 96 and 96 × 5 = 480. ### Recipe Division Our final example of division is based on a recipe. Often when cooking, recipes will tell you how much food they are going to make, enough to feed 6 people, for example. The ingredients below are needed to make 24 fairy cakes, however, we only want to make 8 fairy cakes. We have modified the ingredients slightly for the benefit of this example, original recipe at: http://www.bbc.co.uk/food/recipes/fairycakes_93711). The first thing we need to establish is how many 8's there are in 24 – use the multiplication table above or your memory. 3 × 8 = 24 – if we divide 24 by 8 we get 3. Therefore we need to divide each ingredient below by 3 in order to have to right amount of mixture to make 8 fairy cakes. #### Ingredients • 120g butter, softened at room temperature • 120g caster sugar • 3 free-range eggs, lightly beaten • 1 tsp vanilla extract • 120g self-raising flour • 1-2 tbsp milk The amount of butter, sugar and flour are all the same, 120g it is only necessary to work out 120 ÷ 3 once, the answer will be the same for those three ingredients. 3 1 2 0 As before we start in the right column and divide 3 by 1. However 3 ÷ 1 doesn’t go as 3 is greater than 1. Next we look at how many times 3 goes into 12. Using the multiplication table if needed you can see that 3 goes into 12 exactly 4 times with no remainder. 0 4 0 3 1 2 0 120g ÷ 3 is therefore 40g. We now know that we’ll need 40g of butter, sugar and flour. The next sum is simple. The original recipe calls for 3 eggs, we again divide by 3. So 3 ÷ 3. This is the same as saying how many 3s are there in 3. The answer is obviously 1. Therefore one egg is needed. Next the recipe calls for 1tsp (teaspoon) of vanilla extract. We need to divide one teaspoon by 3. A fraction is written like a division sum, 1 ÷ 3 is the same as ⅓ one third. You’ll need ⅓ of a teaspoon of vanilla extract – in reality it may be difficult to accurately measure ⅓ of a teaspoon! Next the recipe calls for 1-2 tbsp of milk. That is between 1 and 2 tablespoons of milk. We have no definitive amount and how much milk you add will be dependent on your mixture consistency. We already know that 1 ÷ 3 is ⅓; therefore 2 ÷ 3 is ⅔. ⅓ - ⅔ of a tablespoon of milk is needed to make 8 fairy cakes. TOP
# Which Equation is the Inverse of y = 100×2? Source: bing.com ## Understanding Inverse Equations When we talk about inverse equations, we refer to a pair of equations that have opposite operations. In other words, if we apply one equation to a certain variable, applying the inverse equation will reverse the process and bring the variable back to its original value. For instance, if we have an equation that adds 5 to a variable, the inverse equation will subtract 5 from the same variable. Similarly, if we have an equation that multiplies a variable by 2, the inverse equation will divide the same variable by 2. ## Finding the Inverse Equation of y = 100×2 Now, let’s apply this concept to the equation y = 100×2. To find the inverse equation, we need to isolate x and express it in terms of y. Starting with the given equation: y = 100×2 We can divide both sides by 100: y/100 = x2 Source: bing.com To isolate x, we need to take the square root of both sides: sqrt(y/100) = x Source: bing.com Therefore, the inverse equation of y = 100×2 is: x = sqrt(y/100) ## Understanding the Inverse Equation Now that we have found the inverse equation, let’s see what it represents. The equation x = sqrt(y/100) means that if we have a certain value of y, applying this equation to it will give us the corresponding value of x that makes the original equation true. For example, let’s say y = 400. We can plug this value into the inverse equation: x = sqrt(400/100) = sqrt(4) = 2 Source: bing.com Therefore, the x value that corresponds to y = 400 is 2. If we substitute x = 2 into the original equation y = 100×2, we get: y = 100(2)2 = 400 As expected, the original equation is true for x = 2 and y = 400. ## Graphical Representation Another way to understand inverse equations is by looking at their graphical representation. The graph of y = 100×2 is a parabola that opens upwards, as shown below: Source: bing.com On the other hand, the graph of its inverse equation x = sqrt(y/100) is a reflection of the original graph over the line y = x: Source: bing.com Notice that the inverse graph is also a parabola, but it opens to the right instead of upwards. This reflects the fact that applying the inverse equation to the y value will give us two possible values of x, one positive and one negative, due to the square root operation. ## Conclusion Inverse equations are a powerful tool in mathematics that allow us to reverse the effect of a given equation. In the case of y = 100×2, we have found that its inverse equation is x = sqrt(y/100), which represents the value of x that makes the original equation true for a given value of y. Graphically, the inverse equation corresponds to a reflection of the original graph over the line y = x. Understanding inverse equations can help us solve problems and gain a deeper insight into the workings of mathematics.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 3.11: Half Angle Formulas Difficulty Level: At Grade Created by: CK-12 Estimated14 minsto complete % Progress Practice Half Angle Formulas MEMORY METER This indicates how strong in your memory this concept is Progress Estimated14 minsto complete % Estimated14 minsto complete % MEMORY METER This indicates how strong in your memory this concept is After all of your experience with trig functions, you are feeling pretty good. You know the values of trig functions for a lot of common angles, such as \begin{align}30^\circ, 60^\circ\end{align} etc. And for other angles, you regularly use your calculator. Suppose someone gave you an equation like this: \begin{align}\cos 75^\circ\end{align} Could you solve it without the calculator? You might notice that this is half of \begin{align}150^\circ\end{align}. This might give you a hint! ### Half Angle Formulas Here we'll attempt to derive and use formulas for trig functions of angles that are half of some particular value. To do this, we'll start with the double angle formula for cosine: \begin{align}\cos 2 \theta = 1 - 2 \sin^2 \theta\end{align}. Set \begin{align}\theta = \frac{\alpha}{2}\end{align}, so the equation above becomes \begin{align}\cos 2 \frac{\alpha}{2} = 1 - 2 \sin^2 \frac{\alpha}{2}\end{align}. Solving this for \begin{align}\sin \frac{\alpha}{2}\end{align}, we get: \begin{align}\cos 2 \frac{\alpha}{2} & = 1 - 2 \sin^2 \frac{\alpha}{2} \\ \cos \alpha & = 1 - 2 \sin^2 \frac{\alpha}{2} \\ 2 \sin^2 \frac{\alpha}{2} & = 1 - \cos \alpha \\ \sin^2 \frac{\alpha}{2} & = \frac{1 - \cos \alpha}{2} \\ \sin \frac{\alpha}{2} & = \pm \sqrt{\frac{1 - \cos \alpha}{2}}\end{align} \begin{align}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align} if \begin{align}\frac{\alpha}{2}\end{align} is located in either the first or second quadrant. \begin{align}\sin \frac{\alpha}{2} = - \sqrt{\frac{1 - \cos \alpha}{2}}\end{align} if \begin{align}\frac{\alpha}{2}\end{align} is located in the third or fourth quadrant. This formula shows how to find the sine of half of some particular angle. One of the other formulas that was derived for the cosine of a double angle is: \begin{align}\cos 2 \theta = 2 \cos^2 \theta - 1\end{align}. Set \begin{align}\theta = \frac{\alpha}{2}\end{align}, so the equation becomes \begin{align}\cos 2 \frac{\alpha}{2} = - 1 + 2 \cos^2 \frac{\alpha}{2}\end{align}. Solving this for \begin{align}\cos \frac{\alpha}{2}\end{align}, we get: \begin{align} \cos 2 \frac{\alpha}{2} & = 2 \cos^2 \frac{\alpha}{2} - 1 \\ \cos \alpha & = 2 \cos^2 \frac{\alpha}{2} -1 \\ 2 \cos^2 \frac{\alpha}{2} & = 1 + \cos \alpha \\ \cos^2 \frac{\alpha}{2} & = \frac{1 + \cos \alpha}{2} \\ \cos \frac{\alpha}{2} & = \pm \sqrt{\frac{1 + \cos \alpha}{2}}\end{align} \begin{align}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align} if \begin{align}\frac{\alpha}{2}\end{align} is located in either the first or fourth quadrant. \begin{align}\cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \cos \alpha}{2}}\end{align} if \begin{align}\frac{\alpha}{2}\end{align} is located in either the second or fourth quadrant. This formula shows how to find the cosine of half of some particular angle. Let's see some examples of these two formulas (sine and cosine of half angles) in action. 1. Determine the exact value of \begin{align}\sin 15^\circ\end{align}. Using the half angle identity, \begin{align}\alpha = 30^\circ\end{align}, and \begin{align}15^\circ\end{align} is located in the first quadrant. Therefore, \begin{align}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align}. \begin{align}\sin 15^\circ & = \sqrt{\frac{1 - \cos 30^\circ}{2}} \\ & = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}}\end{align} Plugging this into a calculator, \begin{align}\sqrt{\frac{2 - \sqrt{3}}{4}} \approx 0.2588\end{align}. Using the sine function on your calculator will validate that this answer is correct. 2. Use the half angle identity to find exact value of \begin{align}\sin 112.5^\circ\end{align} Since \begin{align}\sin \frac{225^\circ}{2} = \sin 112.5^\circ\end{align}, use the half angle formula for sine, where \begin{align}\alpha = 225^\circ\end{align}. In this example, the angle \begin{align}112.5^\circ\end{align} is a second quadrant angle, and the sin of a second quadrant angle is positive. \begin{align}\sin 112.5^\circ & = \sin \frac{225^\circ}{2} \\ & = \pm \sqrt{\frac{1 - \cos 225^\circ}{2}} \\ & = + \sqrt{\frac{1 - \left (- \frac{\sqrt{2}}{2} \right )}{2}} \\ & = \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}} \\ & = \sqrt{\frac{2 + \sqrt{2}}{4}}\end{align} 3. Use the half angle formula for the cosine function to prove that the following expression is an identity: \begin{align}2 \cos^2 \frac{x}{2} - \cos x = 1\end{align} Use the formula \begin{align}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align} and substitute it on the left-hand side of the expression. \begin{align}2 \left( \sqrt{\frac{1 + \cos \theta}{2}} \right )^2 - \cos \theta & = 1 \\ 2 \left (\frac{1 + \cos \theta}{2} \right ) - \cos \theta & = 1 \\ 1 + \cos \theta - \cos \theta & = 1 \\ 1 & = 1\end{align} ### Examples #### Example 1 Earlier, you were asked you to find \begin{align}\cos 75^\circ\end{align}. If you use the half angle formula, then \begin{align}\alpha = 150^\circ\end{align} Substituting this into the half angle formula: \begin{align}\sin \frac{150^\circ}{2} = \sqrt{\frac{1 - \cos \alpha}{2}} = \sqrt{\frac{1 - \cos 150^\circ}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}\end{align} #### Example 2 Prove the identity: \begin{align}\tan \frac{b}{2} = \frac{\sec b}{\sec b \csc b + \csc b}\end{align} Step 1: Change right side into sine and cosine. \begin{align}\tan \frac{b}{2} & = \frac{\sec b}{\sec b \csc b + \csc b} \\ & = \frac{1}{\cos b} \div \csc b (\sec b + 1) \\ & = \frac{1}{\cos b} \div \frac{1}{\sin b} \left (\frac{1}{\cos b} + 1 \right ) \\ & = \frac{1}{\cos b} \div \frac{1}{\sin b} \left (\frac{1 + \cos b}{\cos b} \right ) \\ & = \frac{1}{\cos b} \div \frac{1 + \cos b}{\sin b \cos b} \\ & = \frac{1}{\cos b} \cdot \frac{\sin b \cos b}{1 + \cos b} \\ & = \frac{\sin b}{1 + \cos b}\end{align} Step 2: At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula. \begin{align}\sqrt{\frac{1 - \cos b}{1 + \cos b}} & = \frac{\sin b}{1 + \cos b} \\ \frac{1 - \cos b}{1 + \cos b} & = \frac{\sin^2 b}{(1 + \cos b)^2} \\ (1 - \cos b)(1 + \cos b)^2 & = \sin^2 b (1 + \cos b) \\ (1 - \cos b)(1 + \cos b) & = \sin^2 b \\ 1 - \cos^2 b & = \sin^2 b\end{align} #### Example 3 Verify the identity: \begin{align}\cot \frac{c}{2} = \frac{\sin c}{1 - \cos c}\end{align} Step 1: change cotangent to cosine over sine, then cross-multiply. \begin{align}\cot \frac{c}{2} & = \frac{\sin c}{1 - \cos c} \\ & = \frac{\cos \frac{c}{2}}{\sin \frac{c}{2}} = \sqrt{\frac{1 + \cos c}{1 - \cos c}} \\ \sqrt{\frac{1 + \cos c}{1 - \cos c}} & = \frac{\sin c}{1 - \cos c} \\ \frac{1 + \cos c}{1 - \cos c} & = \frac{\sin^2 c}{(1 - \cos c)^2} \\ (1 + \cos c)(1 - \cos c)^2 & = \sin^2 c (1 - \cos c) \\ (1 + \cos c)(1 - \cos c) & = \sin^2 c \\ 1 - \cos^2 c & = \sin^2 c\end{align} #### Example 4 Prove that \begin{align}\sin x \tan \frac{x}{2} + 2 \cos x = 2 \cos^2 \frac{x}{2}\end{align} \begin{align}\sin x \tan \frac{x}{2} + 2 \cos x & = \sin x \left(\frac{1 - \cos x}{\sin x} \right ) + 2 \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 1 - \cos x + 2 \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 1 + \cos x \\ \sin x \tan \frac{x}{2} + 2 \cos x & = 2 \cos^2 \frac{x}{2} \end{align} ### Review Use half angle identities to find the exact values of each expression. 1. \begin{align}\sin 22.5^\circ\end{align} 2. \begin{align}\sin 75^\circ\end{align} 3. \begin{align}\sin 67.5^\circ\end{align} 4. \begin{align}\sin 157.5^\circ\end{align} 5. \begin{align}\cos 22.5^\circ\end{align} 6. \begin{align}\cos 75^\circ\end{align} 7. \begin{align}\cos 157.5^\circ\end{align} 8. \begin{align}\cos 67.5^\circ\end{align} 9. Use the two half angle identities presented in this section to prove that \begin{align}\tan(\frac{x}{2})=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}\end{align}. 10. Use the result of the previous problem to show that \begin{align}\tan(\frac{x}{2})=\frac{1-\cos x}{\sin x}\end{align}. 11. Use the result of the previous problem to show that \begin{align}\tan(\frac{x}{2})=\frac{\sin x}{1+\cos x}\end{align}. Use half angle identities to help you find all solutions to the following equations in the interval \begin{align}[0,2\pi)\end{align}. 1. \begin{align}\sin^2 x=\cos^2(\frac{x}{2})\end{align} 2. \begin{align}\tan(\frac{x}{2})=\frac{1-\cos x}{1+\cos x}\end{align} 3. \begin{align}\cos^2 x=\sin^2(\frac{x}{2})\end{align} 4. \begin{align}\sin^2(\frac{x}{2})=2\cos^2 x-1\end{align} To see the Review answers, open this PDF file and look for section 3.11. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Half Angle Identity A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument. Show Hide Details Description Difficulty Level: Tags: Subjects:
# Mathematical Induction #### Introduction In this note, we introduce the proof technique of mathematical induction. Induction is a powerful tool which is used to establish that a statement holds for all natural numbers. Of course, there are infinitely many natural numbers — induction provides a way to reason about them by finite means. Let us demonstrate the intuition behind induction with an example. Suppose we wish to prove the statement: For all natural numbers $$n$$, $$0 + 1 + 2 + 3 + \cdots + n = n(n+1)/2$$. More formally, using the universal quantifier from Note 1, we can write this as: $\forall n \in {\mathbb{N}}, \quad\sum^n_{i=0} i=\frac{n(n+1)}{2}.$(1) How would you prove this? Well, you could begin by checking that it holds for $$n=0,1,2$$, and so forth. But there are an infinite number of values of $$n$$ for which it needs to be checked! Moreover, checking just the first few values of $$n$$ does not suffice to conclude the statement holds for all $$n\in{\mathbb{N}}$$, as the following example demonstrates: Sanity check! Consider the statement: $$\forall n \in {\mathbb{N}}, \; n^2 − n + 41$$ is a prime number. Check that it holds for the first few natural numbers. (In fact, you could check all the way up to $$n = 40$$ and not find a counterexample!) Now check the case of $$n = 41$$. In mathematical induction, we circumvent this problem by making an interesting observation: Suppose the statement holds for some value $$n=k$$, i.e. $$\sum^k_{i=0} i= k(k+1)/2$$. (This is called the induction hypothesis.) Then: $\label{eqn:2_1} \left(\sum_{i=0}^{k} i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2},$ i.e. the claim must also hold for $$n=k+1$$! In other words, if the statement holds for some $$k$$, then it must also hold for $$k+1$$. Let us call the argument above the inductive step. The inductive step is a very powerful tool: If we can show the statement holds for $$k$$, then the inductive step allows us to conclude that it also holds for $$k+1$$; but now that $$k+1$$ holds, the inductive step implies that $$k+2$$ must hold; in fact, we can repeat this argument indefinitely for all $$n\geq k$$! So is that it? Have we proven Equation 1? Almost! The problem is that in order to apply the inductive step, we first have to establish that Equation 1 holds for some initial value of $$k$$. Since our aim is to prove the statement for all natural numbers, the obvious choice is $$k=0$$. We call this choice of $$k$$ the base case. Then, if the base case holds, the axiom of mathematical induction says that the inductive step allows us to conclude that Equation 1 indeed holds for all $$n\in{\mathbb{N}}$$. Let us now formally re-write this proof. Theorem 1 $$\forall n \in {\mathbb{N}}$$, $$\displaystyle\sum_{i=0}^n i = \frac{n(n+1)}{2}$$. Proof. We proceed by induction on the variable $$n$$. Base Case $$(n=0)$$: Here, we have $$\sum_{i=0}^0 i = 0 = 0(0+1)/2$$. Thus, the base case is correct. Inductive Hypothesis: For arbitrary $$n=k\geq 0$$, assume that $$\sum_{i=0}^k i = k(k+1)/2$$. In words, the Inductive Hypothesis says “let’s assume we have proved the statement for an arbitrary value of $$n =k \geq 0$$”. Inductive Step: Prove the statement for $$n=(k+1)$$, i.e. show that $$\sum_{i=0}^{k+1} i = (k+1)(k+2)/2$$: $\displaystyle\sum_{i=0}^{k+1} i = \left(\sum_{i=0}^{k} i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1)+ 2(k+1)}{2} = \frac{(k+1)(k+2)}{2},$(2) where the second equality follows from the Induction Hypothesis. By the principle of mathematical induction, the claim follows. $$\square$$ Sanity check! How exactly did the Induction Hypothesis help us show the second equality in Equation 2? [Hint: We used it to replace $$\sum_{i=0}^k i$$ with something useful.] #### Recap What have we learned so far? Letting $$P(n)$$ denote the statement $$\sum_{i=0}^{n} i = n(n+1)/2$$, our goal was to prove that $$\forall n \in {\mathbb{N}}$$, $$P(n)$$. The principle of induction asserts that to prove this requires three simple steps: 1. Base Case: Prove that $$P(0)$$ is true. 2. Inductive Hypothesis: For arbitrary $$k\geq 0$$, assume that $$P(k)$$ is true. 3. Inductive Step: With the assumption of the Inductive Hypothesis in hand, show that $$P(k+1)$$ is true. Let us visualize how these three steps fit together using dominoes. Let statement $$P(n)$$ be represented by a sequence of dominoes, numbered from $$0, 1, 2, \dotsc, n$$, such that $$P(0)$$ corresponds to the $$0^{\text{th}}$$ domino, $$P(1)$$ corresponds to the $$1^{\text{st}}$$ domino, and so on. The dominoes are lined up so that if the $$k^{\text{th}}$$ domino is knocked over, then it in turn knocks over the $$(k+1)^{\text{st}}$$. Knocking over the $$k^{\text{th}}$$ domino corresponds to proving $$P(k)$$ is true. And the induction step corresponds to the placement of the dominoes to ensure that if the $$k^{\text{th}}$$ domino falls, in turn it knocks over the $$(k+1)^{\text{st}}$$ domino. The base case ($$n=0$$) knocks over the $$0^{\text{th}}$$ domino, setting off a chain reaction that knocks down all the dominoes! Finally, a word about choosing an appropriate base case — in this example, we chose $$k=0$$, but in general, the choice of base case will naturally depend on the claim you wish to prove. #### Another Proof by Induction Let us do another proof by induction. Recall from Note 1 that for integers $$a$$ and $$b$$, we say that $$a$$ divides $$b$$, denoted $$a \mid b$$, if and only if there exists an integer $$q$$ satisfying $$b = aq$$. Sanity check! How would you write the definition of $$a \mid b$$ formally using quantifiers? (Answer: $$\forall a, b \in \mathbb{Z}$$, $$a \mid b$$ iff $$\exists q \in \mathbb{Z}$$ such that $$b = aq$$.) Theorem 2 $$\forall n \in {\mathbb{N}}$$, $$n^3 - n$$ is divisible by 3. Proof. We proceed by induction over $$n$$. Let $$P(n)$$ denote the statement $$\forall n \in {\mathbb{N}}$$, $$n^3 - n$$ is divisible by 3. Base Case ($$n=0$$): $$P(0)$$ asserts that $$3 \mid (0^3 - 0)$$ or $$3 \mid 0$$, which is true since any non-zero integer divides $$0$$. Inductive Hypothesis: Assume for $$n=k\geq 0$$ that $$P(k)$$ is true. That is, $$3 \mid (k^3-k)$$, or $$\exists q \in {\mathbb{Z}}, k^3 - k = 3q$$. Inductive Step: We show that $$P(k+1)$$ is true, i.e. that $$3 \mid ((k+1)^3-(k+1))$$. To show this, we expand the number $$((k+1)^3-(k+1))$$ as follows: \begin{align} (k+1)^3 - (k+1) &= k^3 + 3k^2 + 3k + 1 - (k+1) \\ &= (k^3 - k) + 3k^2 + 3k \\ &= 3q + 3(k^2 + k) \text{ for some } q \in \mathbb{Z} \quad\text{(Induction Hypothesis)}\\ &= 3(q + k^2 + k). \end{align} Sanity check! How exactly did the Induction Hypothesis help us show the third equality above? We conclude that $$3 \mid ((k+1)^3-(k+1))$$. Thus, by the principle of induction, $$\forall n \in {\mathbb{N}}$$, $$3 \mid (n^3-n)$$. $$\square$$ #### Two Color Theorem We now consider a more advanced proof by induction for a simplified version of the famous four color theorem. The four color theorem states that any map can be colored with four colors such that any two adjacent countries (which share a border, but not just a point) have different colors. The four color theorem is very difficult to prove, and several bogus proofs were claimed since the problem was first posed in 1852. In fact, it was not until 1976 that a computer-assisted proof of the theorem was finally given Appel and Haken. (For an interesting history of the problem and state-of-the-art proof, which is nonetheless still very challenging, see www.math.gatech.edu/~thomas/FC/fourcolor.html). In this note, we consider a simpler version of the theorem in which our “map” is given by a rectangle which is divided into regions by drawing lines, such that each line divides the rectangle into two regions. Can we color such a map using no more than two colors (say, red and blue) such that no two bordering regions have the same color? To illustrate, here is an example of a two-colored map: Theorem 3 Let $$P(n)$$ denote the statement “any map with $$n$$ lines is two-colorable”. Then, it holds that $$\forall n\in {\mathbb{N}}\; P(n)$$. Proof. We proceed by induction on $$n$$. Base Case ($$n=0$$): Clearly $$P(0)$$ holds, since if we have $$n=0$$ lines, then we can color the entire map using a single color. Inductive Hypothesis: For any $$n=k\geq 0$$, assume $$P(k)$$. Inductive Step: We prove $$P(k+1)$$. Specifically, we are given a map with $$k+1$$ lines and wish to show that it can be two-colored. Let’s see what happens if we remove a line. With only $$k$$ lines on the map, the Induction Hypothesis says we can two-color the map. Let us make the following observation: Given a valid coloring, if we swap red $$\leftrightarrow$$ blue, we still have a two-coloring. With this in mind, let us place back the line we removed, and leave colors on one side of the line unchanged. On the other side of the line, swap red $$\leftrightarrow$$ blue. This is illustrated by Figure 1. We claim that this is a valid two-coloring for the map with $$k+1$$ lines. Why does this work? Consider two regions separated by a shared border. Then, one of two cases must hold: Case 1 is when the shared border is the line that was removed and replaced, i.e. line $$k+1$$. But by construction, we flipped the colors on one side of this line so that any two regions separated by it have distinct colors. Case 2 is when the shared border is one of the original $$k$$ lines; here, the Induction Hypothesis guarantees that two regions separated by this border have different colors. Thus, in both cases, the regions separated by the shared border have distinct colors, as required. $$\square$$ # Strengthening the Induction Hypothesis When using induction, it can be very important to choose the correct statement to prove. For example, suppose we wish to prove the statement: $$\forall n \geq 1$$, the sum of the first $$n$$ odd numbers is a perfect square. Here is a first proof attempt via induction. Proof Attempt. We proceed by induction on $$n$$. Base Case ($$n = 1$$): The first odd number is $$1$$, which is a perfect square. Inductive Hypothesis: Assume that the sum of the first $$k$$ odd numbers is a perfect square, say $$m^2$$. Inductive Step: The $$(k+1)$$st odd number is $$2k+1$$. Thus, by the induction hypothesis, the sum of the first $$k+1$$ odd numbers is $$m^2 + 2k +1$$. But now we are stuck. Why should $$m^2 + 2k+1$$ be a perfect square? It seems our Induction Hypothesis is too “weak”; it does not give us enough structure to say anything meaningful about the $$(k+1)$$ case. So let’s take a step back for a moment, and do a preliminary check to ensure our claim isn’t obviously false: Let’s compute the values of the first few cases. Perhaps in the process, we can also uncover some hidden structure we have not yet identified. • $$n = 1: 1 = 1^2$$ is a perfect square. • $$n = 2: 1 + 3 = 4 = 2^2$$ is a perfect square. • $$n = 3: 1 + 3 + 5 = 9 = 3^2$$ is a perfect square. • $$n = 4: 1 + 3 + 5 + 7 = 16 = 4^2$$ is a perfect square. It looks like we have good news and even better news: The good news is that we have not yet found a counterexample to our claim. The even better news is that there is a surprising pattern emerging — the sum of the first $$n$$ odd numbers is not just a perfect square, but is equal precisely to $$n^2$$! Motivated by this discovery, let’s try something counterintuitive: Let us try to show the following stronger claim. Theorem 4 For all $$n \geq 1$$, the sum of the first $$n$$ odd numbers is $$n^2$$. Proof. We proceed by induction on $$n$$. Base Case ($$n = 1$$): The first odd number is $$1$$, which is $$1^2$$. Inductive Hypothesis: Assume that the sum of the first $$k$$ odd numbers is $$k^2$$. Inductive Step: The $$(k+1)$$-st odd number is $$2k+1$$. Applying the Induction Hypothesis, the sum of the first $$k+1$$ odd numbers is $$k^2 + (2k +1) = (k+1)^2$$. Thus, by the principle of induction the theorem holds. $$\square$$ So let’s get this straight — we couldn’t prove our original statement, so instead we hypothesized a stronger one and managed to prove that. Why on Earth did this work? The reason is that our original claim did not capture the true structure of the underlying fact we were trying to prove — it was too vague. As a result, our Induction Hypothesis wasn’t strong enough to prove our desired result. In contrast, although our second claim is a priori stronger, it also has more structure to it; this, in turn, makes our Induction Hypothesis stronger — we can use the fact that not only is the sum of the first $$k$$ odd numbers a perfect square, but that it in fact equals $$k^2$$. This additional structure is exactly what we needed to complete the proof. In summary, we have demonstrated an example in which, although a claim was true, the precise formulation of the Induction Hypothesis made the difference between a failing and successful proof. #### Example Let us now try a second example; this time, we’ll let you do some of the work! Suppose we wish to prove the claim: $$\forall n \geq 1$$, $$\sum_{i=1}^n i^{-2}\leq 2$$. Sanity check! The “obvious” choice of Induction Hypothesis says the following: Assume that for $$n = k$$, $$\sum_{i=1}^k i^{-2} \le 2$$. Why does this not suffice to prove the claim for $$n = k+1$$, i.e. to show that $$\sum_{i=1}^k i^{-2} + (k + 1)^{-2} \le 2$$? [Hint: Is it possible that $$\sum_{i=1}^k i^{-2}$$ actually equals $$2$$?] Now let’s again do the unthinkable — let’s prove the following stronger statement, i.e. let’s strengthen our induction hypothesis. Theorem 5 For all $$n \geq 1$$, $$\displaystyle\sum_{i=1}^n \frac{1}{i^2}\leq 2-\frac{1}{n}$$. Proof. We proceed by induction on $$n$$. Base Case ($$n = 1$$): We have $$\sum_{i=1}^1 i^{-2} =1\leq 2-1/1$$, as required. Inductive Hypothesis: Assume that the claim holds for $$n=k$$. Inductive Step: By the Induction Hypothesis, we have that $\sum_{i=1}^{k+1} \frac{1}{i^2}=\sum_{i=1}^{k} \frac{1}{i^2}+ \frac{1}{(k+1)^2}\leq 2-\frac{1}{k} + \frac{1}{(k+1)^2}.$ Thus, to prove our claim, it suffices to show that $-\frac{1}{k} + \frac{1}{(k+1)^2} \leq -\frac{1}{k+1}.$(3) Exercise. Show that Equation 3 holds. [Hint: Multiply both sides of the inequality by $$(k +1)$$.] By the principle of mathematical induction, the claim holds. $$\square$$ # Simple Induction vs. Strong Induction Thus far, we have been using a notion of induction known as simple or weak induction. There is another notion of induction, which we now discuss, called strong induction. The latter is similar to simple induction, except we have a slightly different inductive hypothesis: Instead of just assuming $$P(k)$$ is true (as was the case with simple induction), we assume the stronger statement that $$P(0)$$, $$P(1)$$, …, and $$P(k)$$ are all true (i.e. that $$\bigwedge_{i=0}^{k} P(i)$$ is true). Is there a difference between the power of strong and weak induction, i.e. can strong induction prove statements which weak induction cannot? No! Intuitively, this can be seen by returning to our domino analogy from Mathematical Induction. In this picture, weak induction says that if the $$k$$th domino falls, so does the $$(k+1)^{\text{st}}$$ one, whereas strong induction says that if dominos $$1$$ through $$k$$ fall, then so does the $$(k+1)^{\text{st}}$$ one. But these scenarios are equivalent: If the first domino falls, then applying weak induction repeatedly we conclude that the first domino in turn topples dominos $$2$$ through $$k$$, setting up the case of $$k+1$$, just as is the case with strong induction. With that said, strong induction does have an appealing advantage — it can make proofs easier, since we get to assume a stronger hypothesis. How should we understand this? Consider the analogy of a screwdriver and a power screwdriver — they both accomplish the same task, but the latter is much easier to use! Let’s try a simple example1 of strong induction. As a bonus, this is our first induction proof requiring multiple base cases. Theorem 6 For every natural number $$n\geq 12$$, it holds that $$n=4x+5y$$ for some $$x,y\in{\mathbb{N}}$$. Proof. We proceed by induction on $$n$$. Base Case ($$n = 12$$): We have $$12=4\cdot 3+5\cdot 0$$. Base Case ($$n = 13$$): We have $$13=4\cdot 2+5\cdot 1$$. Base Case ($$n = 14$$): We have $$14=4\cdot 1+5\cdot 2$$. Base Case ($$n = 15$$): We have $$15=4\cdot 0+5\cdot 3$$. Inductive Hypothesis: Assume that the claim holds for all $$12\leq n\leq k$$ for $$k\geq 15$$. Inductive Step: We prove the claim for $$n=k+1\geq 16$$. Specifically, note that $$(k+1)-4\geq 12$$; thus, the Induction Hypothesis implies that $$(k+1)-4=4x'+5y'$$ for some $$x',y'\in{\mathbb{N}}$$. Setting $$x=x'+1$$ and $$y=y'$$ completes the proof. $$\square$$ Sanity check! Why would the proof above fail if we used weak induction instead of strong induction? Let’s try a second, more advanced, example. For this, recall that a number $$n\geq 2$$ is prime if $$1$$ and $$n$$ are its only divisors. Theorem 7 Every natural number $$n > 1$$ can be written as a product of primes. Proof. We proceed by induction on $$n$$. Let $$P(n)$$ be the proposition that $$n$$ can be written as a product of primes. We will prove that $$P(n)$$ is true for all $$n \geq 2$$. Base Case ($$n=2$$): We start at $$n = 2$$. Clearly $$P(2)$$ holds, since 2 is a prime number. Inductive Hypothesis: Assume $$P(n)$$ is true for all $$2 \leq n \leq k$$. Inductive Step: Prove that $$n=k+1$$ can be written as a product of primes. We have two cases: either $$k+1$$ is a prime number, or it is not. For the first case, if $$k+1$$ is a prime number, then we are done. For the second case, if $$k+1$$ is not a prime number, then by definition $$k+1 = xy$$ for some $$x,y \in {\mathbb{Z}}^+$$ satisfying $$1 < x,y < k+1$$. By the Induction Hypothesis, $$x$$ and $$y$$ can each be written as a product of primes (since $$x,y \leq n$$). But this implies that $$k+1$$ can also be written as a product of primes. $$\square$$ Sanity check! Why does this proof fail if we instead use simple induction? [Hint: Suppose $$k = 42$$. Since $$42 = 6 \cdot 7$$, in the Inductive Step we wish to use the facts that $$P(6)$$ and $$P(7)$$ are true. But weak induction does not give us this — which value of P does weak induction allow us to assume is true in this case?] Finally, for those who wish to have a more formal understanding of why weak and strong induction are equivalent, consider the following. Let $$Q(n)=P(0)\wedge P(1)\wedge\cdots\wedge P(n)$$. Then, strong induction on $$P$$ is equivalent to weak induction on $$Q$$. # Recursion, Programming, and Induction There is an intimate connection between induction and recursion, which we explore here via two examples. #### Example 1: Fibonacci’s Rabbits In the 13th century lived a famous Italian mathematician known as Fibonacci2, who in 1202 considered the following rabbit-based puzzle: Starting with a pair of rabbits, how many rabbits are there after a year, if each month each pair begets a new pair which from the second month on becomes productive? This model of population growth can be modeled by recursively defining a function; the resulting sequence of numbers is nowadays referred to as the Fibonacci numbers. To recursively model our rabbit population, let $$F(n)$$ denote the number of pairs of rabbits in month $$n$$. By the rules above, our initial conditions are as follows: Clearly $$F(0)=0$$. In month $$1$$, a single pair of rabbits is introduced, implying $$F(1) = 1$$. Finally, since it takes a month for new rabbits to become reproductive, we also have $$F(2) = 1$$. In month $$3$$, the fun begins. For example, $$F(3)=2$$, since the original pair breeds to produce a new pair. But how about $$F(n)$$ for a general value of $$n$$? It seems difficult to give an explicit formula for $$F(n)$$. However, we can define $$F(n)$$ recursively as follows. In month $$n -1$$, by definition we have $$F(n-1)$$ pairs. How many of these pairs were productive? Well, only those that were already alive in the previous month, i.e. $$F(n-2)$$ of them. Thus, we have $$F(n-2)$$ new pairs in the $$n$$th month in addition to our existing $$F(n-1)$$ pairs. Hence, we have $$F(n) = F(n-1) + F(n-2)$$. To summarize: • $$F(0) = 0$$. • $$F(1) = 1$$. • For $$n \geq 2$$, $$F(n) = F(n-1) + F(n-2)$$. Pretty neat, except you might wonder why we care about rabbit reproduction in the first place! It turns out this simplified model of population growth illustrates a fundamental principle: Left unchecked, populations grow exponentially over time. In fact, understanding the significance of this unchecked exponential population growth was a key step that led Darwin to formulate his theory of evolution. To quote Darwin: “There is no exception to the rule that every organic being increases at so high a rate, that if not destroyed, the earth would soon be covered by the progeny of a single pair." Now, we promised you programming in the title of this section, so here it is — a trivial recursive program to evaluate $$F(n)$$: function F(n) if n = 0 then return 0 if n = 1 then return 1 else return F(n - 1) + F(n - 2) Exercise. How long does it take this program to compute $$F(n)$$, i.e. how many calls to $$F(n)$$ are needed? The exercise above should convince you that this is a very inefficient way to compute the $$n$$th Fibonacci number. Here is a much faster iterative algorithm to accomplish the same task (this should be a familiar example of turning a tail-recursion into an iterative algorithm): function F2(n) if n = 0 then return 0 if n = 1 then return 1 a = 1 b = 0 for k = 2 to n do temp = a a = a + b b = temp return a Exercise. How long does this iterative algorithm take to compute $$F_2(n)$$, i.e. how many iterations of its loop are needed? Can you show by induction that this new function $$F_2(n) = F(n)$$? We next use induction to analyze a recursive algorithm which even your grandmother is likely familiar with — binary search! Let us discuss binary search in the context of finding a word in the dictionary: To find word $$W$$, we open the dictionary to its middle page. If the letter of that page comes after the first letter of $$W$$, we recurse on the first half of the dictionary; else, we recurse on the last half of the dictionary. (For simplicity, we ignore the issue of dividing a dictionary with an odd number of pages into two halves.) Once we’ve narrowed down the dictionary to a single page, we resort to a brute force search to find $$W$$ on that page. In pseudocode, we have: 1. // pre-condition: W is a word and D is a subset of the dictionary with at least 1 page. 2. // post-condition: Either the definition of W is returned, or “W not found” is returned. 3. findWord(W, D) { 4. // Base case 5. If (D has precisely one page) 6. Look for W in D by brute force. If found, return its definition; else, return “W not found”. 7. 8. // Recursive case 9. Let W' be the first word on the middle page of D. 10. If (W comes before W') 11. return findWord(first half of D) 12. Else 13. return findWord(last half of D) 14. } Let us prove using induction that findWord() is correct, i.e. it returns the definition of $$W$$ if $$W$$ is in the dictionary $$D$$. As a bonus, the proof requires us to use strong induction instead of simple induction! Proof of Correctness of findWord(). We proceed by strong induction on the number of pages, $$n$$, in $$D$$. Base Case ($$n = 1$$): If $$D$$ has one page, line $$6$$ searches via brute force for $$W$$. Thus, if $$W$$ is present, it is found and returned; else, we return “$$W$$ not found”, as desired. Inductive Hypothesis: Assume that findWord() is correct for all $$1\leq n\leq k$$. We prove it is correct for $$n=k+1$$. Inductive Step: After Step 9, we know $$W'$$; thus we can determine whether $$W$$ must be in the first or second half of $$D$$. In the first case, we recurse on the first half of $$D$$ in line 11, and in the second case we recurse on the second half of $$D$$ in line 13. By the Induction Hypothesis, the recursive call correctly finds $$W$$ in the first or second half, or returns “$$W$$ not found”. Since we return the recursive call’s answer, we conclude that findWord() correctly finds $$W$$ in $$D$$ of size $$n=k+1$$. By the principle of mathematical induction, findWord() is correct. $$\square$$ Sanity check! Why did we require strong induction in the proof above? # False Proofs It is very easy to prove false things if your proof is incorrect! Let’s illustrate with a famous example. In the middle of the last century, a colloquial expression in common use was “that is a horse of a different color", referring to something that is quite different from normal or common expectation. The reknowned mathematician George Polya, who was also a great expositor of mathematics for the lay public, gave the following proof to show that there is no horse of a different color! Theorem 8 All horses are the same color. Proof. We proceed by induction on the number of horses, $$n$$. Let $$P(n)$$ denote the statement of the claim. Base Case ($$n=1$$): $$P(1)$$ is certainly true, since if you have a set containing just one horse, all horses in the set have the same color. Inductive Hypothesis: Assume $$P(n)$$ holds for any $$n\geq 1$$. Inductive Step: Given a set of $$n+1$$ horses $$\{h_1,h_2,\ldots,h_{n+1}\}$$, we can exclude the last horse in the set and apply the inductive hypothesis just to the first $$n$$ horses $$\{h_1,\ldots,h_n\}$$, deducing that they all have the same color. Similarly, we can conclude that the last $$n$$ horses $$\{h_2,\ldots,h_{n+1}\}$$ all have the same color. But now the “middle” horses $$\{h_2,\ldots,h_n\}$$ (i.e., all but the first and the last) belong to both of these sets, so they have the same color as horse $$h_1$$ and horse $$h_{n+1}$$. It follows, therefore, that all $$n+1$$ horses have the same color. We conclude, by the principle of induction, that all horses have the same color. $$\spadesuit$$ Clearly, it is not true that all horses are of the same color! So, where did we go wrong? Recall that in order for the principle of mathematical induction to apply, the induction step must show the statement: $$\forall n\geq 1$$, $$P(n){\implies}P(n+1)$$. We claim that in the proof above, this statement is false — in particular, there exists a choice of $$n$$ which yields a counterexample to this statement. Exercise. For which value of n is it not true that $$P(n) \implies P(n+1)$$? [Hint: Think about the key property in the proof of having “middle” horses.] # Practice Problems 1. Prove for any natural number $$n$$ that $$1^2 + 2^2 + 3^2 + \cdots + n^2 = n(n+1)(2n+1)/6$$. 2. Prove that $$3^n > 2^n$$ for all natural numbers $$n \geq 1$$. 3. In real analysis, Bernoulli’s Inequality is an inequality which approximates the exponentiations of $$1+x$$. Prove this inequality, which states that $$(1+x)^n \geq 1+nx$$ if $$n$$ is a natural number and $$1+x > 0$$. 4. A common recursively defined function is the factorial, defined for a nonnegative number $$n$$ as $$n! = n(n-1)(n-2) \dotsm 1$$, with base case $$0! = 1$$. Let us reinforce our understanding of the connection between recursion and induction by considering the following theorem involving factorials. Theorem 9 $$\forall n \in {\mathbb{N}}$$, $$n > 1 \Longrightarrow n! < n^n$$. Prove this theorem using induction. [Hint: In the Inductive Step, write $$(n+1)!=(n+1)\cdot n!$$, and use the Induction Hypothesis.] 5. A celebrity at a party is someone whom everyone knows, yet who knows no one. Suppose that you are at a party with $$n$$ people. For any pair of people $$A$$ and $$B$$ at the party, you can ask $$A$$ if they know $$B$$ and receive an honest answer. Give a recursive algorithm to determine whether there is a celebrity at the party, and if so who, by asking at most $$3n - 4$$ questions. (Note: For the purpose of this question you are just visiting the party to ask questions. What you are trying to determine is whether the $$n$$ people actually attending the party include a celebrity.) Prove by induction that your algorithm always correctly identifies a celebrity iff there is one, and that the number of questions is at most $$3n-4$$. 1. This problem also goes under the name of the Postage Stamp Problem, which states that every amount of postage over $$12$$ cents can be paid using $$4$$- and $$5$$-cent stamps. 2. Fibonacci was also known as Leonardo of Pisa. If you ever find yourself in Pisa, Italy, you can visit his grave; his remains are buried at Campo Santo.
## How do you solve a quadratic equation? Solving Quadratic EquationsPut all terms on one side of the equal sign, leaving zero on the other side.Factor.Set each factor equal to zero.Solve each of these equations.Check by inserting your answer in the original equation. ## What is the quadratic equation used for? As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola, and the number of real zeros the quadratic equation contains. ## What is the general formula for quadratic equation? Answer: The general quadratic equation formula is “ax2 + bx + c”. In this formula, a, b, and c are number; they are the numerical coefficient of the quadratic equation and ‘a’ is not zero a 0. ## Why is it called a quadratic equation? We use the word quadratic because “quadra” refers to a square, and the leading term in a quadratic equation is “squared.” This is consistent with calling a degree three polynomial a “cubic” for the leading term represents a cube. The word for an equation with a leading term of x^4 is “quartic.” ## What is a quadratic in math? In mathematics, a quadratic is a type of problem that deals with a variable multiplied by itself — an operation known as squaring. The word “quadratic” comes from quadratum, the Latin word for square. ## What are the 4 ways to solve quadratic equations? The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula. ## Who uses quadratic equations in real life? Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0. ## Why do quadratic equations equal zero? We use the zero product property when we solve quadratic equations. You may have noticed that we always manipulate quadratic equations to ax2+bx+c=0. This is because factoring the equation gives us two expressions that multiply to zero. We can set each factor equal to zero and solve for x. Brahmagupta You might be interested: Harris benedict equation female ## What is a quadratic equation graph? The graph of a quadratic function is a U-shaped curve called a parabola. The sign on the coefficient a of the quadratic function affects whether the graph opens up or down. If a<0 , the graph makes a frown (opens down) and if a>0 then the graph makes a smile (opens up). ## What does 4 mean in math? In mathematics, the number 4 represents a quantity or value of 4. The whole number between 3 and 5 is 4. The number name of 4 is four. ## What is a real life example of a quadratic function? Balls, Arrows, Missiles and Stones. When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster and a Quadratic Equation tells you its position at all times! ### Releated #### Equation of vertical line How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […] #### Bernoulli’s equation example What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
# The Lengths of the Sides of a Triangle Are in the Ratio 3 : 4 : 5 and Its Perimeter is 144 Cm. Find the Area of the Triangle and the Height Corresponding to the Longest Side. - Mathematics The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side. #### Solution Let the sides of a triangle are 3x, 4x and 5x. Now, a = 3x, b = 4x and c = 5x The perimeter 2s = 144 ⇒ 3x + 4x + 5x = 144 [∵ a + b + c = 2s] ⇒ 12x = 144 ⇒ x = 12 ∴ sides of triangle are a = 3(x) = 36cm b = 4(x) = 48 cm c = 5(x) = 60 cm Now semi perimeter s1/2(a+b+c)=1/2(144)=72cm By heron’s formulas ∴ Area of Δle = sqrt(s(s-a)(s-b)(s-c)) =sqrt(72(72-36)(72-48)(72-60) =864cm^2 Let l be the altitude corresponding to longest side,∴1/2xx60xxl=864 ⇒l=(864xx2)/60 ⇒l=28.8cm Hence the altitude one corresponding long side = 28.8 cm Concept: Area of a Triangle by Heron's Formula Is there an error in this question or solution? #### APPEARS IN RD Sharma Mathematics for Class 9 Chapter 17 Heron’s Formula Exercise 17.1 \| Q 9 \| Page 8
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Lesson 3: Adding and subtracting fractions with unlike denominators # Subtracting fractions with unlike denominators To subtract fractions with different denominators, you need to find a common denominator. This can be done by identifying the least common multiple of the two denominators. After rewriting the fractions so they both have the common denominator, you can subtract the numerators as you would with any two fractions. ## Want to join the conversation? • what if they are the same denominaters • If the denominators are the same , you leave the denominators the same and add or subtract the numerators. Make sure you don't add or subtract the denominators. • What do i do when i have this? 14/6 The numerator is bigger than the denominator. • The answer would be 2 2/6 because 6 goes into 14 2 times and 2 would be left and that turns into the numerator, and the denominator would stay the same • how do you subtract 8/2 - 7/12 • create common denominators so set 8/2 to 48/12 (6 x 2=12, 8 x 6=48) then subtract 48 by 7 to get 41 and place that back over 12 to get 41/12 • how do you multiply or divide with unlike denominators • what if there is 2 diffrent denomanonater • You will have to find a multiple that both denominators have. Then, multiply the numerator by the number you multiplied the denominator to get the common multiple. For example: 5/9 - 6/5 You will need to find a common multiple that 9 and 5 both have. 9: 9, 18, 27, 36, 45, 54... 5: 5, 10, 15, 20, 25, 30, 35, 45, 50... In this case 45 is the common multiple. Then, you would have to multiply the numerators by the number you multiplied their denominator to get the common multiple. Since 9 times 5 equals 45, the 5 in 5/9 needs to be multiplied by 5 and the 6 in 6/5 would need to be multiplied by 9. So, 5/9 becomes 25/45 and 6/5 becomes 54/45. Therefore, 25/45 - 54/45 = -29/45 Hope this helped :) • what if there numerators are bigger than there denominators • then you turn the fraction into a mixed number • I was given the question 2/x+4=3/x I'm not sure how to work this out... how wold i make the denominates common? • Your denominator would be either x or what x stands for. (1 vote) • how do I subtract mixed fraction where only one fraction is mixed • You turn both fractions into improper fractions, next you find a like denominator, then you subtract. • what if the numerater is bigger than the denominator
# Volume of a Orthohedron - Examples, Exercises and Solutions Students start learning mathematics as early as elementary school, and as they progress, the subject becomes more and more complicated. Among others, the syllabus devotes a part to geometry and requires students to master different shapes and know how to calculate their area and volume. Are you also studying these days how to calculate the volume of a rectangular prism? ### Suggested Topics to Practice in Advance 1. Parts of a Rectangular Prism ## examples with solutions for volume of a orthohedron ### Exercise #1 Given the cuboid whose length is equal to 7 cm Width is equal to 3 cm The height of the cuboid is equal to 5 cm Calculate the volume of the cube ### Step-by-Step Solution The formula to calculate the volume of a cuboid is: heightlengthwidth We replace the data in the formula: 357 75 = 35 353 = 105 105 cm³ ### Exercise #2 Given the cuboid of the figure: Given: volume of the cuboid is 45 What is the value of X? ### Step-by-Step Solution Volume formula for a rectangular prism: Volume = length X width X height Therefore, first we will place the data we are given into the formula: 45 = 2.54X We divide both sides of the equation by 2.5: 18=4X And now we divide both sides of the equation by 4: 4.5 = X 4.5 ### Exercise #3 Look at the following orthohedron: The volume of the orthohedron is $80~cm^3$. The length of the lateral edge is 4 meters. What is the area of the base of the orthohedron? ### Step-by-Step Solution The formula for the volume of a box is heightlengthwidth In the specific question, we are given the volume and the height, and we are looking for the area of the base, As you will remember, the area is length width If we replace all the data in the formula, we see that: 4 * the area of the base = 80 Therefore, if we divide by 4 we see that Area of the base = 20 20 cm² ### Exercise #4 Given the cuboid of the figure: The area of the base of the cuboid is 15 cm², The length of the lateral edge is 3 cm. what is the volume of the cuboid ### Step-by-Step Solution To calculate the volume of a cuboid, as we mentioned, we need the length, width, and height. It is important to note that in the exercise we are given the height and the base area of the cuboid. The base area is actually the area multiplied by the length. That is, it is the data that contains the two pieces of information we are missing. Therefore, we can calculate the area by height * base area 15*3 = 45 This is the solution! 45 cm² ### Exercise #5 A building is 21 meters high, 15 meters long, and 14+30X meters wide. Express its volume in terms of X. ### Step-by-Step Solution We use a formula to calculate the volume: height times width times length. We rewrite the exercise using the existing data: $21\times(14+30x)\times15=$ We use the distributive property to simplify the parentheses. We multiply 21 by each of the terms in parentheses: $(21\times14+21\times30x)\times15=$ We solve the multiplication exercise in parentheses: $(294+630x)\times15=$ We use the distributive property again. We multiply 15 by each of the terms in parentheses: $294\times15+630x\times15=$ We solve each of the exercises in parentheses to find the volume: $4,410+9,450x$ $4410+9450x$ ### Exercise #6 Look at the cuboid below: What is the volume of the cuboid? 480 cm² ### Exercise #7 Calculate the volume of the rectangular prism below using the data provided. 48 ### Exercise #8 Calculate the volume of the rectangular prism below using the data provided. 180 ### Exercise #9 A rectangular prism has a base measuring 5 units by 8 units. The height of the prism is 12 units. Calculate its volume. 480 ### Exercise #10 Below is a cuboid with a length of 8 cm. Its width is 2 cm and its height is 4 cm. Calculate the volume of the cube. 64 cm³ ### Exercise #11 A cuboid has a length of is 9 cm. It is 4 cm wide and 5 cm high. Calculate the volume of the cube. 180 cm³ ### Exercise #12 Given the cuboid of the figure: What is its volume? 180 ### Exercise #13 The volume of the cuboid is es: ### Video Solution length X widthX height ### Exercise #14 Shown below is a cuboid with a length of 8 cm. Its width is 2 cm and its height is 4 cm. Calculate the volume of the cube. 64 cm³ ### Exercise #15 A cuboid is 9 cm long, 4 cm wide, and 5 cm high. Calculate the volume of the cube.
