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Courses Courses for Kids Free study material Offline Centres More Store # How do you sketch the graph of $y = {\left( {x + 2} \right)^2}$ and describe the transformation? Last updated date: 20th Jun 2024 Total views: 374.7k Views today: 5.74k Verified 374.7k+ views Hint: In this question we need to find the graph of $y = {\left( {x + 2} \right)^2}$ and determine the transformation for it. To obtain the graph find the value of y for the different value of x and then mark the points on the graph to make the graph. Also to determine the transformation, determine the change in the transformation from parent function to given function. Complete Step By Step solution: In this question we have given a function that is $y = {\left( {x + 2} \right)^2}$ and we need to sketch the graph and need to describe the transformation. The above function is the quadratic function. The parent function for the quadratic function is $y = {x^2}$. Consider the table of values for this parent function is, $y$ $x$ $4$ $- 2$ $9$ $- 3$ $25$ $- 5$ $49$ $- 7$ $64$ $- 8$ $4$ $2$ $9$ $3$ $25$ $5$ $49$ $7$ $64$ $8$ From the above table the graph for the parent function is shown below. Now we will consider the table of values for this given function is, $y = {\left( {x + 2} \right)^2}$ $x$ $0$ $- 2$ $1$ $- 3$ $9$ $- 5$ $25$ $- 7$ $36$ $- 8$ $16$ $2$ $25$ $3$ $49$ $5$ $81$ $7$ $100$ $8$ From the above table the graph for the given function is shown in figure below. From the graph for the parent function and the graph for the given equation. It is concluded that the graph of ${\left( {x + 2} \right)^2}$ is shifted $2$ unit to the left from the parent function ${x^2}$. Note: As we know that the quadratic equation is the equation that is of the standard from $a{x^2} + bx + c$. Here, a and b are the coefficients and c is the constant. In the general equation the highest power of the x is $2$ so the equation is called quadratic. The range of all the quadratic functions lies from $- \infty$ to $\infty$.
# 40 percent of six tenths Dollars ### Percentage Calculator What is % of Answer: ### Percentage Calculator 2 is what percent of ? Answer: % ### Percentage Calculator 3 is % of what? Answer: We think you reached us looking for answers to questions like: What is 40 percent of six tenths? Or maybe: 40 percent of six tenths Dollars See the detailed solutions to these problems below. ## How to work out percentages explained step-by-step Learn how to solve percentage problems through examples. In all the following questions consider that: • The percentage figure is represented by X% • The whole amount is represented by W • The portion amount or part is represented by P ### Solution for 'What is 40% of six tenths?' #### Solution Steps The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate". Let's say that you need to find 40 percent of 0.6. What are the steps? Step 1: first determine the value of the whole amount. We assume that the whole amount is 0.6. Step 2: determine the percentage, which is 40. Step 3: Convert the percentage 40% to its decimal form by dividing 40 into 100 to get the decimal number 0.4: 40100 = 0.4 Notice that dividing into 100 is the same as moving the decimal point two places to the left. 40.0 → 4.00 → 0.40 Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount: 0.4 x 0.6 = 0.24 (answer). The steps above are expressed by the formula: P = W × X%100 This formula says that: "To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100". The symbol % means the percentage expressed in a fraction or multiple of one hundred. Replacing these values in the formula, we get: P = 0.6 × 40100 = 0.6 × 0.4 = 0.24 (answer) Therefore, the answer is 0.24 is 40 percent of 0.6. ### Solution for '40 is what percent of six tenths?' The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount". The following problem is of the type "calculating the percentage from a whole knowing the part". #### Solution Steps As in the previous example, here are the step-by-step solution: Step 1: first determine the value of the whole amount. We assume that it is 0.6. (notice that this corresponds to 100%). Step 2: Remember that we are looking for the percentage 'percentage'. To solve this question, use the following formula: X% = 100 × PW This formula says that: "To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100". This formula is the same as the previous one shown in a different way in order to have percent (%) at left. Step 3: replacing the values into the formula, we get: X% = 100 × 400.6 X% = 40000.6 X% = 6,666.67 (answer) So, the answer is 40 is 6,666.67 percent of 0.6 ### Solution for '0.6 is 40 percent of what?' The following problem is of the type "calculating the whole knowing the part and the percentage". ### Solution Steps: Step 1: first determine the value of the part. We assume that the part is 0.6. Step 2: identify the percent, which is 40. Step 3: use the formula below: W = 100 × PX% This formula says that: "To find the whole, divide the part by the percentage then multiply the result by 100". This formula is the same as the above rearranged to show the whole at left. Step 4: plug the values into the formula to get: W = 100 × 0.640 W = 100 × 0.015 W = 1.5 (answer) The answer, in plain words, is: 0.6 is 40% of 1.5.
# Can the common difference of an AP be negative? Progressions are numbers arranged in a particular order such that they form a predictable order. It means that from that series it can be predicted what the next numbers of that series or sequence are going to be. One of the types of progressions is arithmetic progression. Arithmetic Progression is the sequence of the numbers where the difference between any of the two consecutive numbers is the same throughout the sequence. It can also be the common difference of that series. ### Arithmetic Sequence An Arithmetic sequence is a sequence in which every term is created by adding or subtracting a definite number to the preceding number. The definite number in the series is called a Common difference, it is denoted as “d”. Some examples of arithmetic sequence are, • 1, 2, 3, 4, 5, 6,… • 2, 2, 2, 2, 2, 2,… • 22,19,16,13,10 The arithmetic sequence can be of two types, Finite sequence, and infinite sequence. Finite Arithmetic Sequence: 2, 4, 6, 8. Basically, a finite sequence has the finite number of terms. Infinite Arithmetic Sequence 2,4,6,8,10,… The infinite sequence has infinite number of terms, the three dots in the sequence represent that the sequence shall go up to infinity. ### The common difference can be negative Yes, the common difference of an arithmetic sequence can be negative. Lets first learn what is a common difference, a common difference is a difference between two consecutive numbers in the arithmetic sequence. It is simply calculated by taking the difference between the second term and the first term in the arithmetic sequence or the difference between the third term and the second term or any of the two consecutive numbers in the sequence. Common Differences can be positive as well as it can be negative. Example for positive common difference: Sequence â‡¢ 2, 4, 6, 8, 10, 12, Common Difference = Second Term – First Term = 4 – 2 = 2 The common difference of the sequence 2, 4, 6, 8, 10, 12,… is 2 Example for negative Common Difference Sequence â‡¢ 10, 7, 4, 1, -2,… Common Difference = Second