# 3.7 Further applications of newton’s laws of motion (Page 5/6) Page 5 / 6 Solution for (a) We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is $\Delta v=8.00 m/s$ . We are given the elapsed time, and so $\Delta t=2.50 s$ . The unknown is acceleration, which can be found from its definition: $a=\frac{\Delta v}{\Delta t}.$ Substituting the known values yields $\begin{array}{lll}a& =& \frac{8.00 m/s}{2\text{.}50 s}\\ & =& 3\text{.}{\text{20 m/s}}^{2}.\end{array}$ Discussion for (a) This is an attainable acceleration for an athlete in good condition. Solution for (b) Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is, ${F}_{\text{net}}=\text{ma}.$ Substituting the known values of $m$ and $a$ gives $\begin{array}{lll}{F}_{\text{net}}& =& \left(\text{70.0 kg}\right)\left(3\text{.}{\text{20 m/s}}^{2}\right)\\ & =& \text{224 N}.\end{array}$ Discussion for (b) This is about 50 pounds, a reasonable average force. This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. ## Summary • Newton’s laws of motion can be applied in numerous situations to solve problems of motion. • Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether ${F}_{\text{net}}=\text{ma}$ or ${F}_{\text{net}}=0$ . • The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full weight of the object. • Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion. ## Conceptual questions To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at $g$ . Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft? #### Questions & Answers what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej characteristics of micro business Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial how to synthesize TiO2 nanoparticles by chemical methods Zubear how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Berger describes sociologists as concerned with Got questions? 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# Points A and B are at (5 ,9 ) and (8 ,6 ), respectively. Point A is rotated counterclockwise about the origin by pi and dilated about point C by a factor of 4 . If point A is now at point B, what are the coordinates of point C? Jul 5, 2017 The point $C = \left(- \frac{28}{3} , - 14\right)$ #### Explanation: The matrix of a rotation counterclockwise by $\pi$ about the origin is $\left(\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right)$ Therefore, the transformation of point $A$ is $A ' = \left(\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right) \left(\begin{matrix}5 \\ 9\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 9\end{matrix}\right)$ Let point $C$ be $\left(x , y\right)$, then $\vec{C B} = 4 \vec{C A '}$ $\left(\begin{matrix}8 - x \\ 6 - y\end{matrix}\right) = 4 \left(\begin{matrix}- 5 - x \\ - 9 - y\end{matrix}\right)$ So, $8 - x = 4 \left(- 5 - x\right)$ $8 - x = - 20 - 4 x$ $3 x = - 28$ $x = - \frac{28}{3}$ and $6 - y = 4 \left(- 9 - y\right)$ $6 - y = - 36 - 4 y$ $3 y = - 42$ $y = - 14$ Therefore, The point $C = \left(- \frac{28}{3} , - 14\right)$
# Question Video: Completing Symmetrical Patterns Using Mirror Lines Emma is making a symmetrical pattern. She has drawn her mirror line. What is the missing part of her pattern? 02:59 ### Video Transcript Emma is making a symmetrical pattern. She has drawn her mirror line. What is the missing part of her pattern? We’re told that Emma has made a symmetrical pattern and we can see a picture of it. Let’s remind ourselves what the words symmetrical pattern mean. When something symmetrical, it’s the same on both sides. That’s why the idea of a mirror is really useful when we’re thinking about symmetry. A mirror reflects what’s in front of it, and it looks the same on both sides. We’re told in the question that Emma has drawn her mirror line. Can you see where it is? Here it is, let’s label it in a bright color so it stands out for us. Now, this mirror line is important. It’s in the very center of Emma’s pattern. And because Emma’s pattern is symmetrical on either side of the mirror line, the shapes are the same. Up close, next to the mirror line at the top, we have two yellow squares, and the next two shapes along are the same. We have two blue triangles. Now, although Emma is making a symmetrical pattern, she hasn’t quite finished it yet. There’s a part to her pattern that’s missing, and we’re asked, what is the missing part of her pattern? We need to think of the shapes that belong in this space. Now, you may think it’s got something to do with these two shapes that are at the bottom, the pink circle and the blue triangle. And you know what? You’d be right. Now, it would be very easy to look at these shapes and to think, “Well, it must be the same on both sides.” The answer’s got to be a pink circle and a blue triangle. Well, if you think the answer is this, you’re almost there, but the answer is not quite right. And we can see why if we put these shapes into position. Can you spot the mistake? If we start from the mirror line, just like before, and we compare the shapes on either side, we start with a blue triangle on one side, but the shape on the other side doesn’t match. It’s not the same, and neither do the next two shapes along. Can you see what we need to do to make the pattern symmetrical? You’ve got the right shapes, but they’re not in the right order. The nearest shape to the mirror line needs to be a blue triangle and then the pink circle. Can you see how the shapes match now? The pattern is the same on both sides of the mirror line. It’s symmetrical. The missing part of Emma’s pattern is a blue triangle followed by a pink circle.
Second derivative of $\operatorname{asec}{\left(x \right)}$ The calculator will find the second derivative of $\operatorname{asec}{\left(x \right)}$, with steps shown. Related calculators: Derivative Calculator, Logarithmic Differentiation Calculator Leave empty for autodetection. Leave empty, if you don't need the derivative at a specific point. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. Find $\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right)$. Find the first derivative $\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right)$ The derivative of the inverse secant is $\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}$: $${\color{red}\left(\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}\right)}$$ Simplify: $$\frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}} = \frac{\left\|{x}\right\|}{x^{2} \sqrt{x^{2} - 1}}$$ Thus, $\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{\left\|{x}\right\|}{x^{2} \sqrt{x^{2} - 1}}$. Next, $\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{d}{dx} \left(\frac{\left\|{x}\right\|}{x^{2} \sqrt{x^{2} - 1}}\right)$ Apply the quotient rule $\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$ with $f{\left(x \right)} = \left\|{x}\right\|$ and $g{\left(x \right)} = x^{2} \sqrt{x^{2} - 1}$: $${\color{red}\left(\frac{d}{dx} \left(\frac{\left\|{x}\right\|}{x^{2} \sqrt{x^{2} - 1}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\left\|{x}\right\|\right) x^{2} \sqrt{x^{2} - 1} - \left\|{x}\right\| \frac{d}{dx} \left(x^{2} \sqrt{x^{2} - 1}\right)}{\left(x^{2} \sqrt{x^{2} - 1}\right)^{2}}\right)}$$ Apply the product rule $\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$ with $f{\left(x \right)} = x^{2}$ and $g{\left(x \right)} = \sqrt{x^{2} - 1}$: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left\|{x}\right\| {\color{red}\left(\frac{d}{dx} \left(x^{2} \sqrt{x^{2} - 1}\right)\right)}}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left\|{x}\right\| {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) \sqrt{x^{2} - 1} + x^{2} \frac{d}{dx} \left(\sqrt{x^{2} - 1}\right)\right)}}{x^{4} \left(x^{2} - 1\right)}$$ The function $\sqrt{x^{2} - 1}$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \sqrt{u}$ and $g{\left(x \right)} = x^{2} - 1$. Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(x^{2} {\color{red}\left(\frac{d}{dx} \left(\sqrt{x^{2} - 1}\right)\right)} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(x^{2} {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(x^{2} - 1\right)\right)} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ Apply the power rule $\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$ with $n = \frac{1}{2}$: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(x^{2} {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(x^{2} - 1\right) + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(x^{2} {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(x^{2} - 1\right) + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \frac{d}{dx} \left(x^{2} - 1\right)}{2 \sqrt{{\color{red}\left(u\right)}}} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \frac{d}{dx} \left(x^{2} - 1\right)}{2 \sqrt{{\color{red}\left(x^{2} - 1\right)}}} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ The derivative of a sum/difference is the sum/difference of derivatives: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} {\color{red}\left(\frac{d}{dx} \left(x^{2} - 1\right)\right)}}{2 \sqrt{x^{2} - 1}} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{x^{2} - 1}} + \sqrt{x^{2} - 1} \frac{d}{dx} \left(x^{2}\right)\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 2$: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \left({\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(1\right)\right)}{2 \sqrt{x^{2} - 1}} + \sqrt{x^{2} - 1} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \left({\color{red}\left(2 x\right)} - \frac{d}{dx} \left(1\right)\right)}{2 \sqrt{x^{2} - 1}} + \sqrt{x^{2} - 1} {\color{red}\left(2 x\right)}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ The derivative of a constant is $0$: $$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \left(2 x - {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left\|{x}\right\|\right) - \left(\frac{x^{2} \left(2 x - {\color{red}\left(0\right)}\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ The derivative of the absolute value is $\frac{d}{dx} \left(\left\|{x}\right\|\right) = \frac{x}{\left\|{x}\right\|}$: $$\frac{x^{2} \sqrt{x^{2} - 1} {\color{red}\left(\frac{d}{dx} \left(\left\|{x}\right\|\right)\right)} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} {\color{red}\left(\frac{x}{\left\|{x}\right\|}\right)} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)}$$ Simplify: $$\frac{\frac{x^{3} \sqrt{x^{2} - 1}}{\left\|{x}\right\|} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left\|{x}\right\|}{x^{4} \left(x^{2} - 1\right)} = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left\|{x}\right\|}$$ Thus, $\frac{d}{dx} \left(\frac{\left\|{x}\right\|}{x^{2} \sqrt{x^{2} - 1}}\right) = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left\|{x}\right\|}$. Therefore, $\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left\|{x}\right\|}$. $\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left\|{x}\right\|}$A
# Question Video: Proving Polynomial Identities Mathematics Is the equation π₯Β³ β π¦Β³ = (π₯ β π¦)(π₯Β² + π₯π¦ + π¦Β²) an identity? 02:12 ### Video Transcript Is the equation π₯ cubed minus π¦ cubed equals π₯ minus π¦ multiplied by π₯ squared plus π₯π¦ plus π¦ squared an identity? So, if we look and see whether it is an identity, we wanna see whether the left-hand side is identical to the right-hand side. And to do that, what weβre gonna have to do is distribute across the parentheses. And in order to distribute across our parentheses, what weβre gonna do is multiply each of the terms in the left-hand set by each of the terms in the right-hand set. So first to start with, π₯ multiplied by π₯ squared, which is π₯ cubed. Then, weβre gonna have π₯ multiplied by positive π₯π¦, which is gonna give us π₯ squared π¦. And then, weβre gonna have π₯ multiplied by positive π¦ squared, which is gonna give us π₯π¦ squared. Well then, next what weβre gonna have is negative π¦ multiplied by π₯ squared, which gives us negative π₯ squared π¦. And then, weβre gonna have negative π¦ multiplied by positive π₯π¦, which is gonna give us negative π₯π¦ squared. And then finally, weβve got negative π¦ multiplied by positive π¦ squared, which is gonna give us negative π¦ cubed. Okay, great, weβve got to this point, but this doesnβt look like the left-hand side of our equation. But what weβre gonna have to do is tidy up first with some simplifying. Well, first of all, weβve only got one π₯ cubed. Well then, we have π₯ squared π¦ minus π₯ squared π¦, which means these are gonna cancel each other out because theyβre gonna be equal to zero. So then, next weβre gonna have positive π₯π¦ squared minus π₯π¦ squared. So again, these are gonna cancel each other out. And then finally, weβve got minus π¦ cubed. So, weβre now left with π₯ cubed minus π¦ cubed. Well, this is what we started with on the left-hand side of the equation. So therefore, we can say that the left-hand side and the right-hand side are identical. So therefore, the answer is yes. And we can say that the equation π₯ cubed minus π¦ cubed equals π₯ minus π¦ multiplied by π₯ squared plus π₯π¦ plus π¦ squared is an identity.
# The legs of a right triangle are √15 and 1 find the sine of the smallest angle of this triangle. In order for us to find the hypotenuse of the AC, we will use the Pythagorean theorem which is applicable in a right-angled triangle. The Pythagorean theorem sounds like this: the square of the hypotenuse is equal to the sum of the squares of the legs. c ^ 2 = a ^ 2 + b ^ 2, where a and b are the legs of a right triangle, and c is the hypotenuse in a right triangle. Find the hypotenuse AC In our case, the theorem will look like this: AC2 = AB2 + BC2. Let’s substitute the known values into this formula such as the AB leg which is equal to √15 cm and the second BC leg which is 1 cm and find the AC hypotenuse in the right-angled triangle ABC, then we get: AC ^ 2 = AB ^ 2 + BC ^ 2; AC ^ 2 = (√15 cm) ^ 2 + BC ^ 2; AC ^ 2 = (√15 cm) ^ 2 + (1 cm) ^ 2; We bring √15 cm to the square, we get: AC ^ 2 = 15 cm2 + (1 cm) 2; Now let’s bring 1 cm to the square, we get: AC ^ 2 = 15 cm2 + 1 cm2; AC ^ 2 = 16 cm2; Find an AC without a square, we get: AC = √16cm2 = 4 cm. And so we found the hypotenuse AC. Find sinA and sinC sin A = BC / AC = 1/4 = 0.25; sin C = AB / AC = √15 / 4 = 0.968; We get that the sine of angle A is 0.25, which means that its angle is about 14 degrees, and this is the smallest angle in the ABC triangle. Answer: The sine of the smallest angle of this triangle is 0.25. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
# Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Selina Solutions Concise Maths Class 7 Chapter 15 Triangles provides students with a clear idea about the basic concepts covered in this chapter. The solutions are prepared by faculty after conducting research on each topic, keeping in mind the understanding capacity of students. Here, the students can download Selina Solutions Concise Maths Class 7 Chapter 15 Triangles free PDF, from the links which are provided here. Chapter 15 helps students understand the types of triangles based on the angles and length of sides. The solutions improve problem solving and analytical thinking skills among students, which are important from the exam point of view. ## Selina Solutions Concise Maths Class 7 Chapter 15: Triangles Download PDF ### Exercises of Selina Solutions Concise Maths Class 7 Chapter 15 – Triangles Exercise 15A Solutions Exercise 15B Solutions Exercise 15C Solutions ## Access Selina Solutions Concise Maths Class 7 Chapter 15: Triangles #### Exercise 15A page: 176 1. State, if the triangles are possible with the following angles : (i) 20Â°, 70Â° and 90Â° (ii) 40Â°, 130Â° and 20Â° (iii) 60Â°, 60Â° and 50Â° (iv) 125Â°, 40Â° and 15Â° Solution: In a triangle, the sum of three angles is 1800 (i) 20Â°, 70Â° and 90Â° Sum = 200 + 700 + 900 = 1800 Here the sum is 1800 and therefore it is possible. (ii) 40Â°, 130Â° and 20Â° Sum = 40Â° + 130Â° + 20Â° = 1900 Here the sum is not 1800 and therefore it is not possible. (iii) 60Â°, 60Â° and 50Â° Sum = 60Â° + 60Â° + 50Â° = 1700 Here the sum is not 1800 and therefore it is not possible. (iv) 125Â°, 40Â° and 15Â° Sum = 125Â° + 40Â° + 15Â° = 1800 Here the sum is 1800 and therefore it is possible. 2. If the angles of a triangle are equal, find its angles. Solution: In a triangle, the sum of three angles is 1800 So each angle = 1800/3 = 600 3. In a triangle ABC, âˆ A = 45Â° and âˆ B = 75Â°, find âˆ C. Solution: In a triangle, the sum of three angles is 1800 âˆ A + âˆ B + âˆ C = 1800 Substituting the values 450 + 750 + âˆ C = 1800 By further calculation 1200 + âˆ C = 1800 So we get âˆ C = 1800 â€“ 1200 = 600 4. In a triangle PQR, âˆ P = 60Â° and âˆ Q = âˆ R, find âˆ R. Solution: Consider âˆ Q = âˆ R = x âˆ P = 60Â° We can write it as âˆ P + âˆ Q + âˆ R = 1800 Substituting the values 600 + x + x = 1800 By further calculation 600 + 2x = 1800 2x = 1800 â€“ 600 = 1200 So we get x = 1200/2 = 600 âˆ Q = âˆ R = 600 Therefore, âˆ R = 600. 5. Calculate the unknown marked angles in each figure: Solution: In a triangle, the sum of three angles is 1800 (i) From figure (i) 900 + 300 + x = 1800 By further calculation 1200 + x = 1800 So we get x = 1800 â€“ 1200 = 600 Therefore, x = 600. (ii) From figure (ii) y + 800 + 200 = 1800 By further calculation y + 1000 = 1800 So we get y = 1800 â€“ 1000 = 800 Therefore, y = 800. (iii) From figure (iii) a + 900 + 400 = 1800 By further calculation a + 1300 = 1800 So we get a = 1800 â€“ 1300 = 500 Therefore, a = 500. 6. Find the value of each angle in the given figures: Solution: (i) From the figure (i) âˆ A + âˆ B + âˆ C = 1800 Substituting the values 5x0 + 4x0 + x0 = 1800 By further calculation 10x0 = 1800 x = 180/10 = 180 So we get âˆ A = 5x0 = 5 Ã— 180 = 900 âˆ B = 4x0 = 4 Ã— 180 = 720 âˆ C = x = 180 (ii) From the figure (ii) âˆ A + âˆ B + âˆ C = 1800 Substituting the values x0 + 2x0 + 2x0 = 1800 By further calculation 5x0 = 1800 x0 = 1800/5 = 360 So we get âˆ A = x0 = 360 âˆ B = 2x0 = 2 Ã— 360 = 720 âˆ C = 2x0 = 2 Ã— 360 = 720 7. Find the unknown marked angles in the given figure: Solution: (i) From the figure (i) âˆ A + âˆ B + âˆ C = 1800 Substituting the values b0 + 500 + b0 = 1800 By further calculation 2b0 = 1800 â€“ 500 = 1300 b0 = 1300/2 = 650 Therefore, âˆ A = âˆ C = b0 = 650. (ii) From the figure (ii) âˆ A + âˆ B + âˆ C = 1800 Substituting the values x0 + 900 + x0 = 1800 By further calculation 2x0 = 1800 â€“ 900 = 900 x0 = 900/2 = 450 Therefore, âˆ A = âˆ C = x0 = 450. (iii) From the figure (iii) âˆ A + âˆ B + âˆ C = 1800 Substituting the values k0 + k0 + k0 = 1800 By further calculation 3k0 = 1800 k0 = 1800/3 = 600 Therefore, âˆ A = âˆ B = âˆ C = 600. (iv) From the figure (iv) âˆ A + âˆ B + âˆ C = 1800 Substituting the values (m0 â€“ 50) + 600 + (m0 + 50) = 1800 By further calculation m0 â€“ 50 + 600 + m0 + 50 = 1800 2m0 = 1800 â€“ 600Â = 1200 m0 = 1200/2 = 600 Therefore, âˆ A = m0 â€“ 50 = 600 â€“ 50 = 550 âˆ C = m0 + 50 = 600 + 50 = 650 8. In the given figure, show that: âˆ a = âˆ b + âˆ c (i) If âˆ b = 60Â° and âˆ c = 50Â°; find âˆ a. (ii) If âˆ a = 100Â° and âˆ b = 55Â°; find âˆ c. (iii) If âˆ a = 108Â° and âˆ c = 48Â°; find âˆ b. Solution: From the figure AB \|\| CD b = âˆ C and âˆ A = c are alternate angles In triangle PCD Exterior âˆ APC = âˆ C + âˆ D a = b + c (i) If âˆ b = 60Â° and âˆ c = 50Â° âˆ a = âˆ b + âˆ c Substituting the values âˆ a = 60 + 50 = 1100 (ii) If âˆ a = 100Â° and âˆ b = 55Â° âˆ a = âˆ b + âˆ c Substituting the values âˆ c = 100 â€“ 55 = 450 (iii) If âˆ a = 108Â° and âˆ c = 48Â° âˆ a = âˆ b + âˆ c Substituting the values âˆ b = 108 â€“ 48 = 600 9. Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6. Solution: In a triangle, the sum of angles of a triangle is 1800 âˆ A + âˆ B + âˆ C = 1800 It is given that âˆ A: âˆ B: âˆ C = 4: 5: 6 Consider âˆ A = 4x, âˆ B = 5x and âˆ C = 6x Substituting the values 4x + 5x + 6x = 1800 By further calculation 15x = 1800 x = 1800/15 = 120 So we get âˆ A = 4x = 4 Ã— 120 = 480 âˆ B = 5x = 5 Ã— 120 = 600 âˆ C = 6x = 6 Ã— 120 = 720 10. One angle of a triangle is 60Â°. The, other two angles are in the ratio of 5 : 7. Find the two angles. Solution: From the triangle ABC Consider âˆ A = 600, âˆ B: âˆ C = 5:7 In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values 600 + âˆ B + âˆ C = 1800 By further calculation âˆ B + âˆ C = 1800 â€“ 600 = 1200 Take âˆ B = 5x and âˆ C = 7x Substituting the values 5x + 7x = 1200 12x = 1200 x = 1200/12 = 100 So we get âˆ B = 5x = 5 Ã— 100 = 500 âˆ C = 7x = 7 Ã— 100 = 700 11. One angle of a triangle is 61Â° and the other two angles are in the ratio 1 Â½ : 1 1/3. Find these angles. Solution: From the triangle ABC Consider âˆ A = 610 In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values 610 + âˆ B + âˆ C = 1800 By further calculation âˆ B + âˆ C = 1800 â€“ 610 = 1190 âˆ B: âˆ C = 1 Â½: 1 1/3 = 3/2: 4/3 Taking LCM âˆ B: âˆ C = 9/6: 8/ 6 âˆ B: âˆ C = 9: 8 Consider âˆ B = 9x and âˆ C = 8x Substituting the values 9x + 8x = 1190 17x = 1190 x = 1190/ 17 = 70 So we get âˆ B = 9x = 9 Ã— 70 = 630 âˆ C = 8x = 8 Ã— 70 = 560 12. Find the unknown marked angles in the given figures: Solution: In a triangle, if one side is produced Exterior angle is the sum of opposite interior angles (i) From the figure (i) 1100 = x0 + 300 By further calculation x0 = 1100 â€“ 300 = 800 (ii) From the figure (ii) 1200 = y0 + 600 By further calculation y0 = 1200 â€“ 600 = 600 (iii) From the figure (iii) 1220 = k0 + 350 By further calculation k0 = 1220 â€“ 350 = 870 (iv) From the figure (iv) 1350 = a0 + 730 By further calculation a0 = 1350 â€“ 730 = 620 (v) From the figure (v) 1250 = a + c â€¦â€¦ (1) 1400 = a + b â€¦â€¦ (2) a + c + a + b = 1250 + 1400 On further calculation a + a + b + c = 2650 We know that a + b + c = 1800 Substituting it in the equation a + 1800 = 2650 So we get a = 265 â€“ 180 = 850 If a + b = 1400 Substituting it in the equation 850 + b = 1400 So we get b = 140 â€“ 85 = 550 If a + c = 1250 Substituting it in the equation 850 + c = 1250 So we get c = 125 â€“ 85 = 400 Therefore, a = 850, b = 550 and c = 400. #### Exercise 15B page: 180 1. Find the unknown angles in the given figures: Solution: (i) From the figure (i) x = y as the angles opposite to equal sides In a triangle x + y + 800 = 1800 Substituting the values x + x + 800 = 1800 By further calculation 2x = 1800 â€“ 800 = 1000 x = 1000/2 = 500 Therefore, x = y = 500. (ii) From the figure (ii) b = 400 as the angles opposite to equal sides In a triangle a + b + 400 = 1800 Substituting the values a + 400 + 400 = 1800 By further calculation a = 180 â€“ 80 = 1000 Therefore, a = 1000 and b = 400. (iii) From the figure (iii) x = y as the angles opposite to equal sides In a triangle x + y + 900 = 1800 Substituting the values x + x + 900 = 1800 By further calculation 2x = 180 â€“ 90 = 900 x = 90/2 = 450 Therefore, x = y = 450. (iv) From the figure (iv) a = b as the angles opposite to equal sides are equal In a triangle a + b + 800 = 1800 Substituting the values a + a + 800 = 1800 By further calculation 2a = 180 â€“ 80 = 1000 a = 1000/2 = 500 Here a = b = 500 We know that in a triangle the exterior angle is equal to sum of its opposite interior angles x = a + 800 So we get x = 50 + 80 = 1300 Therefore, a = 500, b = 500 and x = 1300. (v) From the figure (v) In an isosceles triangle consider each equal angle = x x + x = 860 2x = 860 So we get x = 860/2 = 430 For a linear pair p + x = 1800 Substituting the values p + 430 = 1800 By further calculation p = 180 â€“ 43 = 1370 Therefore, p = 1370. 2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures: Solution: (i) a = 700 as the angles opposite to equal sides are equal In a triangle a + 700 + x = 1800 Substituting the values 700 + 700 + x = 1800 By further calculation x = 180 â€“ 140 = 400 y = b as the angles opposite to equal sides are equal Here a = y + b as the exterior angle is equal to sum of interior opposite angles 700 = y + y So we get 2y = 700 y = 700/2 = 350 Therefore, x = 400 and y = 350. (ii) From the figure (ii) Each angle is 600 in an equilateral triangle In an isosceles triangle Consider each base angle = a a + a + 1000 = 1800 By further calculation 2a = 180 â€“ 100 = 800 So we get a = 800/2 = 400 x = 600 + 400 = 1000 y = 600 + 400 = 1000 (iii) From the figure (iii) 1300 = x + p as the exterior angle is equal to the sum of interior opposite angles It is given that the lines are parallel Here p = 600 is the alternate angles and y = a In a linear pair a + 1300 = 1800 By further calculation a = 180 â€“ 130 = 500 Here x + p = 1300 Substituting the values x + 600 = 1300 By further calculation x = 130 â€“ 60 = 700 Therefore, x = 700, y = 500 and p = 600. (iv) From the figure (iv) x = a + b Here b = y and a = c as the angles opposite to equal sides are equal a + c + 300 = 1800 Substituting the values a + a + 300 = 1800 By further calculation 2a = 180 â€“ 30 = 1500 a = 150/2 = 750 We know that b + y = 900 Substituting the values y + y = 900 2y = 900 y = 90/2 = 450 where b = 450 Therefore, x = a + b = 75 + 45 = 1200 and y = 450. (v) From the figure (v) a + b + 400 = 1800 So we get a + b = 180 â€“ 40 = 1400 The angles opposite to equal sides are equal a = b = 140/2 = 700 x = b + 400 = 700Â + 400= 1100 Here the exterior angle of a triangle is equal to the sum of its interior opposite angles In the same way y = a + 400 Substituting the values y = 700Â + 400 = 1100 Therefore, x = y = 1100. 3. The angle of vertex of an isosceles triangle is 100Â°. Find its base angles. Solution: Consider âˆ† ABC Here AB = AC and âˆ B = âˆ C We know that âˆ A = 1000 In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values 1000 + âˆ B + âˆ B = 1800 By further calculation 2âˆ B = 1800 â€“ 1000 = 800 âˆ B = 80/2 = 400 Therefore, âˆ B = âˆ C = 400. 4. One of the base angles of an isosceles triangle is 52Â°. Find its angle of vertex. Solution: It is given that the base angles of isosceles triangle ABC = 520 Here âˆ B = âˆ C = 520 In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values âˆ A + 520 + 520 = 1800 By further calculation âˆ A = 180 â€“ 104 = 760 Therefore, âˆ A = 760. 5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle. Solution: Consider the vertical angle of an isosceles triangle = x So the base angle = 4x In a triangle x + 4x + 4x = 1800 By further calculation 9x = 1800 x = 180/9 = 200 So the vertical angle = 200 Each base angle = 4x = 4 Ã— 200 = 800 6. The vertical angle of an isosceles triangle is 15Â° more than each of its base angles. Find each angle of the triangle. Solution: Consider the angle of the base of isosceles triangle = x0 So the vertical angle = x + 150 In a triangle x + x + x + 150 = 1800 By further calculation 3x = 180 â€“ 15 = 1650 x = 165/3 = 550 Therefore, the base angle = 550 Vertical angle = 55 + 15 = 700. 7. The base angle of an isosceles triangle is 15Â° more than its vertical angle. Find its each angle. Solution: Consider the vertical angle of the isosceles triangle = x0 Here each base angle = x + 150 In a triangle x + 150 + x + 150Â + x = 1800 By further calculation 3x + 300 = 1800 3x = 180 â€“ 30 = 1500 x = 150/3 = 500 Therefore, vertical angle = 500 and each base angle = 50 + 15 = 650. 8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle. Solution: Consider each base angle of an isosceles triangle = x Vertical angle = 3 (x + x) = 3 (2x) = 6x In a triangle 6x + x + x = 1800 By further calculation 8x = 1800 x = 180/8 = 22.50 Therefore, each base angle = 22.50 and vertical angle = 3 (22.5 + 22.5) = 3 Ã— 45 = 1350. 9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle. Solution: It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4 Consider base angle = x Vertical angle = 4x In a triangle x + x + 4x = 1800 By further calculation 6x = 1800 x = 180/6 = 300 Therefore, each base angle = x = 300 and vertical angle = 4x = 4 Ã— 300 = 1200. 10. In the given figure, BI is the bisector of âˆ ABC and CI is the bisector of âˆ ACB. Find âˆ BIC. Solution: In âˆ† ABC BI is the bisector of âˆ ABC and CI is the bisector of âˆ ACB Here AB = AC âˆ B = âˆ C as the angles opposite to equal sides are equal We know that âˆ A = 400 In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values 400 + âˆ B + âˆ B = 1800 By further calculation 400Â + 2âˆ B = 1800 2âˆ B = 180 â€“ 40 = 1400 âˆ B = 140/2 = 700 Here BI and CI are the bisectors of âˆ ABC and âˆ ACB âˆ IBC = Â½ âˆ ABC = Â½ Ã— 700 = 350 âˆ ICB = Â½ âˆ ACB = Â½ Ã— 700 = 350 In âˆ† IBC âˆ BIC + âˆ IBC + âˆ ICB = 1800 Substituting the values âˆ BIC + 350 + 350Â = 1800 By further calculation âˆ BIC = 180 â€“ 70 = 1100 Therefore, âˆ BIC = 1100. 11. In the given figure, express a in terms of b. Solution: From the âˆ† ABC BC = BA âˆ BCA = âˆ BAC Here the exterior âˆ CBE = âˆ BCA + âˆ BAC a = âˆ BCA + âˆ BCA a = 2âˆ BCA â€¦â€¦ (1) Here âˆ ACB = 1800 â€“ b Where âˆ ACD and âˆ ACB are linear pair âˆ BCA = 1800 â€“ b â€¦â€¦. (2) We get a = 2 âˆ BCA Substituting the values a = 2 (1800 â€“ b) a = 3600 â€“ 2b 12. (a) In Figure (i) BP bisects âˆ ABC and AB = AC. Find x. (b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects âˆ ABC and âˆ ADB = 70Â°. Solution: (a) From the figure (i) AB = AC and BP bisects âˆ ABC AP is drawn parallel to BC Here PB is the bisector of âˆ ABC âˆ PBC = âˆ PBA âˆ APB = âˆ PBC are alternate angles x = âˆ PBC â€¦.. (1) In âˆ† ABC âˆ A = 600 Since AB = AC we get âˆ B = âˆ C In a triangle âˆ A + âˆ B + âˆ C = 1800 Substituting the values 600 + âˆ B + âˆ C = 1800 We get 600 + âˆ B + âˆ B = 1800 By further calculation 2âˆ B = 180 â€“ 60 = 1200 âˆ B = 120/2 = 600 Â½ âˆ B = 60/2 = 300 âˆ PBC = 300 So from figure (i) x = 300 (b) From the figure (ii) DA = DB = DC Here BD bisects âˆ ABC and âˆ ADB = 700 In a triangle âˆ ADB + âˆ DAB + âˆ DBA = 1800 Substituting the values 700 + âˆ DBA + âˆ DBA = 1800 By further calculation 700 + 2âˆ DBA = 1800 2âˆ DBA = 180 â€“ 70 = 1100 âˆ DBA = 110/2 = 550 Here BD is the bisector of âˆ ABC So âˆ DBA = âˆ DBC = 550 In âˆ† DBC DB = DC âˆ DCB = âˆ DBC Hence, x = 550. 13. In each figure, given below, ABCD is a square and âˆ† BEC is an equilateral triangle. Find, in each case: (i) âˆ ABE (ii) âˆ BAE Solution: The sides of a square are equal and each angle is 900 In an equilateral triangle all three sides are equal and all angles are 600 In figure (i) ABCD is a square and âˆ† BEC is an equilateral triangle (i) âˆ ABE = âˆ ABC + âˆ CBE Substituting the values âˆ ABE = 900 + 600= 1500 (ii) In âˆ† ABE âˆ ABE + âˆ BEA + âˆ BAE = 1800 Substituting the values 1500 + âˆ BAE + âˆ BAE = 1800 By further calculation 2âˆ BAE = 180 â€“ 150 = 300 âˆ BAE = 30/2 = 150 In figure (ii) ABCD is a square and âˆ† BEC is an equilateral triangle (i) âˆ ABE = âˆ ABC – âˆ CBE Substituting the values âˆ ABE = 900 â€“ 600 = 300 (ii) In âˆ† ABE âˆ ABE + âˆ BEA + âˆ BAE = 1800 Substituting the values 300 + âˆ BAE + âˆ BAE = 1800 By further calculation 2âˆ BAE = 180 â€“ 30 = 1500 âˆ BAE = 150/2 = 750 14. In âˆ† ABC, BA and BC are produced. Find the angles a and h. if AB = BC. Solution: In âˆ† ABC, BA and BC are produced âˆ ABC = 540 and AB = BC In âˆ† ABC âˆ BAC + âˆ BCA + âˆ ABC = 1800 Substituting the values âˆ BAC + âˆ BAC + 540 = 1800 2âˆ BAC = 180 â€“ 54 = 1260 âˆ BAC = 126/2 = 630 âˆ BCA = 630 In a linear pair âˆ BAC + b = 1800 Substituting the value 630 + b = 1800 So we get b = 180 â€“ 63 = 1170 In a linear pair âˆ BCA + a = 1800 Substituting the value 630Â + a = 1800 So we get a = 180 â€“ 63 = 1170 Therefore, a = b = 1170. #### Exercise 15C page: 185 1. Construct a âˆ†ABC such that: (i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm (ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm (iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm Solution: (i) Steps of Construction 1. Construct a line segment BC = 4 cm. 2. Taking B as centre and 6 cm as radius construct an arc. 3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A. 4. Now join AB and AC. Therefore, âˆ†ABC is the required triangle. (ii) Steps of Construction 1. Construct a line segment CB = 6.5 cm 2. Taking C as centre and 4.2 cm as radius construct an arc. 3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A. 4. Now join AC and AB. Therefore, âˆ†ABC is the required triangle. (iii) Steps of Construction 1. Construct a line segment BC = 4 cm 2. Taking B as centre and 3.5 cm as radius construct an arc. 3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A. 4. Now join AB and AC. Therefore, âˆ†ABC is the required triangle. 2. Construct a âˆ† ABC such that: (i) AB = 7 cm, BC = 5 cm and âˆ ABC = 60Â° (ii) BC = 6 cm, AC = 5.7 cm and âˆ ACB = 75Â° (iii) AB = 6.5 cm, AC = 5.8 cm and âˆ A = 45Â° Solution: (i) Steps of Construction 1. Construct a line segment AB = 7 cm. 2. At the point B construct a ray which makes an angle 600 and cut off BC = 5cm. 3. Now join AC. Therefore, âˆ†ABC is the required triangle. (ii) Steps of Construction 1. Construct a line segment BC = 6 cm. 2. At the point C construct a ray which makes an angle 750 and cut off CA = 5.7 cm. 3. Now join AB. Therefore, âˆ†ABC is the required triangle. (iii) Steps of Construction 1. Construct a line segment AB = 6.5 cm. 2. At the point A construct a ray which makes an angle 450 and cut off AC = 5.8 cm. 3. Now join CB. Therefore, âˆ†ABC is the required triangle. 3. Construct a âˆ† PQR such that : (i) PQ = 6 cm, âˆ Q = 60Â° and âˆ P = 45Â°. Measure âˆ R. (ii) QR = 4.4 cm, âˆ R = 30Â° and âˆ Q = 75Â°. Measure PQ and PR. (iii) PR = 5.8 cm, âˆ P = 60Â° and âˆ R = 45Â°. Measure âˆ Q and verify it by calculations Solution: (i) Steps of Construction 1. Construct a line segment PQ = 6 cm. 2. At point P construct a ray which makes an angle 450. 3. At point Q construct another ray which makes an angle 600 which intersect the first ray at point R. Therefore, âˆ† PQR is the required triangle. By measuring âˆ R = 750. (ii) Steps of Construction 1. Construct a line segment QR = 4.4 cm. 2. At point Q construct a ray which makes an angle 750. 3. At point R construct another ray which makes an angle 300 which intersect the first ray at point R. Therefore, âˆ† PQR is the required triangle. By measuring the length, PQ = 2.1 cm and PR = 4.4 cm. (iii) Steps of Construction 1. Construct a line segment PR = 5.8 cm. 2. At point P construct a ray which makes an angle 600. 3. At point R construct another ray which makes an angle 450 which intersect the first ray at point Q. Therefore, âˆ† PQR is the required triangle. By measuring âˆ Q = 750. Verification â€“ âˆ P + âˆ Q + âˆ R = 1800 Substituting the values 600Â + âˆ Q + 450 = 1800 By further calculation âˆ Q = 180 â€“ 105 = 750 4. Construct an isosceles âˆ† ABC such that: (i) base BC = 4 cm and base angle = 30Â° (ii) base AB = 6.2 cm and base angle = 45Â° (iii) base AC = 5 cm and base angle = 75Â°. Measure the other two sides of the triangle. Solution: (i) Steps of Construction In an isosceles triangle the base angles are equal 1. Construct a line segment BC = 4 cm. 2. At the points B and C construct rays which makes an angle 300 intersecting each other at the point A. Therefore, âˆ† ABC is the required triangle. By measuring the equal sides, each is 2.5 cm in length approximately. (ii) Steps of Construction In an isosceles triangle the base angles are equal 1. Construct a line segment AB = 6.2 cm. 2. At the points A and B construct rays which makes an angle 450 intersecting each other at the point C. Therefore, âˆ† ABC is the required triangle. By measuring the equal sides, each is 4.3 cm in length approximately. (iii) Steps of Construction In an isosceles triangle the base angles are equal 1. Construct a line segment AC = 5 cm. 2. At the points A and C construct rays which makes an angle 750 intersecting each other at the point B. Therefore, âˆ† ABC is the required triangle. By measuring the equal sides, each is 9.3 cm in length approximately. 5. Construct an isosceles âˆ†ABC such that: (i) AB = AC = 6.5 cm and âˆ A = 60Â° (ii) One of the equal sides = 6 cm and vertex angle = 45Â°. Measure the base angles. (iii) BC = AB = 5-8 cm and ZB = 30Â°. Measure âˆ A and âˆ C. Solution: (i) Steps of Construction 1. Construct a line segment AB = 6.5 cm. 2. At point A construct a ray which makes an angle 600. 3. Now cut off AC = 6.5cm 4. Join BC. Therefore, âˆ† ABC is the required triangle. (ii) Steps of Construction 1. Construct a line segment AB = 6 cm. 2. At point A construct a ray which makes an angle 450. 3. Now cut off AC = 6cm 4. Join BC. Therefore, âˆ† ABC is the required triangle. By measuring âˆ B and âˆ C, both are equal toÂ 67 Â½ 0. (iii) Steps of Construction 1. Construct a line segment BC = 5.8 cm. 2. At point B construct a ray which makes an angle 300. 3. Now cut off BA = 5.8cm 4. Join AC. Therefore, âˆ† ABC is the required triangle. By measuring âˆ C and âˆ A is equal to 750. 6. Construct an equilateral triangle ABC such that: (i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC. (ii) Each side is 6 cm. Solution: (i) Steps of Construction 1. Construct a line segment AB = 5cm. 2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C. 3. Now join AC and BC where âˆ† ABC is the required triangle. 4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p. 5. Join PA, PB and PC. By measuring each is 2.8 cm. (ii) Steps of Construction 1. Construct a line segment AB = 6cm. 2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C. 3. Now join AC and BC Therefore, âˆ† ABC is the required triangle. 7. (i) Construct a âˆ† ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle. (ii) Construct an isosceles âˆ†PQR such that PQ = PR = 6.5 cm and âˆ PQR = 75Â°. Using ruler and compasses only construct a circumcircle to this triangle. (iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm. Construct a circumcircle to this triangle. Solution: (i) Steps of Construction 1. Construct a line segment BC = 4.5 cm. 2. Taking B as centre and 6 cm radius construct an arc. 3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A. 4. Now join AB and AC Therefore, âˆ† ABC is the required triangle. 5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O. 6. Now join OB, OC and OA. 7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C. This is the required circumcircle of âˆ† ABC. (ii) Steps of Construction 1. Construct a line segment PQ = 6.5 cm. 2. At point Q, construct an arc which makes an angle 750. 3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R. 4. Join PR. âˆ† PQR is the required triangle. 4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O. 5. Join OP, OQ and OR. 6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R. This is the required circumcircle of âˆ† PQR. (iii) Steps of Construction 1. Construct a line segment AB = 5.5 cm. 2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C. 3. Now join AC and BC. âˆ† ABC is the required triangle. 4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O. 5. Now join OA, OB and OC. 6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C. This is the required circumcircle. 8. (i) Construct a âˆ†ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius. (ii) Construct an isosceles âˆ† MNP such that base MN = 5.8 cm, base angle MNP = 30Â°. Construct an incircle to this triangle and measure its radius. (iii) Construct an equilateral âˆ†DEF whose one side is 5.5 cm. Construct an incircle to this triangle. (iv) Construct a âˆ† PQR such that PQ = 6 cm, âˆ QPR = 45Â° and angle PQR = 60Â°. Locate its incentre and then draw its incircle. Solution: (i) Steps of Construction 1. Construct a line segment AB = 6 cm. 2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C. 3. Now join AC and BC. 4. Construct the angle bisector of âˆ A and âˆ B which intersect each other at point I. 5. From the point I construct IL which is perpendicular to AB. 6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally. By measuring the required incircle the radius is 1.6 cm. (ii) Steps of Construction 1. Construct a line segment MN = 5.8 cm. 2. At points M and N construct two rays which make an angle 300 each intersecting each other at point P. 3. Construct the angle bisectors of âˆ M and âˆ N which intersect each other at point I. 4. From the point I draw perpendicular IL on MN. 5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ† PMB internally. By measuring the required incircle the radius is 0.6 cm. (iii) Steps of Construction 1. Construct a line segment BC = 5.5 cm. 2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A. 3. Now join AB and AC. 4. Construct the perpendicular bisectors of âˆ B and âˆ C which intersect each other at the point I. 5. From the point I construct IL which is perpendicular to BC. 6. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†ABC internally. This is the required incircle. (iv) Steps of Construction 1. Construct a line segment PQ = 6 cm. 2. At the point P construct rays which make an angle of 450 and at point Q which makes an angle 600Â thats intersects each other at point R. 3. Construct the bisectors of âˆ P and âˆ Q which intersect each other at point I. 4. From the point I construct IL which is perpendicular to PQ. 5. Taking I as centre and IL as radius construct a circle which touches the sides of âˆ†PQR internally. This is the required incircle where the point I is incentre.
Math and Arithmetic Statistics Probability What does the standard deviation of a data set tell us about the data and why is the standard deviation an important measure? Standard Deviation Explained: Standard deviation is a simple measure of width of a distribution of numbers (usually scores or measurements). It is the next high degree of sophistication of characterizing a bunch of number other than just giving the average ( or mean, median, mode). If you know the average, you do not know how tightly the numbers are clustered around the average, that is what the standard deviation tells you. It is one definition, the most common definition, of the width of a distribution. (There are, of course, many other additional characterizations.) It is important because it tells you if the average is a very useful quantity to use to interpret the data. If someone tells you that the average person your age dies in 50 years, that seems important, but if someone says that the average person dies in 50 years, give or take 20 years, suddenly you realize there is more to the story and maybe you should save more money, just in case. Well, the "give or take" part of that statement is very useful, but not well defined. If they say the life expectancy is 50 years with a standard deviation of 20 years, then that is perfectly defined mathematically. Standard deviation is a mathematical measure of the broadness of the distribution of data. The following two data sets, A and B, have the same mean (average): A: 48, 49, 50, 51, 52 B: 30, 40, 50, 60, 70 The distribution of the data about the mean in A is very narrow, whereas the distribution about the mean in B is broad. The S.D. gives us a quantification of the broadness of the distribution. In normal distributions, about 68 percent of the data will fall within one S.D. on either side of the mean. About 96 percent of the data will fall with two S.D. Let's say your teacher gives a test to one hundred kids and the test average is 80 points and the S.D. is 10. If the distribution is "normal," about 34 kids will score between 70 and 80, and about 34 kids will score between 80 and 90. We can also predict that about 14 kids will score between 90 and 100, and 14 will score below 70. That leaves four kids. They fall into two groups: they either totally bombed the test, or they got the extra credit question to boost their score over 100!
Skip to Main Content It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results. # Math help from the Learning Centre This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills. ## Domain and Range Different functions have different domains and ranges. Sometimes there are restrictions on domains and ranges. The domain is the set of all input values. The range is the set of all output values. If we draw a graph with the x-coordinate as the horizontal values, and y-coordinate as the vertical values. Then the domain is where the function lies horizontally and the range is where the function lies vertically. In the graph above, notice there are no values in the domain less than -5 on the left side of the graph. On the right side, the graph keeps going. If we call this graph $$f(x)$$, the domain is stated $$D: \{x\in \mathbb{R}\|x \geq -5\}$$. This means all x values (input values) can be any Real Number greater than or equal to -5. You can also state the domain as the following $$D:[-5, \infty)$$. Notice a [ bracket was used before -5 to state that it is greater than or equal to. A circle bracket, ( ), is used for open ends like infinity, or when it is strictly greater than. The range of the graph $$f(x)$$ above, is stated as $$R: \{f(x)\in \mathbb{R}\|f(x) \leq 5\}$$. You can use $$y$$ instead of $$f(x)$$. This means that all $$y$$ or $$f(x)$$ values are less than or equal to 5. In the other notation, it is stated $$R:(-\infty, 5]$$ ## Finding the Domain and Range of Common Functions Knowing the properties and graphs of functions helps you find the domain and range. For example, we know that $$f(x) = x^2$$ is a parabola opening up with vertex at $$(0,0)$$. The domain of $$f(x)$$ is all the numbers, so $$D:\{x\in\mathbb{R}\}$$. The range is all the numbers greater than or equal to 0, so $$R:\{f(x)\in\mathbb{R}\|f(x) \geq 0\}$$. Other quadratic functions will have similar domains and ranges. For example, if $$f(x) = 3x^2-10$$ is the same graph that has been stretched vertically by a factor of 3 and moved down 10 units. The domain is still $$D:\{x\in\mathbb{R}\}$$. The range is now all the numbers greater than or equal to -10, since the vertex has moved down 10 units, so $$R:\{f(x)\in\mathbb{R}\|f(x) \geq -10\}$$. So knowing the base function properties helps you determine domains and ranges after the graph has been transformed (dilated, moved. reflected). Here are other common functions: The root function, $$f(x) = \sqrt{x}$$: $$D:\{x\in\mathbb{R}\|x \geq 0\}$$ and $$R:\{f(x)\in\mathbb{R}\|f(x) \geq 0\}$$ The cubic function, $$f(x) = x^3$$: $$D:\{x\in\mathbb{R}\}$$ and $$R:\{f(x)\in\mathbb{R}\}$$ The reciprocal function $$f(x) = \frac{1}{x}$$: $$D:\{x\in\mathbb{R}\|x \neq 0\}$$ and $$R:\{f(x)\in\mathbb{R}\|x \neq 0\}$$ Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License. chat loading...