Term – First Term = 7 – 10 = -3 Here the common difference of the sequence 10, 7, 4, 1, -2,… is -3 Note: The common difference changes with the change of the sequence or series ### Sample Problems Question 1: Find the common difference for -1, -2, -3, -4, -5,… Solution: Common Difference = Second Term – First Term = -2 – ( -1 ) = -2 + 1 = -1 The common difference of the sequence -1, -2, -3, -4, -5,… is -1 Question 2: Find the common difference for the sequence, 100, 90, 80, 70, 60,… Solution: Common Difference = Second Term – First Term = 90 – 100 = -10 The common difference of the sequence 100, 90, 80, 70, 60,... is -10 Question 3: Find the 15th term of the sequence 9, 7, 5, 3, 1,… Solution: a = 9 d = 7 – 9 = 5 – 7 = -2 n = 15 a15 = a + ( n – 1 ) Ã— d = 9 + ( 15 – 1 ) Ã— (-2) = 9 + ( 14 ) Ã— ( -2 ) = 9 – 28 a15 = -21 Hence, 15th term in the sequence 9,7,5,3,1,… is -21 Question 4: Find the 50th term of the sequence 293, 290, 287, 284, 281,… Solution: a = 293 d = 290 – 293 = 287 – 290 = -3 n = 50 a50 = a + (n – 1) Ã— d = 293 + ( 50 – 1 ) Ã— (-3) = 293 + ( 49 ) Ã— ( -3 ) = 293 – 147 a50 = 146 Hence, 50th term in the sequence 293, 290, 287, 284, 281,… is 146 Question 5: Find the 100th term of the sequence 100, 50, 0, -50, -100,… Solution: a = 100 d = 50 – 100 = 0 – 50 = -50 n = 100 a100 = a + ( n – 1 ) Ã— d = 100 + ( 100 – 1 ) Ã— (-50) = 100 + ( 99 ) Ã— ( -50 ) = 100 – 4950 a50 = -4850 Hence, 50th term in the sequence 100, 50, 0, -50, -100,... is -4850 Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now! Previous Next Similar Reads Complete Tutorials
# 5.3: Dot Products Difficulty Level: At Grade Created by: CK-12 ## Learning objectives • Calculate the dot product of a pair of vectors. • Determine the angle between a pair of vectors. • Determine the vector and scalar projections of one vector onto another. ## Introduction Previously, we discussed the multiplication of a vector by a scalar. There are actually two ways to multiply two vectors, both of which depend on the relative directions of the two vectors. One of the methods has its maximum when the two vectors are parallel; the other is maximized when the two vectors are perpendicular to one another. In this section we will look at the type of vector multiplication that gives a scalar value as the product. The Dot Product, also known as the scalar product, of two vectors will allow us to determine the angle between the two vectors and therefore will also help us to determine whether two vectors are parallel, perpendicular, or something in between. ## Dot Product The dot product of two vector quantities is a scalar quantity which varies as the angle between the two vectors changes. The dot product is therefore sometimes referred to as the scalar product of the two vectors. The maximum value for the dot product occurs when the two vectors are parallel to one another, but when the two vectors are perpendicular to one another the value of the dot product is equal to 0. Furthermore, the dot product must satisfy several important properties of multiplication. The dot product must be commutative The dot product must also be distributive If we simply multiply the magnitudes of the two vectors together, we would fulfill the commutative property, but not the distributive property, since the magnitude There are two ways to calculate the dot product. One way is to multiply the individual components. Each component of vector is multiplied by the component of vector which is in the same direction. Then we add the results. Another way to describe the process is to say that the dot product is the multiplication of one vector by the component of a second vector which is parallel to the first vector. In the diagram below are two vectors, A and B. A perpendicular line has been drawn radially outward from B towards A to create a right triangle with A as the hypotenuse. The component of which is parallel to is given by A cos θ so the second way to compute the dot product is Likewise, the component of which is parallel to is given by B cos θ , so the dot product No matter which of the two vectors we project onto the other, the value of the dot product is maximized when the two vectors are parallel and zero when the two vectors are perpendicular to one another. When a vector is dotted with itself, the result is the square of the vectors magnitude since, by definition, a vector has the same direction as an equal vector. Also, the dot product of any vector with the zero vector is equal to zero since the magnitude of the zero vector is itself equal to 0. Now that we have two ways to compute the dot product, we can use these two methods to solve problems about vectors. The following examples illustrate one instance of this. Example: (a) Calculate the dot product of the two vectors shown below. (b) Then determine the angle between the two vectors. and . Solution: First we will use the components of the two vectors to determine the dot product. Now that we know the dot product, the alternative definition of the dot product, to find θ, the angle between the vectors. First find the magnitudes of the two vectors: Then use these magnitudes with the cosine version of the dot product to find θ. Example: Calculate the dot product of the two vectors shown below. Then determine the angle between the two vectors. Solution: Use the components of the two vectors to determine the dot product. Here and . Now to find the angle between the vectors, first find the magnitudes of the two vectors: Then use these magnitudes with the cosine version of the dot product to find θ. Notice that these two vectors have these same magnitudes as the vectors in Example 1, but their dot product is not the same because they are placed at different angles relative to one another. ## Scalar Projections A scalar projection of a vector onto a particular direction is given by the dot product of the vector with the unit vector for that direction. For example, the component notations for the vectors shown below are and . The scalar projection of vector AB onto is given by The scalar projection of vector AB onto is given by And the scalar projection of vector AB onto is given by The scalar projections of AB onto the x and y directions are non-zero numbers because the vector is located in the x-y plane. The scalar projection of AB onto the z direction is equal to zero, because the z direction is perpendicular to AB. Example: The diagram below shows both vectors AB and D together on the same grid. Determine the scalar projection of vector AB onto the direction of vector D. Solution: To find the scalar projection onto the direction of another vector we need to know the unit vector in the direction of vector D. First, the components of are Now the magnitude of is . Finally, the direction vector of is . Now we can use the dot-product to calculate the scalar projection of AB onto the direction of vector D. ## Vector Projections The vector projection of a vector onto a given direction has a magnitude equal to the scalar projection. The direction of the vector projection is the same as the unit vector of that given direction. In a previous section, we saw that when a vector is multiplied by a scalar s, its components are given by To calculate the vector projection of AB onto the direction of vector D, use the scalar projection calculated in the previous example and the unit vector . ## Lesson Summary One of the two ways to multiply vector quantities is the Scalar Product. The scalar product, also known as the dot product, multiplies one vector by the component of the second vector which is parallel to the first. The result of a scalar product of two vectors is always a scalar number rather than a vector. There are two ways to calculate the dot product: and. These two versions of the dot product can be use to determine the angle between two vectors. The dot product can also be used to determine the scalar and vector projections of one vector in the direction of the other. The scalar projection of onto the direction of is given by and the vector projection of onto the direction of is given by , where is the unit vector in the direction of . ## Practice Problems 1. Determine the dot product of the two vectors and . 2. Determine the dot product of the two vectors shown in the diagram below. @ and @ . 3. Determine the vector projection of onto the direction of and the vector projection of onto the direction of . @ and @ . 4. Determine the dot product of two vectors and . Then determine the angle between the two vectors. 5. Determine the dot product of the following two vectors 6. Determine the scalar projection of the vector onto the direction of . 7. Determine the vector projection of vector onto the vector . 8. Determine the angle between the two vectors and . 9. Determine the dot product of the two vectors and , then determine the angle between the two vectors. ## Solutions 1. Determine the dot product of the two vectors and . The component form of the dot product is given by . In this case, 2. Determine the dot product of the two vectors shown in the diagram below. @ and @ . The angle form of the dot product is given by . In this case, 3. Determine the vector projection of onto the direction of and the vector projection of onto the direction of . @ and @ . The vector projection of one vector onto the direction of another vector is given by , where is the unit vector in the direction of . Since it is a unit vector has a magnitude of 1 and has the same direction as @ . Therefore, @ @ The vector projection of one vector onto the direction of another vector is given by , where is the unit vector in the direction of . Since it is a unit vector has a magnitude of 1 and has the same direction as , @ . Therefore, @ @ 4. Determine the dot product of two vectors and . Then determine the angle between the two vectors. The dot product of two vectors is defined in two ways: and . We will use the first to calculate the dot product and then we will use that result together with the second definition to determine the angle between the two vectors. To find the angle between the two vectors, we need to know not only the dot product of the two vectors, but also the length of each individual vector. Now use the second definition of the dot product to determine the angle 5. Determine the dot product of the following two vectors. The angle form of the dot product is given by . In this case, 6. Determine the scalar projection of the vector onto the direction of . The scalar projection of one vector onto the direction of the other is the dot product of the first vector with the unit vector representing the direction of the second vector. To calculate the scalar projection, we need to determine the unit vector in the direction of vector . Remember that a unit vector is equal to the ratio of the vector and its magnitude, therefore we first need to calculate the length of vector . Now we can calculate the scalar projection of onto by calculating the dot product 7. Determine the vector projection of vector onto the vector . The vector progression of one vector onto a second vector is the multiplication of the dot products of the two vectors and the unit vector defining the direction of the second vector. In this case, . First we need to identify the components of the two vectors by using the information given on the graph. In this case, and . Then we need to determine the dot product of the two vectors. We also need to determine the unit vector in the direction of .Remember that a unit vector is equal to the ratio of the vector and its magnitude, therefore we first need to calculate the length of vector . Lastly, we multiply the dot product of the two vectors by this unit vector, 8. Determine the angle between the vectors and . We calculated the dot product of and in the previous problem: We can then use the definition to determine the angle between the two vectors. But first we need to determine the magnitudes of the two vectors. By looking at the diagram, we can see that the angle between these two vectors is larger than 90o. Many calculators only give the smaller of the two angles between two lines. As you can see below, both θ and relate the blue line to the red line. For our problem, the calculator returned a value of 53.1o. The actual angle between the two vectors is 180o - 53.1o = 126.9o when we take into account the directions of and . 9. Determine the dot product of the two vectors and , then determine the angle between the two vectors. The component form of the dot product is given by . Now we can find the angle between the two vectors using the other form of the dot-product equation: , but first we need to determine the magnitudes of the two vectors using the Pythagorean Theorem. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# Definite Integrals A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022 ## Definite Integrals A definite integral is an integral with limits. The limits are represented as small numbers at the top and bottom of the integral. We integrate in exactly the same way, except we can leave out the constant of integration $+c$. Then, we get our final answer by substituting the limits into our integrated form, and taking the value for the lower limit away from the value for the upper limit. Definite integrals are useful because they represent the area under the graph. Make sure you are happy with the following topics before continuing. A Level ## Definite Integrals Have Limits The limits of a definite integral are the small numbers at the top and bottom of the integral symbol. Once we have integrated, we substitute the limits in, then subtract the lower limit result from the upper limit result. Example: Find $\int ^{2}_{1}x^{2}dx$ \begin{aligned}\int^{2}_{1}x^{2}dx&=\left[\dfrac{1}{3}x^{3}\right]^{2}_{1}\\[1.2em]&=\left( \dfrac{1}{3}\times2^{3}\right) -\left( \dfrac{1}{3}\times1^{3}\right) \\[1.2em]&=\dfrac{8}{3}-\dfrac{1}{3}\\[1.2em]&=\dfrac{7}{3}\end{aligned} A Level ## Definite Integrals are the Area Under the Graph Definite integrals represent the area under the graph. For example, to find the area under the graph between $-2$ and $2$ of $y=x^{2}-4$, we would do $\int^{2}_{-2}x^{2}-4dx$. Note: Using this metric, area below the $x$ axis is counted as negative. Example: Find the area under the graph $y=x^{3}-6$ between $x=2$ and $x=4$. \begin{aligned}\text{area}&=\int^{4}_{2}x^{3}-6dx\\[1.2em]&=\left[\dfrac{1}{4}x^{4}-6x\right]^{4}_{2}\\[1.2em]&=\left( \dfrac{1}{4}\times4^{4}\right) -\left( 6\times4\right) -\left( \dfrac{1}{4}\times2^{4}\right) +\left( 6\times2\right) \\[1.2em]&=\left( \dfrac{1}{4}\times256\right) -24-\left( \dfrac{1}{4}\times16\right) +12\\[1.2em]&=64-24-4+12\\[1.2em]&=48\end{aligned} A Level Sometimes, the area you have to figure out will not always