Rational Zero Theorem Synthetic & Long Division Using Technology to Approximate Zeros # Rational Zero Theorem Synthetic & Long Division Using Technology to Approximate Zeros Télécharger la présentation ## Rational Zero Theorem Synthetic & Long Division Using Technology to Approximate Zeros - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Rational Zero TheoremSynthetic & Long DivisionUsing Technology to Approximate Zeros Today you will look at finding zeros of higher degree polynomials using the rational zero theorem, synthetic division, and technology. We will also review polynomial long division. 2. Try Factor by Grouping! 3. Factor by grouping worked fine for our first example, but not for the second example so we need another way to find the roots for our second equation. We could just keep trying numbers until we find something that works OR We can use the Rational Zero Theorem to accomplish this more efficiently. 4. Getting to the Root of the (Polynomial) Matter You are often asked to find all the zeros (roots or x-intercepts) of polynomials. To do this in the most efficient way, use the rational zero test. • First, there are some general concepts. • When you FOIL a pair of quadratic binomials with leading coefficients of one, notice that for y = (x – r1)(x – r2) , the constant in the trinomial function is the product (r1 r2). • When we to do the multiplication for y = (x – r1)(x – r2)(x – r3), the constant in the polynomial function is the product (r1 r2 r3). • That means that all rational zeros must be in thefactor listfor the constant in a polynomial function when the leading coefficient is one. 5. Getting to the Root of the (Polynomial) Matter 6. Rational Zero Theorem If f(x) = anxn + . . . + a1x + a0 has integer coefficients, then every rational zero of f has the form: p factor of constant term a0 (c) --- = ----------------------------------------- q factor of leading coefficient an (a) 7. Find all possible rational zeros of the functions below using the Rational Zero Thm. 8. Let’s look at our example again: Let’s apply the Rational Zero Theorem to find all possible rational zeros and then we can use Synthetic Division to test those zeros. 9. Testing the possible rational zeros using Synthetic Division. Find the zeros of • Find all possible rational zeros using Rational Zero Theorem • Use synthetic division to test rational zeros • Factor or use quadratic formula (if you are left with a quadratic function) to find remaining zeros or use synthetic division again until you are left with a quadratic function. 10. More Examples to Try 11. Use Polynomial Long Division to Simplify 12. That means we must have Irrational Zeros. Because they are irrational, we need to use our graphing calculators to approximate the zeros. First, always graph the function. The number of times the function crosses the x-axis gives us the number of real zeros. What happens if we use synthetic division and none of the possible rational zeros work? 13. Zeros of Polynomial Functions Using a Graphing Calculator Approximate the real zeros of 1. Enter equation into Y1 2. Graph the equation 3. Go to CALC: zero 4. Move cursor to left of the zero – hit enter 5. Move cursor to right of zero – hit enter 6. Hit Enter one more time to see the zero. 7. The coordinates of the zero appear. Repeat the process to find all other real zeros. 14. Zeros of Polynomial Functions Using a Graphing Calculator Try this one: Approximate the real zeros of 15. Assignment A 1.9 Sect II & III A 1.10 (Book Reference: Section 2.4 pgs. 214 – 223) See yu tmrrw!
# Finding maximum and minimum When Finding maximum and minimum, there are often multiple ways to approach it. ## Math: How to Find the Minimum and Maximum of a Function There are several methods we can use to find the minimum or maximum values: graphing, calculus, or utilizing the standard form equation. If you are given the graph of a Decide mathematic question What is the value of x in the equation below? Expert instructors will give you an answer in real-time If you're looking for an answer to your question, our expert instructors are here to help in real-time. If you're struggling with math, there are some things you can do to clear up the confusion. ## Introduction to minimum and maximum points For these values, the function f gets maximum and minimum values. f(c) > f(x) > f(d) What is the local minimum of the function as below: f(x) = 2. As the derivative of the function is 0, the local Top Professionals There are few greater professions than being a top professional. Track Progress It's important to track your progress in life so that you can see how far you've come and how far you still have to go. Clarify math Math can be a tricky subject for many people, but with a little bit of practice, it can be easy to understand. Timely deadlines are essential for ensuring that projects are completed on time. • Clear up math questions Math can be a difficult subject for many people, but there are ways to make it easier. ## How to Find Maximum and Minimum Value of a Function Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. These extreme values are obtained, either on a relative ## 3 Ways to Find the Maximum or Minimum Value of a Quadratic The following steps would be useful to find the maximum and minimum value of a function using first and second derivatives. Step 1 : Let f (x) be a function. Find the first derivative of f (x) ## Finding Maxima and Minima using Derivatives Ratio-Test For Maximum Or Minimum Value Use the ratio test to find the maximum value of “f” (“x”) between “x” = 0 and “x” = 1. Draw a horizontal line from 0 to 1, with a slope of 1/2. If “f” (“x”) `
# Whole number – creating 20 Students develop confidence with number sequences to 100 by ones from any starting point. Practical Resource required ## Number and algebra – whole number • describes mathematical situations and methods using everyday and some mathematical language, actions, materials, diagrams and symbols MA1-1WM • uses objects, diagrams and technology to explore mathematical problems MA1-2WM • supports conclusions by explaining or demonstrating how answers were obtained MA1-3WM • applies place value, informally, to count, order, read and represent two- and three-digit numbers MA1-4NA ## Content Develop confidence with number sequences to 100 by ones from any starting point (ACMNA012) ## National Numeracy Learning Progression mapping to the NSW mathematics syllabus When working towards the outcome MA1-4NA the sub-elements (and levels) of Quantifying numbers (QuN6-QuN8) and Understanding money (UnM3-UnM4) describe observable behaviours that can aid teachers in making evidence based decisions about student development and future learning. ## Materials • 1-20 numeral cards ## Teacher instructions The purpose of this task is to gauge students’ understanding of whole number concepts such as: • identify the number before a given two-digit number • identify the number after a given two-digit number • describe a number before as 'one less than' and the number after as 'one more than' a given number Students are presented with shuffled numbers cards with the numbers 1 to 20. Instruct students to place the cards face down in two rows of 10. Students turn over the first card in the top right hand corner and read it aloud. They place the card, face up, where it should go if the cards were in order. They explain the reason for their placement. Students then pick up the card replaced and continue to arrange the numbers in order. For example, if the first card they turn up is the number 5, they say ‘5’, place it where the 5 should go and explain the reason why they placed it in that position. The student takes the card replaced and repeats the process. ## Student instructions Place the cards face down in two rows of 10. Turn over the first card in the top right hand corner. Read the number displayed on the card. Place this card, face up, where it should go if the numbers were in order. Explain why you placed the card in that position. Pick up the card that it replaces and find out where it should be placed. Continue till all the cards are flipped over. If a card is in the right place when turned over, explain how you know. Then select another face down card and continue the task. ## Possible areas for further exploration Is the student reliant on a particular strategy to order the numbers? What reason/s does the student provide? ## Where to next? The task could be repeated using: • a focus on the backward sequence • a different range of numbers
Search 74,212 tutors 0 0 Work with Functions What is the rule for this function table X:0,1,2,3,4 Y:7,8,9,10,11 And after you find the rule tell me whats the Value of y when x = 8. Then i need to know what would be the value of x when y = 35. . One way to tackle this problem is to graph the points to determine the relationship between y and x. If we write the coordinate pairs we have (0, 7), (1, 8), (2, 9), (3, 10), (4, 11). When you plot those you will notice a linear relationship. To get the equation of the line, which will be the function rule, find the y-intercept and the slope. Remember that the y-intercept is where the line crosses the y-axis, so x = 0. One of our points is (0, 7). x = 0 and y = 7. We will use 7 for the y-intercept in the equation. To find the slope we need the coordinates of any two points on the line. Lets use (0, 7) for point 1 and (3, 10) for point 2. We could use any two points and, hopefully, get the same calculated slope. The formula for slope is (y2-y1)/(x2-x1). Point 1 is (0, 7) so x1 = 0 and y1 = 7. Point 2 is (3, 10) so x2 = 3 and y2 = 10. Now substitute into the slope formula --> (10 - 7) / (3 - 0) = 3/3 = 1. So slope = 1 Now write the equation of the line in y = mx + b form. m is the slope and b is the y-intercept. So y = 1x + 7. Now the 1 can be "invisible" so we normally write this y = x + 7. This is the rule, the equation of the line. To find the values for y when x = 8, just substitute for x in the equation above: y = 8 + 7 = 15 when x = 8, y = 15 When y = 35, substitute 35 for y and then solve for x 35 = x + 7 Subtract 7 from both sides --> 28 = x when y = 35, x = 28 If I understand the question correctly - you're looking for a formula to express a graph going through the listed points. Okay... if you plot this on a graph it looks to have a slope of 1/1 (that is, for every one it moves over via 'x' , it moves up one. When x = 0, y = 7, so we know we have to add 7 to the equation. I'm seeing y = x + 7. When x = 8, y would be equal to 15. If you flip the equation (x = y -7), you'll find when y = 35, x will equal 28.
# Learning to Think Mathematically with the Rekenrek - Jeff Frykholm, Ph ## Presentation on theme: "Learning to Think Mathematically with the Rekenrek - Jeff Frykholm, Ph"— Presentation transcript: Learning to Think Mathematically with the Rekenrek - Jeff Frykholm, Ph Learning to Think Mathematically with the Rekenrek - Jeff Frykholm, Ph.d. The Rekenrek ABC MATH What is A more than E? What is B less than AO? There are B ducks on a pond. D more ducks waddle into the pond. How many ducks are there now? Do you have “D-ness”? Do you have “letter sense”? Meet the Rekenrek From start position…..what do you see? Allow students to come up with a discovery about the Rekenrek Cardinality, Subitizing, Decomposing and a understanding of 5 and 10 Rekenrek is a DYNAMIC ten frame The rule is to slide in groups rather than one by one Show Me….1-5 and Make 5 The importance of helping young children anchor on 5 and 10 cannot be stated strongly enough….this helps with mental math strategies for add/sub Show me 1, then 2, then 3, then 4, then 5 Move 2 on top…how many more do we need to move over to make 5? Show me 3…..now make 5 Do this with all combinations of 5 Then try it using two rows Show me 2 on the top row, how many on the bottom row do you need to make 5? Why would students need to move to using the bottom row? Show Me 5-10 This will help students understanding of cardinality. Ex. 5+1….they have to count all to get the answer…they don’t understand to start with 5 and add one more Slide 5 in one move. How did you know to do that? Then continue with this pattern 10, 9, 5, 7, 8, 6, 9, 10, 7 Now use two pushes and the bottom row to show me 8 Make 10, Two Rows Move 5 to the left….then make 10 Students should now know that 5 red beads and 5 white beads make 10 Show me 7….Now what do I need to make ten? You could do the last three on the top….OR…3 on the bottom I will show you a move…show 8 and cover the remaining ones. How many more do I need to make 10? Flash Attack Game This game will wean students from wanting to count every bead Show 6 for 2 seconds, then cover….How many did I show and how did you know? Show 19 for two seconds Then play Flash Attack with top and bottom row Doubles Slide one on top and one on bottom…. Then slide two on top and two on bottom…How many do you see? Demonstrate all the way up to 5. Let’s think about 7 and 7….They may see two sets of doubles. First, they will recognize that two groups of 5 red beads is 10. Next a pair of 2’s is 4. Hence, 7+7= (5+5)+(2+2)= 10+4=14 Almost a Double (Doubles +1) Show three on top and two on bottom…where is the double? 2+2…now what can you tell me about doubles +1? Pose these questions for students to figure out on the rekenrek Does 6+7=12+1? Does 3+2 = 4+1? Part Part Whole Move 6 beads to the left and cover the others…How many are on the top row? What could students say? “5 and 1 more is 6. I counted up to 10….7,8,9,10. 4 are covered” “I know that 6+4 =10. I see 6 so 4 more are covered. “I know that there are 5 red and 5 white beads on a row. I only see one white, so there must be 4 more”. Part Part Whole Continued Move 9 over (6 on top and 3 on bottom) and cover the rest. How many are missing? What could the students say? “5 whites and 2 reds are missing on the bottom. 4 whites are missing on top. 7 and 4 is 11.” “If I slid the 3 red beads under the cover, and slid 3 white beads to the left, then there would be 5 red and 5 white hidden on the bottom, and 1 white hidden on the top. So that is 11 hidden. “Well, 10 and 10 are 20. Right now I can see 9 which is close to 10. Since it is 1 less than 10, then 11 must be covered up.” Rekenrek Recording Paper It’s Automatic…Math Facts Give them 4+5….Show me 4 and then add 5 more beads anyway you want. Without having to count each bead, how do you know that there are 9 beads? You can also use flashcards that show the beads on the rekenrek Subtraction – Take Away Model I have 14 balls. You take away 6. How many are left? How could you show that on a rekenrek? Show 14 and cover the rest…(10 on top and 4 on bottom)…Then take away 6 (4 on bottom and 2 more on top). That leaves 8! Subtraction – Comparison Model I have 8 dollars and you have 6. How much more do I have? How can we show this on the rekenrek? 8 on top and 6 on bottom.. Compare the two rows. Find the beads that are in common on both rows. How many beads are left over? How many more beads on the top row? You can use your pencil to separate them… Game – Roll Three for 20 or Bust! Use 3 dice One red (0,1,2,3,4,5) One blue (2,3,4,5,6,7) One green (4,5,6,7,8,9) Roll the dice to determine the first three numbers to add…”I rolled a 6,7 and 4” Use rekenrek to show how I added these together (Move 6 and then 4 over. That makes 10. Then on the bottom move 5 red and 2 white which is plus 7 equals 17.) Roll 3 for 20 or Bust! Con.. Students must then decide to stay put or roll 1,2,3 dice to add more to the sum they have so far to come close or equal to 20 without going over. “I am going to try again…but I will use the red dice because I am only 3 away from 20” I rolled a 4! Oh that is too much. I went Bust! Play 4 rounds…the person with the highest sum wins…. TENS OR ONES GAME Choose a card from the number cards. “I chose 19” Show that on the rekenrek Then draw a card from the take away tens or run away ones cards. “I chose a take away ten card” Move ten beads back to the right. That leaves is 9! I drew a 16 and a take away ones. I take the 6 ones away and it leaves =10 Brainstorm…what is early number sense? What is Early Number Sense? Counting Cardinality Comparing Relationships Among Numbers (Spatial, 1 more/1 less 2 more 2 less, Anchors of 5 and 10, Part Part Whole) Pre Place Value (tens and ones) Place Value Download ppt "Learning to Think Mathematically with the Rekenrek - Jeff Frykholm, Ph" Similar presentations
St Paul's C of E Primary School Heathside Grove LEARNING to make a difference # Week 11 Year 4 – Maths Home Learning Week 11 Maths Activity 1 Recap: Write these numbers as words: 6,564 2,809 1,499 8,412 1. For the past two weeks we have been using place value grids to add two 4-digit numbers. This week we are going to look at subtracting them. When we subtract numbers using a place value grid and counters, we only make the first number using the counters. Once you have made the first number we subtract the other number by taking them away from each column, starting with the ones. Draw your own place value grids and work these questions out by using/drawing the counters. 5,678 – 2,526 = 8,098 – 6,024 = 3,457 – 2,344 = 9,258 – 4,257 = 1. Now we are going to have a go at some more subtraction questions but without the numbered counters. 7,582 – 2,461 = 6,459 – 2,335 = Maths Activity 2 Recap 1. Answer these questions using the place value grid and the written subtraction method together. 2,348 – 235 = 6,582 – 582 = 4,572 – 2,341 = 7,262 – 7,151 = 2. Maths Activity 3: Recap: Divide these numbers by 10 and 100: 10 100 3,480 5,230 1,700 7,080 1. We are going to have a go at using a bar model to represent our subtractions. We can still work it out using our place value grid but the question will look slightly different. e.g. 2. Can you answer these subtraction questions and represent them as a bar model? There are 3,597 boys and girls in a school. 2,182 are boys. How many are girls? Car A travels 7,653 miles per year. Car B travels 5,612 miles per year. How much further does Car A travel than Car B per year Top
# Unit 4: Ratios, Rates, Proportions, and Percents (January 6-Feb 21)- Math 7th Grade Teacher Carol Brackin Math 7th Grade 19-25 Ratios, Rates, Proportions, and Percents Standard(s) Taught MAFS.7.RP.1.1 Compute unit rates associated with ratios of fractions, including ratios of lengths, areas, and other quantities measured in like or different units. For example, if a person walks 1⁄2 mile in each 1⁄4 hour, compute the unit rate as the complex fraction 1⁄2 miles per 1⁄4 hour, equivalently 2 miles per hour. MAFS.7.RP.1.2 Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table, or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r ) where r is the unit rate. MAFS.7.RP.1.3 Use proportional relationships to solve multi-step ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error. MAFS.7.EE.1.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that “increase by 5%” is the same as “multiply by 1.05.” Learning Targets and Learning Criteria • compute a unit rate by iterating (repeating) or partitioning given rate. • compute a unit rate by multiplying or dividing both quantities by the same factor. • explain the relationship between using composed units and a multiplicative comparison to express a unit rate. • use measures of lengths and areas to calculate unit rates with the given context. a. • analyze ratios in a table to determine if the ratios are equivalent by finding the constant of proportionality. • calculate the cross product to determine if the two ratios are in proportion (equivalent). • graph ratios on a coordinate plane to determine if the ratios are proportional by observing if the graph is a straight line through the origin (y = kx, where k is the slope/constant of proportionality). b. • calculate the constant of proportionality/unit rate from a table or diagram. • calculate the constant of proportionality/unit rate given a verbal description of a proportional relationship. • compute the rate of change/slope from a graph or equation (k in y=kx). c. • solve equations created from proportional relationships • write an equation from a proportional relationship d. • calculate the unit rate by identifying that on a graph when the x-coordinate is 1, the y- coordinate is the unit rate. • define the rate of proportionality from a graph. • explain that the y-coordinate divided by the x-coordinate for every point other than the origin equals the constant of proportionality • explain the meaning of a point on a graph y=kx of a real life situation • use proportional reasoning to solve real- world ratio problems, including those with multiple steps. • use proportional reasoning to solve real- world percent problems, including those with multiple steps, such as markups, markdowns, commissions, and fees • solve percent problems when one quantity is a certain percent more or less than another. • recall conversion of percents from ratio to decimal to percent form. • solve percent of increase/decrease problems. • explain how an equivalent expression relates to the original situation problem • rewrite expressions to help analyze problems. • simplify expressions • translate situation problems to algebraic expressions Classroom Activities Notes Page & Week of Study Topic of Study and Independent Practice in Class Homework 1 – 19 What is a Ratio HW #1 2 – 19 Rates & Unit Rates HW #2 3 – 20 Proportional vs. Nonproportional Relationships HW #3 4 – 20 Solving Proportions HW #4 – 21 Quiz 4-1 None 5 – 21 Proportion Word Problems HW #5 6 – 21 Scale Drawings and Models HW #6 7 – 21 Similar Figures HW #7 – skip Indirect Measurement HW #8 – 22 Quiz 4-2 None 8 – 22 Percent Proportion HW #9 9 – 22 Percent Equation HW #10 10 -23 Computing Percents Mentally (5%, 10%, 15%, 20%) using 10% as a benchmark HW #11 -23 -23 Graphing Proportions Quiz 4-3 None 11- 24 Discount, Mark-Up, Sales Tax, Tip HW #12 12- 24 Percent of Change HW #13 13- 25 Simple Interest HW #14 – 25 Quiz 4-4 None -25 Unit 4 Review – graphing proportions Study for DIA Assignments Due Each study topic above has close notes (fill in the blank) with 6 to 8 guided practice problems done together as a class and corrected in class. Then students do 10 to 12 independent practice problems which are also checked in class. These problems can be used as models to help with the homework problems. Each page of notes has a homework page which is due the next day class meets. On Monday and Tuesday, 2/10, students will be learning to calculate common percentages (5%, 10%, 15%, 20%) using mental math.
# A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. - Mathematics A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes. #### Solution $\text{ We have, }$ $p = \text{ probability of getting an even number on a toss } = \frac{3}{6} = \frac{1}{2} \text{ and }$ $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$ $\text{ Let X denote a success of getting an even number on a toss . Then } ,$ $\text{ X follows binomial distribution with parameters n = 2 and } p = \frac{1}{2}$ $\therefore Var\left( X \right) = \text{ npq } = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}$ Concept: Random Variables and Its Probability Distributions Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 12 Maths Chapter 33 Binomial Distribution Exercise 33.2 \| Q 24 \| Page 25
Find the equation to the straight line (i) cutting off intercepts 3 and 2 from the axes. (ii) cutting off intercepts -5 and 6 from the axes. Asked by Aaryan \| 1 year ago \| 23 ##### Solution :- (i) Cutting off intercepts 3 and 2 from the axes. Given: a = 3, b = 2 Let us find the equation of line cutoff intercepts from the axes. By using the formula, The equation of the line is $$\dfrac{ x}{a} + \dfrac{y}{b} = 1$$ $$\dfrac{ x}{3} + \dfrac{y}{2} = 1$$ By taking LCM, 2x + 3y = 6 ∴ The equation of line cut off intercepts 3 and 2 from the axes is 2x + 3y = 6 (ii) Cutting off intercepts -5 and 6 from the axes. Given: a = -5, b = 6 Let us find the equation of line cutoff intercepts from the axes. By using the formula, The equation of the line is $$\dfrac{ x}{a} + \dfrac{y}{b} = 1$$ $$\dfrac{ x}{-5} + \dfrac{y}{6} = 1$$ By taking LCM, 6x – 5y = -30 The equation of line cut off intercepts 3 and 2 from the axes is 6x – 5y = -30 Answered by Aaryan \| 1 year ago ### Related Questions #### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1. Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1. #### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals. #### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0. Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
# Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 2 Integers Ex 2.1 ## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 2 Integers Ex 2.1 Question 1. Fill in the blanks. (i) The potable water available at 100m below the ground level is denoted as ______ m. Hint: Below ground level – negative; ground level – 0; above ground level – positive. (ii) A swimmer dives to a depth of 7 feet from the ground into the swimming pool. The integer that represents this, is ______ feet. Hint: Below ground level – negative numbers. (iii) -46 is to the _____ of -35 on the number line. Hint: -46 <-35 (iv) There are _____ integers from -5 to +5 (both inclusive). (v) ______ is an integer which is neither positive nor negative. Solutions: (i) -100 (ii) -7 (iii) Left (iv) 11 (v) 0 Question 2. Say True or False. (i) Each of the integers -18, 6, -12, 0 is greater than -20. (ii) -1 is to the right of 0. (iii) -10 and 10 are at equal distance from 1. (iv) All negative integers are greater than zero. (v) All whole numbers are integers. Solution: (i) True (ii) False (iii) False (iv) False (v) True Question 3. Mark the numbers 4, -3, 6, -1 and -5 on the number line. Solution: Question 4. On the number line, which number is i) 4 units to the right of-7? ii) 5 units to the left of 3? Solution: (i) – 3 is 4 units to the right of -7 (ii) – 2 is 5 units to the left of 3 Question 5. Find the opposite of the following numbers. (i) 44 (ii) -19 (iii) 0 (iv) -312 (v) 789 Solution: (i) Opposite of 44 is – 44 (ii) Opposite of-19 is + 19 or 19 (iii) Opposite of 0 is 0 (iv) Opposite of-312 is + 312 or 312 (v) Opposite of 789 is – 789. Question 6. If 15 km east of a place is denoted as + 15 km, what is the integer that represents 15 km west of it? Solution: Opposite of east is west. ∴ If 15 km east is + 15 km, then 15 km west is – 15 km. Question 7. From the following number lines, identify the correct and the wrong representations with reason? Solution: (i) This representation is wrong. Because the numbers are not continuously marked as -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, (ii) This is the correct representation of Integers (iii) The representation is wrong. – 2 is misplaced in the number line. (iv) This representation is correct as the numbers are marked at equal intervals in the correct order. (v) Here negative integers are marked wrongly to the left of 0 Question 8. Write all the integers between the given numbers. (a) 7 and 10 (b) -5 and 4 (c) -3 and 3 (d) -5 and 0 Solution: Question 9. Put the appropriate signs as <, > or = in the box? Solution: (i) We know that -7 is to the left of 8. (ii) -8 is to the left of -7, (iii) -1000 is to the left of -999, (iv) -111 and-111 coincides in the number line. (v) 0 is to the right of -200, Question 10. Arrange the following integers in ascending order. i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20 ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22 iii) -100, 10, -1000, 100,0, -1, 1000, 1, -10 Solution: (i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20 • First separating the positive integers 12, 14, 16, 18 and the negative integers -11,-13,-15,-17,-19,-20. • Then arranging the positive integers in ascending order we get 12, 14, 16, 18 and negative integers in ascending order as -20, -19, -17, -15, -13, -11 4 • Now the ascending order : -20, -19, -17, -15, -13, -11, 12, 14, 16, 18. (ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22 • Positive integers are 6, 8, 12, 4, 22 Negative integers are -28, -5, -40, -1 • Arranging the positive integers in ascending order we get 4, 6, 8, 12, 22 and the negative integers in ascending order -40, -28, -5, -1 • The number 0 is neither positive nor negative and stand in the middle. • In ascending order : -40, -28, -5, -1, 0, 4, 6, 8,12, 22 (iii) -100, 10, -1000, 100, 0, -1, 1000, 1, -10 • Separating positive integers 10, 100, 1000, 1 and negative integers -100, -1000, -1, -10. • Now the positive integers in ascending order 1,10,100,1000 and the negative integers in ascending order. -1000, -100, -10, -1 • Also ‘0’ stand in the middle as its is neither positive nor negative. • ∴ The numbers in ascending order: -1000, -100, -10, -1, 0, 1, 10, 100, 1000. Question 11. Arrange the following integers in descending order. i) 14, 27, 15, -14, -9, 0, 11, -17 ii) -99, -120, 65, -46, 78, 400, -600 iii) 111, -222, 333, -444, 555, -666, 7777, -888 Solution: (i) 14, 27, 15, -14, -9, 0,11, -17 • Separating the positive integers 14, 27, 15, 11 and negative integers -14, -9, -17 • Arranging in descending order we get the positive integers 27,15,14,11 and the negative integers -9, -14, -17. • ‘0’ is neither positive nor negative and so it stand in middle. • ∴ The numbers in descending order : 27, 15, 14, 11, 0, -9, -14, -17 (ii) -99, -120, 65, -46, 78, 400, -600 • Separating the positive integers 65, 78, 400 and negative integers -99, -120, -46, -600 • Arranging the positive integers in descending order as 400, 78, 65 and the negative integers in descending order -46, -99, -120, -600. • The numbers in descending order : 400, 78, 65, -46, -99, -120, -600. (iii) 111, -222, 333, -444, 555, -666, 7777, -888 • Separating the positive integers 111, 333, 555, 7777 and negative integers -222, -444, -666, -888 • Arranging the positive integers in descending order as 7777, 555, 333, 111 and negative integers in descending order as -222, -444, -666, -888 • The numbers in descending order : 7777, 555, 333, 111, -222, -444, -666, -888 Objective Type Questions Question 12. There are _______ positive integers from – 5 to 6. (a) 5 (b) 6 (c) 7 (d) 11 Solution: (b) 6 Hint: Question 13. The opposite of 20 units to the left of 0 is (a) 20 (b) 0 (c) -20 (d) 40 Solution: (a) 20 Hint: Question 14. One unit to the right of -7 is (a) +1 (b) -8 (c) -7 (d) -6 Solution: (d) -6 Hint: Question 15. 3 units to the left of 1 is (a) -4 (b) -3 (c) -2 (d) 3 Solution: (c) -2 Hint: Question 16. The number which determines marking the position of any number to its opposite on a number line is? (a) -1 (b) 0 (c) 1 (d) 10 Solution: (b) 0 Hint:
# BODMAS/PEMDAS Rules - Involving Decimals We will obey the rules for simplifying an expression using BODMAS/PEMDAS rules - involving decimals for solving order of operations. Follow the order of operation as: 1. Bracket Solve inside the Brackets/parenthesis before Of, Multiply, Divide, Add or Subtract. For example: 1.6 × (0.5 + 0.3) = 1.6 × 0.8 = 1.28 2. Of Then, solve Of part (Exponent, Powers, Roots, etc.,) before Multiply, Divide, Add or Subtract. For example: 3.2 + 0.5 of 1.6 - 0.1 = 3.2 + 0.5 × 1.6 - 0.1 = 3.2 + 0.8 - 0.1 = 4 - 0.1 = 3.9 3. Division/Multiplication Then, calculate Multiply or Divide before Add or Subtract start from left to right. For example: 1.2 + 1.5 ÷ 0.3 × 0.3 = 1.2 + 5 × 0.3 = 1.2 + 1.5 = 2.7 4. Addition/Subtraction At last Add or Subtract start from left to right. For example: 1.7 + (4.2 - 0.5) × 0.1 = 1.7 + 3.7 × 0.1 = 1.7 + 0.37 = 2.07 Worked-out problems for solving BODMAS/PEMDAS rules - involving decimals: Simplify using BODMAS/PEMDAS rule: (a) 8 - 4.2 ÷ 6 + 0.3 × 0.4 Solution: 8 - 4.2 ÷ 6 + 0.3 × 0.4 = 8 - 0.7 + 0.3 × 0.4, (Simplifying ‘division’ 4.2 ÷ 6 = 0.7) = 8 - 0.7 + 0.12, (Simplifying ‘multiplication’ 0.3 × 0.4 = 0.12) = 7.3 + 0.12, (Simplifying ‘subtraction’ 8 - 0.7 = 7.3) = 7.42, (Simplifying ‘addition’ 7.3 + 0.12 = 7.42) (b) 7.6 - [3 + 0.5 of (3.1 - 2.3 × 1.02)] Solution: 7.6 - [3 + 0.5 of (3.1 - 2.3 × 1.02)] = 7.6 - [3 + 0.5 of (3.1 - 2.346)], (Simplifying 2.3 × 1.02 = 2.346) = 7.6 - [3 + 0.5 of 0.754], (Simplifying ‘subtraction’ 3.1 - 2.346 = 0.754) = 7.6 - [3 + 0.5 × 0.754], (Simplifying ‘of’) = 7.6 - [3 + 0.377], (Simplifying ‘multiplication’ 0.5 × 0.754 = 0.377) = 7.6 - 3.377, (Simplifying ‘addition’ 3 + 0.377 = 3.377) = 4.223, (Simplifying ‘subtraction’ 7.6 - 3.377 = 4.223) (c) 12.8 - 0.4 of (7.2 - 3.7) + 2.4 × 3.02 Solution: 12.8 - 0.4 of (7.2 - 3.7) + 2.4 × 3.02 = 12.8 - 0.4 of 3.5 + 2.4 × 3.02, (Simplifying 7.2 - 3.7 = 3.5) = 12.8 - 0.4 × 3.5 + 2.4 × 3.02, (Simplifying ‘of’) = 12.8 - 1.4 + 2.4 × 3.02, (Simplifying ‘multiplication’ 0.4 × 3.5 = 1.4) = 12.8 - 1.4 + 7.248, (Simplifying ‘multiplication’ 2.4 × 3.02 = 7.248) = 11.4 + 7.248, (Simplifying ‘subtraction’ 12.8 - 1.4 = 11.4) = 18.648, (Simplifying ‘addition’ 11.4 + 7.248 = 18.648) Related Concept
Top # Subtract Fractions with Different Denominators Fractions are denoted by $\frac{a}{b}$ where a and b are whole numbers and b not equals to 0. For comparing and ordering of unlike fractions, first convert them into like fractions and then compare or rearrange them in the order as desired. When the numerator and denominator of fraction which are multiplied or divided by same number, we get its own equivalent fractions. Related Calculators Subtract Fractions with Unlike Denominators Calculator Adding Fractions with Unlike Denominators Calculator Unlike Fraction Calculator Least Common Denominator Calculator for Fractions ## Like and Unlike Fractions Fractions can be classified on the basis of the denominators. Two fractions may have same or different denominators. • Like fractions • Unlike fractions Like Fractions: The number a is called Fractions having the same denominators are called like fractions Example: $\frac{4}{11}$ , $\frac{6}{11}$ are all like fractions. Unlike Fractions: The fractions which contains different denominator numbers are called unlike fractions. Example: $\frac{3}{5}$ , $\frac{4}{7}$ are all unlike fractions. Let us see how to subtract fractions with unlike denominators in this article. ## How to Subtract Fractions with Unlike Denominators Below are the steps for how to subtract fractions with different denominators - Step 1. Find out Lowest Common Denominator(LCD) of the fractions. Step 2. Rename fractions to have LCD. Step 3. Now subtract the numerators of the given fractions. Step 4. The difference will be the numerator and the LCD will be the denominator of the answer. Step 5. Simplify the Fraction if needed. ## Examples Below are the solved examples on subtracting fractions with different denominators - ### Solved Examples Question 1: Subtract $\frac{5}{12}$ from $\frac{13}{24}$ Solution: Here we have unlike fractions, so we have to take LCM for 12 and 24 is 24 $\frac{5}{12}$ = $\frac{5\times2}{12\times2}$ = $\frac{10}{24}$ $\frac{13}{24}$ - $\frac{5}{12}$ = $\frac{13}{24}$ - $\frac{10}{24}$ = $\frac{3}{24}$ Correct answer is $\frac{1}{8}$ Question 2: $\frac{7}{10}$ - $\frac{8}{15}$ Solution: Here we have unlike fractions, so we have to take LCM for 10 and 15 = 30. $\frac{7}{10}$ - $\frac{8}{15}$ = $\frac{7\times3}{10\times3}$ - $\frac{8\times2}{15\times2}$ = $\frac{21}{30}$ - $\frac{16}{30}$ = $\frac{21-16}{30}$ = $\frac{5}{30}$ Correct answer is $\frac{1}{6}$ Question 3: Subtract 3$\frac{2}{5}$ - 2$\frac{3}{8}$ Solution: Given fraction is unlike mixed fractions take LCM, LCM for 5 and 8 is 40 3$\frac{2}{5}$ - 2$\frac{3}{8}$ = $\frac{17}{5}$ - $\frac{19}{8}$ = $\frac{17\times8}{5\times8}$ - $\frac{19\times5}{8\times5}$ = $\frac{136}{40}$ - $\frac{95}{40}$ = $\frac{41}{40}$ Correct answer is 1$\frac{1}{40}$ ## Practice Problems Problem 1: Subtract $\frac{2}{3}$ from $\frac{6}{11}$. Find the difference between $3\frac{1}{2}$ and $6\frac{1}{3}$.
## Kiss those Math Headaches GOODBYE! ### From GPGCF to GCF … in two easy steps Once you know the GPGCF, you’re two easy steps from finding the GCF. [If you don’t know how, see my last post: “Recent Insight on the GCF (and GPGCF)”] That’s one of the benefits of finding the GPGCF — speed in getting the GCF! Here are the short, sweet steps: 1) Find all factors of the GPGCF, and list them from largest to smallest. 2) Starting with the largest factor and working your way down the list, test to find the first factor that goes into both numbers. The first (largest) to do so is the GCF. You can bet on it! Example 1 (Easy): Find GCF for 30 and 42. 1st) GPGCF is 12. Factors of 12, greatest to least, are 12, 6, 4, 3 and 2. 2nd) Largest factor to go evenly into 30 and 42 is 6. So 6 is GCF. Example 2 (Harder): Find GCF for 72 and 120. 1st) GPGCF is 48. Factors of 48, greatest to least, are 48, 24, 16, 12, 8, 6, 4, 3 and 2. 2nd) Largest factor to go evenly into 72 and 120 is 24. So 24 is GCF. NOW TRY THESE — For each pair: 1) Find GPGCF and say if it is the difference or smaller #. 2) List factors of GPGCF, greatest to least. 3) Find GCF. a) 8 and 12 b) 16 and 40 c) 18 and 63 d) 56 and 140 a) 8 and 12 GPGCF = 4 (difference) Factors of 4: 4 and 2 GCF = 4 b) 16 and 40 GPGCF = 16 (smaller #) Factors of 16: 16, 8, 4 and 2 GCF = 8 c) 18 and 66 GPGCF = 18 (smaller #) Factors of 18: 18, 9, 6, 3 and 2 GCF = 6 d) 56 and 76 GPGCF = 20 (difference) Factors of 20: 20, 10, 5, 4 and 2 GCF = 4 ### Recent insight on the GCF (and GPGCF) A while back I wrote a post about the GCF, and mentioned that there’s a number related to it — a number that I call the GPGCF. “GPGCF” stands for the “Greatest Possible Greatest Common Factor.” In short, the GPGCF is a number that sets an upper limit for the size of the GCF. I’ve seen many students struggle when searching for the GCF, seeking hither and yon for it. I had a sense that students were checking numbers that were too large. That’s what led me to try to figure out what must the the upper limit for the GCF. If you check out that post (10/25/10), you’ll see that, for any two numbers, I said that the difference between those numbers has to be the GPGCF. And I was correct, to a degree. But I recently realized that my little theory needs modifying. For while the difference between any two numbers can be the upper limit for the GCF, that difference is not the only quantity that can set an upper limit for the GCF. There’s another quantity that plays a role. That other quantity, I recently realized, is the size of the smaller of the two numbers. Take the numbers 8 and 24, for example. The difference between these two numbers is 16, so I would have said that 16 is the upper limit for the GCF. But there’s actually another quantity that limits the size of the GCF, and that quantity is 8. For since the GCF of 8 and 24 must by definition fit into both 8 and 24, it must fit into 8. And common sense tells us that there’s no number larger than 8 that can fit into 8! So the size of this number — the smaller of the two numbers — also sets an upper limit for the size of the GCF. So my revised theory about the GPGCF is this: when you need to find the GCF for any two numbers, look at two quantities: 1) the smaller of the two numbers, and 2) the difference between the two numbers. Both of these quantities constrains the size of the GPGCF. So therefore, whichever of these is smaller IS the GPGCF. Once you’ve found the GPGCF, that makes it easier to find the actual GCF. I know this sounds very abstract, so let’s look at a few examples to see what I’m blabbering on about. Example 1: What’s the GPGCF for 6 and 16? Smaller number is 6; difference is 10. 6 and 10 both limit the size of the GCF, but 6 is less than 10, so 6 is the GPGCF. Example 2: What’s the GPGCF for 8 and 12? Smaller number is 8; difference is 4. 4 is less than 8, so 4 is the GPGCF. Example 3: What’s the GPGCF for 30 and 75? Smaller number is 30; difference is 45. 30 is less than 45, so 30 is the GPGCF. Example 4: What’s the GPGCF for 28 and 42? Smaller number is 28; difference is 14. 14 is less than 28, so 14 is the GPGCF. Now, let’s go one step further. From here, how do we figure out the GCF? I’ve done a bit more thinking about this, too, and I’ll share those ideas in my next post.
# How do you solve abs(5-2x)=13? Mar 2, 2018 $x = - 4 \text{ or } x = 9$ #### Explanation: $\text{the value inside the "color(blue)"absolute value bars "" can be}$ $\text{positive or negative }$ $\Rightarrow \text{there are 2 possible solutions}$ $5 - 2 x = 13 \leftarrow \textcolor{red}{\text{positive inside bars}}$ $\text{subtract 5 from both sides}$ $\cancel{5} \cancel{- 5} - 2 x = 13 - 5$ $\Rightarrow - 2 x = 8$ $\text{divide both sides by } - 2$ $\frac{\cancel{- 2} x}{\cancel{- 2}} = \frac{8}{- 2}$ $\Rightarrow x = - 4 \leftarrow \textcolor{b l u e}{\text{first solution}}$ $- \left(5 - 2 x\right) = 13 \leftarrow \textcolor{red}{\text{negative inside bars}}$ $\Rightarrow - 5 + 2 x = 13$ $\text{add 5 to both sides}$ $\cancel{- 5} \cancel{+ 5} + 2 x = 13 + 5$ $\Rightarrow 2 x = 18$ $\text{divide both sides by 2}$ $\frac{\cancel{2} x}{\cancel{2}} = \frac{18}{2}$ $\Rightarrow x = 9 \leftarrow \textcolor{b l u e}{\text{second solution}}$ $\textcolor{b l u e}{\text{As a check}}$ $\text{Substitute these 2 possible solutions into the left side}$ $\text{of the equation and if equal to right side then they are}$ $\text{the solutions}$ $x = - 4 \Rightarrow \| 5 + 8 \| = \| 13 \| = 13 \leftarrow \text{ True}$ $x = 9 \Rightarrow \| 5 - 18 \| = \| = 13 \| = 13 \leftarrow \text{ True}$ $\Rightarrow x = - 4 \text{ or "x=9" are the solutions}$
# What is the standard form of y=(2x + 1) (3x – 4) (2x – 1) ? Jan 27, 2016 $y = 12 {x}^{3} - 16 {x}^{2} - 3 x + 4$ #### Explanation: Visual inspection of the equation shows that it is a cubic function (there are 3 x's all with exponent 1). Hence we know that the standard form of the equation should appear this way: $y = a {x}^{3} + b {x}^{2} + c x + d$ Generally in solving these types of questions, a possible approach would be expanding the equation. Sometimes this may seem tedious especially for longer equations however with a little patience you will be able to reach the answer. Of course it would also help if you know which terms to expand first to make the process less complicated. In this case, you can choose which two terms you wish to expand first. So you can do either of the following Option 1 $y = \left(2 x + 1\right) \left(3 x - 4\right) \left(2 x - 1\right)$ $y = \left(6 {x}^{2} - 8 x + 3 x - 4\right) \left(2 x - 1\right)$ $y = \left(6 {x}^{2} - 5 x - 4\right) \left(2 x - 1\right)$ OR Option 2 $y = \left(2 x + 1\right) \left(2 x - 1\right) \left(3 x - 4\right)$ -> rearranging the terms $y = \left(4 {x}^{2} - 1\right) \left(3 x - 4\right)$ Note that in Option 2 the product of $\left(2 x + 1\right) \left(2 x - 1\right)$ follows the general pattern of $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$. In this case, the product is shorter and simpler than that of the 1st option. Therefore, although both options will lead you to the same final answer, it would be simpler and easier for you to follow the 2nd one. Continuing with the solution from Option 2 $y = \left(4 {x}^{2} - 1\right) \left(3 x - 4\right)$ $y = 12 {x}^{3} - 16 {x}^{2} - 3 x + 4$ But if you still opt to do the 1st solution indicated above... $y = \left(6 {x}^{2} - 5 x - 4\right) \left(2 x - 1\right)$ $y = 12 {x}^{3} - 6 {x}^{2} - 10 {x}^{2} + 5 x - 8 x + 4$ $y = 12 {x}^{3} - 16 {x}^{2} - 3 x + 4$ ...it would still produce the same final answer
# Complete addition or subtraction statements to make a target number Lesson ## Making a whole number When we add one number to another, we make a bigger number. Likewise, if we subtract one number from another, we make a smaller number. If we have $\$22$$22, but need \30$$30, we can work out that we need to add another $\$8$$8 to make \30$$30. ## Making a decimal number by addition With decimals, we are working with tenths and hundredths, rather than tens, hundreds etc., but otherwise the process is very similar. Our numbers can be made up of wholes and parts, or decimals. Let's look through examples of those: • Suppose we have $\$1.20$$1.20, but need \2.00$$2.00. We can make $\$2.00$$2.00 by adding \1.20$$1.20 and $\$0.80$$0.80 In our first video, we make a decimal by adding on tenths, and then move to an example where we add by hundredths. You can see the process is the same, we just need to think about the space, or difference, between the numbers we are adding. A number line can help you see how to add to a decimal number, moving to the right. So now, if you have \3.35$$3.35, and need to calculate how much you need to add to make $\$4.00$$4.00, you can use this method to work out that \3.35$$3.35$+$+$\$0.65$$0.65==\4.00$$4.00. ## Making a decimal by subtraction Imagine that a book you want is on sale for $\$2.20$$2.20, but it was \3.00$$3.00 before the sale. How much will you save if you buy it at the sale price? Well, making a number by subtracting decimals can help you work this out. The amount you would save is the same as the amount you need to subtract from $\$3.00$$3.00 to get to \2.20$$2.20. In our second video, we look at how making a decimal by subtracting can help solve problems such as this. We can use a number line to show that counting back by tenths, or increments of $0.1$0.1, until we reach $\$2.20$$2.20, means we saved \0.80$$0.80. We also look at hundredths, allowing us to calculate how much we need to subtract from $\$5.00$$5.00 to get \3.75$$3.75. If you are paying $\$3.75$$3.75 for a snack, and have \5.00$$5.00, you can work out how much change you should receive, by starting at $\$5.00$$5.00 and subtracting until you reach \3.75$$3.75. Can you work out how much change you should expect? #### Examples ##### Question 1 Which of the options below make $6.8$6.8? There may be more than one correct answer. 1. $6+0.8$6+0.8 A $2.3+4.5$2.3+4.5 B $8.5-1.7$8.51.7 C $10.1-3.9$10.13.9 D ##### Question 2 Which of the options below make $57.88$57.88? There may be more than one correct answer. 1. $28.64+29.04$28.64+29.04 A $46.52+11.36$46.52+11.36 B $72.97-15.09$72.9715.09 C $65.83-6.93$65.836.93 D ##### Question 3 Let’s play a game called target number. Here’s how it works. I give you a target number and starting number and you tell me what I need to subtract to get there. For example, our target number is $0.05$0.05, and if I tell you the number $0.07$0.07 you would say $0.02$0.02, because $0.07-0.02=0.05$0.070.02=0.05. Find the missing numbers below for our target number $0.05$0.05. 1. $0.06-\editable{}=0.05$0.06=0.05 2. $0.08-\editable{}=0.05$0.08=0.05 3. $0.09-\editable{}=0.05$0.09=0.05 ##### Question 4 Let’s play a game called target number. Here’s how it works. I give you a target number and starting number and you tell me what I need to subtract to get there. For example, our target number is $4.57$4.57, and if I tell you the number $6.99$6.99 you would say $2.42$2.42, because $6.99-2.42=4.57$6.992.42=4.57. Find the missing numbers below for our target number $4.57$4.57. 1. $6.69-\editable{}=4.57$6.69=4.57 2. $5.93-\editable{}=4.57$5.93=4.57 3. $7.35-\editable{}=4.57$7.35=4.57
How do you solve the inequality -6abs(2-x)<-12? May 11, 2018 The inequality is fulfilled when $x < 0$ or $x > 4$, i.e. $x \notin \left[0 , 4\right]$ Explanation: Let's draw the graph of this inequality: As we see on the graph x must be outside the area [0, 4] to solve this equation, i.e. $x < 0$ or $x > 4$ To solve this matematically we need to remember that $\| 2 - x \|$ means the absolut value of $\left(2 - x\right)$, i.e. $2 - x$ when $x \le 2$ $x - 2$ when $x \ge 2$ We, therefore must consider two inequalities, one when $x \le 2$, the other when $x \ge 2$ $x \le 2$: $- 6 \left(2 - x\right) = - 12 + 6 x < - 12$ Add 12 on both sides: $- 12 + 6 x + 12 < - 12 + 12$ or $6 x < 0$. Therefore $x < 0$ $x \ge 2$: $- 6 \left(x - 2\right) = - 6 x + 12 < - 12$ Add $6 x + 12$ on both sides: $- 6 x + 12 + 6 x + 12 < - 12 + 6 x + 12$ $24 < 6 x$ If we divide on 6 on both sides, we end up with: $x > 4$ The inequality is fulfilled when $x < 0$ or $x > 4$, in other words $x \notin \left[0 , 4\right]$
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Class 8 (Foundation)>Unit 2 Lesson 2: Multiplying fractions # Multiplying fractions and whole numbers visually Learn the concept of multiplying fractions and whole numbers. Watch how to visually represent this process and practice understanding the relationship between fractions and whole numbers in multiplication. Created by Sal Khan. ## Want to join the conversation? • I'm just wanting to confirm and understand why: 2/5 = 2 x 1/5 and why isn't it like before with mixed numerals where this is done: 2 x 1/5 = ((2 x 5)+1)/5 = 11/5 ? is there a conceptual difference i am missing? thanks in advance !! :) • Yes there's a conceptual difference. I believe you are confusing multiplication of a whole number and a fraction, with addition of a whole number and a fraction. Many students have difficulty in math, including higher levels such as algebra, ultimately because they confuse the fundamental concepts of addition and multiplication. ((2 x 5)+1)/5 is equivalent to the mixed number 2 and 1/5. This mixed number means 2 + 1/5, not 2 x 1/5. 2 x 1/5 can be thought of as the repeated addition 1/5 + 1/5, which is clearly 2/5. • why can't we just add it the normal way.like 1/5+3/4=4/9 • You cant just add the denominators as well as adding the numerators, you need to find a common factor. • If whatever happens to the denominator happens to the numerator why is the whole number only multiplied by the numerator and not the denominator? Are whole numbers +-x/ fractions only between the whole number and numerator? • Technically, A whole number n is the same as n/1. Since anything multiplied by 1 is itself, The denominator doesn't change. You only multiply both numbers by the whole number when you're finding a common denominator. • If 1/2 x 5 is 5/2 does that mean 5 x 1/2 is the same? • yes because 1/2 x 5 is the same as 1/2 x 5/1 (because 5 ones are five) and if you multipy 1 x 5 (the top) and 2 x 1( the bottom) you get 5/2 so you are correct • before i watched the video i already knew that when u multiply a fraction u just get the whole number like this example:5 x3/4 the denominator stays the same and the numerator changes just have to multiply the whle number to the numerator which is 15/? then the denominator stays the same like this 15/4 so yea that what i knew before i watched the vid.then after i watched the vid i taught me the same thing but in a longer form. • how would it be smaller instead of being bigger? • this is because of the fact that the fraction is not equal to one, so it makes the number smaller. Think of these fractions as decimals, when you multiply a number by a decimal it becomes smaller. so 3 x 1/3 is 1, because it becomes 3/3 which is one • why is it that when you add denominators together, they aren't different. For example: 2 fourths + 2 fourths do not equal 4 eighths. it equals 4 fourths instead. Why is that true? • Try imagining that the denominator is telling us how big the parts are. Adding parts together doesn't the size of those parts individually, but changes their amount over-all. Hope this helps. • On the last fraction, why are all the one fifths in parentheses? • I don't understand please help I am trying to get like what to do other than this fifths • multiplying fractions are one of the hardest fractions but sometimes it can be easy for some people (1 vote) • Why is it not 2/10 because 1/5 x 2 = 2/10 cause you cant just add when multiplying we have to multiply the denominator too! ## Video transcript We've already seen that the fraction 2/5, or fractions like the fraction 2/5, can be literally represented as 2 times 1/5, which is the same thing, which is equal to literally having two 1/5s. So 1/5 plus 1/5. And if we wanted to visualize it, let me make a hole here and divide it into five equal sections. And so this represents two of those fifths. This is the first of the fifths, and then this is the second of the fifths, Literally 2/5, 2/5, 2/5. Now let's think about something a little bit more interesting. What would 3 times 2/5 represent? 3 times 2/5. And I encourage you to pause this video and, based on what we just did here, think about what you think this would be equivalent to. Well, we just saw that 2/5 would be the same thing as-- so let me just rewrite this as instead of 3 times 2/5 written like this, let me write 2/5 like that-- so this is the same thing as 3 times 2 times 1/5. And multiplication, we can multiply the 2 times the 1/5 first and then multiply by the 3, or we can multiply the 3 times the 2 first and then multiply by the 1/5. So you could view this literally as being equal to 3 times 2 is, of course, 6, so this is the same thing as 6 times 1/5. And if we were to try to visualize that again, so that's a whole. That's another whole. Each of those wholes have been divided into five equal sections. And so we're going to color in six of them. So that's the first 1/5, second 1/5, third 1/5, fourth 1/5, fifth 1/5-- and that gets us to a whole-- and then we have 6/5 just like that. So literally 3 times 2/5 can be viewed as 6/5. And of course, 6 times 1/5, or 6/5, can be written as-- so this is equal to, literally-- let me do the same color-- 6/5, 6 over 5. Now you might have said, well, what if we, instead of viewing 2/5 as this, as we just did in this example, we view 2/5 as 1/5 plus 1/5, what would happen then? Well, let's try it out. So 3 times 2/5-- I'll rewrite it-- 3 times 2/5, 2 over 5, is the same thing as 3 times 1/5 plus 1/5. 2/5 is the same thing as 1/5 plus 1/5. So 3 times 1/5 plus 1/5 which would be equal to-- well, I just have to have literally three of these added together. So it's going to be 1/5 plus 1/5 plus 1/5 plus 1/5 plus-- I think you get the idea here-- plus 1/5 plus 1/5. Well, what's this going to be? Well, we literally have 6/5 here. We can ignore the parentheses and just add all of these together. We, once again, have 1, 2, 3, 4, 5, 6/5. So once again, this is equal to 6/5. So hopefully this shows that when you multiply-- The 2/5 we saw already represents two 1/5s. We already saw that, or 2 times 1/5. And 3 times 2/5 is literally the same thing as 3 times 2 times 1/5. In this case, that would be 6/5.
Home \| Learning \| History \| Fun \|Real-Life \| Help/Contact \| Index/Site-Map Lessons 1. Basics 2. Deductive Reasoning • 2.1 If-Then Statements • 2.2 How To Prove Theorems • 2.3 Pairs Of Angles • 2.4 Perpendicular Lines 3 - Parallel lines • 3.1 Parallel Lines • 3.2 Angles • 3.3 Inductive Reasoning 4 - Congruent Triangles • 4.1 Congruent Triangles • 4.2 Ways to prove it • 4.3 Bisectors 5 - Quadrilaterals • 5.1 Parallelograms • 5.2 Parallel lines Theorem • 5.3 Special Parallelograms • 5.4 Trapezoids 6 - Inequalities • 6.1 Inequalities • 6.2 Inequalities In A Triangle • 6.3 Inequalities In 2 Triangles 7 - Similar Polygon • 7.1 Ratios and Proportion • 7.2 Similar Polygons • 7.3 Similar Triangles 8 - Rt. Triangles • 8.1 Right Triangle • 8.2 Trig. In Geometry 9 - Circles • 9.1 Tangents, Arcs, and Chords • 9.2 Angles and Segment 10 - Constructions • 10.1 Construction • 10.2 Perpendicular Lines • 10.3 Parallel Lines • 10.4 Concurrent Lines • 10.5 Circles 11 - Areas of 2D objects • 11. 1 Areas Of Polygons • 11. 2 Circles and Similar Figures 12 - Areas and Volumes • 12.1 Prisms • 12.2 Pyramids • 12.3 Cylinders and cones • 12.4 Spheres • 12.5 Similar solids 13 - Coordinates • 13.1 Geometry and Algebra • 13.2 Lines and Coordinates 14 - Reflection/Rotation • 14.1 Basic Mapping • 14.2 Composition and Symmetry Inductive Reasoning Objective: • To be able to distinguish between deductive reasoning and inductive reasoning Lesson 3-3 Inductive Reasoning: Inductive reasoning: (1) Conclusion based on several past observations (2) Conclusion is probably true, but not necessarily true Deductive Reasoning: (1) Conclusion based on accepted statements (definitions, postulates, previous theorems, corollaries, and given information) (2) Conclusion must be true if hypotheses are true Quickie Math Copyright (c) 2000 Team C006354
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # What Is The Prime Factorization Of 25 What is the prime factorization of 25? Answer: 5 * 5 The prime factorization of 25 has 2 prime factors. If you multiply all primes in the factorization together then 25=5 * 5. Prime factors can only have two factors(1 and itself) and only be divisible by those two factors. Any number where this rule applies can be called a prime factor. The biggest prime factor of 25 is 5. The smallest prime factor of 25 is 5. ## How To Write 25 As A Product Of Prime Factors How to write 25 as a product of prime factors or in exponential notation? First we need to know the prime factorization of 25 which is 5 * 5. Next we add all numbers that are repeating more than once as exponents of these numbers. Using exponential notation we can write 25=52 For clarity all readers should know that 25=5 * 5=52 this index form is the right way to express a number as a product of prime factors. ## Prime Factorization Of 25 With Upside Down Division Method Prime factorization of 25 using upside down division method. Upside down division gives visual clarity when writing it on paper. It works by dividing the starting number 25 with its smallest prime factor(a figure that is only divisible with itself and 1). Then we continue the division with the answer of the last division. We find the smallest prime factor for each answer and make a division. We are essentially using successive divisions. This continues until we get an answer that is itself a prime factor. Then we make a list of all the prime factors that were used in the divisions and we call it prime factorization of 25. 5\|25 We divide 25 with its smallest prime factor, which is 5 5 The division of 5/25=5. 5 is a prime factor. Prime factorization is complete The solved solution using upside down division is the prime factorization of 25=5 * 5. Remember that all divisions in this calculation have to be divisible, meaning they will leave no remainder. ## Mathematical Properties Of Integer 25 Calculator 25 is a composite figure. 25 is a composite number, because it has more divisors than 1 and itself. It is not even. 25 is not an even number, because it can't be divided by 2 without leaving a comma spot. This also means that 25 is an odd number. When we simplify Sin 25 degrees we get the value of sin(25)=-0.13235175009777. Simplify Cos 25 degrees. The value of cos(25)=0.99120281186347. Simplify Tan 25 degrees. Value of tan(25)=-0.13352640702154. When converting 25 in binary you get 11001. Converting decimal 25 in hexadecimal is 19. The square root of 25=5. The cube root of 25=2.9240177382129. Square root of √25 simplified is 2√6. All radicals are now simplified and in their simplest form. Cube root of ∛25 simplified is 25. The simplified radicand no longer has any more cubed factors. ## Write Smaller Numbers Than 25 As A Product Of Prime Factors Learn how to calculate factorization of smaller figures like: ## Express Bigger Numbers Than 25 As A Product Of Prime Factors Learn how to calculate factorization of bigger amounts such as: ## Single Digit Properties For 25 Explained • Integer 2 properties: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a number of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2. • Integer 5 properties: 5 is the third the primes, after 3 and before 7. An odd amount and part of the primes of Fermat, Sophie Germain and Eisenstein. It is a prime, which is (5-1)÷2 and still remains one. Five is a pentagonal, square pyramidal, centered square, pentatopic, Perrin, Catalan and a congruent number. The fifth of the Fibonacci sequence, after 3 and before 8. An untouchable amount, not being the sum of the divisors proper to any other. Figures are divisible by 5 if and only if its last digit is 0 or 5. The square of a quantity with the last digit of 5 is equal to a quantity that has the last digits of 25 and as first digits the product of the private starting of 5 for itself increased by one unit. For example, 25²=(2×3)25=625 or 125²=(12×13)25=15625. The total of the first 2 prime numbers(in fact 2+3=5) and the sum of two squares(5=1²+2²). Five is the smallest natural that belongs to 2 Pythagorean triads:(3, 4, 5) and (5, 12, 13). In the binary system a palindrome. In the positional numbering system based on 4 it is a repeated number. In the numerical decimal system a Colombian number, that in addition is integer-free. ## Finding Prime Factorization Of A Number The prime factorization of 25 contains 2 primes. The prime factorization of 25 is and equals 5 * 5. This answer was calculated using the upside down division method. We could have also used other methods such as a factor tree to arrive to the same answer. The method used is not important. What is important is to correctly solve the solution. ## List of divisibility rules for finding prime factors faster Knowing these divisibility rules will help you find primes more easily. Finding prime factors faster helps you solve prime factorization faster. Rule 1: If the last digit of a number is 0, 2, 4, 6 or 8 then it is an even integer. All even integers are divisible by 2. Rule 2: If the sum of digits of a number is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 102 are 1, 0 and 2 so 1+0+2=3 and 3 is divisible by 3, meaning that 102 is divisible by 3). The same logic works also for number 9. Rule 3: If the last two digits of a number are 00 then this number is divisible by 4(example: we know that 212=200+12 and 200 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 12). In order to use this rule to it's fullest it is best to know multiples of 4. Rule 4: If the last digit of a integer is 0 or 5 then it is divisible by 5. We all know that 25=10 which is why the zero is logical. Rule 5: All numbers that are divisible by both 2 and 3 are also divisible by 6. This makes much sense because 23=6. ## What Is Prime Factorization Of A Number? In mathematics breaking down a composite number(a positive integer that can be the sum of two smaller numbers multiplied together) into a multiplication of smaller figures is called factorization. When the same process is continued until all numbers have been broken down into their prime factor multiplications then this process is called prime factorization. Using prime factorization we can find all primes contained in a number.
This chapter can improve student’s math skills, by referring to the Go Math Grade 3 Answer Key Chapter 1 Addition and Subtraction within 1,000 Assessment Test, and with the help of this Go Math Grade 3 Assessment Test Answer Key, students can score good marks in the examination. Go Math Grade 3 Answer Key Chapter 1 contains all the topics of chapter 1 which helps to test the student’s knowledge. Through this assessment test, students can check their knowledge. This assessment test is also helpful for the teachers to know how much a student understood the topics. Chapter 1: Addition and Subtraction within 1,000 Assessment Test ### Test – Page 1 – Page No. 11 Question 1. For numbers 1aâ€“1d, choose Yes or No to tell whether the sum is even. a. 8 + 3 i. yes ii. no Explanation: As 8 + 3 = 11 which is odd number. So the answer is no. Question 1. b. 6 + 6 i. yes ii. no Explanation: As 6 + 6= 12 which is even number. So the answer is true. Question 1. c. 4 + 5 i. yes ii. no Explanation: As 4 + 5 = 9 which is odd number. So the answer is No. Question 1. d. 2 + 6 i. yes ii. no Explanation: As 2 + 6 = 8 which is even number. So the answer is Yes. Question 2. Select the number sentences that show the Commutative Property of Addition. Mark all that apply. Options: a. 9 + 7 = 16 + 0 b. 9 + 7 = 7 + 9 c. (4 + 5) + 7 = (5 + 4) + 7 d. 7 + (4 + 5) = (7 + 4) + 5 Explanation: The “Commutative Laws” say we can swap numbers over and still get the same answer when we add a + b = b+a. Therefore 9 + 7 = 7 + 9 and (4 + 5) + 7 = (5 + 4) + 7 shows the Commutative Property of Addition. Question 3. Select the numbers that round to 500 when rounded to the nearest hundred. Mark all that apply. Options: a. 438 b. 542 c. 450 d. 483 e. 567 Explanation: We know that 542, 450, and 483 are between 400 and 500 and it is closer to 500. So, 483 rounded off to the nearest hundred is 500. Question 4. There are 165 cars in the parking lot. Complete the chart to show 165 rounded to the nearest 10. Explanation: Rounding 165 to nearest 10 = 170. 1 hundred, 7 tens, and 0 ones = Â 170. ### Test – Page 2 – Page No. 12 Question 5. Write each number sentence in the box below the better estimate of the sum. 281 + 125 = â– Â Â Â Â Â 236 + 119 = â– 242 + 128 = â– Â Â Â Â Â 309 + 135 = â– 281 + 125 = 300 + 100 = 400Â Â Â Â 236 + 119 = 200 + 100 = 300 242 + 128 =Â Â 200 + 100 = 300Â Â Â Â 309 + 135 =Â 300 + 100 = 400 Explanation: Question 6. Abby and Cruz are playing a game. Abbyâ€™s score is 586 points. Cruzâ€™s score is 754. Abby estimates she needs about 200 points more to reach Cruzâ€™s score. How did she estimate? Explain. Answer: Cruz rounded the estimates points to the nearest hundred. Explanation: Cruz rounded 586 to 600 and 754 to 800. Then she calculated the difference to estimate points = 800 â€“ 600 = 200. Question 7. The table shows how many shells each person collected. For numbers 7aâ€“7d select True or False for each statement. a. Melba collected about 40 more shells than Pablo. i. True ii. False Explanation: Melba collected 455 shells and Pablo collected 421 shells Difference between both = 455 â€“ 421 = 34 which is near to 40. Question 7. b. Melba and Pablo collected more than 800 shells. i. True ii. False Explanation: Sum of shells collected by Melba and Pablo = 455 + 421 = 876. Question 7. c. Amber collected about 60 fewer shells than Pablo. i. True ii. False Explanation: Difference between Amber and Pablo collected shells = 421 â€“ 382 = 39. Question 7. d. Amber, Melba, and Pablo collected over 1,100 shells. i. True ii. False Explanation: Sum of shells collected all three = 382 + 455 + 421 = 1258. ### Test – Page 3 – Page No. 13 Question 8. Mikio drove 58 miles on Saturday. On Sunday he drove 23 miles. How many miles did he drive on Saturday and Sunday? Explain how you solved the problem. _____ miles Explanation: No of miles drove on Saturday = 58 miles No of miles drove on Sunday = 23 miles Total no of miles he drove on both Saturday and Sunday = 58 + 23 = 81 miles. Question 9. Choose the property that makes the statement true. The Property of Addition describes the number sentence 17 + 1 = 1 + 17. ________ The Property of Addition describes the number sentence 17 + 1 = 1 + 17. Use the table for 10â€“12. Question 10. The table shows the number of students visiting the zoo each day. How many students visited the zoo on Wednesday and Thursday? _____ students Explanation: No of students visited zoo on Wednesday and Thursday = 349 + 508 = 857 students. Question 11. How many more students visited the zoo on Wednesday than on Monday? _____ students Explanation: No of students visited the zoo on Wednesday than on Monday = 349 â€“ 246 = 103. Question 12. How many more students visited the zoo on Monday and Tuesday than on Wednesday? _____ students Explanation: No of students visited the zoo on Monday and Tuesday than on Wednesday = (246+418) â€“ 349 = 315. ### Test – Page 4 – Page No. 14 Question 13. Help Ben find the sum. 2 4 6 3 2 1 +1 2 8 ———- 695 For numbers 13aâ€“13d choose Yes or No to tell Ben when to regroup. a. Regroup the ones. i. yes ii. no Question 13. i. yes ii. no Question 13. c. Regroup the tens. i. yes ii. no Question 13. i. yes ii. no Question 14. Avery sent 58 email invitations to a party. So far, 37 people replied. How many people still need to reply? Draw jumps and label the number line to show your thinking. _____ emails. Explanation: Given that total 58 email invitations are sent to a party No of people replied so far = 37 From the below figure no of people remained to reply =1 + 10 + 10 = 21 people. Question 15. There are 842 seats in the school auditorium. 138 seats need repairs. How many seats do not need repairs? Show your work. _____ seats Explanation: Total seats in school auditorium = 842 No of seats need to be repaired = 138 Therefore no of seats not required to repair = 842 â€“ 138 = 704 seats. Question 16. Madison solves this problem. She says the difference is 419. Explain the mistake Madison made. What is the correct difference? 6 4 5 âˆ’2 3 6 ———– _____ Explanation: When Madison combined the tens and ones, she should have regrouped 1 ten as 10 ones to subtract 36 from 45. Then she would have 0 tens and 9 ones left. The difference is 409, not 419. ### Test – Page 5 – Page No. 15 Question 17. Radburn School recycles aluminum cans to raise money. The third graders have collected 329 cans so far. Their goal is to collect more than 500 cans. What is the least number of cans they need to collect to reach their goal? Complete the bar model and explain how to use it to find the unknown part. _____ cans Explanation: The given model shows a whole of 500 and a part of 329. The unknown part represents the number of cans still to be collected. By solving using subtraction: 500 âˆ’ 329 = 171. So, they need to collect 1 more can than 171, which is 172. Question 18. The Science Center displays 236 butterflies. The number of beetles on display is 89 less than the number of butterflies. Part A About how many beetles are on display at the Science Center? Explain. Explanation: Given that 236 butterflies have displayed in the Science Center, rounding to nearest value = 240 No of displayed beetles are 89 less than the number of butterflies, after rounding = 90 Therefore no of beetles displayed = 240 â€“ 90 = 150. Question 18. Part B How many butterflies and beetles are on display at the Science Center? Show your work. _____ butterflies and beetles Explanation: Given no of butterflies = 236 No of beetles = 236 â€“ 89 = 147 Total no of beetles and butterflies = 236 + 147 = 383. ### Test – Page 6 – Page No. 16 Question 19. Elena used 74 + 37 = 111 to check her subtraction. Which math problem could she be checking? Mark all that apply. Options: a. 74 âˆ’ 37 = â– b. 111 âˆ’ 74 = â– c. 111 + 37 = â– d. 111 âˆ’ 37 = â– Explanation: She could use either option b. 111 â€“ 74 = 37 or option d. 111 â€“ 37 = 74. Question 20. Shawn and Steve are rock hunters. The tables show the kinds of rocks they collected. Part A Who collected more rock samples? How many did he collect? About how many more did he collect? Explain how you solved the problem. __________ Answer: Shawn collected more rock samples which are 288. Explanation: Sum of rock samples collected by Shawn = (127+65+96) = 288 Sum of rock samples collected by Steve = (79 + 109 + 93) = 281 Therefore Shawn collected more rock samples compared to Steve By subtracting 288 â€“ 281 = 9 (after rounding) => 10 Shawn has about 10 more rock samples. Question 20. Part B Shawn and Steve have the greatest number of what kind of rock? How many rocks of that kind do they have? Show your work. Answer: Quartz rocks, 236 rocks; 127 + 109 = 236. Explanation: Shawn collected 127 quartz rocks where Steve collected 109 quartz rocks. Total they both collected 236 Quartz rocks which are greatest in number compared to other types. Conclusion: This assessment test helps students to check their math skills. Go Math Grade 3 Chapter 10 questions are explained in detail that students can understand easily. Scroll to Top
### Chapter 2 in Undergraduate Econometrics ```2.1 Chapter 2: Probability Random Variable (r.v.) is a variable whose value is unknown until it is observed. The value of a random variable results from an experiment. Experiments can be either controlled (laboratory) or uncontrolled (observational). Most economic variables are random and are the result of uncontrolled experiments. Random Variables 2.2 A discrete random variable can take on only a finite number of values such as • • • • The number of visits to a doctor’s office Number of children in a household Flip of a coin Dummy (binary) variable: D=0 if male, D=1 if female A continuous random variable can take any real value (not just whole numbers) in an interval on the real number line such as: • Gross Domestic Product next year • Price of a share in Microsoft • Interest rate on a 30 year mortgage Probability Distributions of Random Variables • All random variables have probability distributions that describe the values the random variable can take on and the associated probabilities of these values. • Knowing the probability distribution of random variable gives us some indication of the value the r.v. may take on. 2.3 2.4 Probability Distribution for Discrete Random Variable Expressed as a table, graph or function 1. Suppose X = # of tails when a coin is flipped twice. X can take on the values 0, 1 or 2. Let f(x) be the associated probabilities: Table Graph X f(x) 0 0.25 f (x) 1 0.50 2 0.25 0.50 Probability is represented as height on this bar graph 0.25 0 1 2 x 2.5 2. Suppose X is a binary variable that can take on two values: 0 or 1. Furthermore, assume P(X=1) = p and P(X=0) = (1-p) Function: P(X=x) = f(x) = px(1-p)1-x for X = 0, 1 Table X f(x) 0 (1-p) 1 p Suppose p = 0.10 Then X takes on 0 with probability 0.90 and X takes on 1 with probability 0.10 Facts about discrete probability distribution functions 1. Each probability P(X=x) = f(x) must lie between 0 and 1: 0  f(x)  1 2. The sum of the probabilities must be 1. If X can take on n different values then: f(x1) + f(x2)+. . .+f(xn) = 1 2.6 2.7 Probability Distribution (Density)for Continuous Random Variables Expressed as a function or graph. Continuous r.v.’s can take on an infinite number of values in a given interval – A table isn’t appropriate to express pdf EX: f(x) = 2x for 0  x  1 =0 otherwise 2.8 Because a continuous random variable has an uncountably infinite number of values, the probability of one occurring is zero. P(X = a) = 0 Instead, we ask “What is the probability that X is between a and b? P[a < X < b] = ? In an experiment, the probability P[a < X < b] is the proportion of the time, in many experiments, that X will fall between a and b. 2.9 Probability is represented as area under the function. Total area must f(x) be 1.0 Area of triangle 2 is 1.0 1 Probability that x lies between 0 and 1/2 P [ 0  X  1/2 ] = 0.25 [Area of any triangle is ½BaseHeight] 1/2 1 x 2.10 Uniform Random Variable: u is distributed uniformly between a and b • p.d.f. is a line between a and b of height 1/(b-a) • f(u) = 1/(b – a) if a  u  b = 0 otherwise EX: Spin a dial on a clock a = 0 and b = 12 Find the probability that u lies between 1 and 2 f(u) 1/12 0 1 2 12 u 2.11 In calculus, the integral of a function defines the area under it: P[aXb]=  b f(x) dx a For continuous random variables it is the area under f(x), and not f(x) itself, which defines the probability of an event. We will NOT be integrating functions; when necessary we use tables and/or computers to calculate the necessary probability (integral). 2.12 Rules of Summation n Rule 1: Rule 2: Rule 3: xi = x1 + x2 + . . . + xn  i=1 n a = na  i=1  axi = a  xi n Rule 4: n n i=1 i=1 xi + yi =  xi +  yi  i=1 2.13 Rules of Summation (continued) n Rule 5: Rule 6: n n i=1 i=1 axi + byi = a  xi + b  yi  i=1 x n = n  xi = i=1 1 x1 + x2 + . . . + xn n From Rule 6, we can prove (in class) that: n xi x) = 0 i=1 2.14 Rules of Summation (continued) n Rule 6: f(xi) = f(x1) + f(x2) + . . . + f(xn)  i=1 Notation: n m Rule 7: n x f(xi) = i f(xi) = i =1 f(xi) n [ f(xi,y1) + f(xi,y2)+. . .+ f(xi,ym)]   f(xi,yj) = i  =1 i=1 j=1 The order of summation does not matter : n m m n f(xi,yj)   f(xi,yj) =j =  1 i=1 i=1 j=1 2.15 The Mean of a Random Variable The mean of a random variable is its mathematical expectation, or expected value. For a discrete random variable, this is: = xif(xi) = x1f(x1) + x2f(x2) + . . . + xnf(xn) where n measures the number of values X can take on E(X) It is a probability-weighted average of the possible values the random variable X can take on. This is a sum for discrete r.v.’s and an integral for continuous r.v.’s 2.16 • E(X) tells us the “long-run” average value for X. It is not the value one would expect X to take on. • If you were to randomly draw values of X from its pdf an infinite number of times and average these values, you would get E(X) • E(X) =  this greek letter “mu” is not used in your text but is commonly used to denote the mean of X. 2.17 Example: Roll a fair die 6 E  X    xi f xi  i 1  1(1 / 6)  2(1 / 6)  3(1 / 6)  4(1 / 6)  5(1 / 6)  6(1 / 6)  21/ 6  3.5 Interpretation: In a large number of rolls of a fair die, onesixth of the values will be 1’s, one-sixth of the values will be 2’s. etc., and the average of these values will be 3.5. Mathematical Expectation • • 2.18 Think of E(.) as an operator that requires you to weight by probabilities any expression inside the parentheses, and then sum E(g(x)) = g(xi)f(xi) = g(x1)f(x1) + g(x2 ) f(x2) + . . . + g(xn ) f(xn) Rules of Mathematical Expectation • E(c) = c where c is a constant • E(cX) = cE(X) where c is a constant and X is a random variable • E(a + cX) = a + cE(X) where a and c are constants and X is a random variable. 2.19 Variance of a Random Variable • • • • Like the mean, the variance of a r.v. is an expected value, but it is the expected value of the squared deviations from the mean Let g(x) = (x – E(x))2 Variance 2 = Var(x) = E(x – E(x))2 = g(xi)f(xi) = (xi – E(xi))2f(xi) It measures the amount of dispersion in the possible values for X. 2.20 • • 2.21 Unit of measurement is X units squared When we create a new random variable as a linear transformation of X: y = a + cx We know that E(y) = a + cE(x) But Var(y) = c2Var(x) (proof in class) This property tells us that the amount of variation in y is determined by: the amount of variation in X and the constant c. The additive constant a in no way alters the amount of variation in the values on x. • E(x – E(x))2 = E[x2 – 2E(x)x + E(x)2] = E(x2) – 2E(x)E(x) + E(x)2 = E(x2) – 2E(x)2 + E(x)2 = E(x2) – E(x)2 • Run the E(.) operator thru, pulling out constants and stopping on random variables. Remember that E(x) is itself a constant, so • E(E(x)) = E(x) 2.22 Standard Deviation • 2.23 Because variance is in squared units of the r.v., we can take the square root of the variance to obtain the standard deviation.  =  2 =  Var(x) Be sure to take the square root after you square and sum the deviations from the mean. Joint Probability 2.24 • An experiment can randomly determine the outcome of more than one variable. • When there are 2 random variables of interest, we study the joint probability density function • When there are more than 2 random variables of interest, we study the multivariate probability density function. For a discrete joint pdf, probability is expressed in a matrix: Let X= return on stocks, Y= return on bonds X Y f(y) -10 0 10 20 6 0 0 0.10 0.10 8 0 0.10 0.30 0.20 10 0.10 0.10 0 0 f(x) P(X=x,Y=y) = f(x,y) e.g. P(X=10,Y=8) = 0.30 2.25 • 2.26 Marginal Probability Distribution: what is the probability distribution for X regardless of what values Y takes on? f(x) = yf(x,y) what is the probability distribution for Y regardless of what values X takes on? f(y) = xf(x,y) 2.27 • Conditional Probability Distribution: What is the probability distribution for X given that Y takes on a particular value? f(x\|y) = f(x,y)/f(y) What is the probability distribution for Y given that X takes on a particular value? f(y\|y\x) = f(x,y)/f(x) 2.28 • Covariance: A measure that summarizes the joint probability distribution between two random variables. cov(x,y) = E[(x – E(x))(y-E(y))] = x y (xi – E(x))(yi – E(y))f(x,y) Ex: 2.29 It measures the joint association between 2 random variables. Try asking: “When X is large, is Y more or less likely to also be large?” If the answer is that Y is likely to be large when X is large, then we say X and Y have a positive relationship. Cov(x,y) > 0 If the answer is that Y is likely to be small when X is large, then we say that X and Y have a negative relationship. Cov(x,y) < 0. cov(x,y) = E[(x – E(x))(y – E(y))] = E[xy – E(x)y – xE(y) + E(x)E(y)] = E(xy) – E(x)E(y) – E(x)E(y) + E(x)E(y) = E(xy) – E(x)E(y)  useful!! 2.30 • Correlation Covariance has awkward units of measurement. Correlation removes all units of measurement by dividing covariance by the product of the standard deviations: xy = Cov(x,y)/(xy) and –1  xy  1 Ex: 2.31 What does correlation look like?? =0 =.7 =.3 =.9 Statistical Independence Two random variables are statistically independent if knowing the value that one will take on does not reveal anything about what value the other may take on: f(x\|y) = f(x) or f(y\|x) = f(y) This implies that f(x,y) = f(x)f(y) if X and Y are independent. If 2 r.v.’s are independent, then their covariance will necessarily be equal to 0. 2.32 Functions of more than one Random Variable 2.33 Suppose that X and Y are two random variables. If we sum them together we create a new random variable that has the following mean and variance: Z = aX + bY  E(Z) = E(aX + bY) = aE(x) + bE(y) Var(Z) = Var(aX + bY) = a2Var(X) + b2Var(Y) + 2abCov(X,Y) If X and Y are independent  Var(Z) = Var(aX + bY) = a2Var(X) + b2Var(Y) see page 31 2.34 Normal Probability Distribution • Many random variables tend to have a normal distribution (a well known bell shape) • Theoretically, x~N(β,2) where E(x) = β and Var(x) = 2 The probability density function is a  ( x  )2  f ( x)  exp  ,  2 2 2  2  1  b  x  x Normal Distribution (con’t) 2.35 • A family of distributions, each with its own mean and variance. The mean anchors the distribution’s center and the variance captures the spread of the bell-shaped curve • To find area under the curve would require integrating the p.d.f – too complicated. Computer generated table gives all the probabilities we need for a normal r.v. that has mean 0 and variance of 1 To use the table (pg. 389), we need to take a normal random variable x~N(,2) and transform it by subtracting the mean and dividing by the standard deviation. This is a linear transformation of X that creates a new random variable that has mean 0 and variance of 1. Z = (x - )/  where z ~N(0,1) Statistical inference: drawing conclusions about a population based on a sample 2.36 T E ( X )     xi f ( xi ) X  Xt t 1 T 2 ( x  x ) s x2   i T 1 Var ( X )   2  E ( X  E ( X )) 2  E ( X   )2  x   x 2  Var ( X ) sx  sx2 Cov ( X ,Y )  E ( X   x )(Y   y ) 1 S xy  ( xt  x )( yt  y )  T 1  E ( XY )   x  y  xy  Cov ( X , Y ) Var ( X ) Var (Y ) r S xy s x2 s 2y   ( xt  x )( yt  y )  ( xt  x )2  ( yt  y )2 ```
Q: # How do you do double digit multiplication? A: To multiply two-digit numbers, write the first number on a piece of paper and write the second number directly underneath it. Multiplying double digits requires breaking the problem into four multiplication equations and one addition calculation. ## Keep Learning Credit: Roland Dan Moment Getty Images 1. Multiply the second digits Multiply the last digits of both numbers together. For example, in 12 x 15, calculate 2 x 5 = 10. Write this answer below the two starting numbers. 2. Multiply the last digit of the second number Replace the second digit of the first number with zero. Multiply the entire top number by the second digit of the second number. So in 12 x 15, calculate 10 x 5 = 50. Write this answer below the number calculated in the first step. 3. Multiply the first digit of the second number Replace the second digit of the second number with zero. Multiply this number by the second digit of the first number. In 12 x 15, multiply 10 x 2 = 20. Write this answer under the number calculated in the second step. 4. Multiply the first digits Replace the second digits of the first and second numbers with zero. Multiply these together. In 12 x 15, multiply 10 x 10 = 100. Write this answer under the number calculated in the third step. Add the four calculated numbers to find the answer to the original multiplication problem. For example, in 12 x 15, add together 10 + 50 + 20 + 100, which equals 180. Sources: ## Related Questions • A: As a general rule of thumb, numbers should be rounded down if the final digit in a number is from 0 to 4, and rounded up if it's 5 to 9. Rounding sometimes... Full Answer > Filed Under: • A: The value of an underlined digit in a given number expresses how much that digit is worth. To determine the digit’s value, multiply it by its place value. ... Full Answer > Filed Under: • A: To tell if a number is evenly divisible by three, add all the digits in it until you arrive at a single digit. For instance, the number 9234 adds up to 18,... Full Answer > Filed Under: • A: Round each number in the question to only have one non-zero digit. For example, 64 becomes 60, and 1,890 becomes 2,000. Then do the calculation required.... Full Answer > Filed Under: PEOPLE SEARCH FOR
# Difference between revisions of "2003 Indonesia MO Problems/Problem 2" ## Problem Given a quadrilateral $ABCD$. Let $P$, $Q$, $R$, and $S$ are the midpoints of $AB$, $BC$, $CD$, and $DA$, respectively. $PR$ and $QS$ intersects at $O$. Prove that $PO = OR$ and $QO = OS$. ## Solution $[asy] pair A=(0,0), B=(60,0), C=(40,80), D=(10,50), P=(30,0), Q=(50,40), R=(25,65), SX=(5,25); draw(A--B--C--D--cycle); draw(P--Q--R--SX--cycle); draw(A--C, dotted); draw(B--D, dotted); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); dot(P); label("P",P,S); dot(Q); label("Q",Q,E); dot(R); label("R",R,N); dot(SX); label("S",SX,W); [/asy]$ Draw lines $AC$ and $BD$. By SAS Similarity, $\triangle DSR \sim \triangle DAC,$ $\triangle CRQ \sim \triangle CDB,$ $\triangle ASP \sim \triangle ADB,$ and $\triangle CRQ \sim \triangle CDB.$ That means $RQ \parallel DB \parallel SP$ and $SR \parallel AC \parallel PQ,$ making $PQRS$ a parallelogram. $[asy] pair P=(30,0), Q=(50,40), R=(25,65), SX=(5,25); draw(P--Q--R--SX--cycle); dot(P); label("P",P,S); dot(Q); label("Q",Q,E); dot(R); label("R",R,N); dot(SX); label("S",SX,W); draw(Q--SX); draw(P--R); dot((27.5,32.5)); label("O",(27.5,32.5),SE); [/asy]$ Since $PQRS$ is a parallelogram, $RS = QP.$ In addition, by the Alternating Interior Angle Theorem, $\angle SRP = \angle RPQ$ and $\angle RSQ = \angle SQP.$ Thus, by ASA Congruency, $\triangle SRO \cong \triangle QPO.$ Finally, using CPCTC shows that $RO = OP$ and $SO = OQ.$ 2003 Indonesia MO (Problems) Preceded byProblem 1 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed byProblem 3 All Indonesia MO Problems and Solutions
#### Transcript Section 3.4 - Shelton State Community College ```Section 3.4 Exponential and Logarithmic Equations Overview • In this section we will solve logarithmic and exponential equations. • Some things we will need: 1. The ability to convert from exponential to logarithmic and vice versa. 2. The properties of logarithms. 3. The change of base formula. 4. The ability to solve linear and quadratic equations 5. Knowledge of the domain of a logarithmic function. A Couple of (New) Things 1. The exponential function is one-to-one: If ax = ay, then x = y. 2. The logarithmic function is one-to-one. In both directions: If x = y, then logax = logay. If logax = logay, then x = y. Case I: Exponential • Exponential equations do not have the word “log” anywhere in the problem. • To solve an exponential equation: 1. Write both sides of the equation as powers of the same base. Then set the exponents equal to each other. 2. Take the natural log of both sides (Know the difference between an exact answer and an • An exact answer will leave the log expressions intact. No decimal approximations will be used for any log expressions. • An approximate answer will involve using a scientific calculator to find approximate values for log expressions. Answers will be rounded to a designated number of decimal places. Examples 8 2 x 1  512 5 e x 7 4 x2  5 1  e 6 4 x 1  5 x 3 Case II: Log = number • These equations will have a log expression on one side and a number on the other. • Solve by converting to exponential form: logax = y is the same as x = ay Examples log3 x  4 ln x  8 6 ln 2 x   18 Case III: Multiple logs = number • These equations will have more than one log expression on one side and a number on the other side. • Use the properties of logarithms to combine the multiple logs into a single logarithmic expression. • Then convert to exponential. Case IV: Logs on both sides • If you have multiple logs on either side, use the properties of logarithms to condense them into single logarithmic expression. • Then set the arguments equal to each other and solve. Examples log2 ( x  3)  log2 ( x  1)  5 logx  6  log x  log 6 3 log x  log125 2 log x  log 9  log 576 ```
Courses Courses for Kids Free study material Offline Centres More Store # How do you factorize the trinomial $3{x^2} - 14x - 5$? Last updated date: 13th Jun 2024 Total views: 384.9k Views today: 4.84k Verified 384.9k+ views Hint: Here we have to find out the roots of the given trinomial expression by the factoring method. On splitting the middle term of the given expression and doing some simplification we get the required answer. Complete step-by-step solution: Given polynomial is $3{x^2} - 14x - 5$ We need to find out the roots for the trinomial polynomial by using factoring. First we need to put it is equal to zero So, we can write it as $3{x^2} - 14x - 5 = 0$ Now solving the given trinomial polynomial as follows: $3{x^2} - 14x - 5$ We need to find the combination of two numbers which when multiply each other it would be $- 15$and $x$ when subtracted both the numbers, so we get We can write the equation as, $\Rightarrow 3{x^2} - 15x + x - 5 = 0$ Taking common the desired number from the above equations in half parts $\Rightarrow 3x(x - 5) + 1(x - 5) = 0$ Arranging the equation as we want as roots $\Rightarrow (x - 5)(3x + 1) = 0$ We put each bracket equal to zero $\Rightarrow x - 5 = 0$ and $3x + 1 = 0$ Hence we get $\Rightarrow x = 5$ and $x = - \dfrac{1}{3}$ Therefore, the roots for the given trinomial $3{x^2} - 14x - 5 = 0$ are $5$ and $- \dfrac{1}{3}$ Note: In algebra, we discuss in the primary level class; A polynomial contains three terms or we can say that any three monomials, is known as a trinomial polynomial, where monomial is the polynomial which has only one term. The equation contains three variables that are $a,b{\text{ and c}}$. Therefore it is trinomial polynomial and Examples as $4{a^4} + 3x - 2,8{x^2} + 2x + 7$. The equations contain only one element that is $z$ therefore, it is not a trinomial. In solving these types of questions we need to understand the key concept of solving and we have to understand all the terms used in the question one by one.
# Probability/The Counting Principle ## The Counting Principle Before we can delve into the properties of probability and odds, we need to understand the Counting Principle. We use the Counting Principle to determine how many different ways one can choose/do certain events. It is easier to define the Principle through examples: ### Independent Events Let's say that John is at a deli sub restaurant. There are 3 different breads, 3 different cheeses, 3 different condiments, and 3 different vegetables he can place on his sandwich, assuming that he can only place one from each category on to his sandwich. How many different ways can he set up his sandwich? Since choosing a cheese doesn't affect the number of choices of vegetables, condiments, or bread, these events are called independent events. For this problem, we will multiply ${\displaystyle 3\times 3\times 3\times 3}$ , so ${\displaystyle 3^{4}}$ , which is 81. So there are 81 different possible combinations to form this sandwich. #### Practice Problems 1) Thomas goes to a McDonalds' restaurant and chooses to create his own burger. He has 2 different kinds of cheese, 3 different breads, and 3 different sauces he can choose from, but he can only choose one of each category. How many different ways can he create this burger? 2) Diane is ordering pizza for her family. There are 4 different possible sizes of the pizza. Also, she has to choose one of 5 toppings to place on the pizza and one of 3 different types of cheese for the pizza. In addition, she must choose one of 3 different kinds of crust. How many different ways can she have her pizza? 3) a) How many 3-digit numbers can be formed from the digits 2, 3, 4, 5, 7 and 9? b) How many of these numbers are less than 400? 1) ${\displaystyle (2)(3)(3)=18}$ 2) ${\displaystyle (4)(5)(3)(3)=180}$ 3) a) Since there are six available digits, the answer is ${\displaystyle (6)(6)(6)=216}$ b) For the value of the 3-digit number to be less than 400, we only have two choices for the first digit, namely 2 or 3. After that we can choose the other two digits freely. The answer is thus ${\displaystyle (2)(6)(6)=72}$ . ### Dependent Events Assume that John is now working at a library. He must put 5 books on a shelf in any order. How many different ways can he order the books on the shelf? Unlike the independent events, when John puts a book on the shelf, that eliminates one book from the remaining choices of books to put next on the shelf; thus these are referred to as dependent events. At first, he has 5 different choices, so our first number in the multiplication problem will be 5. Now that one is missing, the number is reduced to 4. Next, it's down to 3, and so on. So, the problem will be ${\displaystyle (5)(4)(3)(2)(1)}$ However, there is a symbol for this very idea. A ${\displaystyle !}$ represents the term factorial. So, for example, ${\displaystyle 3!=(3)(2)(1)}$ . Factorials are very useful in statistics and probability. Therefore, the problem can be rewritten as ${\displaystyle 5!}$ , which ends up being equal to 120. However, not all dependent event problems are that simple. Let's say that there are 10 dogs at a dog competition. How many different ways can one select the Champion AND the Runner-Up? This problem could actually be considered simpler than the last one, but it doesn't involve a factorial. So, how many different ways can the judge determine the Champion? Of course, there are 10 different dogs, so 10 different ways. Next, how many dogs are left to select the Runner-Up from? Well, you removed one, so it's down to 9. Instead of placing a factorial, though, you will only multiply 10 and 9, resulting in 90. ### Independent Or Dependent? To help you differentiate between the two, we will do a few more examples, but we will have to decide if it is dependent or independent before solving the problem. Let's say that you are creating a 5-digit garage door opener code (the digits would include 0-9). If there were absolutely no restrictions, would the events be independent of each other or dependent on each other? Of course, there are no restrictions, since you could have five 4's for the code, so to solve the problem, you multiply 10 by itself 5 times, resulting in 100000. Alternatively, suppose that the first number of the code cannot be 0, and that there can be no repeated numbers whatsoever. Obviously these events are dependent on each other, because there cannot be any repetitions. Let's look at this problem one number at a time. The first number can be all the numbers except 0, reducing the possible numbers to 9. The second number can be 0 this time, so the possible numbers returns to 10. However, it cannot be a repeat of the previous number, so there are 9 possible choices again. After that, the numbers will reduce by one each time, due to the fact that there cannot be repeats, so the problem would look like this ${\displaystyle (9)(9)(8)(7)(6)={\frac {(9)(9!)}{(5!)}}=27216}$ Now, just one more example. Let's say that you were choosing your schedule for school. There are 8 periods each day, and there are 7 classes to choose from. Nonetheless, you must have a lunch period during the 4th period. We can think of 4th period as non existent because it is in a constant position and therefore does not affect the possible choices. With 7 slots and 7 options, the answer is simply ${\displaystyle 7!}$ . ${\displaystyle (7!)=5040}$ ### Review Of The Counting Principle So, we use the Counting Principle to determine the different unique ways we can do something, such as a sandwich or a selection of classes. Sometimes, these events will affect each other, such as when you can't choose the same number twice for your garage door code, so they are dependent events. However, other times, one event has no effect on the next event, such as when you have different cheeses and breads to choose for your sandwich, so they are independent events. The Counting Principle is a fundamental mathematical idea and an essential part of probability. ## Counting Rules Rule 1: If any one of ${\displaystyle k}$ mutually exclusive and exhaustive events can occur on each of ${\displaystyle n}$ trials, there are ${\displaystyle k^{n}}$ different sequences that may result from a set of such trials. • Example: Number of possible sequences resulted from flipping a coin three times is ${\displaystyle k^{n}=2^{3}=8}$ . Rule 2: If ${\displaystyle k_{1},\dots ,k_{n}}$ are the numbers of distinct events that can occur on trials ${\displaystyle 1,\dots ,n}$ in a series, the number of different sequences of ${\displaystyle n}$ events that can occur is ${\displaystyle k_{1}\times \cdots \times k_{n}}$ . • Example: Number of possible sequences resulted from flipping a coin and roll a six-faced die is ${\displaystyle (k_{1})(k_{2})=(2)(6)=12}$ . Rule 3: The number of different ways that ${\displaystyle n}$ distinguishable things may be arranged in order is ${\displaystyle n!=(1)(2)(3)\times \cdots \times (n-1)(n)}$ , in which ${\displaystyle 0!{\overset {\text{def}}{=}}1}$ . An arrangement in order is called a permutation. The total number of permutations of ${\displaystyle n}$ objects is ${\displaystyle n!}$ (the symbol ${\displaystyle n!}$ Is called n-factorial). • Example: Number of possible ways to arrange 10 items in order is ${\displaystyle 10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=3,628,800}$ Rule 4: The number of ways of selecting and arranging ${\displaystyle k}$ distinguishable objects from among ${\displaystyle n}$ distinct objects is: ${\displaystyle {\frac {n!}{(n-k)!}}}$ , or as seen on calculators, [nPr]. • Example: Number of possible ways to pick 3 things from 10 items, and arrange them in order is ${\displaystyle {\frac {10!}{(10-3)!}}={\frac {10!}{7!}}=720}$ . Rule 5: The total number of ways of selecting ${\displaystyle k}$ distinct combinations of ${\displaystyle n}$ distinguishable objects, irrespective of order (i.e. order is NOT important), is: ${\displaystyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}}$ or as seen on calculators, [nCr]. • Example: Number of possible ways to pick 3 items from 10 items irrespective of order is ${\displaystyle {\binom {10}{3}}={\frac {10!}{3!(7!)}}={\frac {720}{6}}=120}$ .
# Use proportional relationships to solve multistep ratio and percent problems. ### Popular Tutorials in Use proportional relationships to solve multistep ratio and percent problems. #### How Do You Solve a Word Problem Using a Percent Proportion? Word problems allow you to see the real world uses of math! This tutorial shows you how to take a words problem and turn it into a percent proportion. Then see how to solve for the answer using the mean extremes property of proportions. Take a look! #### How Do You Set Up a Proportion from a Word Problem? Sometimes the hardest part of a word problem is figuring out how to turn the words into an equation you can solve. This tutorial let's you see the steps to take in order to turn a word problem involving a blueprint into a proportion. Take a look! #### How Do You Use a Proportion to Find a Part of a Whole? Taking a percent of a number? Trying to figure out the result? Use a percent proportion to solve! This tutorial will show you how! #### How Do You Use the Formula for Simple Interest? If you already have a bank account or if you plan to have one in the future, then this tutorial is a must see! Follow along as this tutorial goes through a word problem involving simple interest. #### What is the Formula for Simple Interest? Interest is found in a bunch of places: savings accounts, mortgages, loans, investments, credit cards, and more! Watch this tutorial and learn how to calculate simple interest! #### How Do You Figure Out a Percent of Change? Word problems allow you to see the real world uses of math! In this tutorial, learn how to calculate the percent of increase using the percent of change formula. #### What's a Percent of Change? Lots of things in this world change their value such as cars, video games, and computers. When something either increases or decreases in value, it can be useful to know the percent of that change in value. To figure out that percent, you'll need the percent of change formula. Learn it with this tutorial! #### How Do You Figure Out How Much Something is Marked Down? Going shopping? Is something you want on sale? Trying to figure out the sale price of that item? Follow along with this word problem and you'll see how to calculate that price! #### How Do You Set Up a Percent Proportion from a Word Problem? Sometimes the hardest part of a word problem is figuring out how to turn the words into an equation you can solve. This tutorial let's you see the steps to take in order to do just that! Take a look! You'll be glad you did! #### How Do You Figure Out Sales Tax? Going shopping can be tons of fun, but things can go sour when you get to the register and realize that the sales tax puts you over your budget. Always stay under budget by figuring out your total cost BEFORE you hit the check out. Watch this tutorial and learn how to calculate sales tax! #### How Do You Use an Equation to Find a Part of a Whole? Taking a percent of a number? Trying to figure out the result? Convert the percent to a decimal and multiply it by the number! This tutorial will show you how! #### How Do You Solve a Proportion Using Cross Products? Want to solve a percent proportion? Just use the means extremes property of proportions to cross multiply! Solve for the variable, and you have your answer! Learn how with this tutorial. #### How Do You Use a Proportion to Find What Percent a Part is of a Whole? A part is some percent of a whole. Trying to calculate the percent? Use a percent proportion to solve! This tutorial will show you how! #### What's the Means-Extremes Property of Proportions? The means-extremes property of proportions allows you to cross multiply, taking the product of the means and setting them equal to the product of the extremes. This property comes in handy when you're trying to solve a proportion. Watch this tutorial to learn more! #### What are the Means and Extremes of Proportions? A proportion is just an equation where two ratios are equal, and each piece of the proportion has a special name. This tutorial will teach you those names, and this will help you understand cross multiplication when you learn it later! #### How Do You Solve a Word Problem Using a Proportion? This tutorial provides a great real world application of math. You'll see how to use the scale from a blueprint of a house to help find the actual height of the house. This tutorial shows you how to use a proportion to solve! #### How Do You Figure Out a Tip? If you need to leave a tip at a restaurant, you can quickly estimate the amount in your head! This tutorial shows you how to use estimation and mental math to estimate a tip! #### How Do You Estimate a Sale Price? Sales are great, but how much are you really saving? This tutorial shows you how to estimate the sales price of an item. #### How Do You Figure Out a Percent From a Part to Part Ratio? Word problems and percents can be a fun combination! This tutorial shows you how to find the percent of something in a basket using ratios! #### How Do You Figure Out a Percent From a Part to Whole Ratio? If you want to find a percent in a word problem, you may be able to use a ratio to help you! This tutorial shows you how to do exactly that! #### How Do You Solve a Proportion by Finding an Equivalent Ratio? Trying to find a missing value in order to create a proportion with two ratios? Take the ratios in fraction form and identify their relationship. Use that relationship to find your missing value. This tutorial will show you how! #### How Do You Solve a Proportion Using the Multiplication Property of Equality? Trying to find a missing value in a ratio to create proportional ratios? You could use the multiplication property of equality! In this tutorial, see how to use this property to find a missing value in a ratio. Take a look! #### How Do You Solve a Word Problem Using Ratios? This tutorial shows you how to use a ratio to create equivalent ratios. Then, use a multiplier to find a missing value and solve the word problem. Take a look! #### How Do You Figure Out the Price of a Marked Up Item? The price of items is always changing. You've probably went to the store to buy an item and found that its price has been marked up. In this tutorial, learn how to figure out the new price of an item that was marked up. Take a look! #### How Do You Figure Out Whether a Percent of Change is an Increase or a Decrease? Word problems are a great way to see the real world applications of math! In this tutorial, you'll see how the percent of change can be found from the information given in a word problem. Check it out!