be bounded by one function to integrate. In this case, you often have to add two integrals together. Example: Find the area between the $x$ axis, the graph $y=x^{3}$ and $y=2-x$. The area runs from $x=0$ to $x=2$. The graphs intersect at $x=1$, so the area changes from being under $y=x^{3}$ to being under $y=2-x$ at $x=1$. Hence, we can find the area: \begin{aligned}\text{area}&=\int^{1}_{0}x^{3}dx+\int^{2}_{1}\left( 2-x\right) dx\\[1.2em]&=\left[\dfrac{1}{4}x^{4}\right]^{1}_{0}+\left[2x-\dfrac{1}{2}x^{2}\right]^{2}_{1}\\[1.2em]&=\dfrac{1}{4}-0+\left( 2\times2\right) -\left( \dfrac{1}{2}\times2^{2}\right) -\left( 2\times1\right) +\left( \dfrac{1}{2}\times1^{2}\right) \\[1.2em]&=\dfrac{1}{4}+4-\left( \dfrac{1}{2}\times4\right) -2+\left( \dfrac{1}{2}\times1\right) \\[1.2em]&=\dfrac{1}{4}+4-2-2+\dfrac{1}{2}\\[1.2em]&=\dfrac{3}{4}\end{aligned} A Level ## Subtracting Integrals Some problems involving areas on graphs require you to subtract integrals. Generally, these questions take the form of asking you to find the area between two curves. We solve them by finding the entire area under the upper curve, then subtracting the area under the lower curve. Example: Find the area between the curves $y=x^{2}$ and $y=2-x^{2}$. The area runs from $x=-1$ to $x=1$. The upper curve is $y=2-x^{2}$. Integrating this gives the entire area under this curve. The lower curve is $y=x^{2}$. Subtracting the integral of this gets rid of the area we don’t want. \begin{aligned}\text{area}&=\int^{1}_{-1}\left( 2-x^{2}\right) dx-\int^{1}_{-1}x^{2}dx\\[1.2em]&=\left[2x-\dfrac{1}{3}x^{3}\right]^{1}_{-1}-\left[\dfrac{1}{3}x^{3}\right]\\[1.2em]&=\left( 2\times1\right) -\left( \dfrac{1}{3}\times1^{3}\right) -\left( 2\times(-1)\right) +\left( \dfrac{1}{3}\times(-1)^{3}\right) \\[1.2em]&\left( -\dfrac{1}{3}\times1^{3}\right) +\left( \dfrac{1}{3}\times(-1)^{3}\right) \\[1.2em]&=2-\left( \dfrac{1}{3}\times1\right) +2-\left( \dfrac{1}{3}\times1\right) -\left( \dfrac{1}{3}\times1\right) -\left( \dfrac{1}{3}\times1\right) \\[1.2em]&=4-\dfrac{4}{3}\\[1.2em]&=\dfrac{8}{3}\end{aligned} A Level ## Example Questions i)\begin{aligned}\int^{3}_{1}x^{3}dx&=\left[\dfrac{1}{4}x^{4}\right]^{3}_{1}\\[1.2em]&=\left( \dfrac{1}{4}\times3^{4}\right) -\left( \dfrac{1}{4}\times1^{4}\right) \\[1.2em]&=\left( \dfrac{1}{4}\times81\right) -\left( \dfrac{1}{4}\times1\right) \\[1.2em]&=\dfrac{81}{4}-\dfrac{1}{4}\\[1.2em]&=\dfrac{80}{4}\\[1.2em]&=20\end{aligned} ii)\begin{aligned}\int^{4}_{0}2x+1dx&=[x^{2}+x]^{4}_{0}\\[1.2em]&=4^{2}+4-0^{2}+0\\[1.2em]&=16+4\\[1.2em]&=20\end{aligned} iii)\begin{aligned}\int^{1}_{-1}x^{-3}dx&=\left[-\dfrac{1}{2}x^{-2}\right]^{1}_{-1}\\[1.2em]&=\left( -\dfrac{1}{2}\times1^{-2}\right) +\left( \dfrac{1}{2}\times(-1)^{-2}\right) \\[1.2em]&=-\dfrac{1}{2}+\dfrac{1}{2}\\[1.2em]&=0\end{aligned} i) \begin{aligned}&\int^{3}_{-3}\left( x^{2}+4x+8\right) dx=\left[\dfrac{1}{3}x^{3}+2x^{2}+8x\right]^{3}_{-3}\\[1.2em]&=\left( \dfrac{1}{3}\times3^{3}\right) +\left( 2\times3^{2}\right) +\left( 8\times3\right) -\left( \dfrac{1}{3}\times(-3)^{3}\right) -\left( 2\times(-3)^{2}\right) -\left( 8\times(-3)\right) \\[1.2em]&=\left( \dfrac{1}{3}\times27\right) +\left( 2\times9\right) +24+\left( \dfrac{1}{3}\times27\right) -\left( 2\times9\right) +24\\[1.2em]&=9+18+24+9-18+24\\[1.2em]&=66\end{aligned} ii) \begin{aligned}&\int^{5}_{1}\left( x^{\frac{6}{7}}-\sqrt{x}\right) dx=\int^{5}_{1}\left( x^{\frac{6}{7}}-x^{\frac{1}{2}}\right) dx\\[1.2em]&=\left[\dfrac{7}{13}x^{\frac{13}{7}}-\dfrac{2}{3}x^{\frac{3}{2}}\right]^{5}_{1}\\[1.2em]&=\left( \dfrac{7}{13}\times5^{\frac{13}{7}}\right) -\left( \dfrac{2}{3}\times5^{\frac{3}{2}}\right) -\left( \dfrac{7}{13}\times1^{\frac{13}{7}}\right) +\left( \dfrac{2}{3}\times1^{\frac{3}{2}}\right) \\[1.2em]&=\left( \dfrac{7}{13}\times5^{\frac{13}{7}}\right) -\left( \dfrac{2}{3}\times5^{\frac{3}{2}}\right) -\dfrac{7}{13}+\dfrac{2}{3}\\[1.2em]&=3.37\end{aligned} iii) \begin{aligned}&\int^{200}_{1}\left( 15x^{\frac{1}{3}}+10x^{\frac{1}{4}}-3x^{\frac{5}{12}}\right) dx\\[1.2em]&=\left[\left( 15\times\dfrac{3}{4}x^{\frac{4}{3}}\right) +\left( 10\times\dfrac{4}{5}x^{\frac{5}{4}}\right) -\left( 3\times\dfrac{12}{17}x^{\frac{17}{12}}\right) \right]^{200}_{1}\\[1.2em]&=\left[\dfrac{45}{4}x^{\frac{4}{3}}+8x^{\frac{5}{4}}-\dfrac{36}{17}x^{\frac{17}{12}}\right]^{200}_{1}\\[1.2em]&=\left( \dfrac{45}{4}\times200^{\frac{4}{3}}\right) +\left( 8\times200^{\frac{5}{4}}\right) -\left( \dfrac{36}{17}\times200^{\frac{17}{12}}\right) \\[1.2em]&\left( -\dfrac{45}{4}\times1^{\frac{4}{3}}\right) -\left( 8\times1^{\frac{5}{4}}\right) +\left( \dfrac{36}{17}\times1^{\frac{17}{12}}\right) \\[1.2em]&=\left( \dfrac{45}{4}\times200^{\frac{4}{3}}\right) +\left( 8\times200^{\frac{5}{4}}\right) -\left( \dfrac{36}{17}\times200^{\frac{17}{12}}\right) \\[1.2em]&-\dfrac{45}{4}-8+\dfrac{36}{17}\\[1.2em]&=15300\end{aligned} Graphs intersect at $(4,2)$, so we want the area under $y=\sqrt{x}$ from $0$ to $4$ added to the area under $y=10-x^{\frac{3}{2}}$ from $4$ to where this graph touches the $x$ axis. Find where $y=10-x^{\frac{3}{2}}$ touches the $x$ axis. $0=10-x^{\frac{3}{2}}$ $x^{\frac{3}{2}}=10$ $x=10^{\frac{2}{3}}$ Putting this all together: \begin{aligned}\text{area}&=\int^{4}_{0}\sqrt{x}dx+\int^{10^{\frac{2}{3}}}_{4}\left( 10-x^{\frac{3}{2}}\right) dx\\[1.2em]&=\int^{4}_{0}x^{\frac{1}{2}}dx+\left[10x-\dfrac{2}{5}x^{\frac{5}{2}}\right]^{10^{\frac{2}{3}}}_{4}\\[1.2em]&=\left[\dfrac{2}{3}x^{\frac{3}{2}}\right]^{4}_{0}+\left( 10\times10^{\frac{2}{3}}\right) -\left( \dfrac{2}{5}\times(10^{\frac{2}{3}})^{\frac{5}{2}}\right)-\left( 10\times4\right) +\left( \dfrac{2}{5}\times4^{\frac{5}{2}}\right) \\[1.2em]&=\left( \dfrac{2}{3}\times4^{\frac{3}{2}}\right) -\left( \dfrac{2}{3}\times0^{\frac{3}{2}}\right) +10^{\frac{5}{3}}-\left( \dfrac{2}{5}\times10^{\frac{5}{3}}\right) -40+\left( \dfrac{2}{5}\times32\right) \\[1.2em]&=\left( \dfrac{2}{3}\times8\right) +\left( \dfrac{3}{5}\times10^{\frac{5}{3}}\right) -40+\dfrac{64}{5}\\[1.2em]&=5.98\end{aligned} From the graph we can see that we want the area under $y=x^{2}-4x+36$ minus the area under $y=x^{5}$, and the graphs intersect at $(2,32)$ so our area runs from $0$ to $2$. \begin{aligned}\text{area}&=\int^{2}_{0}\left( x^{2}-4x+36\right) dx-\int^{2}_{0}x^{5}dx\\[1.2em]&=\left[\left( \dfrac{1}{3}x^{3}\right) -2x^{2}+36x\right]^{2}_{0}-\left[\dfrac{1}{6}x^{6}\right]^{2}_{0}\\[1.2em]&=\left( \dfrac{1}{3}\times2^{3}\right) -\left( 2\times2^{2}\right) +\left( 36\times2\right) -\left( \dfrac{1}{3}\times0^{3}\right) +\left( 2\times0^{2}\right) -\left( 36\times0\right) -\left( \dfrac{1}{6}\times2^{6}\right) +\left( \dfrac{1}{6}\times0^{6}\right) \\[1.2em]&=\left( \dfrac{1}{3}\times8\right) -\left( 2\times4\right) +72-\left( \dfrac{1}{6}\times64\right) \\[1.2em]&=\dfrac{8}{3}-8+72-\dfrac{32}{3}\\[1.2em]&=64-\dfrac{24}{3}\\[1.2em]&=64-8\\[1.2em]&=56\end{aligned} A Level A Level A Level A Level ## You May Also Like... ### A Level Maths Revision Cards The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams. £14.99 ### A Level Maths – Cards & Paper Bundle A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers. From: £22.99 ### Transition Maths Cards The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths! £8.99
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. Zeroes of Rational Functions Values where the numerator equals zero but the denominator doesn't. 0% Progress Practice Zeroes of Rational Functions