## Video and Examples Learn how to find the circumference of a circle if you know the measure of the circle's diameter. $C\,=\,\pi~\cdot~d$ If you measure the distance around a circle and divide it by the distance across the circle through its center, you will always come close to a particular value, depending upon the accuracy of your measurement. This value is approximately 3.14159265358979323846... We use the Greek letter (Pi) to represent this value. Using computers, mathematicians have been able to calculate the value of to billions of places. The distance around a circle is called its circumference. The distance across a circle through its center is called its diameter. Pi ($\large\pi$) is the ratio of the circumference of a circle to its diameter. For any circle, if you divide its circumference by its diameter, you get a value close to $\pi$. This relationship is expressed in the following formula: $\frac{C}{d}\,=\,\pi$, where C is the circumference and D is the diameter. You can test this formula at home with a dinner plate. If you measure the circumference and the diameter of the plate and then divide the circumference by the diameter, your quotient should come close to $\pi$. Another way to write this formula is: $\large~C\,=\,\pi\cdot~d$ Using $\pi~\,=\,3.14$ Using $\pi~\,=\,3\frac{1}{7}$ Find the circumference of this circle with a diameter of 8 inches. $\mbox{Use}\,\pi~\,=\,3.14$ $C\,=\,\pi~\,\cdot~\,d$ $C\,=\,(3.14)\,\cdot~\,(8)$ $C\,=\,25.12\,inches$ Find the circumference of this circle with a diameter of 35 inches. $\mbox{Use}\,\pi~\,=\,3\frac{1}{7}$ $C\,=\,\pi~\,\cdot~\,d$ $C\,=\,3\frac{1}{7}\,\cdot~\,35$ $C\,=\,\frac{22}{7}\,\cdot~\,\frac{35}{1}$ $C\,=\,110\,cm$ Find the circumference of this circle with a diameter of 8 inches in terms of $\large~\pi$. $C\,=\,\pi~\,\cdot~\,d$ $C\,=\,\pi~\,\cdot~\,(8)$ $C\,=\,8\pi~\,inches$ Use this applet to observe how to find the circumference of any circle using either 3.14 or $3\frac{1}{7}$ and when given the radius or the diameter. Move the blue and green points to create the circle of your choice. Then use the slider on the left to choose whether the radius or the diameter is given. The slider on the right selects which version of $\pi$ is being used. # Self-check Q1: What is the circumference of a circle with a diameter of 10 cm? (Use π = 3.14) [show answer] Q2: What is the circumference of a circle with a diameter of 14 cm? (Use π = 3 1/7) [show answer] Q3: In terms of π, what is the circumference of a circle with a diameter of 8 cm? [show answer]
A statistical hypothesis is an assumption about a population parameter. This assumption may or may not be true Presentation on theme: "A statistical hypothesis is an assumption about a population parameter. This assumption may or may not be true"— Presentation transcript: A statistical hypothesis is an assumption about a population parameter. This assumption may or may not be true http://stattrek.com/Lesson5/HypothesisTesting.aspx?Tutorial=Stat The Null Hypothesis (H 0 ) Sample observations are resulting purely from chance H 0 always contains an “equals” statement The null hypothesis is assumed to be true We need strong statistical evidence to reject it The Alternative Hypothesis (H a ) This is usually what we’re trying to prove Everything that’s not the null hypothesis Sample observations are influenced by some non- random cause For example Null Hypothesis Alternative Hypothesis H 0 : µ=\$10H a : µ≠\$10µ is \$10 or it isn’t H 0 : µ≥\$10H a : µ<\$10µ is at least \$10, or it is less H 0 : µ≤\$10H a : µ>\$10 µ is no more than \$10, or it is more We want to know if a coin is fair and balance. H 0 : Probability of each coin toss coming up heads = 0.5. H a : Probability of each coin toss coming up heads ≠ 0.5. Suppose we test this hypothesis by flipping the coin 50 times. What would we conclude? Consider the following hypotheses. The director of a city’s transit system claims that 35% of the system’s riders consists of senior citizens. In a recent study, researchers found that only 23% of the riders were seniors. Is the director’s claim wrong? (Transit example) A parts supplier claims that no more than 20% of the parts he delivers are defective. But a random sample of a recent delivery shows that 25% of the parts were defects. Is the suppliers claim wrong? (Parts example) Transit example Assertion: 35% of the riders are senior citizens Null hypothesis: H 0 : π = 0.35 The null hypothesis is identical to the transit manager’s statement since he claimed an exact value for the parameter. Alternative hypothesis: H a : π ≠ 0.35 If the null is incorrect, the proportion must be something other than 0.35. Parts example Assertion: Not more than 20% of the parts are defective Null hypothesis: H 0 : π ≤ 0.20 Alternative hypothesis: H a : > 0.20 1. State the hypotheses Null Alternate 2. Formulate an analysis plan How to use the sample data Significance level Test method Mean, proportion, difference between means, z-score, t- score, etc. A research firm claims that 62% of women age 40-49 participate in a 401(k) retirement account. We want to test this percentage by randomly selecting a group of 300 women. What are the null and alternative hypotheses? Type I Error A Type I error occurs when the researcher rejects a null hypothesis when it is true. The probability of committing a Type I error = the significance level (α). Type II Error A Type II error occurs when the researcher fails to reject a null hypothesis that is false. The probability of committing a Type II error is called Beta, and is often denoted by β. The probability of not committing a Type II error is called the power of the test. When a robot welder is in adjustment, its mean time to perform a task is 1.325 minutes. Past experience tells us the standard deviation for the cycle time is 0.0396 minutes. Using a recent random sample of 80 jobs, the mean cycle time for the welder was 1.3229 minutes. Do we need to adjust the machine? Our company produces light bulbs with a mean life of 1030.0 hours and a standard deviation of 90.0 hours. We’re approached by a salesman who says he can sell us a process that will extend the life of our bulbs. We decide to test this product and, using a sample of 40 bulbs, we discover that the mean life for our bulbs using the new process is 1061.6 hours. Should we invest in this new process? When set properly, a machine produces nails that are 2 inches long with a standard deviation of 0.070 inches. We took a sample of 35 nails and found that their mean length was 2.025 inches. Using a significance level of 0.01, is the machine properly adjusted? The p-value is the probability that we would get a test statistic as extreme as the one observed, if the null hypothesis was true. In practice, all statistical software packages calculate a p-value and return that value as part of the data. If p-value is ≤ our level of significance, we have evidence to reject H 0 I took a random sample (n=30) from a population with a known standard deviation (σ=1) and a known mean (μ=0) -1.62390.35581.335410.23904-0.35570.03744 -0.08742-0.6836-2.031710.05390.30906-2.37686 -1.59076-0.582991.71301-1.587930.40806-0.2797 -0.58236-1.723331.21996-0.797860.773081.34005 -1.07465-1.15515-0.98725-1.711280.58343-1.74421 I then used a statistical package (Minitab®) to run a 1-sample z-test. Here are my results: Test of mu = 0 vs not = 0 The assumed standard deviation = 1 Variable N Mean StDev SE Mean 95% CI Z C1 30 -0.420281 1.113508 0.182574 ( -0.778120, -0.062442) -2.30 Variable P C1 0.021 What happened? The credit manager of a department store claims that the mean balance for the store’s charge account customers is \$410. An independent auditor selects 18 accounts at random and finds a mean balance of \$511.33 and a standard deviation of \$183.75. The population of account balances is assumed to be normally distributed. Is the credit manager correct? An inventor developed an, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours on one gallon of regular gasoline. Suppose a simple random sample of 50 engines is tested. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. We want to test this claim (α =.05) State the hypotheses Formulate an analysis plan Analyze sample data Interpret results Bon Air Elementary School has 300 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01. State the hypotheses Formulate an analysis plan Analyze sample data Interpret results The sampling method is simple random sampling. Each sample point can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. np ≥ 5 and n(1-p) ≥ 5 The population size is at least 20 times as big as the sample size. The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisfied. Based on these findings, what can we conclude about the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance. Two sample t-test The sampling method for each sample is simple random sampling The samples are independent. Each population is at least 20 times larger than its respective sample. Each sample is drawn from a normal or near-normal population. Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students. At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15. Can we determine if Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.) State the hypotheses Formulate an analysis plan Analyze sample data Interpret results Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 30 78.0 10.0 1.8 2 25 85.0 15.0 3.0 Difference = mu (1) - mu (2) Estimate for difference: -7.00 90% CI for difference: (-12.91, -1.09) T-Test of difference = 0 (vs not =): T-Value = -1.99 P-Value = 0.053 DF = 40 Download ppt "A statistical hypothesis is an assumption about a population parameter. This assumption may or may not be true" Similar presentations
# Get most out of 25 divided by 6 25 divided by 6 numbers 25 divided by 6 may seem like a daunting task, but fear not! Today, we’re here to demystify the process and help you get the most out of one particular division: 25 divided by 6. Whether you’re a student struggling with math homework or simply looking to sharpen your skills, this blog post is for you. We’ll explore the basics of division, share tips and tricks specifically for dividing by 6, dive into real-world applications, highlight common mistakes to avoid, provide practice exercises – all so that you can conquer division like a pro! So grab your pencils and let’s embark on this mathematical journey together. Let’s unlock the secrets hidden within those sixes! ## Understanding Basic Division Understanding Basic Division When it comes to division, it’s essential to have a clear grasp of the fundamentals. At its core, division is simply the process of sharing or distributing a quantity equally among a given number of groups. Think of it as splitting up a pie into equal slices for your friends. To perform basic division, you’ll need two main components: the dividend and the divisor. The dividend represents the total amount you want to divide, while the divisor tells us how many groups we’re dividing it into. In our case, 25 is our dividend and 6 is our divisor. So, what exactly does it mean when we say “25 divided by 6”? It means that we’re dividing twenty-five items into six equal parts or groups. Each group will receive an equal share – but what about those extra pieces? That’s where remainders come in handy! Remainders occur when we can’t distribute everything evenly among all groups. In this case, there will be one leftover item that cannot be divided equally among six groups. By understanding these basic concepts of division – dividends, divisors, quotient (the answer), and remainder – you’ll have a solid foundation for tackling more complex mathematical problems involving division. Now that we’ve covered the basics let’s dive deeper into why having knowledge of division tables is so important! Stay tuned! ## The Importance of Knowing Division Tables Knowing division tables is an essential skill that can have a significant impact on our everyday lives. Whether we realize it or not, division plays a crucial role in various aspects of our daily routine. One of the main reasons why knowing division tables is important is because it helps us solve problems quickly and efficiently. Imagine having to divide 25 by 6 without knowing the division table – it would take much longer to find the answer! By memorizing these tables, we can save time and effort when faced with similar calculations. Additionally, having a solid understanding of division tables allows us to grasp more complex mathematical concepts. Division lays the foundation for higher-level math skills such as fractions, decimals, and ratios. Without a strong knowledge of basic division, it becomes much harder to comprehend and solve advanced mathematical problems. Moreover, knowing division tables can enhance our problem-solving abilities in real-world scenarios. From dividing grocery expenses among friends to calculating cooking measurements or determining how many days’ worth of supplies are left during camping trips – being able to divide quickly and accurately proves invaluable in practical situations. By familiarizing ourselves with division tables early on, we develop mental agility and numeracy skills that extend beyond mathematics alone. These skills include critical thinking, logical reasoning, pattern recognition, and even estimation abilities – all highly transferable across different disciplines and areas of life. In conclusion (as per your instruction), mastering division tables has wide-ranging benefits that extend far beyond simple arithmetic calculations. It enhances problem-solving capabilities both inside and outside the classroom while fostering valuable cognitive skills that serve us well throughout our lives. So let’s embrace this fundamental aspect of mathematics – because knowing how to divide pays dividends! ## Tips and Tricks for Dividing by 6 Tips and Tricks for Dividing by 6 Dividing numbers can sometimes be a challenging task, especially when dealing with unfamiliar divisors like 6. However, there are several tips and tricks that can help make dividing by 6 much easier. One helpful tip is to recognize that any number multiplied by 6 will always end in either 0, 2, 4, 6 or 8. This means that when dividing a number by 6, you can quickly determine if the result will have a remainder or not based on its last digit. For example, if the last digit of the dividend is even (0), then the division will be exact; otherwise, there will likely be a remainder. Another useful trick is to break down larger numbers into smaller factors of both the dividend and divisor. For instance, instead of trying to divide a large number like 48 directly by 6, you can first divide it by smaller factors such as dividing it by both 2 and then again by another factor until you reach an easier calculation. Additionally, knowing your multiplication tables well can greatly assist in dividing efficiently. By memorizing multiples of six up to at least twelve times six (72), you can easily calculate divisions involving six using mental math. These tips and tricks may take some practice to master but utilizing them regularly will undoubtedly improve your ability to divide numbers effectively. So next time you encounter division problems involving six as the divisor – don’t fret! Implement these strategies and watch how quickly and effortlessly you conquer those calculations! ## Real World Applications of Dividing by 6 Real World Applications of Dividing by 6 Dividing by 6 may seem like a basic math skill, but it has practical applications in the real world that we encounter every day. One such application is when dividing resources evenly among a group of people or objects. For example, if you have 25 cookies and want to divide them equally among 6 friends, knowing how to divide by 6 will help you determine that each person gets 4 cookies with one remaining. Another situation where dividing by 6 comes in handy is when calculating rates or proportions. Let’s say you’re planning a road trip and need to know how many gallons of gas your car will require for every six hours of driving. By dividing the total distance you plan to travel by the number of hours per interval (in this case, 6), you can easily calculate your fuel consumption rate. Furthermore, dividing by 6 can be useful in financial calculations. If you receive \$150 as part of your monthly allowance and want to save one-sixth of it, knowing division skills allows you to quickly determine that \$25 should be set aside for savings. In the field of cooking, understanding division tables can be beneficial as well. Recipes often call for specific measurements divided into portions suitable for serving sizes. When scaling recipes up or down according to the number of servings needed, being able to divide accurately helps ensure proper ingredient amounts are used. Moreover, businesses frequently use division when analyzing data or calculating sales figures. Retailers may calculate average revenue per day over a six-day period using this fundamental operation. These are just a few examples showcasing how valuable it is to know how to divide numbers effectively not only within an academic setting but also in everyday situations we encounter regularly. ## Common Mistakes to Avoid When Dividing by 6 Common Mistakes to Avoid When Dividing by 6 When it comes to dividing by 6, there are a few common mistakes that people often make. By being aware of these errors and learning how to avoid them, you can improve your division skills and get the most out of this mathematical operation. One common mistake is forgetting to simplify the fraction when dividing. Sometimes, the quotient may be expressed as a fraction instead of a whole number. To prevent this error, always simplify the result if possible. Another mistake to watch out for is accidentally switching the dividend and divisor. Remember that in division, you divide the dividend (the number being divided) by the divisor (the number doing the division). Double-checking your calculations can help prevent this slip-up. It’s also important not to ignore any remainders when dividing by 6. If there is a remainder after dividing, it should be included in your answer or converted into a decimal or fraction depending on what is appropriate for the situation. Some people may struggle with keeping track of their place value while performing long division with numbers divisible by 6. This can lead to errors in both digit placement and calculation accuracy. Take your time and ensure each step is done correctly. Rounding errors can occur if you round too early during division calculations involving decimals or fractions. It’s best practice to work with exact values until reaching an appropriate point for rounding. By being mindful of these common pitfalls when dividing by 6, you’ll enhance your understanding of this operation and minimize errors along the way! So keep practicing and soon enough you’ll become proficient at divvying up numbers evenly using this method! ## Practice Exercises to Improve Division Skills If you want to improve your division skills and get the most out of 25 divided by 6, practice is key! By regularly engaging in division exercises, you can sharpen your mathematical abilities and become more comfortable with dividing numbers. One effective exercise to enhance division skills is practicing long division problems. Start by solving simple equations, gradually progressing to more complex ones. This will allow you to develop a systematic approach and gain confidence in tackling any division problem that comes your way. Another useful exercise is using manipulatives or visual aids when dividing. For example, you can use blocks or counters to represent numbers and physically divide them into groups of six. This hands-on approach helps reinforce the concept of division and makes it easier to visualize the process. Additionally, timed drills can be an excellent way to build speed and accuracy in division. Set a timer for a specific duration (such as five minutes) and challenge yourself to solve as many division problems as possible within that time frame. Furthermore, incorporating real-life scenarios into your practice sessions can make learning more enjoyable. Try applying division skills while splitting up quantities of food or sharing items among friends or family members. Don’t forget about online resources! There are numerous websites offering interactive games and quizzes specifically designed for practicing divisions at different levels of difficulty. By regularly engaging in these practice exercises, you’ll not only strengthen your understanding of dividing by 6 but also improve overall math proficiency. So grab a pencil or fire up that computer – it’s time to conquer those divisions! ## Conclusion Conclusion In this article, we have explored the world of division and specifically focused on dividing by 6. Understanding basic division is essential for developing strong mathematical skills, and knowing the division tables can greatly enhance your ability to solve problems quickly and efficiently. We have also discussed various tips and tricks that can help you divide by 6 more effectively. Whether it’s recognizing patterns or using mental math strategies, these techniques can make the process much simpler. Furthermore, we have highlighted real-world applications where dividing by 6 comes in handy. From calculating ratios to solving everyday problems involving 25 divided by 6 sharing or distributing items, this skill has practical uses beyond just numbers on a page. To truly master dividing by 6 (and any other number), it’s important to be aware of common mistakes to avoid. By double-checking calculations and being mindful of potential pitfalls like not reducing fractions properly, you can ensure accurate results every time. Practice exercises are crucial for improving your division skills. The more you practice dividing by 6 with different numbers and scenarios, the more comfortable you will become with the process. So don’t shy away from challenging yourself! Remember that becoming proficient in dividing by 6 takes time and effort but is well worth it in the end. With consistent practice and application in real-life situations, you will soon find yourself effortlessly performing divisions involving this number. So go ahead – dive into the world of division! Strengthen your mathematical abilities through mastering how to divide efficiently by exploring its 25 divided by 6 many facets – including diving into other divisors as well! Happy practicing!
You are on page 1of 10 # Solving Compound Interest Problems What is Compound Interest? If you walk into a bank and open up a savings account you will earn interest on the money you deposit in the bank. If the interest is calculated once a year then the interest is called simple interest. If the interest is calculated more than once per year, then it is called compound interest. Compound Interest Formula The mathematical formula for calculating compound interest depends on several factors. These factors include the amount of money deposited called the principal, the annual interest rate (in decimal form), the number of times the money is compounded per year, and the number of years the money is left in the bank. These factors lead to the formula FV = future value of the deposit P = principal or amount of money deposited r = annual interest rate (in decimal form) n = number of times compounded per year t = time in years. Example 1: If you deposit \$4000 into an account paying 6% annual interest compounded quarterly, how much money will be in the account after 5 years? Plug in the giving information, P = 4000, r = 0.06, n = 4, and t = 5. Use the order or operations to simplify the problem. If the problem has decimals, keep as many decimals as possible until the final step. FV = 5387.42 Round your final answer to two decimals places. After 5 years there will be \$5387.42 in the account. n t r FV P 1 n = + 4(5) 0.06 FV 4000 1 4 = + 20 FV 4000(1.015) FV 4000(1.346855007) Example 2: If you deposit \$6500 into an account paying 8% annual interest compounded monthly, how much money will be in the account after 7 years? Plug in the giving information, P = 6500, r = 0.08, n = 12, and t = 7. Use the order or operations to simplify the problem. If the problem has decimals, keep as many decimals as possible until the final step. FV = 11358.24 Round your final answer to two decimals places. After 7 years there will be \$11358.24 in the account. Example 3: How much money would you need to deposit today at 9% annual interest compounded monthly to have \$12000 in the account after 6 years? Plug in the giving information, FV = 12000, r = 0.09, n = 12, and t = 6. Use the order or operations to simplify the problem. If the problem has decimals, keep as many decimals as possible until the final step. P = 7007.08 Divide and round your final answer to two decimals places. You would need to deposit \$7007.08 to have \$12000 in 6 years. Example 4: If you deposit \$5000 into an account paying 6% annual interest compounded monthly, how long until there is \$8000 in the account? Plug in the giving information, FV = 8000, P = 5000, r = 0.06, and n = 12. Use the order or operations to simplify the problem. Keep as many decimals as possible until the final step. Divide each side by 5000. Take the logarithm of each side. Then use Property 5 to rewrite the problem as multiplication. Divide each side by log 1.005. 94.23553232 12t Use a calculator to find log 1.6 divided by log 1.005. t 7.9 Finish solving the problem by dividing each side by 12 and round your final answer. It will take approximately 7.9 years for the account to go from \$5000 to \$8000. Example 5: If you deposit \$8000 into an account paying 7% annual interest compounded quarterly, how long until there is \$12400 in the account? Plug in the giving information, FV = 12400, P = 8000, r = 0.07, and n = 4. Use the order or operations to simplify the problem. Keep as many decimals as possible until the final step. Divide each side by 8000. Take the logarithm of each side. Then use Property 5 to rewrite the problem as multiplication. Divide each side by log 1.0175. 25.26163279 4t Use a calculator to find log 1.55 divided by log 1.0175. t 6.3 Finish solving the problem by dividing each side by 4 and round your final answer. It will take approximately 6.3 years for the account to go from \$8000 to \$12400. Example 6: At 3% annual interest compounded monthly, how long will it take to double your money? At first glance it might seem that this problem cannot be solved because we do not have enough information. It can be solved as long as you double whatever amount you start with. If we start with \$100, then P = \$100 and FV = \$200. Plug in the giving information, FV = 200, P = 100, r = 0.03, and n = 12. Use the order or operations to simplify the problem. Keep as many decimals as possible until the final step. Divide each side by 100. Take the logarithm of each side. Then use Property 5 to rewrite the problem as multiplication. Divide each side by log 1.0025. 277.6053016 12t Use a calculator to find log 2 divided by log 1.0025. t 23.1 Finish solving the problem by dividing each side by 12 and round your final answer. At 3% annual interest it will take approximately 23.1 years to double your money Practice Problems Now it is your turn to try a few practice problems on your own. Work on each of the problems below and then click on the link at the end to check your answers. Problem 1: If you deposit \$4500 at 5% annual interest compounded quarterly, how much money will be in the account after 10 years? Problem 2: If you deposit \$4000 into an account paying 9% annual interest compounded monthly, how long until there is \$10000 in the account? Problem 3: If you deposit \$2500 into an account paying 11% annual interest compounded quarterly, how long until there is \$4500 in the account? Problem 4: How much money would you need to deposit today at 5% annual interest compounded monthly to have \$20000 in the account after 9 years? Problem 5: If you deposit \$6000 into an account paying 6.5% annual interest compounded quarterly, how long until there is \$12600 in the account? Problem 6: If you deposit \$5000 into an account paying 8.25% annual interest compounded semiannually, how long until there is \$9350 in the account Problem If you deposit \$4500 into an account paying 7% annual interest compounded semi anualy , how much money will be in the account after 9 years? Result The amount is \$8358.7 and the interest is \$3858.7. Explanation STEP 1: To find amount we use formula: A = total amount P = principal or amount of money deposited, A=P(1+rn)nt ## r = annual interest rate n = number of times compounded per year t = time in years ## P=\$4500 , r=7% , n=2 and t=9 years After plugging the given information we have AAAA=4500(1+0.072)29=45001.03518=45001.857489=8358.7 STEP 2: To find interest we use formula A=P+I, since A=8358.7 and P = 4500 we have: A8358.7II=P+I=4500+I=8358.74500=3858.7 Problem If you deposit \$4500 into an account paying 7% annual interest compounded semi anualy , how much money will be in the account after 9 years? Result The amount is \$8358.7 and the interest is \$3858.7. Explanation STEP 1: To find amount we use formula: A=P(1+rn)nt A = total amount P = principal or amount of money deposited, ## r = annual interest rate n = number of times compounded per year t = time in years In this example we have ## P=\$4500 , r=7% , n=2 and t=9 years After plugging the given information we have AAAA=4500(1+0.072)29=45001.03518=45001.857489=8358.7 STEP 2: To find interest we use formula A=P+I, since A=8358.7 and P = 4500 we have: A8358.7II=P+I=4500+I=8358.74500=3858.7 Problem How much money would you need to deposit today at 8% annual interest compounded monthly to have \$1200 in the account after 12 years? Result The principal is \$460.72. Explanation To find amount we use formula: A = total amount P = principal or amount of money deposited, A=P(1+rn)nt ## r = annual interest rate n = number of times compounded per year t = time in years ## A=\$1200 , r=8% , n=12 and t=12 years After plugging the given information we have 120012001200PP=P(1+0.0812)1212=P1.00667144=P2.604631=12002.604631=460. 72 Problem How much money would you need to deposit today at 12% annual interest compounded quarterly to have \$1000 in the account after 2 years? Result The principal is \$789.41. Explanation To find amount we use formula: A=P(1+rn)nt A = total amount P = principal or amount of money deposited, r = annual interest rate n = number of times compounded per year t = time in years In this example we have ## A=\$1000 , r=12% , n=4 and t=2 years After plugging the given information we have 100010001000PP=P(1+0.124)42=P1.038=P1.26677=10001.26677=789.41 Problem Suppose that a savings account is compounded yearly with a principal of \$1700. After 2 years years, the amount increased to\$1910. What was the per annum interest rate? Result Interest rate per anum was 6%. Explanation To find interest rate we use formula: A = total amount P = principal or amount of money deposited, A=P(1+rn)nt ## r = annual interest rate n = number of times compounded per year t = time in years ## A=\$1910 , P=\$1700,t=2 years and n=1 After plugging the given information we have 19101910(1+r1)2(1+r1)2ln(1+r1)22ln(1+r1)ln(1+r1)ln(1+r1)1+r11+r1r1 r=1700(1+r1)12=1700(1+r1)2=19101700=1.12353 of each ## Take the natural logarithm side=ln(1.12353)=ln(1.12353)=ln(1.12353)2=0.05824=e0.05824=1.05996935783=0.0599 693578315=0.05996935783156%
# What is 2/3 as a decimal? How to show 2/3 as a decimal? It’s is very easy. Follow this article to know more. The process of conversion of the fraction to a decimal is a very important step and you must learn about it. Here we shall take the simple example of the fraction 2/3 and convert it to its decimal value. Let us learn the method of converting a given fraction to its decimal form. This method can be replicated to convert any fractions to decimals. ## Basic concepts of converting a fraction to a decimal 1) A fraction is made of two parts, the numerator, and the denominator. 2) The numerator refers to the digits that are present above the line of division. 3) A denominator is a digit that is present below the line of the fraction. 4) When we are converting a fraction to a decimal, we need to initiate the process of division where the numerator becomes the dividend and the denominator becomes the divisor. 5) So in the given fraction, 2 is the numerator and hence it becomes the dividend, and 3 the divisor. What is 2/3 as a Decimal? The answer is 2/3 = 0.66667 ## Calculations to show 2/3 as a decimal The calculations of converting a fraction to decimal are very simple and involve the basic step of the division of the numerator by the denominator. Thus, 2÷3 =0.66667 #### 2/3 = 0.66667 Hence we obtain the value as a recurring decimal digit. There are no complex steps that are involved and thus with some practice, you will be able to solve the problems of decimal conversions easily. Also read: What is 3/4 as a Decimal? ## What is the need for you to convert a fraction to a decimal? 1) The process of converting a fraction to a decimal is a better way of representing the answer in a better way. 2) The idea of decimals is very important in daily life as you might be presented with an item in the shop and the quantity might be measured or expressed in decimals. Thus you need to know about decimals. Further, the price may also be given decimal, in such case, we can say that both currency and weight might be expressed in decimals and you should be clear about the concept. 3) When you are adding a certain ingredient to a particular recipe you are cooking, the amount might be mentioned in the form of a fraction in the cookbook. In such a case you need to know about how to change the fractional value to decimals. 4) Further, even while comparing the value of two fractions and determining which one is greater or smaller you need to know how to first change it to decimals and then compare the values of the digit. Hence we can say that the knowledge of converting a fraction to a decimal is very important and you must make sure that you practice it well to learn the steps. Also read: What is 1/2 as a decimal?
Forming Linear Equations - Two digit number Chapter 2 Class 8 Linear Equations in One Variable Concept wise To do this question, first check Two digit number and its reverse . ### Transcript Question 3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?Let original number be ab Let the digit at tens place = a & digit at units place = b Given that Sum of digits of a 2-digit number = 9 a + b = 9 b = 9 a Original Number ab Digit at units place = b = 9 a Digit at tens place = a Original number = (10 Digit at tens place) + (1 Digit at units place) = (10 a) + (1 (9 a)) = 10a + 9 a = 9a + 9 Reverse number ba Digit at units place = a Digit at tens place = b = 9 a Resulting number = (10 Digit at tens place) + (1 Digit at units place) = 10 (9 a) + (1 a) = 90 10a + a = 90 9a Hence, Original number = 9a + 9 Resulting number = 90 9a Given that, Resulting number is greater than original number by 27. Resulting number = Original number + 27 (90 9a) = (9a + 9) + 27 90 9a = 9a + 9 + 27 90 27 9 = 9a + 9a 54 = 18a 54/18 = a 3 = a a = 3 Therefore, Digit at tens place = a = 3 & Digit at units place = b = 9 a = 9 3 = 6 Hence, Original number = 36
Get Dream IIT in Drop Year \| Up to 70% OFF \| Limited Seats # NCERT Solutions For Class 9 Maths Chapter 3 Exercise 3.2 Coordinate Geometry - PDF Download Class 9 NCERT solutions for class 9 maths chapter 3 ex 3.2 Coordinate Geometry is very helpful for class 9 students to learn the cartesian system. It provides you an in-depth exploration to learn more about the Cartesian plane, quadrants, coordinate axes etc. Our subject matter experts have prepared these NCERT solutions for class 9 maths based on extensive research on the subject.The best way to study for the exam is to use these NCERT solutions. NCERT solutions class 9 maths chapter 3 ex 3.2 is composed of two questions that are related to the plotting of points on the cartesian plane. NCERT solutions make it easy for students to understand this concept by giving them a thorough overview of all the main topics and their related subtopics. Download NCERT solutions for class 9 maths chapter 3 ex 3.2 free PDF from eSaral’s expert teacher. ## Topics Covered in Exercise 3.2 class 9 Mathematics Questions Ex 3.2 discusses the important topic of the cartesian system and its related concepts. You can check the detailed explanations mentioned below. 1 Cartesian System 2 Important Points Origin Coordinate Axes and Quadrants Point in Different Quadrants 1. Cartesian System - The Cartesian system describes a point’s location in a plane. In a cartesian system, a point is defined by reference to two perpendicular lines. The horizontal line in the X-axis is called XX’ and the vertical line is called YY’. 2. Important Points • Origin - The point of origin, denoted by the letter 'O', is the point where the vertical and horizontal lines intersect. Positive directions correspond to OX and OY, while negative directions correspond to OX' and OY'. • Coordinate Axes and Quadrants - The plane is divided into four parts by the axes X and Y. The four quadrants of the plane are numbered as I, II, III, IV anti-clockwise from OX. The Cartesian plane, also called coordinate plane or XY plane, is made up of these axes and quadrants. These axes are also referred to as coordinate axes. • Points in Different Quadrants 1. The first quadrant - coordinates of a point are of the form (+ , +) 2. The second quadrant - coordinates of a point are of the form (–, +) 3. The third quadrant - coordinates of a point are of the form (–, –) 4. The fourth quadrant - coordinates of a point are of the form (+, –) where ‘+’ denotes a positive real number and ‘–’ denotes a negative real number. ## Tips for Solving Exercise 3.2 class 9 chapter 3 Coordinate Geometry NCERT solutions class 9 maths chapter 3 ex 3.2, students can easily learn how to plot points and objects on the cartesian plane. You can go through these tips to get a better understanding of concepts and questions. 1. These NCERT solutions cover both theory and practice questions, making it easy for students to learn this important subject. 1. Once students are familiar with the concepts of the cartesian system, they can understand the other difficult concepts using the NCERT solutions class 9 maths chapter 3 ex 3.2. 2. Students can also download the solutions PDF of ex 3.2 provided on eSaral website. This will help them to gain a better comprehension of the coordinate system as well as other related terms. 3. By solving the questions of ex 3.2, you will get an amazing score in exams. Students should make sure to check out the solutions and solve the examples provided in NCERT solutions to get a good grasp of the concepts. ## Importance of Solving Ex 3.2 class 9 Maths chapter 3 Coordinate Geometry By solving these questions, you will be able to gain a better understanding of the concepts. You will also get to know some other advantages include: 1. In NCERT solutions provided by our subject specialists, you will be able to understand the various types of questions asked in ex 3.2. 2. NCERT solutions give a comprehensive understanding to learn cartesian system concepts with ease. 3. You will get enough practice to solve the questions. 4. The solutions are accessible in PDF format, allowing students to access them anytime. #### Frequently Asked Questions Question 1. What is the cartesian plane in NCERT solutions class 9 maths chapter 3 ex 3.2? Answer 1. The cartesian plane is also known as a two dimensional coordinate plane that is defined as the plane where the x-axis is perpendicular to the y-axis. This point is called the origin of the cartesian plane. Question 2. What are the components of a Cartesian plane? Answer 2. The cartesian plane is composed of three different components. These components are essential when attempting to locate a specific point on the plane or construct a graph for a function. Here, we provide a comprehensive list of the components of a Cartesian plane. Origin - Origin is defined as the point where the vertical and horizontal lines intersect. The coordinate points of origin, denoted by (o,o). Axes - The axes are the horizontal and vertical lines that intersect to form the cartesian coordinate system. The horizontal line is the x-axis and the vertical line is the y-axis. Quadrants - When the axes of x and y intersect, the cartesian plane divides into four parts called quadrants. These quadrants can be infinite. Click here to get exam-ready with eSaral
Grade 5 Math Operations in Base Ten \| Student Handouts Grade 5: Math - Operations in Base Ten www.studenthandouts.com > Grade 5 > Grade 5 Math > Grade 5: Number & Operations: Base Ten CCSS.MATH.CONTENT.5.NBT.A.1 - Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. CCSS.MATH.CONTENT.5.NBT.A.2 - Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. CCSS.MATH.CONTENT.5.NBT.A.3 - Read, write, and compare decimals to thousandths. CCSS.MATH.CONTENT.5.NBT.A.3.A - Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). CCSS.MATH.CONTENT.5.NBT.A.3.B - Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. CCSS.MATH.CONTENT.5.NBT.A.4 - Use place value understanding to round decimals to any place. CCSS.MATH.CONTENT.5.NBT.B.5 - Fluently multiply multi-digit whole numbers using the standard algorithm. CCSS.MATH.CONTENT.5.NBT.B.6 - Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. CCSS.MATH.CONTENT.5.NBT.B.7 - Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
### Transcript Now we'll talk about the Quadratic Formula. Now, of course, this is out of place in some sense because, mostly when we're talking about quadratics, that was back in the algebra section. We were talking about factoring. Usually, if you have to solve a quadratic, the best method will be factoring it, but, sometimes, quadratic equations cannot be factored, and at least one option for these is to use the Quadratic Formula. There's another method called completing the square, which I have demonstrated in a few lessons. If the quadratic resembles the square of a binomial, and you should know those patterns, the square of the sum, the square of a difference, if it's close to one of those patterns, it's often pretty easy to solve it without the Quadratic Formula. However, there are some quadratics that are not factorable, and you can't really fit it very easily into one of those neat algebraic patterns, so then the Quadratic Formula is often your best bet. So this is the Quadratic Formula. First of all, keep in mind, it is an If, Then statement. If ax squared + bx + c = 0, so notice that is a quadratic equation set equal to 0, so that's in standard form. If we put that quadratic into standard form and read off the coefficients, then the solution for x will follow that familiar pattern, that familiar formula. Notice, also, there's a +/- sign in that formula, typically, we'll get two roots. Remember that the graph of a quadratic is a parabola, and many parabolas intersect the x-axis twice, so that's why you'd get two roots. You do not, do not need to memorize this formula. The test will always provide this formula if you need it, and, once again, most of the time, with a quadratic, your best bet is to factor it. You don't need this formula. The only time the test will give you this formula if something is utterly unfactorable, so just keep that in mind. In particular, notice the expression under the radical in the quadratic formula, b squared- 4ac. This is sometimes called the discriminant. That's a term you do not need to know for the ATC, but that little expression, b squared- 4ac, that's very important. The reason is, if it's positive, then we get two real square roots, and then we're gonna wind up with two different values for x. It's gonna be negative b plus something and negative b minus something over 2a. We're gonna get two roots. In rare cases, when b squared- 4ac= 0, then the quadratic has one real root. And you'll notice, for any perfect square, if you go back and look at the square of a sum or square of a difference from the algebra section, all of these obey this condition that b squared- 4ac = 0. So this would be a parabola that is simply tangent to the x-axis at its vertex, so it has only one solution, and, of course, this expression can also be negative. Well, think about that. If it's negative, this is something under the square root, so we have a square root of a negative. That would be an imaginary number. We'd get two imaginary solutions, and, of course, as always, what you'd get is some real number plus or minus some imaginary number. The two roots would be two complex conjugates. It is very unlikely that the ACT is gonna give you a quadratic formula that's going to wind up having an imaginary root, but it could happen. So it's just something to keep in mind. Here's a very simple example. So there's a quadratic equation. It's already in standard form. It's already set equal to zero, and we're gonna solve for x. So it's not immediately obvious how we would factor that or use completing the square, so the quadratic formula's actually not a bad choice with this equation. So we can see that a = 1, b and c = -1. Very important to remember those negative signs when you're reading off a, b and c for the quadratic formula. So the quadratic formula, which the test would give us, we'd plug in these values. Under the radical, we get a root 5, and so we have two roots here. The two roots, one of these 1 + root 5 over 2. The other is 1- root 5 over 2. Incidentally, that first root is the golden ratio, and the second root is the reciprocal of the golden ratio, but you do not need to know that for the ACT. Here's another example. We're gonna solve this. Now, notice this one is not in standard form, so step one is always put things in standard form, so we have to put it in standard form. We subtract 12 from both sides, we subtract 4. As it turns out, if I were gonna solve this, I would actually say that this is very, very close to a completing the square problem, and I would it solve it that way, but let's solve it with the quadratic formula. So, we get a = 1, b = -12, c = 31. Plug in all those numbers, and 4 times 31, well, 4 times 30 is 120, so 4 times 31 is 124. We subtract that. We get 12 + root 20 over 2. Now remember the lessons that we had on radicals. We can simplify 20 because 20 is 4 times 5, and 4 is a perfect square, so we can separate that into square root of 4 times square root of 5, and the square root of 4 is 2. Well, now everything in the numerator's divisible by 2, so we can cancel the 2, and we get 6 plus or minus root 5, and that is the solution to this equation. 6 + root 5 and 6 + root 5, those are the two roots. Here's a practice problem, pause the video, and then we'll talk about this. Okay, so this is in the form that the ACT would state it, they give us the quadratic formula, they state all that very clearly, and then they ask us to solve the problem. Now, of course, they did slip us the trick here. They gave us an equation that's not in standard form, it's not equal to 0, so we just have to subtract 3 from both sides and get it equal to 0. So now we have something in standard form. Incidentally, if you wanted to solve this with completing the square, if you wanted to add whatever you needed to add to get the perfect square, the square of a difference. That's also a perfectly valid way to solve it. I'll just point out, don't feel compelled to use the quadratic formula if another method of solution is easier for you, but here I'll demonstrate the quadratic formula. We plug everything in. Of course, 4(7) is 28, 36- 28 is 8. The square root of 8 can be simplified because 8 is 4 x 2, so the square root of 8 is the square root of 4 times square root of 2, or in other words 2 root 2. Then we can divide everything by 2, and we get 3 +/- root 2. So that's the solution to this particular equation. We go back to the problem, and we select answer choice B. In summary, we can find the solution of an unfactorable quadratic using the quadratic formula. Now I wanna emphasis once again, it's not your only option. In fact, completing the square is often a much quicker, more efficient option, but you certainly can use the quadratic formula. When you need to use the quadratic formula, the ACT will always supply it. You do not need to memorize it. You must make sure that the quadratic equation you are solving is in standard form before you read a, b,and c to plug into the quadratic formula.
Two things: the somewhat random-looking picture above is from The Moon Is Blue, a 1953 film that’s the first known reference to the ‘dating rule’ discussed here. Secondly, I don’t make any judgement about the validity of the ‘dating rule’ - you might find it a useful rule of thumb or a ludicrous restriction; I find it a nice thing to do some algebra on. There is a so-called rule about dating: the youngest age you are supposed to date is half your age plus seven. So, if you’re 16, the youngest age you should consider is 15 – because 16 divided by two is eight and 8+7 equals 15. We can write the dating rule as an equation: $y=x \div 2 +7$. In this equation, $x$ is your age and $y$ is the minimum age you are supposed to date. Why would we want to do this? Well, it means we can work out some other implications of the rule. For instance, you might want to know what the oldest age you could reasonably date would be. The rule doesn’t say anything about that, but it does also apply to the other person. So if you’re 13, you can date anyone over $13 \frac{1}{2}$. But the $13 \frac{1}{2}$ year old can only date people who are at least $13 \frac{3}{4}$ – so you can date them, but they can’t date you. How do we work that out? Well, we need to rearrange the equation. In this case, we want to solve for $x$. We take away seven from each side to get $y-7=x \div 2$, and then double both sides to get $2y-14=x$. That means, to find the oldest age you can date, you have to double your age and take away 14. So, if you’re 16, the minimum age you’re supposed to date is 15; the oldest is 18. If you’re 13, you’re out of luck. The youngest you can date is 13 1/2, but the oldest is 12. So even somebody your own age is simultaneously too old and too young for you. Nightmare! It turns out that the minimum age this rule allows dating is 14 – for extra credit, can you explain why?
# What does mean mode and range mean Mean, Median, Mode, and Range # What does mean mode and range mean Mean, Median, Mode, Range Calculator The mode is the number that is repeated more often than any other, so 13 is the mode. The largest value in the list is 21, and the smallest is 13, so the range is 21 – 13 = 8. mean: Range is equal to maximum value minus minimum value which gives us: 12 ? 2 = Example 3: Find the mean, median, mode and range for the following list of values. To determine the value of the mean, obtain the total of all the numbers and then divide by the number of numbers in the list. The word mean, which is a homonym for multiple other words in the What does mean mode and range mean language, is similarly ambiguous even in the area of mathematics. Depending on the context, whether mathematical or statistical, what is meant by the "mean" changes. In its simplest mathematical definition regarding data sets, the mean used is the arithmetic mean, also referred to as mathematical expectation, or average. In this form, the mean refers to an intermediate value between a discrete set of numbers, namely, the sum of all values in the data set, divided by the total number of values. The equation for calculating an arithmetic mean is virtually identical to that for calculating the statistical concepts of population and sample mean, with slight variations in the variables used:. Given the data set 10, 2, 38, 23, 38, 23, 21, applying the summation above yields:. As previously mentioned, this is one of the simplest definitions of the mean, and some others include the weighted arithmetic mean which only differs in that what does mean mode and range mean values in the data set contribute more value than othersand geometric mean. Proper understanding of given situations and contexts can often provide a person with the tools necessary to determine what statistically relevant method to use. In general, mean, median, mode and range should ideally all be computed and analyzed for a given sample or data set since they elucidate different aspects of the given data, and if considered alone, can lead to misrepresentations of the data, as will be demonstrated in the what happens after summary judgement sections. The statistical concept of the median is a value that divides a data sample, population, or probability distribution into two halves. Finding the median essentially involves finding the value in what airlines fly to santorini greece data sample that has a physical location between the rest of the numbers. Note that when calculating the median of how to draw kuchiki byakuya finite list of numbers, the order of the data samples is important. Conventionally, the values are listed in ascending order, but there is no real reason that listing the values in descending order would provide different results. In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values. While this can be confusing, simply remember that even though the median sometimes involves the computation of what is early intrauterine pregnancy mean, when this case arises, it will involve only the two what are common tax deductions values, while a mean involves all the values how to get password from computer the data sample. In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same. Given the same data set as before, the median would be acquired in the following manner:. After listing the data in ascending order, and determining that there are an odd number of values, it is clear that 23 is the median given this case. If there were another value added to the data set:. Since there are an even number of values, the median will be the average of the two middle numbers, in this case 23 and 23, the mean of which is Note that in this particular data set, the addition of an outlier a value well outside the expected range of valuesthe value 1,, has no real effect on the data set. If however the mean is computed for this data set, the result isThis value is clearly not a good representation of the seven other values in the data set that are far smaller and closer in value than the average and the outlier. This is the main advantage of using the median in describing statistical data when compared to the mean. While both, as well as other statistical values, should be calculated when describing data, if only one can be used, the median can provide how to get cbse 10th marksheet duplicate better estimate of a typical value in a given data set when there are extremely large variations between values. In statistics, the mode is the value in a data set that has the highest number of recurrences. It is possible for a data set to be multimodal, meaning that it has more than one mode. For example:. Similarly to mean and median, the mode is used as a way to express information about random variables and populations. Unlike mean and median however, the mode is a concept that can be applied to non-numerical values such as the brand of tortilla chips most commonly purchased from a grocery store. For example, when comparing the brands Tostitos, Mission, and XOCHiTL, if it is found that in the sale of tortilla chips, XOCHiTL is the mode and sells in a ratio compared to Tostitos and Mission brand tortilla chips respectively, the ratio could be used to determine how many bags of each brand to stock. In the case where 24 bags of tortilla chips sell during a given period, the store would stock 12 bags of XOCHiTL chips, 8 of Tostitos, and 4 of Mission if using the mode. If however the store simply used an average and sold 8 bags of each, it could potentially lose 4 sales if a customer desired only XOCHiTL chips and not any other brand. As is evident from this example, it is important to take all manners of statistical values into account when attempting to draw conclusions about any data sample. The range of a data set in statistics is the difference between the largest and the smallest values. While range does have different meanings within different areas of statistics and mathematics, this is its most basic definition, and is what is used by the provided calculator. Using the same example:. The range in this example is Similarly to the mean, range can be significantly affected by extremely large or small values. Using the same example as previously:. The range in this case would what does mean mode and range mean 1, compared what does mean mode and range mean 36 in the previous case. As such, it is important to extensively analyze data sets to ensure that outliers are accounted for. Financial Fitness and Health Math Other. Purplemath Dec 22, · The terms mean, median, mode, and range describe properties of statistical distributions. In statistics, a distribution is the set of all possible values for terms that represent defined events. The value of a term, when expressed as a variable, is called a random variable. There are two major types of statistical distributions. Sep 21, · Mean, median and mode are different ways of determining the average from a set of numbers. Range gives the difference between the highest and lowest values. The mean takes the sum of all the values, then divides the total by the number of values given. The resulting average will not necessarily be equal to any of the values previously given and may contain a decimal. Mean, median, and mode are all types of averages. Together with range, they help describe the data. Mean, median and mode are different ways of determining the average from a set of numbers. Range gives the difference between the highest and lowest values. The mean takes the sum of all the values, then divides the total by the number of values given. The resulting average will not necessarily be equal to any of the values previously given and may contain a decimal. The mean is commonly taught as being the standard average. The median is the middle value within a set of values arranged in numerical order. When a set has an even number of values, the median is determined by calculating the mean between the two most central values. The mode is the number that is repeated most often within a set of values. If no numbers are repeated, then the set is determined to have no mode. A set can also have multiple modes if multiple numbers are equally repeated. The range subtracts the smallest value from the largest value within the set. Range is useful to determine how spread out all the variables are. However, a single outlier within the set can affect the result in a way that does not accurately reflect the data. More From Reference. What Are the Different Departments of a Bank? More articles in this category: <- How to get rid of blackheads with glue stick - What is work cited page-> ## 2 thoughts on “What does mean mode and range mean” 1. Vudobar: FBE the next Mortal Combat reactors tournament if it happens.