Progress 0% Zeroes of Rational Functions The zeroes of a function are the collection of $x$ values where the height of the function is zero. How do you find these values for a rational function and what happens if the zero turns out to be a hole Watch This Focus on the portion of this video discussing holes and $x$ -intercepts. http://www.youtube.com/watch?v=UnVZs2EaEjI James Sousa: Find the Intercepts, Asymptotes, and Hole of a Rational Function Guidance Zeroes are also known as $x$ -intercepts, solutions or roots of functions. They are the $x$ values where the height of the function is zero. For rational functions, you need to set the numerator of the function equal to zero and solve for the possible $x$ values. If a hole occurs on the $x$ value, then it is not considered a zero because the function is not truly defined at that point. Example A Identify the zeroes and holes of the following rational function. $f(x)=\frac{(x-1)(x+3)(x+3)}{x+3}$ Solution: Notice how one of the $x+3$ factors seems to cancel and indicate a removable discontinuity. Even though there are two $x+3$ factors, the only zero occurs at $x=1$ and the hole occurs at (-3, 0). Example B Identify the zeroes, holes and $y$ intercepts of the following rational function without graphing. $f(x)=\frac{x(x-2)(x-1)(x+1)(x+1)(x+2)}{(x-1)(x+1)}$ Solution: The holes occur at $x=-1, 1$ . To get the exact points, these values must be substituted into the function with the factors canceled. $f(x) &= x(x-2)(x+1)(x+2)\\f(-1) &= 0, f(1)=-6$ The holes are (-1, 0); (1, 6). The zeroes occur at $x=0, 2, -2$ . The zero that is supposed to occur at $x=-1$ has already been demonstrated to be a hole instead. Example C Create a function with holes at $x=1,2,3$ and zeroes at $x=0, 4$ Solution: There are an infinite number of possible functions that fit this description because the function can be multiplied by any constant. One possible function could be: $f(x)=\frac{(x-1)(x-2)(x-3)x(x-4)}{x(x-4)}$ Concept Problem Revisited To find the zeroes of a rational function, set the numerator equal to zero and solve for the $x$ values. When a hole and a zero occur at the same point, the hole wins and there is no zero at that point. Vocabulary A zero is where a function crosses the $x$ -axis. It is also known as a root, solution or $x$ -intercept. A rational function is a function with at least one rational expression. A rational expression is a ratio of two polynomial expressions. Guided Practice 1. Create a function with holes instead of zeroes. 2. Identify the $y$ intercepts, holes, and zeroes of the following rational function. $f(x)=\frac{6x^3-7x^2-x+2}{x-1}$ 3. Identify the zeroes and holes of the following rational function. $f(x)=\frac{2(x+1)(x+1)(x+1)}{2(x+1)}$ 1. There are an infinite number of functions that meet the requirements. An illustrative example would be: $f(x)=(x-1)(x+1) \cdot \frac{(x-1)(x+1)}{(x-1)(x+1)}$ The two $x$ values that are holes are identical to the two $x$ values that would be zeroes. Therefore, this function has no zeroes because holes exist in their place. 2. After noticing that a possible hole occurs at $x=1$ and using polynomial long division on the numerator you should get: $f(x)=(6x^2-x-2) \cdot \frac{x-1}{x-1}$ A hole occurs at $x=1$ which turns out to be the point (1, 3) because $6 \cdot 1^2-1-2=3$ The $y$ -intercept always occurs where $x=0$ which turns out to be the point (0, -2) because $f(0)=-2$ . To find the $x$ -intercepts you need to factor the remaining part of the function: $(2x+1)(3x-2)$ Thus the zeroes ( $x$ -intercepts) are $x=-\frac{1}{2}, \frac{2}{3}$ . 3. The hole occurs at $x=-1$ which turns out to be a double zero. The hole still wins so the point (-1, 0) is a hole. There are no zeroes. The constant 2 in front of the numerator and the denominator serves to illustrate the fact that constant scalars do not impact the $x$ values of either the zeroes or holes of a function. Practice Identify the intercepts and holes of each of the following rational functions. 1. $f(x)=\frac{x^3+x^2-10x+8}{x-2}$ 2. $g(x)=\frac{6x^3-17x^2-5x+6}{x-3}$ 3. $h(x)=\frac{(x+2)(1-x)}{x-1}$ 4. $j(x)=\frac{(x-4)(x+2)(x+2)}{x+2}$ 5. $k(x)=\frac{x(x-3)(x-4)(x+4)(x+4)(x+2)}{(x-3)(x+4)}$ 6. $f(x)=\frac{x(x+1)(x+1)(x-1)}{(x-1)(x+1)}$ 7. $g(x)=\frac{x^3-x^2-x+1}{x^2-1}$ 8. $h(x)=\frac{4-x^2}{x-2}$ 9. Create a function with holes at $x=3, 5, 9$ and zeroes at $x=1, 2$ . 10. Create a function with holes at $x=-1, 4$ and zeroes at $x=1$ 11. Create a function with holes at $x=0, 5$ and zeroes at $x=2, 3$ . 12. Create a function with holes at $x=-3, 5$ and zeroes at $x=4$ 13. Create a function with holes at $x=-2, 6$ and zeroes at $x=0, 3$ . 14. Create a function with holes at $x= 1, 5$ and zeroes at $x=0,6$ . 15. Create a function with holes at $x=2, 7$ and zeroes at $x=3$ . Vocabulary Language: English Hole Hole A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero. Rational Expression Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator. Rational Function Rational Function A rational function is any function that can be written as the ratio of two polynomial functions. Zero Zero The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
# Inscribed Angles in Circles ## Angle with its vertex on a circle and sides that contain chords. Estimated8 minsto complete % Progress Practice Inscribed Angles in Circles MEMORY METER This indicates how strong in your memory this concept is Progress Estimated8 minsto complete % Inscribed Angles Point \begin{align}A\end{align} is the center of the circle below. What can you say about \begin{align}\Delta CBD\end{align} ? ### Inscribed Angles An inscribed angle is an angle with its vertex on the circle. The sides of an inscribed angle will be chords of the circle. Below, \begin{align}\angle CED\end{align} is an inscribed angle. Inscribed angles are inscribed in arcs. You can say that \begin{align}\angle CED\end{align} is inscribed in \begin{align}\widehat{CED}\end{align}. You can also say that \begin{align}\angle CED\end{align} intercepts \begin{align}\widehat{CD}\end{align}. The measure of an inscribed angle is always half the measure of the arc it intercepts. You will prove and then use this theorem in the problems below. #### Proving the Inscribed Angle Theorem 1. Consider the circle below with center at point \begin{align}A\end{align}. Prove that \begin{align}m \angle AEC = \frac{1}{2} m \angle CAD = \frac{1}{2} m \widehat{CD}\end{align}. \begin{align}\overline{CA}\end{align} and \begin{align}\overline{EA}\end{align} are both radii of the circle and are therefore congruent. This means that \begin{align}\Delta CAE\end{align} is isosceles. The base angles of isosceles triangles are congruent so \begin{align}m \angle ACE = m \angle AEC\end{align}. \begin{align}\angle CAD \end{align} is an exterior angle of \begin{align}\Delta CAE\end{align}, so its measure must be the sum of the measures of the remote interior angles. Therefore, \begin{align}m \angle CAD = m \angle ACE + m \angle AEC\end{align}. By substitution, \begin{align}m \angle CAD = m \angle AEC + m \angle AEC\end{align}. This means \begin{align}m \angle CAD = 2m \angle AEC\end{align} and \begin{align}\frac{1}{2} m \angle CAD = m \angle AEC\end{align}. A central angle has the same measure as its intercepted arc, so \begin{align}m \angle CAD = m \widehat{CD}\end{align}. Therefore by substitution, \begin{align}\frac{1}{2} m \widehat{CD} = m \angle AEC\end{align}. This proves that when an inscribed angle passes through the center of a circle, its measure is half the measure of the arc it intercepts. 2. Use the result from the previous problem to prove that \begin{align}m \angle CED = \frac{1}{2}m \widehat{CD}\end{align}. Draw a diameter through points \begin{align}E\end{align} and \begin{align}A\end{align}. From #1, you know that \begin{align}m \angle FEC = \frac{1}{2} m \angle FAC\end{align}. You also know that \begin{align}m \angle FED = \frac{1}{2} m \angle FAD\end{align}. Because \begin{align}m \angle FED = m \angle FEC + m \angle CED\end{align}, by substitution \begin{align}\frac{1}{2} m \angle FAD = \frac{1}{2} m \angle FAC + m \angle CED\end{align}. This means \begin{align}m \angle CED = \frac{1}{2} (m \angle FAD - m \angle FAC)\end{align}. \begin{align}m \angle FAD - m \angle FAC = m \angle CAD\end{align}, so \begin{align}m \angle CED = \frac{1}{2} m \angle CAD\end{align}. Because \begin{align}m \angle CAD = m \widehat{CD}, m \angle CED = \frac{1}{2} m \widehat{CD} \end{align}. This proves in general that the measure of an inscribed angle is half the measure of its intercepted arc. Now let's find the measure of an angle. Find \begin{align}m \angle CED\end{align}. Notice that both \begin{align}\angle CED\end{align} and \begin{align}\angle CBD\end{align} intercept \begin{align}\widehat{CD}\end{align}. This means that their measures are both half the measure of \begin{align}\widehat{CD}\end{align}, so their measures must be equal. \begin{align}m \angle CED=38^\circ\end{align}. ### Examples #### Example 1 Earlier, you were asked what can you say about \begin{align}\Delta CBD\end{align} Point \begin{align}A\end{align} is the center of the circle below. What can you say about \begin{align}\Delta CBD\end{align} ? If point \begin{align}A\end{align} is the center of the circle, then \begin{align}\overline{CB}\end{align} is a diameter and it divides the circle into two equal halves. This means that \begin{align}m\widehat{CB}=180^\circ\end{align}\begin{align}\angle CDB\end{align} is an inscribed angle that intercepts \begin{align}\widehat{CB}\end{align}, so its measure must be half the measure of \begin{align}\widehat{CB}\end{align}. Therefore, \begin{align}m \angle CDB=90^\circ\end{align} and \begin{align} \Delta CBD\end{align} is a right triangle. In general, if a triangle is inscribed in a semicircle then it is a right triangle. In #2-#3, you will use the circle below to prove that when two chords intersect inside a circle, the products of their segments are equal. #### Example 2 Prove that \begin{align}\Delta EFC \sim \Delta BFD\end{align}. Hint: Look for congruent angles! \begin{align}\angle CED \cong \angle CBD\end{align} because both are inscribed angles that intercept the same arc \begin{align}(\widehat{CD})\end{align}\begin{align}\angle EFC \cong \angle BFD \end{align} because they are vertical angles. Therefore, \begin{align}\Delta EFC \sim \Delta BFD\end{align} by \begin{align}AA \sim\end{align}. #### Example 3 Prove that \begin{align}EF \cdot FD = BF \cdot FC\end{align}. Because \begin{align}\Delta EFC \sim \Delta BFD\end{align}, its corresponding sides are proportional. This means that \begin{align}\frac{EF}{BF}=\frac{FC}{FD}\end{align}. Multiply both sides of the equation by \begin{align}BF \cdot FD\end{align} and you have \begin{align}EF \cdot FD = BF \cdot FC\end{align}. This proves that in general, when two chords intersect inside a circle, the products of their segments are equal. #### Example 4 For the circle below, find \begin{align}BF\end{align}. Based on the result of #2, you know that \begin{align}15 \cdot 6=BF \cdot 9\end{align}. This means that \begin{align}BF=10\end{align}. ### Review 1. How are central angles and inscribed angles related? In the picture below, \begin{align}\overline{BD} \ \\| \ \overline{EC}\end{align}. Use the picture below for #2-#6. 2. Find \begin{align}m \widehat{BD}\end{align}. 3. Find \begin{align}m \angle ABD\end{align}. 4. Find \begin{align}m \widehat{EB}\end{align}. 5. Find \begin{align}m \angle ECD\end{align}. 6. What type of triangle is \begin{align}\Delta ACD\end{align} ? Solve for \begin{align}x\end{align} in each circle. If \begin{align}x\end{align} is an angle, find the measure of the angle. 7. 8. 9. 10. \begin{align}m \widehat{DC} = 95^\circ\end{align} 11. 12. 13. 14. 15. In the picture below, \begin{align}\overline{BD}\ \|\| \ \overline{EC}\end{align}. Prove that \begin{align}\widehat{BC} \cong \widehat{ED}\end{align}. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Arc An arc is a section of the circumference of a circle. Intercepts The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis. Inscribed Angle Theorem The Inscribed Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc. Semicircle Theorem The Semicircle Theorem states that any time a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter and the diameter is the hypotenuse. Inscribed Angle An inscribed angle is an angle with its vertex on the circle. The measure of an inscribed angle is half the measure of its intercepted arc.
BREAKING NEWS Elementary algebra ## Summary ${\displaystyle {\overset {}{\underset {}{x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}}}}$ The quadratic formula, which is the solution to the quadratic equation ${\displaystyle ax^{2}+bx+c=0}$ where ${\displaystyle a\neq 0}$. Here the symbols a, b, and c represent arbitrary numbers, and x is a variable which represents the solution of the equation. Two-dimensional plot (red curve) of the algebraic equation ${\displaystyle y=x^{2}-x-2}$. Elementary algebra encompasses some of the basic concepts of algebra, one of the main branches of mathematics. It is typically taught to secondary school students and builds on their understanding of arithmetic. Whereas arithmetic deals with specified numbers,[1] algebra introduces quantities without fixed values, known as variables.[2] This use of variables entails use of algebraic notation and an understanding of the general rules of the operations introduced in arithmetic. Unlike abstract algebra, elementary algebra is not concerned with algebraic structures outside the realm of real and complex numbers. The use of variables to denote quantities allows general relationships between quantities to be formally and concisely expressed, and thus enables solving a broader scope of problems. Many quantitative relationships in science and mathematics are expressed as algebraic equations. ## Algebraic notation Algebraic notation describes the rules and conventions for writing mathematical expressions, as well as the terminology used for talking about parts of expressions. For example, the expression ${\displaystyle 3x^{2}-2xy+c}$ has the following components: A coefficient is a numerical value, or letter representing a numerical constant, that multiplies a variable (the operator is omitted). A term is an addend or a summand, a group of coefficients, variables, constants and exponents that may be separated from the other terms by the plus and minus operators.[3] Letters represent variables and constants. By convention, letters at the beginning of the alphabet (e.g. ${\displaystyle a,b,c}$) are typically used to represent constants, and those toward the end of the alphabet (e.g. ${\displaystyle x,y}$ and z) are used to represent variables.[4] They are usually written in italics.