# Line Graphs ## Display given data as linear change on coordinate graphs. Estimated7 minsto complete % Progress Practice Line Graphs MEMORY METER This indicates how strong in your memory this concept is Progress Estimated7 minsto complete % Line Graphs ### Line Graphs Before you continue to explore the concept of representing data graphically, it is very important to understand the meaning of some basic terms that will often be used in this concept. The first such definition is that of a variable. In statistics, a variable is simply a characteristic that is being studied. This characteristic assumes different values for different elements, or members, of the population, whether it is the entire population or a sample. The value of the variable is referred to as an observation, or a measurement. A collection of these observations of the variable is a data set. Variables can be quantitative or qualitative. A quantitative variable is one that can be measured numerically. Some examples of a quantitative variable are wages, prices, weights, numbers of vehicles, and numbers of goals. All of these examples can be expressed numerically. A quantitative variable can be classified as discrete or continuous. A discrete variable is one whose values are all countable and does not include any values between 2 consecutive values of a data set. An example of a discrete variable is the number of goals scored by a team during a hockey game. A continuous variable is one that can assume any countable value, as well as all the values between 2 consecutive numbers of a data set. An example of a continuous variable is the number of gallons of gasoline used during a trip to the beach. A qualitative variable is one that cannot be measured numerically but can be placed in a category. Some examples of a qualitative variable are months of the year, hair color, color of cars, a person’s status, and favorite vacation spots. The following flow chart should help you to better understand the above terms. Variables can also be classified as dependent or independent. When there is a linear relationship between 2 variables, the values of one variable depend upon the values of the other variable. In a linear relation, the values of \begin{align}y\end{align} depend upon the values of \begin{align}x\end{align}. Therefore, the dependent variable is represented by the values that are plotted on the \begin{align}y\end{align}-axis, and the independent variable is represented by the values that are plotted on the \begin{align}x\end{align}-axis. Linear graphs are important in statistics when several data sets are used to represent information about a single topic. An example would be data sets that represent different plans available for cell phone users. These data sets can be plotted on the same grid. The resulting graph will show intersection points for the plans. These intersection points indicate a coordinate where 2 plans are equal. An observer can easily interpret the graph to decide which plan is best, and when. If the observer is trying to choose a plan to use, the choice can be made easier by seeing a graphical representation of the data. #### Describing Variables Select the best descriptions for the following variables and indicate your selections by marking an ‘\begin{align}x\end{align}’ in the appropriate boxes. Variable Quantitative Qualitative Discrete Continuous Number of members in a family A person’s marital status Length of a person’s arm Color of cars Number of errors on a math test The variables can be described as follows: Variable Quantitative Qualitative Discrete Continuous Number of members in a family \begin{align}x\end{align} \begin{align}x\end{align} A person’s marital status \begin{align}x\end{align} Length of a person’s arm \begin{align}x\end{align} \begin{align}x\end{align} Color of cars \begin{align}x\end{align} Number of errors on a math test \begin{align}x\end{align} \begin{align}x\end{align} #### Creating a Table of Values Sally works at the local ballpark stadium selling lemonade. She is paid $15.00 each time she works, plus$0.75 for each glass of lemonade she sells. Create a table of values to represent Sally’s earnings if she sells 8 glasses of lemonade. Use this table of values to represent her earnings on a graph. The first step is to write an equation to represent her earnings and then to use this equation to create a table of values. \begin{align}y=0.75x+15\end{align}, where \begin{align}y\end{align} represents her earnings and \begin{align}x\end{align} represents the number of glasses of lemonade she sells. Number of Glasses of Lemonade Earnings 0 $15.00 1$15.75 2 $16.50 3$17.25 4 $18.00 5$18.75 6 $19.50 7$20.25 8 21.00 The dependent variable is the money earned, and the independent variable is the number of glasses of lemonade sold. Therefore, money is on the \begin{align}y\end{align}-axis, and the number of glasses of lemonade is on the \begin{align}x\end{align}-axis. From the table of values, Sally will earn21.00 if she sells 8 glasses of lemonade. Now that the points have been plotted, the decision has to be made as to whether or not to join them. Between every 2 points plotted on the graph are an infinite number of values. If these values are meaningful to the problem, then the plotted points can be joined. This type of data is called continuous data. If the values between the 2 plotted points are not meaningful to the problem, then the points should not be joined. This type of data is called discrete data. Since glasses of lemonade are represented by whole numbers, and since fractions or decimals are not appropriate values, the points between 2 consecutive values are not meaningful in this problem. Therefore, the points should not be joined. The data is discrete. #### Interpreting a Graph The following graph represents 3 plans that are available to customers interested in hiring a maintenance company to tend to their lawn. Using the graph, explain when it would be best to use each plan for lawn maintenance. From the graph, the base fee that is charged for each plan is obvious. These values are found on the \begin{align}y\end{align}-axis. Plan A charges a base fee of $200.00, Plan C charges a base fee of$100.00, and Plan B charges a base fee of 50.00. The cost per hour can be calculated by using the values of the intersection points and the base fee in the equation \begin{align}y=mx+b\end{align} and solving for \begin{align}m\end{align}. Plan B is the best plan to choose if the lawn maintenance takes less than 12.5 hours. At 12.5 hours, Plan B and Plan C both cost150.00 for lawn maintenance. After 12.5 hours, Plan C is the best deal, until 50 hours of lawn maintenance is needed. At 50 hours, Plan A and Plan C both cost $300.00 for lawn maintenance. For more than 50 hours of lawn maintenance, Plan A is the best plan. All of the above information was obvious from the graph and would enhance the decision-making process for any interested client. --> ### Example #### Example 1 The local arena is trying to attract as many participants as possible to attend the community’s “Skate for Scoliosis” event. Participants pay a fee of$10.00 for registering, and, in addition, the arena will donate 3.00 for each hour a participant skates, up to a maximum of 6 hours. Create a table of values and draw a graph to represent a participant who skates for the entire 6 hours. How much money can a participant raise for the community if he/she skates for the maximum length of time? The equation for this scenario is \begin{align}y=3x+10\end{align}, where \begin{align}y\end{align} represents the money made by the participant, and \begin{align}x\end{align} represents the number of hours the participant skates. Numbers of Hours Skating Money Earned 010.00 1 $13.00 2$16.00 3 $19.00 4$22.00 5 $25.00 6$28.00 The dependent variable is the money made, and the independent variable is the number of hours the participant skated. Therefore, money is on the \begin{align}y\end{align}-axis, and time is on the \begin{align}x\end{align}-axis as shown below: A participant who skates for the entire 6 hours can make $28.00 for the "Skate for Scoliosis" event. The points are joined, because the fractions and decimals between 2 consecutive points are meaningful for this problem. A participant could skate for 30 minutes, and the arena would pay that skater$1.50 for the time skating. The data is continuous. ### Review 1. What term is used to describe a data set in which all points between 2 consecutive points are meaningful? 1. discrete data 2. continuous data 3. random data 4. fractional data 2. What type of variable is represented by the number of pets owned by families? 1. qualitative 2. quantitative 3. independent 4. continuous 3. What type of data, when plotted on a graph, does not have the points joined? 1. discrete data 2. continuous data 3. random data 4. independent data 4. Select the best descriptions for the following variables and indicate your selections by marking an ‘\begin{align}x\end{align}’ in the appropriate boxes. Variable Quantitative Qualitative Discrete Continuous Men’s favorite TV shows Salaries of baseball players Number of children in a family Favorite color of cars Number of hours worked weekly You are selling your motorcycle, and you decide to advertise it on the Internet on Walton’s Web Ads. He has 3 plans from which you may choose. The plans are shown on the following graph. Use the graph and explain when it is best to use each plan. 1. When would it be best to use Plan A? 2. When would it be best to use Plan B? 3. When would it be best to use Plan C? 4. What is the dependent variable in the following relationship? The time it takes to run the 100 yard dash and the fitness level of the runner. 1. fitness level 2. time 3. length of the track 4. age of the runner 5. If the relationship in question 8 were graphed on a coordinate grid, what variable would be on the x-axis? 6. If the relationship in question 8 were graphed on a coordinate grid, what variable would be on the y-axis? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition bar chart A bar chart is a graphic display of categorical variables that uses bars to represent the frequency of the count in each category. broken line graph A broken line graph is a graph that is used to show changes over time. A line is used to join the values but the line has no defined slope. continuous variables A continuous variable is a variable that takes on any value within the limits of the variable. data set A collection of these observations of the variable is a data set. dependent variable The dependent variable is the output variable in an equation or function, commonly represented by $y$ or $f(x)$. discrete random variables Discrete random variables represent the number of distinct values that can be counted of an event. independent variable The independent variable is the input variable in an equation or function, commonly represented by $x$. qualitative variable A qualitative variable is one that cannot be measured numerically but can be placed in a category. quantitative variable A quantitative variable is a variable that takes on numerical values that represent a measurable quantity. Examples of quantitative variables are the height of students or the population of a city. variable In statistics, a variable is simply a characteristic that is being studied.
# Shear Force and Bending Moments.docx ##### Citation preview Shear Force and Bending Moments Part 1 Example Problem A beam is loaded and supported as shown in Fig. 1. For this beam a. Draw complete shear force and bending moment diagrams. b. Determine the equations for the shear force and the bending moment as functions of x. Fig. 1 Solution Overall Equilibrium We start by drawing a free-body diagram (Fig. 2) of the beam and determining the support reactions. Summing moments about the left end of the beam MA = 7RC - 2 [ 4 × 10 ] gives - 4(16) - 9(19) = 0 (1a) Fig. 2 RC = 45 kN Then, summing forces in the vertical direction F = RA + RC - 4 × 10 - 16 - 19 = 0 gives RA = 30 kN (1b) (2a) (2b) Drawing the Shear Force Diagram Sometimes we are not so much interested in the equations for the shear force and bending moment as we are in knowing the maximum and minimum values or the values at some particular point. In these cases, we want a quick and efficient method of generating the shear force and bending moment diagrams (graphs) so we can easily find the maximum and minimum values. That is the subject of this first part of the problem. Concentrated Force The 30-kN concentrated force (support reaction) at the left end of the beam causes the shear force graph to jump up (in the direction of the force) by 30 kN (the magnitude of the force) from 0 kN to 30 kN. Fig. 3 Distributed Load The downward distributed load causes the shear force graph to slope downward (in the direction of the load). Since the distributed load is constant, the slope of the shear force graph is constant (dV/dx = w = constant). The total change in the shear force graph between points A and B is 40 kN (equal to the area under the distributed load between points A and B) from +30 kN to -10 kN. Fig. 4 We also need to know where the shear force becomes zero. We know that the full 4 m of the distributed load causes a change in the shear force of 40 kN. So how much of the distributed load will it take to cause a change of 30 kN (from +30 kN to 0 kN)? Since the distributed load is uniform, the area (change in shear force) is just 10 × b = 30, which gives b = 3 m. That is, the shear force graph becomes zero at x = 3 m (3 m from the beginning of the uniform distributed load). Concentrated Force Fig. 5 The 16-kN concentrated force at B causes the shear force graph to jump down (in the direction of the force) by 16 kN (the magnitude of the force) from -10 kN to -26 kN. No Loads Since there are no loads between points B and C, the shear force graph is constant (the slope dV/dx = w = 0) at -26 kN. Fig. 6 Concentrated Force The 45-kN concentrated force (support reaction) at C causes the shear force graph to jump up (in the direction of the force) by 45 kN (the magnitude of the force) from -26 kN to +19 kN. Fig. 7 No Loads Since there are no loads between points C and D, the shear force graph is constant (the slope dV/dx = w = 0) at +19 kN. Fig. 8 Concentrated Force The 19-kN concentrated force at D causes the shear force graph to jump down (in the direction of the force) by 19 kN (the magnitude of the force) from +19 kN to 0 kN. Fig. 9 Drawing the Bending Moment Diagram Since there are no concentrated moments acting on this beam, the bending moment diagram (graph) will be continuous (no jumps) and it will start and end at zero. Decreasing Shear Force The bending moment graph starts out at zero and with a large positive slope (since the shear force starts out with a large positive value and dM/dx = V ). As the shear force decreases, so does the slope of the bending moment graph. At x = 3 m the shear force becomes zero and the bending moment is at a local maximum (dM/dx = V = 0 ) For values of x greater than 3 m (3 < x < 4 m) the shear force is negative and the bending moment decreases (dM/dx = V < 0). The shear force graph is linear (1st order function of x ), so the bending moment graph is a parabola (2nd order function of x ). Fig. 10 The change in the bending moment between x = 0 m and x = 3 m is equal to the area under the shear graph between those two points. The area of the triangle is M = (1/2)(30 × 3) = 45 kN·m So the value of the bending moment at x = 3 m is M = 0 + 45 = 45 kN·m. The change in the bending moment between x = 3 and x = 4 m is also equal to the area under the shear graph M = (1/2)(-10 × 1) = -5 kN·m So the value of the bending moment at x = 4 m is M = 45 - 5 = 40 kN·m. Constant Shear Force Although the bending moment graph is continuous at x = 4 m, the jump in the shear force at x = 4 m causes the slope of the bending moment to change suddenly from dM/dx = V = -10 kN·m/m to dM/dx = -26 kN·m/m. Since the shear force graph is constant between x = 4 m and x = 7 m, the bending moment graph has a constant slope between x = 4 m and x = 7 m (dM/dx = V = -26 kN·m/m). That is, the bending moment graph is a straight line. The change in the bending moment between x = 4 m and x = 7 m is equal to the Fig. 11 area under the shear graph between those two points. The area of the rectangle is just M = (-26 × 3) = -78 kN·m. So the value of the bending moment at x = 7 m is M = 40 - 78 = -38 kN·m. Constant Shear Force Again the bending moment graph is continuous at x = 7 m. The jump in the shear force at x = 7 m causes the slope of the bending moment to change suddenly from dM/dx = V = -26 kN·m/m to dM/dx = +19 kN·m/m. Since the shear force graph is constant between x = 7 m and x = 9 m, the bending moment graph has a constant slope between x = 7 m and x = 9 m (dM/dx = V = +19 kN·m/m). That is, the bending moment graph is a straight line. The change in the bending moment between x = 7 m and x = 9 m is equal to the Fig. 12 area under the shear graph between those two points. The area of the rectangle is just M = (+19 × 2) = +38 kN·m. So the value of the bending moment at x = 7 m is M = -38 + 38 = 0 kN·m. Determining the Shear Force and Bending Moment Equations Sometimes we are not so much interested in the graphs of the shear force and bending moment as we are in knowing the equations. In particular, we need to integrate the equation for the bending moment to determine the shape of beam and how much the beam will bend as a result of the loads. That is the subject of the second part of this problem. Shear Force and Bending Moments Part 2 Solution (Cont.) Determining the Shear Force and Bending Moment Equations The easiest way to get the equations for the shear force and bending moment as functions of the position x is to use equilibrium. 0 m < x < 4 m Figure 13 shows a free-body diagram of the left end of the beam to an arbitrary position, 0 m < x < 4 m. The righthand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction gives Fig. 13 F = 30 - (10x ) - V = 0 V = 30 - (10x ) kN 3a (0 m < x < 4 m) Summing moments about a point on the "cut end" of the beam Mcut = M + (10x )(x/2) - 30x = 0 gives M = 30x - (5x 2) kN·m (0 m < x < 4 m) 3b 4a 4b 4 m < x < 7 m Figure 14 shows a free-body diagram of the left end of the beam to an arbitrary position, 4 m < x < 7 m. The right-hand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction gives F = 30 - (10 × 4) - 16 - V = 0 V = -26 kN (4 m < x < 7 m) Fig. 14 5a 5b Summing moments about a point on the "cut end" of the beam Mcut = M + (10 × 4)(x - 2) + 16(x - 4) - 30x = 0 gives M = 144 - 26x kN·m (4 m < x < 7 m) 6a 6b 7 m < x < 9 m Figure 15 shows a free-body diagram of the left end of the beam to an arbitrary position, 7 m < x < 9 m. The right-hand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction gives Fig. 15 F = 30 - (10 × 4) - 16 - 45 - V = 0 V = 19 kN (7 m < x < 9 m) Summing moments about a point on the "cut end" of the beam Mcut = M + (10 × 4)(x - 2) + 16(x - 4) - 30x - 45(x - 7) = 0 gives M = 171 - 19x kN·m (7 m < x < 9 m) 7a 7b 8a 8b It is easily verified that these equations have the appropriate character to match the shear force and bending moment diagrams developed in the first part of this problem. It is also easily verified that these equations match the previous graphs at the points x = 0 m, x = 3 m, x = 4 m, x = 7 m, and x = 9 m. Finally, note that these equations satisfy the load-shear force-bending moment relationships dV/dx = w dM/dx = V
# Angles ## Basics ### Key Idea Angle measures a turn. The angle between two line segments? measures how far one turns to go from one line to the other. There are two main ways of measuring angles. • Degrees KS4? and KS5? When measuring in degrees, we effectively measure in fractions? of a full turn. Because it is easier to work with integers? rather than fractions?, we divide a full turn into 360 parts and count how many parts we turn. One degree is therefore one 360th of a full turn. The symbol for degrees is ${}^\circ$. When measuring in radians, we measure how far we travel when we travel around the circumference of a unit circle? when making the turn. There are therefore $2\pi$ radians in a full turn. Radians are usually written without a symbol. When a symbol is used, it is ${}^c$. As there are $360$ degrees and $2\pi$ radians in a full turn, the conversion is: To ConvertMultiply By Degrees to Radians$\frac{\pi}{180}$ Radians to Degrees$\frac{180}{\pi}$ ## Results All results about angles hinge on the following fact. Angles are preserved by translation?, rotation?, reflection?, and scaling?. ### Basic Results 1. Angles around a point add up to $360^\circ$. 2. Angles at a point on a straight line add up to $180^\circ$. 3. Opposite angles are the same. 4. Corresponding angles are the same. ### Dependent Results All of the following results can be deduced from the basic results. 1. Angles in a triangle? add up to $180^\circ$. 2. Alternate angles are the same. 3. Co-interior angles add up to $180^\circ$. ### Circle Theorems The key to the circle theorems is the abundance of isosceles triangles. Any triangle? which is formed using two radii? of the same circle? must be isosceles?. 1. The angle in a semicircle? is a right angle?. 2. The angle at the centre is twice the angle at the circumference?. 3. Angles in the same segment? are equal. category: geometry
# Unlocking the Power of 17/22 as a Percentage: Discover the Benefits and Applications ## Understanding the Importance of 17/22 as a Percentage 17/22 as a Percentage: In mathematical terms, 17/22 as a percentage refers to expressing the fraction 17/22 as a decimal multiplied by 100. This calculation allows us to understand the proportion of 17 out of 22 as a fraction of 100. By converting it to a percentage, we can easily compare and analyze the data in various contexts. Application in Statistics: Understanding 17/22 as a percentage is particularly significant in statistics. It enables us to grasp the relative occurrence or frequency of an event within a given sample size. For example, if we were conducting a survey and found that 17 out of 22 respondents preferred a certain product, we could express this result as a percentage to provide a clearer understanding of the preferred choice among the respondents. Interpreting Percentages: Interpreting percentages is crucial in various areas, including finance, economics, and business. A percentage indicates a relative or proportional value, making it easier to compare data across different contexts. When expressing 17/22 as a percentage, we can quickly identify the significance of the proportion and relate it to other data points. Additionally, percentages are useful for visualizing and communicating data effectively through charts, graphs, and infographics. Therefore, understanding the importance of expressing 17/22 as a percentage allows us to analyze data within a larger context, make informed decisions, and effectively communicate statistical information. Stay tuned for more insights on mathematical concepts and their applications. ## Breaking Down 17/22: Converting Fractions to Percentages ### Understanding Fractions To convert fractions to percentages, it is crucial to have a clear understanding of what a fraction represents. In the context of mathematics, a fraction represents a part of a whole. The top number, known as the numerator, indicates the number of parts we have, while the bottom number, called the denominator, represents the total number of equal parts that make up the whole. For example, in the fraction 17/22, 17 is the numerator, indicating that we have 17 parts, and 22 is the denominator, representing the total number of equal parts that make up the whole. ### The Process of Converting Fractions to Percentages To convert fractions to percentages, we can follow a simple process. First, we divide the numerator by the denominator. In the case of 17/22, we would divide 17 by 22. This calculation gives us a decimal value. To convert this decimal to a percentage, we then multiply it by 100. Finally, we add the percentage symbol (%) to indicate that the number represents a percentage. For example: 17 ÷ 22 = 0.77272727 0.77272727 x 100 = 77.272727% When converting fractions to percentages, it is important to remember that in some cases, the resulting decimal may be long and repeating. In such instances, it is common practice to round the decimal to a desired number of decimal places, such as two or three, depending on the required level of accuracy. For instance: 2/3 = 0.66666666 (rounded to two decimal places: 0.67) 5/8 = 0.625 (rounded to three decimal places: 0.625) Understanding how to convert fractions to percentages can be helpful in various real-life scenarios, such as calculating discounts, analyzing statistical data, or solving mathematical problems involving proportions. By mastering this skill, we can enhance our mathematical proficiency and improve our problem-solving abilities. So, buckle up and get ready to conquer the art of converting fractions to percentages! ## Mastering the Calculation of 17/22 as a Percentage ### Understanding the Basics When it comes to calculating percentages, it’s essential to grasp the fundamentals. For instance, to convert a fraction to a percentage, you need to divide the numerator by the denominator and multiply the result by 100. In the case of 17/22, divide 17 by 22 and multiply the resulting decimal by 100. This will yield the percentage equivalent of the fraction. To make the process clearer, let’s break it down step by step: Step 1: Divide 17 by 22: 17 ÷ 22 = 0.772727… Step 2: Multiply by 100 to get the percentage: 0.772727… × 100 = 77.27… Therefore, 17/22 can be expressed as approximately 77.27%. ### Round It Off In some cases, it may be necessary to round off the resulting percentage to make it more concise or suitable for a specific context. For example, rounding 77.27% to the nearest whole number would give us 77%. However, keep in mind that rounding may introduce some level of approximation and may not always be appropriate, depending on the desired level of accuracy. ### Application in Real-Life Scenarios Knowing how to calculate percentages is not only helpful in math exercises but also in real-life scenarios. For instance, when calculating discounts during a sale or determining the proportion of a budget allocated to different expenses, understanding how to convert fractions to percentages is crucial. By mastering the calculation of 17/22 as a percentage, you’ll have the necessary skills to analyze and interpret data more effectively, both in school and everyday life. This knowledge allows you to understand the significance of percentages and make informed decisions based on numerical information. Remember, practice makes perfect when it comes to calculations, so keep working on your skills to become more proficient at converting fractions to percentages. ## Key Concepts: Exploring 17/22 as a Percentage in Depth ### What is 17/22 as a percentage? Understanding how to express fractions as percentages is an essential concept in mathematics. In this article, we will delve deep into the key concept of converting 17/22 into a percentage. When a fraction like 17/22 is converted to a percentage, it represents the fraction as a proportion of 100. In other words, we want to find the equivalent value of 17/22 out of 100. Calculating 17/22 as a percentage: To convert 17/22 into a percentage, we divide the numerator (17) by the denominator (22) and multiply the result by 100. Performing this calculation, we get 77.27%. Therefore, 17/22 is approximately equal to 77.27% as a percentage. ### Application and significance of 17/22 as a percentage: The concept of expressing fractions as percentages is widely used in various fields, including finance and statistics. When dealing with data or ratios, it is often more convenient to work with percentages rather than fractions. Understanding the percentage value of 17/22 can help in various scenarios, such as analyzing data, calculating proportions, and making comparisons. Example: Let’s say we have a group of 22 students, and 17 of them prefer basketball. By converting this fraction to a percentage, we can determine that 77.27% of the students in the group prefer basketball. This information can be useful for statistical analysis or making decisions related to sports activities. Converting fractions to percentages is a fundamental skill that helps us express proportions in a more relatable way. It allows for easier comparison and interpretation of data. Understanding the concept of converting 17/22 to a percentage provides a solid foundation for further exploration of fractions and their applications in various fields. You may also be interested in: Unveiling the Truth: How Many Days is 9 Months? A Comprehensive Guide Explained! ## Unlocking the Secrets Behind 17/22 as a Percentage ### Understanding 17/22 as a Fraction When we see the fraction 17/22, it represents a part of the whole. In this case, the numerator, 17, represents the part we want to find out of the total, which is represented by the denominator, 22. To convert this fraction to a percentage, we need to unlock the secrets behind the calculation. Conversion to Decimal and Multiplication by 100 To convert 17/22 to a percentage, we first need to convert it to a decimal. To do this, we divide the numerator (17) by the denominator (22). The result is 0.7727272727 (rounded to several decimal places). To convert this decimal to a percentage, we multiply it by 100. Thus, 0.7727272727 x 100 equals 77.27%. You may also be interested in: Converting 5.5 oz to g: A Complete Guide to Gram-Ounce Conversion ### Real-World Application Understanding how to convert 17/22 to a percentage can be valuable in various situations. For example, imagine you are a salesperson and want to determine your sales conversion rate. If you made 17 sales out of 22 leads, you can calculate that your conversion rate is 77.27%. Calculating Success Rates and Goal Achievements Knowing how to unlock the secrets behind 17/22 as a percentage allows you to calculate success rates in any scenario. Whether you are measuring the completion rate of tasks, the accuracy of test scores, or the achievement of personal goals, converting fractions to percentages enables you to assess progress and identify areas for improvement. This skill empowers you to make informed decisions and set realistic targets for future endeavors.
### Chapter 3 - Kawameeh Middle School ```S Chapter 3 Lesson 1 Energy S The ability of a system to do work. Two types of energy: • Kinetic Energy • Potential Energy States of Energy All forms of energy can be in either of two states: What’s the difference? S Kinetic Energy is the energy of MOTION S Potential Energy is STORED energy. Kinetic Energy – what does it depend on? The faster an object moves, the more kinetic energy it has.  The greater the mass of a moving object, the more kinetic energy it has.  Kinetic energy depends on both mass and velocity.  Calculating Kinetic Energy KE  0.5  mass  velocity 2 What has a greater affect of kinetic energy, mass or velocity? Why? What is the unit for Kinetic Energy? S Unit: Joule S Named after: James Prescott Joule S He discovered the relationship between heat (energy) and mechanical work which led to the law of conservation of energy. S How do we derive this unit? S 1 Joule = 1kg ∙ m2/s2 S KE = ½ ∙ m(kg) ∙ v(m/s) 2 Some types of Potential Energy include…  Gravitational potential energy - stored energy due to an objects position (height) - depends on mass of the object and its distance from earth.  Elastic potential energy - Stored energy due to - Stored energy due to compression or expansion of an elastic object Calculating Potential Energy PE  mass  gravity  height S OR you could multiply weight (in Newton’s) by height  What unit do we use for Potential Energy? S Unit: Joule S How do we derive this unit? S 1 Joule = 1kg ∙ m2/s2 S PE = m(kg) ∙ g (m/s2 ) ∙ height (m) 1. You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has ___________ energy. Calculate it. 2. A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 N. The carriage has ____________ energy. Calculate it. 3. A car is traveling with a velocity of 40 m/s and has a mass of 1120 kg. The car has ___________energy. Calculate it. 4. A cinder block is sitting on a platform 20 m high. It weighs 79 N. The block has _____________ energy. Calculate it. 5. A roller coaster is at the top of a 72 m hill and weighs 966 N. The coaster (at this moment) has ____________ energy. Calculate it. 6. Two objects were lifted by a machine. One object had a mass of 2 kilograms, and was lifted at a speed of 2 m/sec. The other had a mass of 4 kilograms and was lifted at a rate of 3 m/sec. a. Which object had more kinetic energy while it was being lifted? b. Which object had more potential energy when it was lifted to a distance of 10 meters? Show your calculation.  As a player throws the ball into the air, various energy conversions take place. Lesson 2 Law of Conservation of Energy S Energy can be neither created nor destroyed by ordinary means. transformed or transferred one form to another. S Energy can be from is the same before and after any transformation or transfer. S The total amount of energy Energy Transfer S Energy TRANSFER is the passing of energy from one object to another object. Example: A cup of hot tea has thermal energy. Some of this thermal energy is transferred to the particles in cold milk, in which you put to make the coffee cooler. Energy Transformation S A change from one form of energy to another. S Single Transformations S Occur when one form of energy into another to get work done. needs to be transformed S Multiple Transformations S Occur when a series of energy transformations are needed to do work Energy Transformation S An objects energy can be: 1. All Kinetic Energy 2. All Potential Energy 3. A combination of both increases and potential energy decreases S As velocity decreases kinetic energy decreases and potential energy increases WHAT IS THE TYPE OF RELATIONSHIP KE AND PE HAVE? INVERSE S As velocity increases kinetic energy Roller Coasters Does energy get transferred or transformed? • As you move up to the first hill on a roller coaster the distance between the coaster and the Earth increases, resulting in an increase of Gravitational Potential Energy. • At the top of the first hill you have the most Gravitational Potential Energy • As you begin your trip down the hill you increase your speed resulting in a transformation from GPE to KE. • At the bottom of the hill right before it goes back upward the GPE is small, but the KE is Large. • As it starts to move up the next hill or loop KE is transformed back into GPE CjyjA ```
Percentiles: Measure of Relative Standing Percentiles are a measure of the relative standing of observation within a data. Percentiles divide a set of observations into 100 equal parts, and percentile scores are frequently used to report results from national standardized tests such as NAT, GAT, and GRE, etc. The $p$th percentile is the value $Y_{(p)}$ in order statistic such that $p$ percent of the values are less than the value $Y_{(p)}$ and $(100-p)$ (100-p) percent of the values are greater $Y_{(p)}$. The 5th percentile is denoted by $P_5$, the 10th by $P_{10}$ and 95th by $P_{95}$. Percentiles for the Ungrouped data To calculate percentiles (a measure of the relative standing of an observation) for the ungrouped data, adopt the following procedure: 1. Order the observation 2. For the $m$th percentile, determine the product $\frac{m.n}{100}$. If $\frac{m.n}{100}$ is not an integer, round it up and find the corresponding ordered value and if $\frac{m.n}{100}$ is an integer, say k, then calculate the mean of the $K$th and $(k+1)$th ordered observations. Example: For the following height data collected from students find the 10th and 95th percentiles. 91, 89, 88, 87, 89, 91, 87, 92, 90, 98, 95, 97, 96, 100, 101, 96, 98, 99, 98, 100, 102, 99, 101, 105, 103, 107, 105, 106, 107, 112. Solution: The ordered observations of the data are 87, 87, 88, 89, 89, 90, 91, 91, 92, 95, 96, 96, 97, 98, 98, 98, 99, 99, 100, 100, 101, 101, 102, 103, 105, 105, 106, 107, 107, 112. $P_{10}= \frac{10 \times 30}{100}=3$ So the 10th percentile i.e. $P_{10}$ is the 3rd observation in sorted data is 88, which means that 10 percent of the observations in the data set are less than 88. $P_{95}=\frac{95 \times 30}{100}=28.5$ The 29th observation is our 95th Percnetile i.e., $P_{95}=107$ Percentiles for the Frequency Distribution Table (Grouped data) The $m$th percentile (a measure of the relative standing of an observation) for the Frequency Distribution Table (grouped data) is $P_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)$ Like median, $\frac{m.n}{100}$ is used to locate the $m$th percentile group. $l$ is the lower class boundary of the class containing the $m$th percentile $h$ is the width of the class containing $P_m$ $f$ is the frequency of the class containing $n$ is the total number of frequencies $P_m$ $c$ is the cumulative frequency of the class immediately preceding the class containing $P_m$ Note that the 50th percentile is the median by definition as half of the values in the data are smaller than the median and half of the values are larger than the median. Similarly, the 25th and 75th percentiles are the lower ($Q_1$) and upper quartiles ($Q_3$) respectively. The quartiles, deciles, and percentiles are also called quantiles or fractiles. Example: For the following grouped data compute $P_{10}$, $P_{25}$, $P_{50}$, and $P_{95}$ given below.Solution: 1. Locate the 10th percentile (lower deciles i.e. $D_1$)by $\frac{10 \times n}{100}=\frac{10 \times 3o}{100}=3$ observation. so, $P_{10}$ group is 85.5–90.5 containing the 3rd observation \begin{align} P_{10}&=l+\frac{h}{f}\left(\frac{10 n}{100}-c\right)\\ &=85.5+\frac{5}{6}(3-0)\\ &=85.5+2.5=88 \end{align} 2. Locate the 25th percentile (lower quartiles i.e. $Q_1$) by $\frac{10 \times n}{100}=\frac{25 \times 3o}{100}=7.5$ observation. so, $P_{25}$ group is 90.5–95.5 containing the 7.5th observation \begin{align} P_{25}&=l+\frac{h}{f}\left(\frac{25 n}{100}-c\right)\\ &=90.5+\frac{5}{4}(7.5-6)\\ &=90.5+1.875=92.375 \end{align} 3. Locate the 50th percentile (Median i.e. 2nd quartiles, 5th deciles) by $\frac{50 \times n}{100}=\frac{50 \times 3o}{100}=15$ observation. so, P50 group is 95.5–100.5 containing the 15th observation \begin{align} P_{50}&=l+\frac{h}{f}\left(\frac{50 n}{100}-c\right)\\ &=95.5+\frac{5}{10}(15-10)\\ &=95.5+2.5=98 \end{align} 4. Locate the 95th percentile by $\frac{95 \times n}{100}=\frac{95 \times 30}{100}=28.5$th observation. so, $P_{95}$ group is 105.5–110.5 containing the 3rd observation \begin{align} P_{95}&=l+\frac{h}{f}\left(\frac{95 n}{100}-c\right)\\ &=105.5+\frac{5}{3}(28.5-26)\\ &=105.5+4.1667=109.6667 \end{align} The percentiles and quartiles may be read directly from the graphs of the cumulative frequency function.
## Intermediate Algebra: Connecting Concepts through Application $m \ge-\dfrac{68}{5}$ $\bf{\text{Solution Outline:}}$ To solve the given inequality, $\dfrac{1}{4}m+\dfrac{2}{3}(m-5)\le\dfrac{1}{3}(4m+7) ,$ use the Distributive Property and the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Distributive Property, which is given by $a(b+c)=ab+ac,$ the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{4}m+\dfrac{2}{3}(m-5)\le\dfrac{1}{3}(4m+7) \\\\ \dfrac{1}{4}m+\dfrac{2}{3}(m)+\dfrac{2}{3}(-5)\le\dfrac{1}{3}(4m)+\dfrac{1}{3}(7) \\\\ \dfrac{1}{4}m+\dfrac{2}{3}m-\dfrac{10}{3}\le\dfrac{4}{3}m+\dfrac{7}{3} .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{4}m+\dfrac{2}{3}m-\dfrac{10}{3}\le\dfrac{4}{3}m+\dfrac{7}{3} \\\\ 12\left( \dfrac{1}{4}m+\dfrac{2}{3}m-\dfrac{10}{3} \right)\le12\left( \dfrac{4}{3}m+\dfrac{7}{3} \right) \\\\ 3m+8m-40 \le16m+28 \\\\ 3m+8m-16m \le28+40 \\\\ -5m \le68 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol) results to \begin{array}{l}\require{cancel} -5m \le68 \\\\ m \ge\dfrac{68}{-5} \\\\ m \ge-\dfrac{68}{5} .\end{array}
# Unit 2: The Derivative Much of the work we are going to do in this course consists of taking one or more functions, and producing a new function. The first of these methods is differentiation: we take a function, call it f(x) for convenience, and produce a new function which we call the derivative of f(x). We have several notations for this process. (In fact, we probably have too many notations. Such is mathematics.) If we again denote our function by f(x), its derivative can be denoted by If we express our function in the x-y plane as y=f(x), then we gain the notations There is, of course, a reason we want the derivative of a function. The derivative is the slope of the original function at any given point. This will be explained in the reading along with pretty pictures of tangent lines. The method used to define the tangent is to approach the tangent line through a sequence of secants; again, refer to pictures in the various texts. Following the secant lines, we arrive at our definition of the derivative: (Sometimes , ``Delta -x'', is used instead of h. The triangle symbol is the Greek letter "Delta".) This definition raises several interesting questions: what is this limit? (This problem was brushed off for a few centuries. Its eventual answer forms the subject material of analysis.) How do we know it exists? (Sometimes it doesn't. See discussion of \|x\| in the text.). We follow a long tradition of beginning calculus courses in firmly brushing these questions under the rug. In most of the cases of interest, we are dealing with a ``smooth'' function where we may take the derivative with impunity. There is another interpretation of the derivative that you should also understand: the derivative indicates the rate of change of the original function. This interpretation is especially important when the function is expressed in terms of time. For example we may have the height of a rocket expressed as h = f(t), giving the height for every instant. The upward component of the velocity of the rocket, i.e., the rate of change of the height, is given by Become accustomed to this idea now; you will see it again and again. Now that we've developed a feel for the derivative, we need method for calculating the derivative without applying the definition over and over again. Using the definiton, the following rules of differentiation can be derived. You should make sure to go over the derivation of these rules: 1. The constant rule: if c is a constant, then 2. The sum and difference rules: 3. The product rule: 4. The quotient rule: 5. The power rule (first version): Note that this is only the first version of the power rule; we will generalize this rule in the next unit, when we have the chain rule. The proofs of the various rules follow from the definition of the derivative and some algebraic manipulation. You should be able to follow the proofs, but you needn't worry about reproducing them. The power rule is an interesting case in that it can be proven using induction for positive integral powers. (The following discussion will be extended when the World Wide Math Integer Arithmetic page is completed.) To over-generalize madly, induction arguments can come up whenever the natural numbers are involved in a formula. Both texts have further examples of induction proofs in their respective appendices. While induction proofs aren't vital to your understanding of calculus, they are a piece of mathematical culture you should learn. The general form is to prove a base case, either n = 0 or n = 1 or wherever you want to start: for the case of the power rule, Next is the inductive step: assume that the proposition is true for all values less than n, then prove it for n+1: Note that the product rule was used in the above. Mastering these rules for differentiation is important; work enough practice problems so that you can differentiate blindfolded!! Now is a convenient time as any to mention that you can take derivatives multiples times. The derivative of a function is a perfectly good function in its own right: it can be graphed, and one can ask for its slope at any point. Relative to the function we started with, this is a second order derivative. We, of course, have lots of fun notation for this notion, and for the obvious next steps: ### Objectives: After completing this unit you should be able to: 1. Understand the derivative as expressing the slope of a curve at each point on the curve. 2. Understand the derivative as expressing a rate of change. 3. Be able to use to set up derivatives, and to calculate derivatives in some cases. 4. Derive the rules of differentiation. 5. Calculate the derivative of any rational function. 6. Calculate higer order derivatives. ### Suggested Procedure: 1. Read Simmons, first edition chapter 2 and sections 3.1, 3.2, and 3.5, or 2. Read Simmons, second edition chapter 2 and sections 3.1, 3.2, and 3.6. 3. Read the following World Web Math Pages: 4. Do some problems in Simmons • 2.2 : # 1, 3, 5, 7, 8 • 2.3 : # 1, 4, 12 • 2.4 : # 1, 4, 8, 10, 11 • 3.1 : # 1, 9 - 12 • 3.2 : # 1, 2, 5, 7, 8, 9 5. Take the Practice Unit Test, Xdvi or PDF
or Find what you need to study Light Find what you need to study # 4.4 Parametrically Defined Circles and Lines Jesse Jesse A parametrically defined circle is a circle that is defined by a set of parametric equations, rather than by its center and radius. These equations use a parameter, typically denoted by t, to specify the position of a point on the circle as it moves in a counterclockwise direction. ๐ Similarly, a parametrically defined line is a line that is defined by a set of parametric equations, rather than by its two endpoints. โ ## Fun with Circles A complete counterclockwise revolution around the unit circle that starts and ends at (1,0) and is centered at the origin can be modeled by the parametric equations (x(t), y(t)) = (cos(t), sin(t)) with the domain 0 โค t โค 2ฯ. Source: Math Easy Solutions โ YouTube The equation x(t) = cos(t) represents the x-coordinate of the point on the circle as it moves in a counterclockwise direction. The value of cos(t) ranges from -1 to 1, with cos(0) = 1 and cos(2ฯ) = 1, which corresponds to the starting and ending point (1, 0) on the x-axis. Similarly, the equation y(t) = sin(t) represents the y-coordinate of the point on the circle as it moves in a counterclockwise direction. The value of sin(t) ranges from -1 to 1, with sin(0) = 0 and sin(2ฯ) = 0, which corresponds to the starting and ending point (1, 0) on the x-axis. By defining the domain of t to be 0 โค t โค 2ฯ, we ensure that the point traces out a complete revolution around the unit circle, starting at (1, 0) and ending at (1, 0) again. ๐ ### Transformations The parametric function (x(t), y(t)) = (cos(t), sin(t)) is known as the standard parametric equation for a unit circle centered at the origin. By applying different transformations to this equation, we can model any circular path traversed in the plane. โป๏ธ For example, to model a circle centered at the point (a, b) with radius r, we can apply the following transformations to the standard equation: (x(t), y(t)) = (a + rcos(t), b + rsin(t)) This transformation shifts the center of the circle from the origin to the point (a,b) and scales the radius from 1 to r. 1๏ธโฃ We can also apply rotation transformation to the standard equation to model a circular path traversed in a different direction. (x(t), y(t)) = (cos(t + c), sin(t + c)), where c is the angle of rotation. ## Fun with Lines A linear path along the line segment from the point (x1, y1) to the point (x2, y2) can be parametrized in many ways, including using an initial position (x1, y1) and rates of change for x with respect to t and y with respect to t. โ๏ธ One common way to parametrize a linear path is to use the equation of a line in slope-intercept form: y = mx + b, where m is the slope of the line and b is the y-intercept. We can use the two points on the line segment to find the slope (m) of the line and the y-intercept (b). ๐ค We can also parametrize the line segment by using a point on the line (x1, y1) and the direction vector of the line segment, which is given by: (x2 - x1, y2 - y1) By using this method, we can represent the line segment as the set of all points that can be written as: (x1 + k(x2 - x1), y1 + k(y2 - y1)) Where k is a scalar that varies between 0 and 1. ๐ซก In addition, we can also use the parametric equations of the line: x = x1 + k(x2 - x1) y = y1 + k(y2 - y1) Where k is a scalar that varies between 0 and 1. So, there are many ways to parametrize a linear path along the line segment from the point (x1, y1) to the point (x2, y2) and each one of them has its own advantages and disadvantages! ๐ # 4.4 Parametrically Defined Circles and Lines Jesse Jesse A parametrically defined circle is a circle that is defined by a set of parametric equations, rather than by its center and radius. These equations use a parameter, typically denoted by t, to specify the position of a point on the circle as it moves in a counterclockwise direction. ๐ Similarly, a parametrically defined line is a line that is defined by a set of parametric equations, rather than by its two endpoints. โ ## Fun with Circles A complete counterclockwise revolution around the unit circle that starts and ends at (1,0) and is centered at the origin can be modeled by the parametric equations (x(t), y(t)) = (cos(t), sin(t)) with the domain 0 โค t โค 2ฯ. Source: Math Easy Solutions โ YouTube The equation x(t) = cos(t) represents the x-coordinate of the point on the circle as it moves in a counterclockwise direction. The value of cos(t) ranges from -1 to 1, with cos(0) = 1 and cos(2ฯ) = 1, which corresponds to the starting and ending point (1, 0) on the x-axis. Similarly, the equation y(t) = sin(t) represents the y-coordinate of the point on the circle as it moves in a counterclockwise direction. The value of sin(t) ranges from -1 to 1, with sin(0) = 0 and sin(2ฯ) = 0, which corresponds to the starting and ending point (1, 0) on the x-axis. By defining the domain of t to be 0 โค t โค 2ฯ, we ensure that the point traces out a complete revolution around the unit circle, starting at (1, 0) and ending at (1, 0) again. ๐ ### Transformations The parametric function (x(t), y(t)) = (cos(t), sin(t)) is known as the standard parametric equation for a unit circle centered at the origin. By applying different transformations to this equation, we can model any circular path traversed in the plane. โป๏ธ For example, to model a circle centered at the point (a, b) with radius r, we can apply the following transformations to the standard equation: (x(t), y(t)) = (a + rcos(t), b + rsin(t)) This transformation shifts the center of the circle from the origin to the point (a,b) and scales the radius from 1 to r. 1๏ธโฃ We can also apply rotation transformation to the standard equation to model a circular path traversed in a different direction. (x(t), y(t)) = (cos(t + c), sin(t + c)), where c is the angle of rotation. ## Fun with Lines A linear path along the line segment from the point (x1, y1) to the point (x2, y2) can be parametrized in many ways, including using an initial position (x1, y1) and rates of change for x with respect to t and y with respect to t. โ๏ธ One common way to parametrize a linear path is to use the equation of a line in slope-intercept form: y = mx + b, where m is the slope of the line and b is the y-intercept. We can use the two points on the line segment to find the slope (m) of the line and the y-intercept (b). ๐ค We can also parametrize the line segment by using a point on the line (x1, y1) and the direction vector of the line segment, which is given by: (x2 - x1, y2 - y1) By using this method, we can represent the line segment as the set of all points that can be written as: (x1 + k(x2 - x1), y1 + k(y2 - y1)) Where k is a scalar that varies between 0 and 1. ๐ซก In addition, we can also use the parametric equations of the line: x = x1 + k(x2 - x1) y = y1 + k(y2 - y1) Where k is a scalar that varies between 0 and 1. So, there are many ways to parametrize a linear path along the line segment from the point (x1, y1) to the point (x2, y2) and each one of them has its own advantages and disadvantages! ๐
Question Video: Identifying Geometric Sequences Mathematics • 9th Grade Determine π and π that make the following a geometric sequence, starting at πβ, of the form π_(π) = ππ^(π β 1): 2, 6, 18, 54, 162, ... 02:34 Video Transcript Determine π and π that make the following a geometric sequence, starting at π sub one, of the form π sub π equals π times π to the power of π minus one: two, six, 18, 54, 162, and so on. Letβs start by recalling that a geometric sequence is a sequence which has a fixed ratio between successive terms. In other words, if you perform the same multiplication on any term, you will get the next term. So letβs consider how we would go from two to six. Well, we could do this by multiplying by three. Notice that if we were adding or subtracting values to two, then this wouldnβt be a geometric sequence. It would in fact be an arithmetic sequence. So if we also multiplied six by three, would we get 18? Yes, we would. In the same way, if we multiply 18 by three, we get 54. And 54 multiplied by three does give us 162. So it looks as though we do have a geometric sequence. So the next thing to do is see if we can write it in this form of the πth term, which involves π and π. We should remember that we can write any term π sub π in a geometric sequence as π times π to the power of π minus one, where π is the index, π or π sub one represents the first term, and π is the fixed ratio between terms. Itβs worth noting that sometimes we have sequences that begin with index zero. However, here, weβre given that this sequence should start with an index of one with the term π sub one. So letβs compare this standard way of writing the πth term with the form in which we were given, which involves π and π. When we look at the equivalent values in the general πth term formula, we could see that π must be equivalent to the first term. π would be equivalent to the common ratio or fixed ratio between terms. So letβs consider what the value of π would be for this given sequence. Itβs quite easy to spot. Itβs the first term. And so π must be equal to two. We then determined that π must be equivalent to the fixed ratio between terms. And weβve already worked out that the fixed ratio must be three. We can therefore give the answer that π is equal to two and π is equal to three.