[5] Algebraic operations work in the same way as arithmetic operations,[6] such as addition, subtraction, multiplication, division and exponentiation.[7] and are applied to algebraic variables and terms. Multiplication symbols are usually omitted, and implied when there is no space between two variables or terms, or when a coefficient is used. For example, ${\displaystyle 3\times x^{2}}$ is written as ${\displaystyle 3x^{2}}$, and ${\displaystyle 2\times x\times y}$ may be written ${\displaystyle 2xy}$.[8] Usually terms with the highest power (exponent), are written on the left, for example, ${\displaystyle x^{2}}$ is written to the left of x. When a coefficient is one, it is usually omitted (e.g. ${\displaystyle 1x^{2}}$ is written ${\displaystyle x^{2}}$).[9] Likewise when the exponent (power) is one, (e.g. ${\displaystyle 3x^{1}}$ is written ${\displaystyle 3x}$).[10] When the exponent is zero, the result is always 1 (e.g. ${\displaystyle x^{0}}$ is always rewritten to 1).[11] However ${\displaystyle 0^{0}}$, being undefined, should not appear in an expression, and care should be taken in simplifying expressions in which variables may appear in exponents. ### Alternative notation Other types of notation are used in algebraic expressions when the required formatting is not available, or can not be implied, such as where only letters and symbols are available. As an illustration of this, while exponents are usually formatted using superscripts, e.g., ${\displaystyle x^{2}}$, in plain text, and in the TeX mark-up language, the caret symbol "^" represents exponentiation, so ${\displaystyle x^{2}}$ is written as "x^2".,[12][13] as well as some programming languages such as Lua. In programming languages such as Ada,[14] Fortran,[15] Perl,[16] Python[17] and Ruby,[18] a double asterisk is used, so ${\displaystyle x^{2}}$ is written as "x*2". Many programming languages and calculators use a single asterisk to represent the multiplication symbol,[19] and it must be explicitly used, for example, ${\displaystyle 3x}$ is written "3x". ## Concepts ### Variables Example of variables showing the relationship between a circle's diameter and its circumference. For any circle, its circumference c, divided by its diameter d, is equal to the constant pi, ${\displaystyle \pi }$ (approximately 3.14). Elementary algebra builds on and extends arithmetic[20] by introducing letters called variables to represent general (non-specified) numbers. This is useful for several reasons. 1. Variables may represent numbers whose values are not yet known. For example, if the temperature of the current day, C, is 20 degrees higher than the temperature of the previous day, P, then the problem can be described algebraically as ${\displaystyle C=P+20}$.[21] 2. Variables allow one to describe general problems,[22] without specifying the values of the quantities that are involved. For example, it can be stated specifically that 5 minutes is equivalent to ${\displaystyle 60\times 5=300}$ seconds. A more general (algebraic) description may state that the number of seconds, ${\displaystyle s=60\times m}$, where m is the number of minutes. 3. Variables allow one to describe mathematical relationships between quantities that may vary.[23] For example, the relationship between the circumference, c, and diameter, d, of a circle is described by ${\displaystyle \pi =c/d}$. 4. Variables allow one to describe some mathematical properties. For example, a basic property of addition is commutativity which states that the order of numbers being added together does not matter. Commutativity is stated algebraically as ${\displaystyle (a+b)=(b+a)}$.[24] ### Simplifying expressions Algebraic expressions may be evaluated and simplified, based on the basic properties of arithmetic operations (addition, subtraction, multiplication, division and exponentiation). For example, • Added terms are simplified using coefficients. For example, ${\displaystyle x+x+x}$ can be simplified as ${\displaystyle 3x}$ (where 3 is a numerical coefficient). • Multiplied terms are simplified using exponents. For example, ${\displaystyle x\times x\times x}$ is represented as ${\displaystyle x^{3}}$ • Like terms are added together,[25] for example, ${\displaystyle 2x^{2}+3ab-x^{2}+ab}$ is written as ${\displaystyle x^{2}+4ab}$, because the terms containing ${\displaystyle x^{2}}$ are added together, and, the terms containing ${\displaystyle ab}$ are added together. • Brackets can be "multiplied out", using the distributive property. For example, ${\displaystyle x(2x+3)}$ can be written as ${\displaystyle (x\times 2x)+(x\times 3)}$ which can be written as ${\displaystyle 2x^{2}+3x}$ • Expressions can be factored. For example, ${\displaystyle 6x^{5}+3x^{2}}$, by dividing both terms by ${\displaystyle 3x^{2}}$ can be written as ${\displaystyle 3x^{2}(2x^{3}+1)}$ ### Equations Animation illustrating Pythagoras' rule for a right-angle triangle, which shows the algebraic relationship between the triangle's hypotenuse, and the other two sides. An equation states that two expressions are equal using the symbol for equality, = (the equals sign).[26] One of the best-known equations describes Pythagoras' law relating the length of the sides of a right angle triangle:[27] ${\displaystyle c^{2}=a^{2}+b^{2}}$ This equation states that ${\displaystyle c^{2}}$, representing the square of the length of the side that is the hypotenuse, the side opposite the right angle, is equal to the sum (addition) of the squares of the other two sides whose lengths are represented by a and b. An equation is the claim that two expressions have the same value and are equal. Some equations are true for all values of the involved variables (such as ${\displaystyle a+b=b+a}$); such equations are called identities. Conditional equations are true for only some values of the involved variables, e.g. ${\displaystyle x^{2}-1=8}$ is true only for ${\displaystyle x=3}$ and ${\displaystyle x=-3}$. The values of the variables which make the equation true are the solutions of the equation and can be found through equation solving. Another type of equation is inequality. Inequalities are used to show that one side of the equation is greater, or less, than the other. The symbols used for this are: ${\displaystyle a>b}$ where ${\displaystyle >}$ represents 'greater than', and ${\displaystyle a where ${\displaystyle <}$ represents 'less than'. Just like standard equality equations, numbers can be added, subtracted, multiplied or divided. The only exception is that when multiplying or dividing by a negative number, the inequality symbol must be flipped. #### Properties of equality By definition, equality is an equivalence relation, meaning it has the properties (a) reflexive (i.e. ${\displaystyle b=b}$), (b) symmetric (i.e. if ${\displaystyle a=b}$ then ${\displaystyle b=a}$) (c) transitive (i.e. if ${\displaystyle a=b}$ and ${\displaystyle b=c}$ then ${\displaystyle a=c}$).