# Lesson 16 Solving Systems by Elimination (Part 3) ## 16.1: Multiplying Equations By a Number (10 minutes) ### Warm-up In this warm-up, experiment with the graphical effects of multiplying both sides of an equation in two variables by a factor. Students are prompted to multiply one equation in a system by several factors to generate several equivalent equations. They then graph these equations on the same coordinate plane that shows the graphs of the original system. Students notice that no new graphs appear on the coordinate plane and reason about why this might be the case. The work here reminds students that equations that are equivalent have all the same solutions, so their graphs are also identical. Later, students will rely on this insight to explain why we can multiply one equation in a system by a factor—which produces an equivalent equation—and solve a new system containing that equation instead. ### Launch Display the equation $$x+5 = 11$$ for all to see. Ask students: • "What is the solution to this equation?" ($$x=6$$) • "If we multiply both sides of the equation by a factor, say, 4, what equation would we have?" ($$4x + 20 = 44$$) "What is the solution to this equation?" ($$x=6$$) • "What if we multiply both sides by 100?" ($$100x + 500 = 1,\!100$$) "By 0.5?" ($$0.5x + 2.5 = 5.5$$) • "Is the solution to each of these equations still $$x=6$$?" (Yes) Remind students that these equations are one-variable equations, and that multiplying both sides of a one-variable equation by the same factor produces an equivalent equation with the same solution. Ask students: "What if we multiply both sides of a two-variable equation by the same factor? Would the resulting equation have the same solutions as the original equation?" Tell students that they will now investigate this question by graphing. Arrange students in groups of 2–4 and provide access to graphing technology to each group. To save time, consider asking group members to divide up the tasks. (For example, one person could be in charge of graphing while the others write equivalent equations, and everyone analyze the graphs together.) ### Student Facing Consider two equations in a system: \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1 &\quad&\text{Equation A}\\ x + 2y &= \hspace {2mm} 9&\quad&\text{Equation B} \end{align} \end{cases} 1. Use graphing technology to graph the equations. Then, identify the coordinates of the solution. 2. Write a few equations that are equivalent to equation A by multiplying both sides of it by the same number, for example, 2, -5, or $$\frac12$$. Let’s call the resulting equations A1, A2, and A3. Record your equations here: 1. Equation A1: 2. Equation A2: 3. Equation A3: 3. Graph the equations you generated. Make a couple of observations about the graphs. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Activity Synthesis Invite students to share their observations about the graphs they created. Ask students why the graphs of equations A1, A2, and A3 all coincide with the graphs of the original equation A. Discuss with students: • "How can we explain the identical graphs?" (Equations A1, A2, and A3 are equivalent to equation A, so they all have the same solutions as equation A, and their graphs are the same line as the graph of A.) • "What move was made to generate A1, A2, and A3? Why did it create equations that are equivalent to A?" (The two sides of equation A was multiplied by the same factor, which keeps the two sides equal.) To further illustrate that equations A1, A2, and A3 are equivalent to equation A, and if time permits, consider: • Using tables to visualize the identical $$(x,y)$$ pairs. For example: $$4x+y=1$$ x y 0 1 1 -3 2 -7 3 -11 $$20x+5y=5$$ x y 0 1 1 -3 2 -7 3 -11 $$2x+\frac12y=\frac12$$ x y 0 1 1 -3 2 -7 3 -11 • Reminding students that, earlier in the unit, they saw that isolating one variable is a way to see if two equations are equivalent. If we isolate $$y$$ in equations A, A1, A2, and A3, the rearranged equation will be identical: $$y=\text-4x+1$$ ## 16.2: Writing a New System to Solve a Given System (15 minutes) ### Activity In an earlier lesson, students learned that adding the two equations in a system creates a new equation that shares a solution with the system. In the warm-up, they saw that multiplying an equation by a factor creates an equivalent equation that shares all the same solutions as the original equation. Here students learn that each time we perform a move that creates one or more new equations, we are in fact creating a new system that is equivalent to the original system. Equivalent systems are systems with the same solution set, and we can write a series of them to help us get closer to finding the solution of an original system. For instance, if $$4x+y=1$$ and $$x+2y=9$$ form a system, and $$4x+8y=36$$ is a multiple of the second equation, the equations $$4x+y=1$$ and $$4x+8y=36$$ form an equivalent system that can help us eliminate $$x$$ and make progress toward finding the value of $$y$$. Students also learn to make an argument that explains why each new system is indeed equivalent to the one that came before it (MP3), building on the work of justifying equivalent equations in earlier lessons. ### Launch Arrange students in groups of 2. Give students 2–3 minutes of quiet work time, and then 1–2 minutes to discuss their thinking with their partner. Follow with a whole-class discussion. ### Student Facing Here is a system you solved by graphing earlier. \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1 &\quad&\text{Equation A}\\ x + 2y &= \hspace {2mm} 9&\quad&\text{Equation B} \end{align} \end{cases} To start solving the system, Elena wrote: \begin {align} 4x + \hspace{2.2mm} y &= \hspace {2mm}1\\ 4x + 8y &= 36 \end{align} And then she wrote: \begin {align} 4x + \hspace{2.2mm} y &= \hspace {3mm}1\\ 4x + 8y &= \hspace{1mm}36 \hspace{1.5mm}- \\ \overline {\hspace{8mm}\text-7y} &\overline{\hspace{1mm}=\text-35 \hspace{5mm}}\end{align} 1. What were Elena's first two moves? What might be possible reasons for those moves? 2. Complete the solving process algebraically. Show that the solution is indeed $$x=\text-1, y=5$$. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Anticipated Misconceptions Some students may be confused by the subtraction symbol after the second equation, wondering if some number is supposed to appear after the sign. Encourage them to ignore the sign at first, find the relationship between the three equations, and then think about what the sign might mean. ### Activity Synthesis Invite students to share their analyses of Elena's moves. Highlight responses that point out that Elena's moves enabled her to eliminate the $$x$$-variable. (In other words, multiplying equation B by 4 gives $$4x + 8y=36$$ and subtracting this equation from equation A removes the $$x$$-variable.) Display for all to see the two original equations in the system and the new equations Elena wrote ($$4x + 8y=36$$ and $$\text-7y=\text-35$$). Then, ask students to predict what the graphs of all four equations might look like. Next, use graphing technology to display all four graphs. Invite students to share their observations about the graphs. Students are likely to observe that the graphs all intersect at the same point, $$(\text-1,5)$$ and that there are only three lines, instead of 4. Discuss why only three lines are visible. Make sure students understand that this is because the equations $$x+2y =9$$ (equation B) and $$4x+8y=36$$ are equivalent, so they share all the same solutions. Then, focus students' attention on two things: the series of systems that came into play in solving the original system, and the explanations that justify each step along the way. Display the following systems and sequence the discussion as follows: \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ x + 2y &=9&\quad&\text{B} \end{align} \end{cases} • "Here are the two equations in the original system. In solving the system, what do we assume about the $$x$$- and $$y$$-values in the equations?" (We assume that there is a pair of $$x$$- and $$y$$-values that make both equations true.) \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ 4x + 8y &= 36&\quad&\text{B1} \end{align} \end{cases} • "We didn't use the original two equations to solve. Instead, we multiplied each side of equation B by 4 to get equation B1. How do we know that the same $$(x,y)$$ pair is also a solution to equation B1?" (Multiplying each side of equation B by the same number gives an equation that is equivalent to B. This means it has all the same solutions as B, including the pair that made the original system true.) \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ \text-7y &= \text-35&\quad&\text{C} \end{align} \end{cases} • "We couldn't yet solve the system with equations A and B1, so we subtracted B1 from A and got equation C. How do we know that the same $$(x,y)$$ pair from earlier is also a solution to equation C?" (When we subtracted $$4x+8y$$ from $$4x+1$$ and subtracted 36 from 1, we subtracted equal amounts from each side of a true equation, which kept the two sides equal. Even though the $$x$$-value was eliminated in the result, the $$y$$-value that makes the original equations true hasn't changed and is also a solution to C.) \begin {cases}\begin {align} 4x + \hspace{2.2mm} y &=1 &\quad&\text{A}\\ y &= 5&\quad&\text{Solution} \end{align} \end{cases} • Solving equation C gives us $$y=5$$. How do we find the $$x$$-value?" (Substituting this value into equation A or B and solving it gives us the $$x$$-value.) \begin {cases}\begin {align} x&= \text-1 &\quad&\text{Solution}\\ y &= \hspace {1.5mm}5 &\quad&\text{Solution}\end{align} \end{cases} • "If we substitute this pair of values for $$x$$ and $$y$$ in equations A, B, B1, and C and evaluate the expressions, can we expect to find true statements?" (Yes. For A, it will be $$1=1$$. For B, it will be $$9=9$$. For B1, it will be 36=36. For C, it will be $$\text-35=\text-35$$.) Explain that what we have done was to create equivalent systems—systems with the exact same solution set—to help us get closer and closer to the solution of the original system. One way to create an equivalent system is by multiplying one or both equations by a factor. It helps to choose the factor strategically—one that would allow one variable to be eliminated when the two equations in the new system are added or subtracted. Elena chose to multiply equation B by 4 so that the $$x$$-variable could be eliminated. Ask students: • "Suppose we want to solve the system by first eliminating the $$y$$-variable. What factor should we choose? Which equation should we apply it to?" (We could multiply equation A by 2, or multiply equation B by $$\frac12$$.) • "Could we multiply equation A by 6 and multiply equation B by -3 as a way to eliminate the $$y$$-variable?" (Yes. Each new equation ($$24x+6y=6$$ and $$\text-3x -6y = \text-27$$) would be equivalent to the original. Adding the two new equations would eliminate the $$y$$-variable.) ## 16.3: What Comes Next? (10 minutes) ### Activity Earlier, students encountered the idea of equivalent systems. They described the moves that lead to equivalent equations and explained why the solution to those equations was the same as the solution to the original system. This activity reinforces that work. Given an unordered set of equivalent systems, students work with a partner to arrange the systems such that the order constitutes logical steps toward solving an original system. Along the way, they take turns describing how each system came from the previous one and explaining how they know it has the same solution as its predecessor. The work allows students to practice constructing logical arguments and critiquing those of others (MP3). Here is an image of the cards for reference and planning. ### Launch Arrange students in groups of 2. Give one set of pre-cut slips or cards from the blackline master to each group. Ask students to find the slip or card that shows a system of equations and is labeled “Start here.” Explain that all the other slips contain equivalent systems that represent steps in solving the starting system. Ask students to arrange the slips in the order that would lead to its solution, and to make sure they can explain what moves take each system to the next system and why each system is equivalent to the one before it. Partners should take turns finding the next step in the solving process and explaining their reasoning. As one student explains, the partner's job is to listen and make sure they agree and the explanation makes sense. If they disagree, the partners should discuss until they reach an agreement. Give students 5 minutes to arrange the systems. Follow with a whole-class discussion. Speaking, Representing: MLR8 Discussion Supports. Use this routine to support student conversation as they take turns selecting cards and explaining how each system came from the previous one. Display the following prompts for students to use as they respond to reasoning shared by the partner who selected the card: “I agree because….” or “I disagree because….” Encourage students not only to challenge each other if they disagree with the card selected, but also to press for clear explanations that use mathematical language. Design Principle(s): Support sense-making Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organization and problem solving. For example, guide students in a strategy by prompting them to begin by identifying the first and last cards. Next, ask students to identify the card where the solution for $$y$$ first appears, and to explain how they know. Encourage students to use these three cards as benchmarks for sorting the remaining cards. Supports accessibility for: Organization; Attention ### Student Facing Your teacher will give you some slips of paper with systems of equations written on them. Each system represents a step in solving this system: $$\begin {cases}\frac45 x + 6y = 15\\ \text-x + 18y = 11 \end{cases}$$ Arrange the slips in the order that would lead to a solution. Be prepared to: • Describe what move takes one system to the next system. • Explain why each system is equivalent to the one before it. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Student Facing #### Are you ready for more? This system of equations has solution $$(5,\text-2)$$$$\begin {cases}Ax - By = 24\\ Bx + Ay = 31 \end{cases}$$ Find the missing values $$A$$ and $$B$$. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response. ### Anticipated Misconceptions Students who are thinking algorithmically about solving systems of equations may think that the first step should involve multiplying each side of the second equation by $$\frac45$$, becoming frustrated when no cards show that step. Encourage them to compare both the first and the second equations in the starting card to the first and second equations on other cards to gain some ideas about what steps might have been taken. ### Activity Synthesis Ask students to share the order in which the systems should be arranged to lead to the solution. Display the ordered systems for all to see. Point out to students that this particular solution path involves multiplying each of the two equations by a factor in order to eliminate the $$x$$-variable. Ask students if it's possible to eliminate a variable by multiplying only one equation by a factor. (Yes, we could multiply the first equation by $$\frac54$$ to eliminate $$x$$, or by 3 to eliminate $$y$$. Or we could multiply the second equation by $$\frac45$$ to eliminate $$x$$, or by $$\frac13$$ to eliminate $$y$$.) See Lesson Synthesis for discussion questions and ways to help students connect the ideas in the lesson. ## 16.4: Build Some Equivalent Systems (15 minutes) ### Optional activity This optional activity gives students another opportunity to identify moves that lead to equivalent systems and to explain why the resulting systems have the same solution (MP3). It also prompts students to build their own equivalent systems, which encourages them to think strategically about what to do to reach the goal of solving the system, rather than on a particular solution path. As students work, prompt students to articulate the reasoning and assumptions. Ask questions such as: • "How do you know the factor to use when multiplying an equation so a variable can be eliminated?" • "Can the factor be a fraction? A negative number? Why or why not?" • "Why doesn't adding one equation to another equation change the solution of the system?" ### Launch Keep students in groups of 2, if desired. ### Student Facing Here is a system of equations: \begin {cases} \begin {align}12a + 5b &= \text-15\\8a + \hspace{2mm}b &= \hspace{1.5mm}11 \end{align} \end {cases} 1. To solve this system, Diego wrote these equivalent systems for his first two steps. Step 1: \begin {cases} \begin {align}12a + \hspace{1.5mm}5b &= \text-15\\\text-40a + \text-5b &= \text-55 \end{align} \end {cases} Step 2: \begin {cases} \begin {align}12a + 5b &= \text-15\\\text-28a \hspace{8.5mm}&= \text-70 \end{align} \end {cases} Describe the move that Diego made to get each equivalent system. Be prepared to explain how you know the systems in Step 1 and Step 2 have the same solution as the original system. 2. Write another set of equivalent systems (different than Diego's first two steps) that will allow one variable to be eliminated and enable you to solve the original system. Be prepared to describe the moves you make to create each new system and to explain why each one has the same solution as the original system. 3. Use your equivalent systems to solve the original system. Then, check your solution by substituting the pair of values into the original system. ### Student Response Teachers with a valid work email address can click here to register or sign in for free access to Student Response. ### Anticipated Misconceptions If students struggle to create an equivalent system of their own, ask them to start by deciding on a variable they'd like to eliminate. Then, ask them to think about a factor that, when multiplied to one equation, would produce the same or opposite coefficients for that variable. If they are uncomfortable using a fractional factor, ask if they could find a factor to apply to each equation such that the resulting equations have the same or opposite coefficients for the variable they wish to eliminate. ### Activity Synthesis Invite students with different first steps to display their equivalent systems and solution paths. Prompt them (or others students) to explain why each system generated share the same solution as the original system or the system before it. Verify that, regardless of the moves made, the different paths all led to the same pair of values. ## Lesson Synthesis ### Lesson Synthesis Refer to the systems that students have ordered in the "What Comes Next?" activity, Invite students to explain what move takes one system to the next and how they know the new system is equivalent to the one before it (even though one equation has been transformed). As students share their responses, record the descriptions of moves and the justifications for all to see. Consider using a graphic organizer, as shown here. (A completed organizer could be preserved and used as a reference by students later.) step system what move was made? why is the new system equivalent to the previous one? 0 $$\begin {cases}\frac45 x + 6y = 15\\ \text-x + 18y = 11 \end{cases}$$ 1 \begin {cases} \begin{align}4x + 30y = 75\\ \text-x + 18y = 11 \end{align} \end{cases} 2 . . . If time is limited, focus on discussing the justifications on why one system in the sequence is equivalent to the before it (the last column in the graphic organizer). Highlight statements such as: "Adding equal amounts to the two sides of an equation keeps the two sides equal, so the same $$x$$- and $$y$$-values that make the first equation true also makes this new equation true. This means that same pair of values is also a solution to the new system." If no students explained in this way, demonstrate it. ## 16.5: Cool-down - Make Your Move (5 minutes) ### Cool-Down Teachers with a valid work email address can click here to register or sign in for free access to Cool-Downs. ## Student Lesson Summary ### Student Facing We now have two algebraic strategies for solving systems of equations: by substitution and by elimination. In some systems, the equations may give us a clue as to which strategy to use. For example: $$\begin{cases} y=2x-11 \\ 3x+2y=18 \\ \end{cases}$$ In this system, $$y$$ is already isolated in one equation. We can solve the system by substituting $$2x-11$$ for $$y$$ in the second equation and finding $$x$$. \begin{cases} \begin {align} 3x-y&=\text-17 \\ \text-3x+4y&=23 \\ \end {align} \end{cases} This system is set up nicely for elimination because of the opposite coefficients of the $$x$$-variable. Adding the two equations eliminates $$x$$ so we can solve for $$y$$. In other systems, which strategy to use is less straightforward, either because no variables are isolated, or because no variables have equal or opposite coefficients. For example: \begin{cases} \begin {align} 2x+3y&=15 \quad&\text{Equation A}\\ 3x-9y&=18 \quad&\text{Equation B} \\ \end{align}\end{cases} To solve this system by elimination, we first need to rewrite one or both equations so that one variable can be eliminated. To do that, we can multiply both sides of an equation by the same factor. Remember that doing this doesn't change the equality of the two sides of the equation, so the $$x$$- and $$y$$-values that make the first equation true also make the new equation true. There are different ways to eliminate a variable with this approach. For instance, we could: • Multiply Equation A by 3 to get $$6x +9y = 45$$. Adding this equation to Equation B eliminates $$y$$. \displaystyle \begin{cases} \begin {align} 6x+9y&=45 &\quad&\text{Equation A1} \\ 3x-9y&=18 &\quad&\text{Equation B}\end{align}\end{cases} • Multiply Equation B by $$\frac23$$ to get $$2x - 6y = 12$$. Subtracting this equation from Equation A eliminates $$x$$. \begin{cases} \begin {align} 2x+3y&=15 &\quad&\text{Equation A}\\ 2x - 6y &=12 &\quad&\text{Equation B1} \\ \end{align}\end{cases} • Multiply Equation A by $$\frac12$$ to get $$x+\frac32y = 7\frac12$$ and multiply Equation B by $$\frac13$$ to get $$x - 3y = 6$$. Subtracting one equation from the other eliminates $$x$$. \begin{cases} \begin {align} x+\frac32y&= 7\frac12 &\quad&\text{Equation A2}\\ x - 3y &= 6 &\quad&\text{Equation B2} \\ \end{align}\end{cases} Each multiple of an original equation is equivalent to the original equation. So each new pair of equations is equivalent to the original system and has the same solution. Let’s solve the original system using the first equivalent system we found earlier. \displaystyle \begin{cases} \begin {align} 6x+9y&=45 &\quad&\text{Equation A1} \\ 3x-9y&=18 &\quad&\text{Equation B}\end{align}\end{cases} • Adding the two equations eliminates $$y$$, leaving a new equation $$9x=63$$, or $$x=7$$. \displaystyle \begin {align} 6x+9y&=45 \\ 3x-9y&=18 \quad+\\ \overline {9x + 0\hspace{2mm}}&\overline{= 63}\\x &=7 \end{align} • Putting together $$x=7$$ and the original $$3x-9y=18$$ gives us another equivalent system. \displaystyle \begin{cases} \begin {align} x&=7 \\ 3x-9y&=18 \end{align}\end{cases} • Substituting 7 for $$x$$ in the second equation allows us to solve for $$y$$. \displaystyle \begin{align} 3(7) - 9y &=18\\ 21- 9y&=18\\ \text-9y &= \text-3\\ y&=\frac13 \end{align} When we solve a system by elimination, we are essentially writing a series of equivalent systems, or systems with the same solution. Each equivalent system gets us closer and closer to the solution of the original system. \begin{cases} \begin {align} 2x+3y&=15\\ 3x-9y&=18\\ \end{align}\end{cases} \displaystyle \begin{cases} \begin {align} 6x+9y&=45\\ 3x-9y&=18 \end{align}\end{cases} \displaystyle \begin{cases} \begin {align} x&=7 \\ 3x-9y&=18 \end{align}\end{cases} \displaystyle \begin{cases} \begin {align} x&=7 \\ y&=\frac13\end{align}\end{cases}
2009 AMC 8 Problems/Problem 21 Problem Andy and Bethany have a rectangular array of numbers with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$. Bethany adds the numbers in each column. The average of her $75$ sums is $B$. What is the value of $\frac{A}{B}$? $\textbf{(A)}\ \frac{64}{225} \qquad \textbf{(B)}\ \frac{8}{15} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{225}{64}$ Solution First, note that $40A=75B=\text{sum of the numbers in the array}$. Solving for $\frac{A}{B}$, we get $\frac{75}{40} =\frac{15}{8} =\frac{A}{B}$. $\boxed{\text{D}}$.
# Eureka Math Algebra 1 Module 5 Lesson 2 Answer Key ## Engage NY Eureka Math Algebra 1 Module 5 Lesson 2 Answer Key ### Eureka Math Algebra 1 Module 5 Lesson 2 Example Answer Key Example 1. Noam and Athena had an argument about whether it would take longer to get from NYC to Boston and back by car or by train. To settle their differences, they made separate, nonstop round trips from NYC to Boston. On the trip, at the end of each hour, both recorded the number of miles they had traveled from their starting points in NYC. The tables below show their travel times, in hours, and the distances from their starting points, in miles. The first table shows Noamâ€™s travel time and distance from the starting point, and the second represents Athenaâ€™s. Use both data sets to justify your answers to the questions below. a. Who do you think is driving, and who is riding the train? Explain your answer in the context of the problem. This is an open – ended question with no right or wrong answer. Check for mathematically sound reasoning based on the data. Sample Response: It appears that Athena is riding the train since she was able to go 81 miles in the first hour, and Noam was able to go only 55, which is a typical highway speed. Also, Athena made the round trip in 10 hours, while Noam made it in 8, so it looks like the train may have had to stop at stations when it was near Boston, slowing its progress considerably during the 2 hours when the train was outside of Boston. b. According to the data, how far apart are Boston and New York City? Explain mathematically. This is an open – ended question with no right or wrong answer. Check for mathematically sound reasoning based on the data. Sample Response: Based on the symmetry of the values in the table, Noamâ€™s maximum distance was 220 miles (at 4 hours). Using the same symmetry, Athenaâ€™s maximum distance was 225 miles. They may have started or ended at different places, but it is more likely that the route for the train was slightly different from that of the car. c. How long did it take each of them to make the round trip? Letâ€™s call Noamâ€™s distance N and Athenaâ€™s distance A. Noam was traveling for 8 hours and Athena for 10 hours. The zeros for N are (0,0) and (8,0) and for A are (0,0) and (10,0). It took Noam 8 hours; it took Athena 10 hours. d. According to their collected data, which method of travel was faster? If we assume that Noam was traveling by car, then the car was faster by 2 hours, overall. However, the speed of the train was faster for the first and the last hour and then slowed. e. What was the average rate of change for Athena for the interval from 3 to 4 hours? How might you explain that in the context of the problem? She only traveled 27 miles per hour: $$\frac{216 – 189}{4 – 3}$$. Since she was likely on the train, there may have been stops during that time period. f. Noam believes a quadratic function can be used as a model for both data sets. Do you agree? Use and describe the key features of the functions represented by the data sets to support your answer. The two data sets have several things in common. Both are symmetric, both have a vertex, and both have (0,0) as one of the x – intercepts and the y – intercept. The other x – intercept is (8,0) for Noam and (10,0) for Athena. However, Noamâ€™s data set has a positive constant difference of + 55 on one side of the vertex, which then changes to – 55 on the other side. Noamâ€™s data set can be best modeled by an absolute value or other piecewise function. Athenaâ€™s data does not have a constant first difference, so it is not linear. When we check the second differences, we find it to be constant ( – 18). Therefore, Athenaâ€™s trip can be modeled with a quadratic function. ### Eureka Math Algebra 1 Module 5 Lesson 2 Exercise Answer Key Opening Exercise When tables are used to model functions, we typically have just a few sample values of the function and therefore have to do some detective work to figure out what the function might be. Look at these three tables: â†’ What do you notice about the three sets of data? Can you identify the type of function they represent? Students may observe some of the examples listed below. It is also possible that they will not notice much of anything, or they may have ideas that are incorrect. Just let them brainstorm, and then come back to their ideas later to see how many were helpful. â†’ The orange set is growing much faster than the first two. Many of the function values in the blue and green are the same. It is difficult to know the function by looking at only these numbers for the first two. However, I notice that the function values in the green set were growing steadily and then decreased at the end. It might help if we had more data points. â†’ Have students make conjectures about the type of function they believe each data set represents and support their conjectures with evidence, which might include a graph. Now show a data plot for each: â†’ Now can you see the trends more clearly? â†’ Yes, it is obvious that the first is linear. It is clearer that the second is quadratic. And the third looks to be exponential. â†’ We learned to determine the type of function by looking at the shapes of graphs. Can we also determine the type of function by just using its table of values? Analyze each table and find any special patterns that may help determine the type of function it represents. â†’ The table of values for the linear function has a constant first difference of 6, indicating the graph will have a slope of 6. The quadratic function does not have a constant difference like a linear function. However, the difference of differences (second difference) is constant. The exponential function does not have constant first or second differences. However, the y – value is multiplied by a constant value of 3. It should be noted that the x – values are all increasing by one. Not all data are presented where the x – values are increasing by a constant value, so it is necessary to interpolate values to produce equal – sized intervals. Exercises Exercise 1. Explain why each function can or cannot be used to model the given data set. a. f(x) = 3x + 5 This function cannot be used to model the data set. The y – intercept is 5, but the first difference is not constant, and the data set is not a linear function. b. f(x) = – (x – 2)2 + 9 This function can be used to model the data set. The second difference has a constant value of – 2; therefore, it is a quadratic function. The vertex is (2,9), and it is a maximum since the leading coefficient is negative. c. f(x) = – x2 + 4x â€“5 This function cannot be used to model the data set. The y – intercept for this equation is – 5 instead of 5. d. f(x) = 3x + 4 This function cannot be use used to model the data set. The y – intercept is 5, and f(x) values are not being multiplied by a constant value. It is not an exponential function. e. f(x) = (x – 2)2 + 9 This function cannot be used to model the data set. The vertex is (2,9); however, it is a minimum in this function. f. f(x) = – (x + 1)(x – 5) This function can be used to model the data set. One of the x – intercepts is x = 5, and the second x – intercept is x = – 1 by following the pattern of the data. The function equation indicates x – intercepts of – 1 and 5, where the vertex is a maximum value of the function. Exercise 2. Match each table below to the function and the context, and explain how you made your decision. Equations: f(x) = 12x h(x) = – 9\|x – 3\| + 27 g(x) = – (x)(x – 6) p(x) = 2x q(x) = – 16x2 + 30x + 160 Contexts: 1. The population of bacteria doubled every month, and the total population vs. time was recorded. 2. A ball was launched upward from the top of a building, and the vertical distance of the ball from the ground vs. time was recorded. 3. The height of a certain animalâ€™s vertical leap was recorded at regular time intervals of one second; the animal returned to ground level after six seconds. 4. Melvin saves the same amount of money every month. The total amount saved after each month was recorded. 5. Chris ran at a constant rate on a straight – line path and then returned at the same rate. His distance from his starting point was recorded at regular time intervals. ### Eureka Math Algebra 1 Module 5 Lesson 2 Problem Set Answer Key Question 1. a. Determine the function type that could be used to model the data set at the right, and explain why. b. Complete the data set using the special pattern of the function you described in part (a). See the completed table for answers. c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why. Minimum value occurs at (4, – 8). Question 2. a. Determine the function type that could be used to model the data set, and explain why. Exponentialâ€”y – value is being multiplied by constant value 4. b. Complete the data set using the special pattern of the function you described in part (a). See responses in the completed table. c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why. Since this is an exponential function, the y – values increase as x – values increase, and there is no maximum value. As x gets smaller and smaller (i.e., moves further to the left on the number line), the y – values approach 0 but never reach 0 and never become negative. Therefore, there is no minimum value of this function. Question 3. a. Determine the function type that could be used to model the data set, and explain why. Linearâ€”the first difference is 6. b. Complete the data set using the special pattern of the function you described in part (a). See responses in the completed table. c. If it exists, find the minimum or maximum value for the function model. If there is no minimum or maximum, explain why. There is no minimum or maximum value for a linear function (except for a horizontal line). For this function, the y – values decrease as the x – values decrease, and the y – values increase as the x – values increase. Question 4. Circle all the function types that could possibly be used to model a context if the given statement applies. a. When x – values are at regular intervals, the first difference of y – values is not constant. Linear FunctionÂ Â Â Â Â Â Quadratic FunctionÂ Â Â Â Â Exponential FunctionÂ Â Â Â Â Â Absolute Value Function b. When x – values are at regular intervals, the second difference of y – values is not constant. Linear FunctionÂ Â Â Â Â Â Quadratic FunctionÂ Â Â Â Â Exponential FunctionÂ Â Â Â Â Absolute Value Function c. When x – values are at regular intervals, the quotient of any two consecutive y – values is a constant that is not equal to 0 or 1. Linear FunctionÂ Â Â Â Â Quadratic FunctionÂ Â Â Â Â Exponential FunctionÂ Â Â Â Â Â Absolute Value Function d. There may be up to two different x – values for y = 0. Linear FunctionÂ Â Â Â Â Quadratic FunctionÂ Â Â Â Â Â Exponential FunctionÂ Â Â Â Â Absolute Value Function ### Eureka Math Algebra 1 Module 5 Lesson 2 Exit Ticket Answer Key Question 1. Analyze these data sets, recognizing the unique pattern and key feature(s) for each relationship. Then use your findings to fill in the missing data, match to the correct function from the list on the right, and describe the key feature(s) that helped you choose the function. Equations: f(x) = 6x h(x) = – 3(x – 2)2 + 18 g(x) = – 2(x + 1)(x – 3) r(x) = 4x + 6 Table A: __________________ Key Feature(s): ________________________________________________________ Table B: __________________ Key Feature(s): ________________________________________________________ Table C: __________________ Key Feature(s): ________________________________________________________ Table D: __________________ Key Feature(s): ________________________________________________________
### The Determinant There is another way to solve systems of equations with three variables. It involves a quantity called the determinant. Every m×m matrix has a unique determinant. The determinant is a single number. To find the determinant of a 2×2matrix, multiply the numbers on the downward diagonal and subtract the product of the numbers on the upward diagonal: A = detA = a1b2 - a2b1. For example, det = 4(6) - (- 1)(- 2) = 24 - 2 = 22 To find the determinant of a 3×3 matrix, copy the first two columns of the matrix to the right of the original matrix. Next, multiply the numbers on the three downward diagonals, and add these products together. Multiply the numbers on the upward diagonals, and add these products together. Then subtract the sum of the products of the upward diagonals from the sum of the product of the downward diagonals (subtract the second number from the first number): A = Example: Find the determinant of: Solution: Step 1 Step 2 Step 3 Step 4 10 - 80 = -70. detA = - 70. ### Cramer's Rule Recall the general 3×4 matrix used to solve systems of three equations: This matrix will be used to solve systems by Cramer's Rule. We divide it into four separate 3×3 matrices: D = Dx = Dy = Dz = D is the 3×3 coefficient matrix, and Dx, Dy, and Dz are each the result of substituting the constant column for one of the coefficient columns in D. Cramer's Rule states that: x = y = z = Thus, to solve a system of three equations with three variables using Cramer's Rule, 1. Arrange the system in the following form: a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 2. Create D, Dx, Dy, and Dz. 3. Find detD, detDx, detDy, and detDz. 4. x = , y = , and z = . Note: If detD = 0 and detDx, detDy, or detDz≠ 0, the system is inconsistent. If detD = 0 and detDx = detDy = detDz = 0, the system has multiple solutions.
# SAT Math : How to find the nth term of an arithmetic sequence ## Example Questions ← Previous 1 ### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence 2, 8, 14, 20 The first term in the sequence is 2, and each following term is determined by adding 6. What is the value of the 50th term? 302 320 300 296 296 Explanation: We start by multiplying 6 times 46, since the first 4 terms are already listed. We then add the product, 276, to the last listed term, 20. This gives us our answer of 296. ### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence Which of the following could not be a term in the sequence 5, 10, 15, 20...? 3751 10005 2500 35 3751 Explanation: All answers in the sequence must end in a 5 or a 0. ### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence In an arithmetic sequence, each term is two greater than the one that precedes it. If the sum of the first five terms of the sequence is equal to the difference between the first and fifth terms, what is the tenth term of the sequence? 10 –0.6 15.6 0.6 2.4 15.6 Explanation: Let a1 represent the first term of the sequence and an represent the nth term. We are told that each term is two greater than the term that precedes it. Thus, we can say that: a2 = a1 + 2 a3 = a1 + 2 + 2 = a1 + 2(2) a4 = a1 + 3(2) a5 = a1 + 4(2) an = a1 + (n-1)(2) The problem tells us that the sum of the first five terms is equal to the difference between the fifth and first terms. Let's write an expression for the sum of the first five terms. sum = a1 + (a1 + 2) + (a1 + 2(2)) + (a1 + 3(2)) + (a1 + 4(2)) = 5a1 + 2 + 4 + 6 + 8 = 5a1 + 20 Next, we want to write an expression for the difference between the fifth and first terms. a5 - a1 = a1 + 4(2) – a1 = 8 Now, we set the two expressions equal and solve for a1. 5a1 + 20 = 8 Subtract 20 from both sides. 5a1 = –12 a1 = –2.4. The question ultimately asks us for the tenth term of the sequence. Now, that we have the first term, we can find the tenth term. a10 = a1 + (10 – 1)(2) a10 = –2.4 + 9(2) = 15.6 ### Example Question #4 : How To Find The Nth Term Of An Arithmetic Sequence In a certain sequence, an+1 = (an)2 – 1, where an represents the nth term in the sequence. If the third term is equal to the square of the first term, and all of the terms are positive, then what is the value of (a2)(a3)(a4)? 6 63 72 24 48 48 Explanation: Let a1 be the first term in the sequence. We can use the fact that an+1 = (an)2 – 1 in order to find expressions for the second and third terms of the sequence in terms of a1. a2 = (a1)2 – 1 a3 = (a2)2 – 1 = ((a1)2 – 1)2 – 1 We can use the fact that, in general, (a – b)2 = a2 – 2abb2 in order to simplify the expression for a3. a= ((a1)2 – 1)2 – 1 = (a1)4 – 2(a1)2 + 1 – 1 = (a1)4 – 2(a1)2 We are told that the third term is equal to the square of the first term. a3 = (a1)2 We can substitute (a1)4 – 2(a1)for a3. (a1)4 – 2(a1)= (a1)2 Subtract (a1)2 from both sides. (a1)4 – 3(a1)= 0 Factor out (a1)from both terms. (a1)2 ((a1)2 – 3) = 0 This means that either (a1)= 0, or (a1)2 – 3 = 0. If (a1)= 0, then a1 must be 0. However, we are told that all the terms of the sequence are positive. Therefore, the first term can't be 0. Next, let's solve (a1)2 – 3 = 0. (a1)= 3 Take the square root of both sides. a1 = ±√3 However, since all the terms are positive, the only possible value for a1 is √3. Now, that we know that a1 = √3, we can find a2, a3, and a4. a2 = (a1)2 – 1 = (√3)2 – 1 = 3 – 1 = 2 a3 = (a2)2 – 1 = 2– 1 = 4 – 1 = 3 a4 = (a3)2 – 1 = 32 – 1 = 9 – 1 = 8 The question ultimately asks for the product of the a2, a3, and a4, which would be equal to 2(3)(8), or 48. ### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence In the given sequence, the first term is 3 and each term after is one less than three times the previous term. What is the sixth term in the sequence? Explanation: The fourth term is: 3(23) – 1 = 69 – 1 = 68. The fifth term is: 3(68) – 1 = 204 – 1 = 203. The sixth term is: 3(203) – 1 = 609 – 1 = 608. ### Example Question #1 : How To Find The Nth Term Of An Arithmetic Sequence Consider the following sequence of numbers: What will be the 8th term in the sequence? 49 58 60 56 51 51 Explanation: Each number in the sequence in 7 more than the number preceding it. The equation for the terms in an arithmetic sequence is an = a1 + d(n-1), where d is the difference. The formula for the terms in this sequence is therefore an = 2 + 7(n-1). Plug in 8 for n to find the 8th term: a8 = 2 + 7(8-1) = 51 ### Example Question #7 : How To Find The Nth Term Of An Arithmetic Sequence Find the seventh term in the following sequence: Explanation: The difference between each term can be found through subtraction. For example the difference between the first and the second term can be found as follows: One can check and see that this is the case for the other given numbers in the sequence as well. In order to find the seventh term, expand the sequence by adding 14 to the last given number (4th number) and all of the following numbers until the 7th number in the sequence is reached. This gives the sequence: As seen above the seventh number in the sequence is 87 and the correct answer. ### Example Question #8 : How To Find The Nth Term Of An Arithmetic Sequence What is the tenth number in the sequence: Explanation: The purpose of this question is to understand the patterns of sequences. First, an equation for the term in the sequence must be determined (). This is true because will create , will create , will create , will create . Then, the eqution must be applied to find the specified term. For the tenth term, the expression must be evaluated, yielding 103. ### Example Question #1 : Nth Term Of An Arithmetic Sequence You are given a sequence with the same difference between consecutive terms. We know it starts at and its 3rd term is . Find its 10th term. Explanation: From the given information, we know , which means each consecutive difference is 3. ### Example Question #10 : How To Find The Nth Term Of An Arithmetic Sequence An arithmetic sequence begins as follows: Give the sixteenth term of this sequence. None of the other responses give the correct answer.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy. You are viewing an older version of this Concept. Go to the latest version. # Alternate Exterior Angles ## Angles on opposite sides of a transversal, but outside the lines it intersects. Estimated6 minsto complete % Progress Practice Alternate Exterior Angles Progress Estimated6 minsto complete % Alternate Exterior Angles What if you were presented with two angles that are on opposite sides of a transversal, but outside the lines? How would you describe these angles and what could you conclude about their measures? After completing this Concept, you'll be able to answer these questions using your knowledge of alternate exterior angles. ### Watch This Watch the portions of this video dealing with alternate exterior angles. ### Guidance Alternate Exterior Angles are two angles that are on the exterior of \begin{align}l\end{align} and \begin{align}m\end{align}, but on opposite sides of the transversal. \begin{align}\angle 1\end{align} and \begin{align}\angle 8\end{align} are alternate exterior angles. Alternate Exterior Angles Theorem: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent. The proof of this theorem is very similar to that of the Alternate Interior Angles Theorem. Converse of the Alternate Exterior Angles Theorem: If two lines are cut by a transversal and the alternate exterior angles are congruent, then the lines are parallel. #### Example A Using the picture above, list all the pairs of alternate exterior angles. Alternate Exterior Angles: \begin{align}\angle 2\end{align} and \begin{align}\angle 7\end{align}, \begin{align}\angle 1\end{align} and \begin{align}\angle 8\end{align}. #### Example B Find \begin{align}m \angle 1\end{align} and \begin{align}m \angle 3\end{align}. \begin{align}m \angle 1 = 47^\circ\end{align} because they are vertical angles. Because the lines are parallel, \begin{align}m \angle 3 = 47^\circ\end{align} by the Corresponding Angles Theorem. Therefore, \begin{align}m \angle 2 = 47^\circ\end{align}. \begin{align}\angle 1\end{align} and \begin{align}\angle 3\end{align} are alternate exterior angles. #### Example C The map below shows three roads in Julio’s town. Julio used a surveying tool to measure two angles at the intersections in this picture he drew (NOT to scale). Julio wants to know if Franklin Way is parallel to Chavez Avenue. The labeled \begin{align}130^\circ\end{align} angle and \begin{align}\angle a\end{align} are alternate exterior angles. If \begin{align}m \angle a = 130^\circ\end{align}, then the lines are parallel. To find \begin{align}m \angle a\end{align}, use the other labeled angle which is \begin{align}40^\circ\end{align}, and its linear pair. Therefore, \begin{align}\angle a + 40^\circ = 180^\circ\end{align} and \begin{align}\angle a = 140^\circ\end{align}. \begin{align}140^\circ \neq 130^\circ\end{align}, so Franklin Way and Chavez Avenue are not parallel streets. Watch this video for help with the Examples above. ### Vocabulary Alternate Exterior Angles are two angles that are on the exterior of \begin{align}l\end{align} and \begin{align}m\end{align}, but on opposite sides of the transversal. ### Guided Practice 1. Find the measure of each angle and the value of \begin{align}y\end{align}. 2. Give THREE examples of pairs of alternate exterior angles in the diagram below: Answers: 1. The given angles are alternate exterior angles. Because the lines are parallel, we can set the expressions equal to each other to solve the problem. \begin{align}(3y+53)^\circ & = (7y-55)^\circ\\ 108^\circ & = 4y\\ 27^\circ & = y\end{align} If \begin{align}y = 27^\circ\end{align}, then each angle is \begin{align}3(27^\circ) + 53^\circ\end{align}, or \begin{align}134^\circ\end{align}. 2. There are many examples of alternate exterior angles in the diagram. Here are some possible answers: • \begin{align} \angle 1 \end{align} and \begin{align} \angle 14\end{align} • \begin{align} \angle 2 \end{align} and \begin{align} \angle 13\end{align} • \begin{align} \angle 12 \end{align} and \begin{align} \angle 13\end{align} ### Practice 1. Find the value of \begin{align}x\end{align} if \begin{align}m\angle 1 = (4x + 35)^\circ, \ m\angle 8 = (7x - 40)^\circ\end{align}: 2. Are lines 1 and 2 parallel? Why or why not? For 3-6, what does the value of \begin{align}x\end{align} have to be to make the lines parallel? 1. \begin{align}m\angle 2 = (8x)^\circ\end{align} and \begin{align}m\angle 7 = (11x-36)^\circ\end{align} 2. \begin{align}m\angle 1 = (3x+5)^\circ\end{align} and \begin{align}m\angle 8 = (4x-3)^\circ\end{align} 3. \begin{align}m\angle 2 = (6x-4)^\circ\end{align} and \begin{align}m\angle 7 = (5x+10)^\circ\end{align} 4. \begin{align}m\angle 1 = (2x-5)^\circ\end{align} and \begin{align}m\angle 8 = (x)^\circ\end{align} 5. \begin{align}m\angle 2 = (3x+50)^\circ\end{align} and \begin{align}m\angle 7 = (10x+1)^\circ\end{align} 6. \begin{align}m\angle 1 = (2x-12)^\circ\end{align} and \begin{align}m\angle 8 = (x+1)^\circ\end{align} For 9-12, determine whether the statement is true or false. 1. Alternate exterior angles are always congruent. 2. If alternate exterior angles are congruent then lines are parallel. 3. Alternate exterior angles are on the interior of two lines. 4. Alternate exterior angles are on opposite sides of the transversal. For questions 13-15, use the picture below. 1. What is the alternate exterior angle with \begin{align}\angle 2\end{align}? 2. What is the alternate exterior angle with \begin{align}\angle 7\end{align}? 3. Are the two lines parallel? Explain. ### Vocabulary Language: English alternate exterior angles Alternate exterior angles are two angles that are on the exterior of two different lines, but on the opposite sides of the transversal. ### Explore More Sign in to explore more, including practice questions and solutions for Alternate Exterior Angles. Please wait... Please wait...
# Probability Probability is a way of describing uncertainty in numerical terms. In this lesson, we review some of the terminology used in elementary probability theory: Probability of simple events, Rules of Probability, Mutually Exclusive and Independent Events. A probability experiment, also called a random experiment, is an experiment for which the result, or outcome, is uncertain. We assume that all of the possible outcomes of an experiment are known before the experiment is performed, but which outcome will actually occur is unknown. The set of all possible outcomes of a random experiment is called the sample space, and any particular set of outcomes is called an event. For example, consider a cube with faces numbered 1 to 6, called a 6-sided die. Rolling the die once is an experiment in which there are 6 possible outcomeseither 1, 2, 3, 4, 5, or 6 will appear on the top face. The sample space for this experiment is the set of numbers 1, 2, 3, 4, 5, and 6. Two examples of events for this experiment are (i) rolling the number 4, which has only one outcome, and (ii) rolling an odd number, which has three outcomes. The probability of an event is a number from 0 to 1, inclusive, that indicates the likelihood that the event occurs when the experiment is performed. The greater the number, the more likely the event. In general, for a random experiment with a finite number of possible outcomes, if each outcome is equally likely to occur, then the probability that an event A, written P(A), occurs is defined by the ratio Calculating the Probability of Simple Events. This video shows the basic idea and a few simple examples of calculating the probability of simple events. 1. What is the probability of the next person you meeting having a phone number that ends in 5? 2. What is the probability of getting all heads if you flip three coins? 3. What is the probability that the person you meet next has a birthday in February) (Assume non-leap year) This video introduces probability and determine the probability of basic events. 1. A bag contains 8 marbles numbered 1 to 8. a) What is the probability of selecting 2 from the bag? b) What is the probability of selecting an odd number? c) What is the probability of selecting a number greater than 6? 2. Using a standard deck of cards, determine each probability a) P(face card) b) P(5) c) P(non face card) ## Rules of Probability The following are general facts about probability. If an event E is certain to occur, then P E( ) = 1. If an event E is certain not to occur, then P E( ) = 0. If an event E is possible but not certain to occur, then 0 < P(E) <1. The probability that an event E will not occur is equal to 1 - P E( ). If E is an event, then the probability of E is the sum of the probabilities of the outcomes in E. The sum of the probabilities of all possible outcomes of an experiment is 1. If E and F are two events of an experiment, we consider two other events related to E and F. The event that both E and F occur, that is, all outcomes in the set E ∩ F. The event that E or F, or both, occur, that is, all outcomes in the set E ∪ F. This video explains the rules of probability. ## Mutually Exclusive and Independent Events Events that cannot occur at the same time are said to be mutually exclusive. For example, if a 6-sided die is rolled once, the event of rolling an odd number and the event of rolling an even number are mutually exclusive. But rolling a 4 and rolling an even number are not mutually exclusive, since 4 is an outcome that is common to both events. For events E and F, we have the following rules. P(E or F) = P(E) + P(F) - P(E and F) which is the inclusion-exclusion principle applied to probability. If E and F are mutually exclusive, then P(E and F) = 0 and therefore, P(E or F) = P(E) + P(F) E and F are said to be independent if the occurrence of either event does not affect the occurrence of the other. If two events E and F are independent, then P(E and F) = P(E)P(F) Probability - P(A ∪ B) and Mutually Exclusive Events. Calculating Probability - " And " statements, independent events. This video shows the basic idea, formula, and two examples. How not to confuse the notions of mutually exclusive events and independent events This video shows a way to get the definitions for mutually exclusive, and independent events right. For two events, P(A) > 0 and P(B) > 0 • The events can't be both mutually exclusive and independent • If A and B are independent then A and B are not mutually exclusive • If A and B are mutually exclusive then A and B can't be independent. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solving Equations with Variables on Both Sides Pre-Algebra. ## Presentation on theme: "Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solving Equations with Variables on Both Sides Pre-Algebra."— Presentation transcript: Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solving Equations with Variables on Both Sides Pre-Algebra 10-3 Solving Equations with Variables on Both Sides To solve multistep equations with variables on both sides: 1.First combine like terms. 2.Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. 3.Then use properties of equality to isolate the variable. Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 4x + 6 = x Example 1: 4x + 6 = x – 4x 6 = –3x Subtract 4x from both sides. Divide both sides by –3. –2 = x 6 –3 –3x –3 = Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. B. 9b – 6 = 5b + 18 Example 2: 9b – 6 = 5b + 18 – 5b 4b – 6 = 18 4b4b 4 24 4 = Subtract 5b from both sides. Divide both sides by 4. b = 6 + 6 4b = 24 Add 6 to both sides. Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. C. 9w + 3 = 5w + 7 + 2w Example 3: 9w + 3 = 5w + 7 + 2w 2w + 3 = 7 9w + 3 = 7w + 7Combine like terms. – 7w Subtract 7w from both sides. 2w = 4 – 3 Subtract 3 from both sides. Divide by 2 on both sides. w = 2 Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 10z – 15 – 4z = 8 – 2z - 15 Example 4: 10z – 15 – 4z = 8 – 2z – 15 + 15 +15 6z – 15 = –2z – 7Combine like terms. + 2z Add 2z to both sides. 8z – 15 = – 7 8z = 8 z = 1 Add 15 to both sides. Divide both sides by 8. 8z 8 8 8 = Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Xavier & Jonathon have the same number of super hero action figures in their collections. Xavier has 5 complete sets and 4 individual figures. Jonathon has 3 complete sets and 14 individual figures. How many figures make up a complete set? Let x equal the number of figures in a complete set. 5x + 4 = 3x + 14 Now solve… Pre-Algebra 10-3 Solving Equations with Variables on Both Sides 5x + 4 = 3x + 14 – 3x 2x + 4 = 14 Subtract 3x from both sides. – 4 Subtract 4 from both sides. 2x = 10 2x 2 10 2 = Divide both sides by 2. x = 5Each complete set has 5 super hero action figures. Download ppt "Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solving Equations with Variables on Both Sides Pre-Algebra." Similar presentations
Methods of Integration The standard formulae for integration are only useful when the integrand is given in the ‘standard’ form. For most physical applications or analysis purposes, advanced techniques of integration are required, which reduce the integrand analytically to a suitable solvable form. Two such methods – Integration by Parts, and Reduction to Partial Fractions are discussed here. Both have limited applicability, but are of immense use in solving integrands which may seem unsolvable otherwise. So let’s begin! Integration by Parts Another useful technique for evaluating certain integrals is integration by parts. Here, the integrand is usually a product of two simple functions (whose integration formula is known beforehand). One of the functions is called the ‘first function’ and the other, the ‘second function’.Â The general formula for the Integration by parts method then can be given as – $$\int uv dx = u\int v dx – \int[\frac{d}{dx}(u)\int vdx]dx + c$$ where both u and v are functions of x. u(x) – the first function v(x) – the second function The preference for deciding the first and second functions is usually as follows – Remember the code ILATE for the precedence, where I – Inverse functions L – Logarithmic Functions A – Algebraic Functions T – Trigonometric Functions E – Exponential Functions Application –Â Suppose your integrand is a product of two functions – exponential and logarithmic. On comparing with the ILATE form of precedence, the logarithmic function will be chosen as the first function and the exponential function can then be taken as the second function for easy evaluation. Thus, for solvingÂ âˆ«(ex)log(4x2)dx, we can choose log(4x2) as the first function, and the ex as the second function to easily obtain the result. Method of Partial Fractions This method relies on the fact that the integration of functions of the form $$\frac{1}{f(x)}$$, where f(x) is a linear function with some exponent, can be done quite easily. Thus, the integrands involving polynomial functions in their numerator and denominator are reduced to partial fractions first, to ease the process of integration. Different types of integrands need to be handled differently under this method. Therefore, we have the following three general cases – Assume that the integrand is of the form $$\frac{P(x)}{Q(x)}$$ with both P(x) and Q(x) being polynomials that may be factorized into multiple polynomials. Under every case, we’ll evaluate the partial fractions only under the condition that the degree of P(x) is less than the degree of Q(x). Case I: Q(x) contains non-repeated linear factors Basically, Q(x) in this case should be of the form (x – a)(x – b)(x – c)… while P(x) may be any polynomial that satisfies the condition of its degree being less than that of Q(x). Example – To find the partial fractions corresponding to $$\frac{x}{(2 – x)(x + 3)}$$: Let $$\frac{x}{(2 – x)(x + 3)} = \frac{A}{(2 – x)} + \frac{B}{(x + 3)}$$ $$= \frac{A(x + 3) + B(2 – x)}{(2 – x)(x + 3)}$$ To findÂ AÂ andÂ B, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get – x =Â A(x + 3) + B(2 – x) Now we can proceed by equating the coefficients of equal powers of x on both sides, which is the standard method of solving such algebraic equations; or, we may use a shortcut. Note that on substituting x = -3 in the equation, we get – -3 = A(-3 + 3) + B(2 – (-3)) -3 = 0 + 5B B = -3/5 Similarly, on substituting x = 2; 2 = A(2 + 3) + B(2 – 2) 2 = 5A + 0 A = 2/5 Clearly, all we did was to subsitute the values of x that made the coefficient of A and B = 0 separately. Our final result then can be written as -$$\frac{x}{(2 – x)(x + 3)} = \frac{A}{(2 – x)} + \frac{B}{(x + 3)}$$ $$\frac{x}{(2 – x)(x + 3)} = \frac{2}{5(2 – x)} – \frac{5}{3(x + 3)}$$ You can also see from the shortcut that you may directly substitute the values of x for which the denominators of our partial fractions tend to go to 0 as well. For eg – To make the denominator of $$\frac{B}{(x + 3)}$$ = 0, we substitute x = -3 throughout the equation, after multiplying the entire equation by the denominator ofÂ Â $$\frac{B}{(x + 3)}$$ i.e. (x + 3). We’ll then be left with – $$\frac{x}{(2 – x)(x + 3)}.(x + 3) = \frac{A}{(2 – x)}.(x + 3) + \frac{B}{(x + 3)}.(x + 3)$$ $$\frac{x}{(2 – x)} = \frac{A}{(2 – x)}.(x + 3) + B$$ Now put x = -3 to get – $$\frac{3}{(2 – (-3))} = \frac{A}{(2 – x)}.(-3 + 3) + B$$ $$B = -\frac{5}{3}$$, which is the same result as before. A similar analysis can be done to find A as well. This way, all the partial fractions of an integrand of the form $$\frac{P(x)}{Q(x)}$$ can be found, where the degree of P(x) is less than that of Q(x). Case II: Q(x) contains one or more repeated linear factors Here, Q(x) may have repeated roots and could be of the form (x – a)(x – b)3(x – c)2 In such a case, if a linear factor is repeatedÂ Â times in the denominator, there will beÂ Â corresponding partial fractions with degree 1 toÂ . Example – To find the partial fractions corresponding to $$\frac{1}{(x – 1)^2(x + 1)}$$: Let $$\frac{1}{(x – 1)^2(x + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x – 1)^2} + \frac{C}{(x + 1)}$$ $$= \frac{A(x – 1)(x + 1) + B(x + 1) + C(x – 1)^2}{(x – 1)^2(x + 1)}$$ To findÂ A,B and C, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get – 1 =Â A(x – 1)(x + 1) + B(x + 1) + C(x – 1)2 Now we can proceed by equating the coefficients of equal powers of x on both sides or, once again we may proceed by direct substitution. Note that on substituting x = 1 in the equation, we get – 1 = A(1 – 1)(1 + 1) + B(1 + 1) + C(1 – 1)2 1 = 0 + 2B + 0 B = 1/2 Similarly, on substituting x = -1; 1 = A(-1 – 1)(-1 + 1) + B(-1 + 1) + C(-1 -1)2 1 = 0 + 0 + 4C C = 1/4 Now to get A, we may substitute the values of B and C in the numerator balance equation, along with any arbitrary value of x (besides 1 and -1) to get – 1 =Â A(x – 1)(x + 1) + B(x + 1) + C(x – 1)2 Put B = 1/2 and C = 1/4 1 =Â A(x – 1)(x + 1) + $$\frac{1}{2}$$(x + 1) + $$\frac{1}{4}$$(x – 1)2 Then let us put x = 0 in this case because it seems like an easy value to evaluate the equation for. : 1 = A(0 – 1)(0 + 1)Â + $$\frac{1}{2}$$(0 + 1) + $$\frac{1}{4}$$(0 – 1)2Â : 1 = -A + $$\frac{1}{2}$$ + $$\frac{1}{4}$$ : A = -$$\frac{1}{4}$$ Our final result then can be written as -$$\frac{1}{(x – 1)^2(x + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x – 1)^2} + \frac{C}{(x + 1)}$$ $$Â = – \frac{1}{4(x – 1)} + \frac{1}{2(x – 1)^2} + \frac{1}{4(x + 1)}$$ Similarly, all other partial fractions in the case of a repeated linear root may be found. Â Case III: Q(x) contains a higher order polynomial factor Under this case, some factors of Q(x) might be quadratic, cubic or higher order polynomials in nature. Let us understand this method with respect to a quadratic term in the denominator – Example – To find the partial fractions corresponding to $$\frac{x^3 – 2}{x^4 – 1}$$: First let us factorize the denominator – $$x^4 – 1 = (x^2 – 1)(x^2 + 1) = (x – 1)(x + 1)(x^2 + 1)$$. Now, let $$\frac{x^3 – 2}{(x – 1)(x + 1)(x^2 + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x + 1)} + \frac{Cx + D}{(x^2 + 1)}$$ $$= \frac{A(x + 1)(x^2 + 1) + B(x – 1)(x^2 + 1) + (Cx + D)(x – 1)(x + 1)}{(x – 1)(x + 1)(x^2 + 1)}$$ Basically, we have chosen the numerator for the partial fraction involving the quadtratic term as a linear function in x: Cx + D (i.e. one degree less than the denominator). To findÂ A,B,CÂ andÂ D, now we will equate the coefficients of the numerators on both sides of the equation, since the fraction represented by them is the same. We then get – x3 – 2 =Â A(x + 1)(x2 + 1) + B(x – 1)(x2 + 1) + (Cx + D)(x – 1)(x + 1) Now we can proceed by equating the coefficients of equal powers of x on both sides; or once again proceed with direct substitution. Note that on substituting x = 1 in the equation, we get – 13 – 2 =A(1 + 1)(12 + 1) + B(1 – 1)(12 + 1) + (C.1 + D)(1 – 1)(1 + 1) -1 = 4A + 0 + 0 A = -1/4 Similarly, on substituting x = -1; (-1)3 – 2 =A(-1 + 1)((-1)2 + 1) + B(-1 – 1)((-1)2 + 1) + (C.(-1) + D)(-1 – 1)(-1 + 1) -3 = 0 – 4B + 0 B = 3/4 Now we have no option but to compare the coefficients of same powers of x on both sides. Coefficient of x3 on LHSÂ =Â Coefficient of x3 on RHS 1 = A + B + C Use values of A and B to get – C = 1/2 Constant term on LHSÂ =Â Constant term on RHS 1 = A – B – D Use values of A and B to get – D = 1 Our final result then can be written as -$$\frac{x^3 – 2}{(x – 1)(x + 1)(x^2 + 1)} = \frac{A}{(x – 1)} + \frac{B}{(x + 1)} + \frac{Cx + D}{(x^2 + 1)}$$ $$Â = -\frac{1}{4(x – 1)} + \frac{3}{4(x + 1)} + \frac{x/2 + 1}{(x^2 + 1)}$$ $$Â = -\frac{1}{4(x – 1)} + \frac{3}{4(x + 1)} + \frac{x + 2}{2(x^2 + 1)}$$ Similarly, all other partial fractions in the case of a higher polynomial function may be found. Definite Integration All of the integration fundamentals that you have studied so far have led up to this point, wherein now you can apply theÂ integral evaluation techniques to more practical situations by incorporating the boundaries (or the limits) to which your integrand actually holds true. This is that branch of integration which is of widespread use in Economics, Mathematics, Engineering, and many other disciplines. Its utilization in the task of finding the area under a curve finds great application in estimating the total revenues from related marginal functions, or the total growth from the growth rates trend graph etc. So let’s understand its working formula! Given an integral $$\int f(x)dx$$. We call the evaluation of such an integral asÂ Indefinite Integration. Formula The solution of this integration is a resultant function in x plus some arbitrary constant. Let us represent the solution in this form – $$\int f(x)dx = F(x) + c$$ In the method of definite integration, the integral actually has to evaluated in some domain of the variable x. Therefore, we represent it by $$\int_{x_1}^{x_2}$$. What this actually means is that the integrand f(x) now will be bound over the values which the variable of integration i.e. x assumes here, which is x1 to x2. Then we can give the end result by – $$\int_{x_1}^{x_2} f(x)dx = F(x_2) – F(x_a)$$ Here, x2 – The Upper Limit x1 – The Lower Limit Note that there is no arbitrary constant of integration in the solution of a definite integration. Also, you need to know beforehand the result of the indefinite integration of f(x), then use its value at the limits of integration to get this solution. Properties For a,b,c as real numbers – $$\int_a^b f(x)dx = \int_a^b f(t)dt$$ {Change of variable of integration} â‡’ $$\int_a^b f(x)dx = – \int_b^a f(x)dx$$ {Reversal of Lower and Upper Limits} â‡’ $$\int_a^b f(x)dx =Â \int_a^c f(x)dx +Â \int_c^b f(x)dx, where a<c<b â‡’ \(\int_0^a f(x)dx =Â \int_0^a f(a – x)dx$$ â‡’ If f(x) = f(a + x) {Periodic Function}; $$\int_0^{na} f(x)dx =Â n\int_0^a f(x)dx$$ â‡’ $$\int_{-a}^a f(x)dx = 2\int_{0}^a f(x)dx$$Â if f(-x) = f(x) {Even Function} = 0Â if f(-x) = -f(x) {Odd Function} These properties are very useful for changing the limits, or simplifying the integral given in the problem. Thus, we must understand and memorize them well. Now let us look at some solved examples on what we have learned so far. Solved Examples for You Question: Solve $$Â \int{\frac{1}{(x – 1)^2(x + 1)}}dx$$ Solution:Â We can evaluate this type of integrand easily by partial fractions method. In the example for Case 2) above, we have expressed exactly this integrand in terms of partial fractions. Now let us proceed further to find the complete solution. $$Â \int \frac{1}{(x – 1)^2(x + 1)} dx$$ $$= \int[- \frac{1}{4(x – 1)} + \frac{1}{2(x – 1)^2} + \frac{1}{4(x + 1)}]dx$$ $$\text{Using the standard formulae and the variable substitution method -}$$ $$= -\frac{1}{4}ln\|x – 1\|Â – \frac{1}{2(x – 1)} +Â \frac{1}{4}ln\|x + 1\| + c$$ $$= \frac{1}{4}ln \|\frac{x + 1}{x – 1}\|Â – \frac{1}{2(x – 1)} + c$$ Question:Â Solve $$\int{x^2lnx dx}$$ Solution: We’ll use integration by parts here with x2 (algebraic) as the second function, and ln x (logarithmic) as the first function. Using the formula, we get – $$Â \int{x^2lnx dx}=ln x\int x^2 dx – \int[\frac{d}{dx}(ln x)\int x^2 dx]dx + c$$ $$= ln x. \frac{x^3}{3} – \int[\frac{1}{x}.\frac{x^3}{3}]dx + c$$ $$= \frac{x^3 ln x}{3} – \int{\frac{x^2}{3} dx}+c$$ $$=Â \frac{x^3 ln x}{3} – \frac{x^3}{9} + c$$Â Â Question:Â Solve $$\int_{-2}^{+2} x^2 dx$$ Solution:Â First, you must note that for this definite integral, the integrand x3 is an even function, that is – (-x)2 = (x)2 Thus, by the properties of such definite integrals we have – $$\int_{-2}^{+2} x^2 dx = 2\int_{0}^{2} x^2dx$$ $$= 2[\frac{x^3}{3}]_0^2$$ $$= 2.[\frac{2^3}{3} – \frac{0^3}{3}]$$ $$= \frac{16}{3}$$ Share with friends Customize your course in 30 seconds Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. 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# How To Solve Algebra Problems With Fractions $The best way to approach word problems is to “divide and conquer”.HOW TO: SOLVE EQUATIONS WITH FRACTION COEFFICIENTS BY CLEARING THE FRACTIONS Step 1. This kind of equation will occur when we solve problems dealing with money and percent.Find the least common denominator of all the fractions in the equation. But decimals are really another way to represent fractions. So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator. Solution Look at the decimals and think of the equivalent fractions.We changed the problem to one we already knew how to solve! Solve using the General Strategy for Solving Linear Equations.We then used the General Strategy for Solving Linear Equations. $$\begin x \frac &= \frac x - \frac \ (\textcolor) \frac &\stackrel \frac (\textcolor) - \frac \ (-1) \frac &\stackrel - \frac - \frac \ - \frac \frac &\stackrel - \frac - \frac \ - \frac &\stackrel - \frac \ - \frac &= - \frac\; \checkmark \end$$ Some equations have decimals in them.👍If you like this resource, then please rate it and/or leave a comment💬.If the rate-resource button on this page does not work, then go to your ratings page by clicking here 👉tes.com/.../rate-resources…They may involve a single person, comparing his/her age in the past, present or future.They may also compare the ages involving more than one person.-- This worksheet contains the full range of exam-type questions that require students to solve equations that involve algebraic fractions.Questions gradually increase in difficulty with the last row being particularly challenging and involve solving quadratics. .06 = \frac, \qquad 0.02 = \frac, \qquad 0.25 = \frac, \qquad 1.5 = 1 \frac$$Notice, the LCD is 100.By multiplying by the LCD we will clear the decimals. Tags: Database Of ThesisEssays On American SlaveryCeelphone Description EssayScience Homework IdeasA Day I Will Always Remember EssayHamlet Critcal EssaysResearch Paper About Study HabitsCover Page For Term PaperEnglish Critical Essay Int 2Extended Definition Essay About Beauty Consecutive Integer Problems deal with consecutive numbers. The number sequences may be Even or Odd, or some other simple number sequences. Ideal for GCSE revision, this is one of a collection of worksheets which contain exam-type questions that gradually increase in difficulty. These review sheets are great to use in class or as a homework. Many students find solving algebra word problems difficult. [[ The best way to approach word problems is to “divide and conquer”. HOW TO: SOLVE EQUATIONS WITH FRACTION COEFFICIENTS BY CLEARING THE FRACTIONS Step 1. This kind of equation will occur when we solve problems dealing with money and percent. Find the least common denominator of all the fractions in the equation. But decimals are really another way to represent fractions. So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator. Solution Look at the decimals and think of the equivalent fractions. We changed the problem to one we already knew how to solve! Solve using the General Strategy for Solving Linear Equations. We then used the General Strategy for Solving Linear Equations.$$\begin x \frac &= \frac x - \frac \\ (\textcolor) \frac &\stackrel \frac (\textcolor) - \frac \\ (-1) \frac &\stackrel - \frac - \frac \\ - \frac \frac &\stackrel - \frac - \frac \\ - \frac &\stackrel - \frac \\ - \frac &= - \frac\; \checkmark \end$$Some equations have decimals in them. \|\| The best way to approach word problems is to “divide and conquer”.HOW TO: SOLVE EQUATIONS WITH FRACTION COEFFICIENTS BY CLEARING THE FRACTIONS Step 1. This kind of equation will occur when we solve problems dealing with money and percent.Find the least common denominator of all the fractions in the equation. But decimals are really another way to represent fractions. So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator. Solution Look at the decimals and think of the equivalent fractions.We changed the problem to one we already knew how to solve! Solve using the General Strategy for Solving Linear Equations.We then used the General Strategy for Solving Linear Equations.$$\begin x \frac &= \frac x - \frac \\ (\textcolor) \frac &\stackrel \frac (\textcolor) - \frac \\ (-1) \frac &\stackrel - \frac - \frac \\ - \frac \frac &\stackrel - \frac - \frac \\ - \frac &\stackrel - \frac \\ - \frac &= - \frac\; \checkmark \end$\$ Some equations have decimals in them.👍If you like this resource, then please rate it and/or leave a comment💬.If the rate-resource button on this page does not work, then go to your ratings page by clicking here 👉tes.com/.../rate-resources…They may involve a single person, comparing his/her age in the past, present or future.They may also compare the ages involving more than one person.-- This worksheet contains the full range of exam-type questions that require students to solve equations that involve algebraic fractions.Questions gradually increase in difficulty with the last row being particularly challenging and involve solving quadratics. ]] ## Comments How To Solve Algebra Problems With Fractions • ###### How to solve algebra problems with fractions - Patatoast è il Toast a Torino. Patatoast, la Toasteria a Torino è anche street food. Il Toast non è mai stato così buono e fresco.… • ###### Equations Containing Fractions Solving equations containing fractions by finding the lowest common denominator LCD.… • ###### SOLVING EQUATIONS INVOLVING FRACTIONS - SOS Math The first step is solving a rational equation is to convert the equation to an equation without denominators. This new equation may be equivalent same solutions.… • ###### How Do You Solve an Equation Where You're Multiplying. Solving an equation with multiple fractions in different forms isn't so bad. This tutorial shows you how to convert a mixed fraction to an improper fraction in order.… • ###### Manipulating Algebraic Equations and Expressions with. Sometimes the answer explanation suggests that I clear fractions by multiplying an entire equation by a common denominator. Sometimes the.… • ###### Algebraic fractions - Edexcel - Revision 1 - GCSE Maths - BBC. Revise how to add, subtract, multiply and divide algebraic fractions as well as simplify algebraic fractions with. expressions or algebraic fractions works in the same way as simplifying normal fractions. Solving quadratic equations - Edexcel.… • ###### GCSE Revision Algebraic Fractions Solving Equations by. Ideal for GCSE revision, this is one of a collection of worksheets which contain exam-type questions that gradually increase in difficulty. These review sheets are.… • ###### Solve Equations with Fraction or Decimal Coefficients. The General Strategy for Solving Linear Equations can be used to solve for equations with fraction or decimal coefficients. Clearing the.…
HANDS-ON ACTIVITY 3.1: WHAT IS A LIMIT? - Limits and Continuity - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC ## Master AP Calculus AB & BC Part II. AP CALCULUS AB & BC REVIEW CHAPTER 3. Limits and Continuity OVERVIEW • Hands-On Activity 3.1: What is a limit? • Evaluating limits analytically • Continuity • Hands-On Activity 3.2: The extreme value theorem • Hands-On Activity 3.3: The intermediate value theorem • Limits involving infinity • Special limits • Technology: Evaluating limits with a graphing calculator • Summing it up The concepts of limits stymied mathematicians for a long, long time. In fact, the discovery of calculus hinged on these wily little creatures. Limits allow us to do otherwise illegal things like divide by zero. Since, technically, it is never acceptable to divide by zero, limits allow uptight math people to say that they are dividing by “basically” zero or “essentially” zero. Limits are like fortune tellers—they know where you are heading, even though you may not ever get there. Unlike fortune tellers, however, the advice of limits is always free, and limits never have bizarre names like “Madame Vinchense.” HANDS-ON ACTIVITY 3.1: WHAT IS A LIMIT? By completing this activity, you will discover what a limit is, when it exists, and when it doesn’t exist. As in previous Hands-On Activities, spend quality time trying to answer the questions before you break down and look up the answers. 1. Let What is the domain of f(x)? Graph f(x). 2. The table below gives x-values that are less than but increasingly closer and closer to —1. These values are said to be approaching —1 from the left. Use your calculator to fill in the missing values of f(x) for each x. TIP. Limits can help you understand the behavior of points not in the domain of a function, like the value you described in Number 1. 3. The y-value (or height) you are approaching as you near the x value of —1 in the table above is called the left-hand limit of — 1 and is written What is the left-hand limit of f(x)? _____________________________________________________ 4. The table below gives x-values that are greater than but increasingly closer to —1. These values are approaching —1 from the right. Use your calculator to fill in the missing values, as you did in Number 2. 5. The y-value (or height) you are approaching as you near the x value of —1 in the table above is called the right-hand limit of —1 and is written What is the right-hand limit of f(x)? _____________________________________________________ ALERT! Just because the general limit may exist at one x value, that does not guarantee that it exists for all x in the function! 6. Graph f(x) below, and draw the left- and right-hand limits as arrows on the graph. 7. When the left- and right-hand limits as x approaches —1 both exist and are equal, the general limit at x = —1 exists and is written Does the general limit exist at x = —1? If so, what is it? 8. Write a few sentences describing what a limit is and how it is found. _____________________________________________________ _____________________________________________________ _____________________________________________________ 9. Each of the following graphs has no limit at the indicated point. Use a graphing calculator and your knowledge of limits to determine why the limits do not exist. 10. Complete this statement: A limit does not exist if... A. _____________________________________________________ B. _____________________________________________________ C. _____________________________________________________ SELECTED SOLUTIONS TO HANDS-ON ACTIVITY 3.1 1. The domain is (—∞,—1) u (—1,∞), or all real numbers excluding x = —1, as this makes the denominator zero and the fraction undefined. The graph looks like y = x — 4 with a hole at point (—1, —5). 3. The graph seems to be heading toward a height of —5, so that is the left-hand limit. 5. The right-hand limit also appears to be —5. 6. The graph is identical toy — x — 4, except/(x) is undefined at the point (-1,-5). Even though the function is undefined there, the graph is still “headed” toward the height (or limit) of —5 from the left and right of the point. 7. The general limit does exist at x = —1, and 8. A general limit exists on f(x) at x = c (c is a constant) if the left- and right-hand limits exist and are equal at x = c. Mathematically, if and only if 9. (a) Because the general limit does not exist, according to your conclusion from problem 8. (b) As you get closer to x = 0 from the left or the right, the function does not approach any one height—it oscillates infinitely between heights, as demonstrated in its graph on the following page. (c) As you approach x = 0 from the left or the right, the function grows infinitely large, never reaching any specified height. Functions that increase or decrease without bound, such as this one, have no general limit. EXERCISE 1 Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book. For problems 1 through 6, determine if the following limits exist, based on the graph below of p(x). If the limits do exist, state them. ALERT! You can write Technically, however, a limit of ∞ meanst here is no limit, since a limit must be a real number. For problems 7 through 9, evaluate (if possible) the given limits, based on the graphs below of f(x) and g(x). 1. This limit does not exist, as p increases without bound as x2+. You can also say which means there is no limit. 2. You approach a height of —1. In this instance, 3. As stated in Number 2, the general limit exists at x = 3 and is equal to —1 (since the left- and right-hand limits are equal to —1). Notice that cannot exist. Thus, the general limit, Even though the function is undefined at x = —1, p is still headed for a height of 0, and that’s what’s important. 9. Although appears to be approximately 1/4, does not exist. Therefore, cannot exist.
Name # MATH 204 PRACTICE FINAL Please work out each of the given problems on your own paper. Credit will be based on the steps that you show towards the final answer. Show your work. Problem 1 Solve the following differential equations. A. (x2 + 1)y' + 4xy = x y(0) = 2 Solution This equation is separable (also linear, but solving using separation of variables is usually easier). We have x - 4xy x y' = = (1 - 4y) x2 + 1 x2 + 1 Separating, we get dy x = dx 1 - 4y x2 + 1 Now integrate both sides to get -1/4 ln\|1 - 4y\| = 1/2 ln(x2 + 1) + C1 ln\|1 - 4y\| = -2 ln(x2 + 1) + C1 = ln(x2 + 1) -2 + C1 \|1 - 4y\| = C(x2 + 1)2 B. (yex + 2ex + y2)dx + (ex + 2xy)dy = 0 Solution We have My = ex + 2y = Nx Hence this is an exact differential equation. We have fx = yex + 2ex + y2 Integrating with respect to x, we get f(x,y) = yex + 2ex + xy2 + C(y) Taking the derivative with respect to y and setting it equal to N gives fy(x,y) = ex + 2xy + C'(y) = ex + 2xy Hence C'(y) = 0 We can conclude that yex + 2ex + x y2 = C is the solution. C. y(iv) + 4y'' = 2 Solution First we find the homogeneous solution by finding the roots of the characteristic polynomial r4 + 4r2 = r2(r2 + 4) = 0 r = 0 (repeated twice) and r = 2i, r = -2i The homogeneous solution is yh = c1 + c2t + c3 cos(2t) + c4 sin(2t) Next, we seek a particular solution. We use the method of undetermined coefficients. The UC set for g(t) is {1} Since this intersects with the homogeneous solution, we multiply by t2 (note that multiplying by t does not get it out of the homogeneous solution.) Let yp = At2 yp' = 2At yp'' = 2A yp''' = yp(iv) = 0 Substituting back into the differential equation gives 0 + 4(2A) = 2 A = 1/4 The solution is y = yh + yp = c1 + c2t + c3 cos(2t) + c4 sin(2t) + 1/4 D. 1 y'' + 3y' + 2y = (You may leave this as an integral) 1 + et Solution We first find the homogeneous solution by finding the roots of the characteristic polynomial r2 + 3r + 2 = (r + 2)(r + 3) = 0 r = -2 or r = -3 The homogeneous solution is yh = c1e-2t + c2e-3t Since g(t) is not a UC function, we use the method of variation of parameters. We let yp = v1e-2t + v2e-3t Now find the Wronskian matrix and determinant. This matrix has determinant W = -3e-5t + 2e-5t = -e-5t The inverse of the Wronskian matrix is We get Now integrate and add to get y = yh + yp E. (x2 + 1)y'' + xy' + y = 0 y(0) = 1, y'(0) = 0 Solution We solve this differential equation using the method of power series. Let Plug these into the differential equation to get Next, change shift the index of summation in the second sum to get an xn. Now pull out terms so that the summations all begin at n = 2. From the initial conditions, we have a0 = 1 a1 = 0 The pulled out terms from the series gives 2a2 + 1 = 0 a2 = -1/2 and 6a3 + 0 = 0 a3 = 0 Putting the sums together and setting the coefficients equal to zero gives n(n - 1)an + (n + 2)(n + 1)an+2 + nan + an = 0 [n(n - 1) + n + 1]an + (n + 2)(n + 1)an+2 = 0 (n2 + 1)an + (n + 2)(n + 1)an+2 = 0 -(n2 + 1)an an+2 = (n + 2)(n + 1) We see that all the odd terms will be 0. We use the recursion relationship to find the even terms. 5 a4 = 4 .3.2 -37.5 a6 = 6 .5. 4.3.2 (-1)n5.37 . ... . ((2n)2 + 1) a2n = (2n)! F. x2y'' + 2xy' + y = 0 Solution This is an Euler equation. Assume y = xr We find F(r) = r2 + r + 1 This has complex roots The general solution is G. Solution We take the Laplace transform of both sides L{y''} + 2L{y'} + 5L{y} = L{1 - up(t)} s2L{y} + 2sL{y} + 5L{y} = 1/s - e-ps / s (s2 + 2s + 5)L{y} = 1/s - e-ps / s 1 L{y} = (1 - e-ps) s(s2 + 2s + 5) Now use partial fractions A Bs + C 1 + = s s2 + 2s + 5 s(s2 + 2s + 5) A(s2 + 2s + 5) + (Bs + C)s = 1 Let s = 0: 5A = 1 A = 1/5 We have 1/5 s2 + 2/5 s + 1 + Bs2 + Cs = 1 Hence B = -1/5 and C = -2/5 We have 1 s + 1 3 5L{y} = - - s (s + 1)2 + 1 (s + 1)2 + 1 1 s + 1 3 + e-ps - - ) s (s + 1)2 + 1 (s + 1)2 + 1 y = 1/5(1 - etcos t - 3etsin t + up(t)(1 - et -pcos(t - p)- 3et - psin(t - p)) H. x1' = 4x1 + x2 x2' = 3x1 + 2x2 Solution This is the system Find the eigenvalues by finding the determinant of The determinant is (4 - r)(2 - r) - 3 = r2 - 6r + 5 = (r - 5)(r - 1) The roots are r = 1 and r = 5 Now plug in to find the eigenvectors. Plugging in 1 gives which gives 3x1 + x2 = 0 So an eigenvalue is Plugging in 5 gives which gives -x1 + x2 = 0 So an eigenvalue is The solution is or x1 = c1et + c2e5t x2 = -3c1et + c2e5t Problem 2 Lake Tahoe holds 30 cubic kilometers and Fallen Leaf Lake holds 0.5 cubic kilometers. It is estimated that there are currently 20 tons of trout in Lake Tahoe and 1 ton of trout in Fallen Leaf Lake. Water flows from Fallen Leaf Lake to Lake Tahoe at 0.1 cubic kilometers per day. Through an underground channel, water flows from Lake Tahoe to Fallen Leaf Lake at 0.01 cubic kilometers per day. Water with 2 tons of trout per cubic kilometer flows into Fallen Leaf Lake from the desolation wilderness at a rate of 0.09 cubic kilometers per day. Water flows out of Lake Tahoe into the Truckee River at a rate of 0.09 cubic kilometers per day. Assume that when water flows out of either lake the density of the trout in the outflow is equal to the density of the trout in the lake that is the source of the outflow. Set up a system of differential equations that model this situation. A diagram is shown below We use the equation Rate = Rate In - Rate Out For Fallen leaf (x tons of trout), we have x' = (2(.09) + .01y/30) - (.1x / .5) For Lake Tahoe, we have y' = (.1x / .5) - (.01y / 30 + .09y / 30) The differential equations reduce to x' = .18 + y/3000 - x/5 y' = x/5 - y/300 x(0) = 1 y(0) = 20 Problem 3 Given that y = x is a solution of (x2 + 1)y'' - 2xy' + 2y = 0 find a linearly independent solution by reducing the order. Solution We use reduction of order y = vx y' = v'x + v y'' = v''x + 2v' Substituting into the original equation gives (x2 + 1)(v''x + 2v') - 2x(v'x + v) + 2vx = x3v'' + xv'' + 2x2v' + 2v' - 2x2v' - 2xv + 2xv (x3 + x)v''+ 2v' = 0 Now let w = v' to get (x3 + x)w'+ 2w = 0 We have 2w w' = x(x2 + 1) dw 2 = dx w x(x2 + 1) Use partial fractions for the right hand side to get A Bx + C 2 + = x x2 + 1 x(x2 + 1) A(x2 + 1) + (Bx + C)x = 2 Setting x = 0 gives A = 2 We have 2x2 + 2 + Bx2 + Cx = 2 Hence B = 2 and C = 0 This gives ln(w) = ln(x) + ln(x2 + 1) or w = x3 + x Integrate to get v = 1/4 x4 + 1/2 x2 Hence, a second solution is y = vx = 1/4 x5 + 1/2 x3 Problem 4 A six Newton weight is attached to the lower end of a coil spring suspended from the ceiling, the spring constant of the spring being 27 Newtons per meter. The weight comes to rest in its equilibrium position, and beginning at t = 0 and external force given by F(t) = 12 cos(20t) is applied to the system. Determine the resulting displacement as a function of time, assuming damping is negligible. We have the differential equation 6y'' + 27y = 12cos(20t) We have F0 = 12, w0 = 3 /, w = 20 The equation is 24 (20 - 3 /)t 20 + 3 /t y = sin sin 6(400 - 9/4) 2 2 Problem 5 Please answer the following true or false. If false, explain why or provide a counter-example. If true, explain why. A. If the lion population P(t) of Africa follows the differential equation P'(t) = -10P(10,000 - P)(30,000 - P) and there are 20,000 lions in Africa today, then the lions will eventually become extinct in Africa. Solution We graph P' vs. P below We see that 10,000 is an unstable equilibrium point and 30,000 is a stable equilibrium point. Hence if we begin with a population of 20,000, it will rise towards 30,000 and the population will not become extinct. The statement is false. B. (15 Points) Let f(t) = sin(t2)e10t . Then the Laplace transform of f(t) is defined for all s > 10 . Solution True, by the existence theorem of Laplace Transforms, \|sin(t2)e10t\| < e10t C. (15 Points) Suppose that y1(x) and y2(x) are differentiable functions and y'' + p(x)y' + q(x)y = 0 is a differential equation such that p(x) and q(x) are continuous functions. If the Wronskian of y1(x) and y2(x) is e3x W = 1 + x2 then y1(x) and y2(x) cannot be solutions to the differential equation. Solution False, W is never zero. D. (15 Points) Suppose that two-dimensional system x' = Ax of differential equations has the phase portrait pictured below. Then the eigenvalues of A are real and distinct. False, the origin is a spiral node. Hence the eigenvalues of A are complex.
Select Page One of the most fundamental and widely applicable skills for GMAT quant is the ability to break down integers to their prime factors. Integers are a class of numbers sometimes referred to as “whole numbers.” Integers can be positive, negative, or 0 in value, and they can be written without the use of fractions or decimals. Integers Non-Integers 17 83.35 -460 76/45 0 -⅙ ## Prime number divisibility An integer’s prime factors are the prime numbers that, when multiplied together, produce the integer. A prime number is an integer that is not divisible by any integers other than itself and 1. Divisibility means that when an integer is divided by another integer, the result (called the quotient) is also an integer. Here are some examples: 56 / 7 = 8 → 56 is divisible by 7 56 / 6 = 9.33 → 56 is not divisible by 6 132 / 12 = 11 → 132 is divisible by 11 132 / 15 = 8.8 → 132 is not divisible by 15 65 / 13 = 5 → 65 is divisible by 13 65 / 8 = 8.125 → 65 is not divisible by 8 If integer n is divisible by integer x, then integer x is called a divisor of integer n. Integer x may also be called a factor of integer n. The term “divisor” is, of course, related to the operation of division, and the term “factor” is related to the operation of multiplication. Relationships between integers may be spoken of in terms of either division or multiplication. These represent two interchangeable ways of speaking about integers. If we want to speak in terms of multiplication, then instead of saying that integer n is divisible by integer x, we say that integer n is a multiple of integer x. This means that there is some integer by which x can be multiplied to produce n. Here are the same examples from before, this time in terms of multiples. 56 / 7 = 8 → 56 is a multiple of 7 (and a multiple of 8) 56 / 6 = 9.33 → 56 is not a multiple of 6 132 / 12 = 11 → 132 is a multiple of 12 (and a multiple of 11) 132 / 15 = 8.8 → 132 is not a multiple of 15 65 / 13 = 5 → 65 is a multiple of 13 (and a multiple of 5) 65 / 8 = 8.125 → 65 is not a multiple of 8 Note the parentheticals: we could rearrange the values to say 56 / 8 = 7 or 132 / 11 = 12 or 65 / 5 = 13. To generalize this principle, if integer n is a multiple of integer x, then integer n is also a multiple of whatever integer, when multiplied by x, produces n ### Here are five different ways to ask the same question about two integers, n and x: Is n divisible by x? Is n a multiple of x? Is x a divisor of n? Is x a factor of n? Is n/x an integer? As you prepare to take the GMAT, you should see so many of these questions that you understand what is being asked without really noticing the changes in terminology. The goal is for your brain to develop a concept of divisibility that is readily identified by any of the above ways of speaking. The frequency of this concept on GMAT quant makes it hard to get a great score without knowing your multiplication table like the back of your hand. If there are any two integers between 1 and 12, inclusive, whose product (the result when these integers multiply each other) you don’t know instantly, it’s time to review your multiplication table and possibly whip out the flash cards. As elementary as this feels, it will serve you well. ## The first 15 prime numbers Now to return to prime factors. (We could call these prime divisors if we liked, but to do so would be rather unconventional.) Again, a prime number is an integer that is not divisible by any integers other than itself and 1. It’s worth pointing out that every integer is divisible by itself and 1. Every number, whether it’s an integer or not, is divisible by itself, since x/x = 1. (The case of 0/0 is best left to professional mathematicians; it won’t appear on the GMAT.) And dividing any number, whether it’s an integer or not, by 1, effectively does nothing, just as multiplying any number by 1 effectively does nothing. x/1 = x, and x1 = x. Here are the first 15 prime numbers, the ones occurring between 1 and 50: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 Try dividing any of these numbers by an integer other than itself or 1, and your quotient won’t be an integer. (If you’re curious why these and not any other integers represent the first 15 primes, you’ll have a hard time finding an explanation, because a mathematically provable model for describing or predicting the series of prime numbers remains one of the most elusive mysteries in mathematics.) Note that every prime number except for 2 is odd. With this understanding of prime numbers in place, it follows that the prime factors of an integer are the integer’s “building blocks” or fundamental elements, which cannot themselves be further broken down. Let’s use some official GMAT problems to practice the process of reducing integers to their prime factors. We refer both to this process and to the resulting list of primes as the “prime factorization” of the integer. The “prime sum” of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example, the prime sum of 12 is 7, since 12 = 2 2 * 3 and 2 + 2 + 3 = 7. For which of the following integers is the prime sum greater than 35? (A) 440 (B) 512 (C) 620 (D) 700 (E) 750 ### Factor Tree This problem is helpful because it includes an example of prime factorization with the integer 12. There are two possible “routes” for us to perform the prime factorization of 12. We can work through and display these “routes” with a factor tree tool. When we perform a prime factorization of an integer, we “break down” the number into sets of factors until all of the factors are prime and cannot be further broken down. With the integer 12, we may begin with the factor pair of 2 * 6 or with the factor pair of 3 * 4. In the first case, we finish by breaking the 6 into 2 * 3. In the second case, we finish by breaking the 4 into 2 * 2. Either way, the resulting prime factorization of 12 is 2 * 2 * 3. (I like to underline the prime factors as I find them in order to keep track.) No matter which “route” is taken in the prime factorization of an integer, the resulting list of prime factors is always the same. It is often helpful to group repetitions of the same prime factor using exponents. So for 12, instead of “2 * 2 * 3,” we can express the prime factorization as “22 * 3.” In this problem, we are looking for the number whose “prime sum” – or, as we are told, the sum of all its prime factors – is greater than 35. If you know your first 15 prime numbers, a certain answer choice here jumps out. But for the sake of learning and practice, we will prime factorize each answer choice and check its prime sum. Here’s answer choice A: Prime factorization: 23 * 5 * 11 Prime sum = 2 + 2 + 2 + 5 + 11 = 24 When an integer ends with one or more 0s (as 440 does), it’s best to begin by isolating the factors of 10, each of which breaks into the prime factors of 2 and 5. Here’s how this would work for a larger integer: The prime sum for 440 was 24, which is too low, so we’ll have to move on to answer choice B: This is a prime example (pun intended) of why you should memorize some powers of 2, 3, 5, and 6 along with your multiplication table! When you see a number like 512, it can be a big time-saver to know immediately, “Oh, that’s 29.” The prime factorization of 512 is 29, and its prime sum is 18. This is too low, so we’ll move on to answer choice C: Prime factorization: 22 * 5 * 31 Prime sum: 2 + 2 + 5 + 31 = 40 C is the correct answer, because 620 contains the prime factor 31, virtually guaranteeing that the prime sum is greater than 35. (Given the importance of primes on GMAT quant, perhaps it is no coincidence that the section contains 31 questions.) ## Official GMAT problem for practice Here are the prime factorizations of answer choices D and E, just for practice: Prime factorization is one of the easiest GMAT quant skills to practice because you can use any integer that comes to mind! Let’s try another official GMAT problem: How many prime numbers between 1 and 100 are factors of 7,150? (A) One (B) Two (C) Three (D) Four (E) Five This is a very straightforward problem. Just prime factorize 7,150 and see how many different prime factors there are. Straightforward, but easier said than done. If you got stuck trying to divide 715, you can always use an “obvious factor” like 5. Since 715 has a units digit of 5, we know that it is divisible by 5. Then all you need is the quotient of 715 / 5. Dividing 143 is a little trickier, since it has no obvious factors. If you mistakenly conclude that 143 is itself a prime factor, you will choose incorrect answer choice B for this problem. Sharp factoring eyes will notice that 143 = 130 + 13, or 11 * 13. We can count four different prime factors of 7,150 (2, 5, 11, and 13), so the correct answer choice is D. If this was challenging for you, remember, you can practice with any integer. If 3 < x < 100, for how many values of x is x/3 the square of a prime number? (A) Two (B) Three (C) Four (D) Five (E) Nine The initial step to solving this problem is less obvious. We want to know how many integers greater than 3 and less than 100 are divisible by 3 and then by the square of a prime number, with no further prime factors. The only values that can work are the multiples of 3 (since these are the only values divisible by 3), but there are too many multiples of 3 between 3 and 100 to check! The best way to proceed is to reverse the process. Square a prime, multiply your result by 3, and see if this final product is less than 100. 22 * 3 = 12 Hey, we know this one from before. Let’s keep going: 32 * 3 = 9 * 3 = 27 52 * 3 = 25 * 3 = 75 72 * 3 = 49 * 3 = 147 This is already too high, so only three values (12, 27, and 75) satisfy the conditions specified in the problem. The correct answer choice is B. This concludes our introduction to number properties and prime factors. In the next article, we’ll learn about the factors of perfect squares. If you are looking for extra help in preparing for the GMAT, we offer extensive one-on-one GMAT tutoring. You can schedule a complimentary 30-minute consultation call with one of our tutors to learn more! Contributor: Elijah Mize (Apex GMAT Instructor)
# Explain Adding and Subtracting Fractions Adding or subtracting like fractions -- fractions with the same denominator -- is easy. Just add or subtract the numerators, and write the sum or difference over the common denominator. ## More Fraction Resources * AAA Math * Brain Pop * The Learning Equation: Math * Who Wants Pizza? To add or subtract fractions with different denominators, you first must find the least common denominator -- the smallest number that can be divided evenly by both denominators. The easiest way to find the least common denominator is to write the multiples of both denominators until you find a number they have in common. For example, to add the fractions 1/3 and 2/5, begin with some multiples of 3 (3, 6, 9, 12, 15) and some multiples of 5 (5, 10, 15). Youll find that 15 is the lowest number that both numbers divide evenly into. So,15 is the least common denominator of 1/3 and 2/5. The next step is to rewrite those fractions as like fractions. To re-write fractions as like fractions using the least common denominator, 1. Divide the least common denominator (15 in the example above) by the denominator of the first fraction. (15 divided by 3 = 5) 2. Multiple the answer times the numerator of the fraction. (5 x 1 = 5) 3. Re-write the fraction using the answer from step 2 above (5) as the numerator and the least common denominator (15) as the denominator. (1/3 = 5/15) 4. Repeat the steps above for all additional fractions in the equation. (15 divided by 5 = 3; 3 x 2 = 6; 2/5 = 6/15) Finally, use the like fractions to solve the equation by adding or subtracting the numerators and writing the sum or difference above the common denominator: 5/15 + 6/15 = 11/15. Fraction Lesson Plans Fraction Worksheets Fraction Interactive Activities
### Module 3 L6 ```Module 3 Lesson 6 Objective • Write base ten numbers in expanded form Warm up • Write 832 in place value and unit form • 8 hundreds • 3 tens • 2 ones 832 800 30 2 Rules: Drag your finger from zero to 27cm. Then slide up 35 more centimeters. You might first skip-count by ten three times, then go up 5 ones. Count up • If I say 324 you say 3 hundreds 2 tens and 4 ones • Let’s give it a try • 398 • 890 • 174 • Timmy the Monkey picked 46 bananas from the tree. When he was done, there were 50 bananas left. How many bananas were on the tree at first? • Close your eyes and visualize Timmy and all of the bananas. • Whisper to your partner what you could draw to show what you see. • What is the question asking? Reread. • Share your thinking • Is there another way to solve it? • Timmy the Monkey picked 46 bananas from the tree. When he was done, there were 50 bananas left. How many bananas were on the tree at first? • The tree had 96 bananas before Timmy picked them. Expanded form in Unit Order Hundreds Tens Ones 2 4 3 100 100 10 10 10 10 1 1 1 number to me in unit forms. • 2 hundreds • 4 tens • 3 ones • Great! • Let’s count to 243 in bundles Expanded form in Unit Order • Each time we Hundreds Tens Ones • Explain to your count a new unit, neighbor whyit this is the to what wesame as 243 before. 100 +100 +10 +10+10 +10+1+1 +1 •• Let’s together symbols 2 200 4 + 40 3 + 3 = Let’s add the total of each unit Hundreds Tens Ones 2 4 3 200 + 40 + 3= 243 Now again give me the addition sentence that adds up the total of each unit. What happens if we move the units around? +1+1+1 +100+100 10+10+10+10 243 100+100 + 10+10+10+10 + 1+1+1= 243 40 + 3 + 200 = 243 Can someone explain what they understand about the order of the units and the total value? Review • What is 2+4+3? • What is 3+4+2? • Explain to your neighbors why these totals are equal. • When you add it doesn’t matter if the parts are switched around. So 2+4+3 = 3+4+2 And 200+40+3 = 3+40+200 Now you do it! • Practice the Skills we just learned in your problem set. • Rotate for your math groups to practice skills more math skills. ```
# Lesson 4All Things Being EqualDevelop Understanding ## Learning Focus Solve quadratic equations graphically and algebraically. How can we use graphs to solve quadratic equations? ## Open Up the Math: Launch, Explore, Discuss As we have learned, quadratic functions can be very useful models for a lot of real situations. They are part of understanding the motion of objects, business and economic models, and many other applications. Using quadratic functions often requires solving a quadratic equation. This can be a pretty straightforward process, or it can be a little complicated. The good news is that we know a lot about quadratic functions that we can apply to solving equations. Let’s get started. ### 1. If you were given the quadratic equation, , how would you solve it? What solution(s) would you get? ### 2. Graph the function: . #### b. For what values of does ? How did you find the values? How do they compare to the solutions of the equation? ### 3. Given the equation : #### a. Solve the equation algebraically. #### b. Graph the function and mark the points where . ### 4. Given the equation: #### a. Graph the function and find where . (Just to be efficient, use technology and record your graph here with the solution points marked.) #### b. Solve the equation algebraically. Don’t feel bad if you couldn’t solve the equation algebraically. It’s one of those complicated equations that can’t be solved directly with inverse operations. Knowing that the graph of an equation is the set of all of its solutions, we can solve any quadratic equation graphically using technology. Now we’re going to use our understanding of the graph of a quadratic function and its symmetry to develop other algebraic techniques for solving quadratic equations. Let’s start in a familiar place. ### 5. If you’re given the function and asked to find the -intercepts, what quadratic equation would you be solving? ### 6. How could you use this kind of thinking to solve: ? ### 7. Now you can try a few. Solve each quadratic equation by factoring. #### c. When we were graphing, we saw that some functions are easier to factor, and others are easier to complete the square. Let’s start with one where the work has been done for us so that we can start seeing some relationships. ### 8. Consider the function: and the equation . #### a. Graph the function: #### b. Line of symmetry: Vertex: #### d. What are the -intercepts of ? #### e. How far are the -intercepts from the line of symmetry? (Think of the distance left and right.) #### f. Use inverse operations to solve the equation . #### g. How does solving the equation relate to the graph and the -intercepts? Don’t be mad, but that equation would have factored easily if it were given in standard form. That’s why the solutions were integers. Not all equations factor easily, and the relationship that we are beginning to see can be very helpful. Let’s look at a function and equation that do not factor easily from standard form, and see how vertex form can help us solve the equation or find the -intercepts. ### 9. #### a. Graph the function by putting it in vertex form. Vertex: #### c. Line of symmetry: #### d. What do you estimate the -intercepts to be? #### e. What do you estimate the distance from each -intercept to the line of symmetry to be? #### f. Starting with in vertex form, write and solve the equation using inverse operations. #### g. How do your solutions compare with your estimates of the -intercepts and the distance from the line of symmetry and the -intercepts? ### 10. Given: #### a. Solve the equation by completing the square and using inverse operations. #### b. Let . What is the line of symmetry? #### c. What is the distance from line of symmetry to an -intercept? ### 11. Given: #### a. Solve the equation by completing the square and using inverse operations. #### b. Let . What is the line of symmetry? #### c. What is the distance from line of symmetry to an -intercept? Find two methods for solving this equation graphically: ## Takeaways Solving quadratic equations using inverse operations: Example Procedure Given Example Procedure Given Example Procedure Given – The equation must be equal to 0. Rearrange the terms if necessary. Solving quadratic equations by completing the square: Example Procedure Given – Arrange the terms so the equation equals 0, if it isn’t already. ## Lesson Summary In this lesson, we learned methods for solving quadratic equations. Some quadratic equations can be solved using inverse operations and taking the square root of both sides of the equations. Some quadratic equations can be solved by factoring and using the zero product property. Some quadratic equations can be solved by completing the square and then using inverse operations. Quadratic equations that have real solutions can also be solved by graphing, and each of these algebraic methods has connections to graphing. ## Retrieval ### 1. Find the features for the function represented in the graph. Intervals of Increase: Intervals of Decrease: Maximum: Minimum: Domain: Range : ### 2. Use the function to find the indicated values.
Chapter 1 Class 7 Integers Concept wise ### Transcript Ex .1, 10 A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step. (i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level? Question 10 A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step. (i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level? We represent, Going down by a negative number. Going up by a positive number. Since monkey starts from first step, his current position is 1st step And he has to go to 9th step So, his final position will be 9th step Let’s check how he goes So, it took 11 jumps, to go to the last step Question 10 (ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step? We represent, Going down by a negative number. Going up by a positive number. Since monkey starts from last step, his current position is 9th step And he has to go to 1st step So, his final position will be 1st step Let’s check how he goes Question 10 (iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) – 3 + 2 – ... = – 8 (b) 4 – 2 + ... = 8. In (a) the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent? In part (i) He goes down 8 steps by going down 3 steps, and coming up 2 steps. The sum is –3 + 2 –3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 –3 = – 8 Here, –8 represents going down 8 steps In part (ii) He goes up 8 steps by going up 4 steps, and going down 2 steps. The sum is 4 – 2 + 4 – 2 + 4 = 8 Here, 8 represents going up 8 steps

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