[28] It also satisfies the important property that if two symbols are used for equal things, then one symbol can be substituted for the other in any true statement about the first and the statement will remain true. This implies the following properties: • if ${\displaystyle a=b}$ and ${\displaystyle c=d}$ then ${\displaystyle a+c=b+d}$ and ${\displaystyle ac=bd}$; • if ${\displaystyle a=b}$ then ${\displaystyle a+c=b+c}$ and ${\displaystyle ac=bc}$; • more generally, for any function f, if ${\displaystyle a=b}$ then ${\displaystyle f(a)=f(b)}$. #### Properties of inequality The relations less than ${\displaystyle <}$ and greater than ${\displaystyle >}$ have the property of transitivity:[29] • If ${\displaystyle a and ${\displaystyle b then ${\displaystyle a; • If ${\displaystyle a and ${\displaystyle c then ${\displaystyle a+c;[30] • If ${\displaystyle a and ${\displaystyle c>0}$ then ${\displaystyle ac; • If ${\displaystyle a and ${\displaystyle c<0}$ then ${\displaystyle bc. By reversing the inequation, ${\displaystyle <}$ and ${\displaystyle >}$ can be swapped,[31] for example: • ${\displaystyle a is equivalent to ${\displaystyle b>a}$ ### Substitution Substitution is replacing the terms in an expression to create a new expression. Substituting 3 for a in the expression a5 makes a new expression 35 with meaning 15. Substituting the terms of a statement makes a new statement. When the original statement is true independently of the values of the terms, the statement created by substitutions is also true. Hence, definitions can be made in symbolic terms and interpreted through substitution: if ${\displaystyle a^{2}:=a\times a}$ is meant as the definition of ${\displaystyle a^{2},}$ as the product of a with itself, substituting 3 for a informs the reader of this statement that ${\displaystyle 3^{2}}$ means 3 × 3 = 9. Often it's not known whether the statement is true independently of the values of the terms. And, substitution allows one to derive restrictions on the possible values, or show what conditions the statement holds under. For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0. Then we can substitute again, letting x = b and y = c, to show that if bc = 0 then b = 0 or c = 0. Therefore, if abc = 0, then a = 0 or (b = 0 or c = 0), so abc = 0 implies a = 0 or b = 0 or c = 0. If the original fact were stated as "ab = 0 implies a = 0 or b = 0", then when saying "consider abc = 0," we would have a conflict of terms when substituting. Yet the above logic is still valid to show that if abc = 0 then a = 0 or b = 0 or c = 0 if, instead of letting a = a and b = bc, one substitutes a for a and b for bc (and with bc = 0, substituting b for a and c for b). This shows that substituting for the terms in a statement isn't always the same as letting the terms from the statement equal the substituted terms. In this situation it's clear that if we substitute an expression a into the a term of the original equation, the a substituted does not refer to the a in the statement "ab = 0 implies a = 0 or b = 0." ## Solving algebraic equations A typical algebra problem. The following sections lay out examples of some of the types of algebraic equations that may be encountered. ### Linear equations with one variable Linear equations are so-called, because when they are plotted, they describe a straight line. The simplest equations to solve are linear equations that have only one variable. They contain only constant numbers and a single variable without an exponent. As an example, consider: Problem in words: If you double the age of a child and add 4, the resulting answer is 12. How old is the child? Equivalent equation: ${\displaystyle 2x+4=12}$ where x represent the child's age To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable.[32] This problem and its solution are as follows: Solving for x 1. Equation to solve: ${\displaystyle 2x+4=12}$ 2. Subtract 4 from both sides: ${\displaystyle 2x+4-4=12-4}$ 3. This simplifies to: ${\displaystyle 2x=8}$ 4. Divide both sides by 2: ${\displaystyle {\frac {2x}{2}}={\frac {8}{2}}}$ 5. This simplifies to the solution: ${\displaystyle x=4}$ In words: the child is 4 years old. The general form of a linear equation with one variable, can be written as: ${\displaystyle ax+b=c}$ Following the same procedure (i.e. subtract b from both sides, and then divide by a), the general solution is given by ${\displaystyle x={\frac {c-b}{a}}}$ ### Linear equations with two variables Solving two linear equations with a unique solution at the point that they intersect. A linear equation with two variables has many (i.e. an infinite number of) solutions.[33] For example: Problem in words: A father is 22 years older than his son. How old are they? Equivalent equation: ${\displaystyle y=x+22}$ where y is the father's age, x is the son's age. That cannot be worked out by itself. If the son's age was made known, then there would no longer be two unknowns (variables). The problem then becomes a linear equation with just one variable, that can be solved as described above. To solve a linear equation with two variables (unknowns), requires two related equations. For example, if it was also revealed that: Problem in words In 10 years, the father will be twice as old as his son. Equivalent equation {\displaystyle {\begin{aligned}y+10&=2\times (x+10)\\y&=2\times (x+10)-10&&{\text{Subtract 10 from both sides}}\\y&=2x+20-10&&{\text{Multiple out brackets}}\\y&=2x+10&&{\text{Simplify}}\end{aligned}}} Now there are two related linear equations, each with two unknowns, which enables the production of a linear equation with just one variable, by subtracting one from the other (called the elimination method):[34] ${\displaystyle {\begin{cases}y=x+22&{\text{First equation}}\\y=2x+10&{\text{Second equation}}\end{cases}}}$ {\displaystyle {\begin{aligned}&&&{\text{Subtract the first equation from}}\\(y-y)&=(2x-x)+10-22&&{\text{the second in order to remove }}y\\0&=x-12&&{\text{Simplify}}\\12&=x&&{\text{Add 12 to both sides}}\\x&=12&&{\text{Rearrange}}\end{aligned}}} In other words, the son is aged 12, and since the father 22 years older, he must be 34. In 10 years, the son will be 22, and the father will be twice his age, 44. This problem is illustrated on the associated plot of the equations. For other ways to solve this kind of equations, see below, System of linear equations. Quadratic equation plot of ${\displaystyle y=x^{2}+3x-10}$ showing its roots at ${\displaystyle x=-5}$ and ${\displaystyle x=2}$, and that the quadratic can be rewritten as ${\displaystyle y=(x+5)(x-2)}$ A quadratic equation is one which includes a term with an exponent of 2, for example, ${\displaystyle x^{2}}$,[35] and no term with higher exponent. The name derives from the Latin quadrus, meaning square.[36] In general, a quadratic equation can be expressed in the form ${\displaystyle ax^{2}+bx+c=0}$,[37] where a is not zero (if it were zero, then the equation would not be quadratic but linear). Because of this a quadratic equation must contain the term ${\displaystyle ax^{2}}$, which is known as the quadratic term. Hence ${\displaystyle a\neq 0}$, and so we may divide by a and rearrange the equation into the standard form ${\displaystyle x^{2}+px+q=0}$ where ${\displaystyle p={\frac {b}{a}}}$ and ${\displaystyle q={\frac {c}{a}}}$. Solving this, by a process known as completing the square, leads to the quadratic formula ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}},}$ where the symbol "±" indicates that both ${\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}}$ are solutions of the quadratic equation. Quadratic equations can also be solved using factorization (the reverse process of which is expansion, but for two linear terms is sometimes denoted foiling). As an example of factoring: ${\displaystyle x^{2}+3x-10=0,}$ which is the same thing as ${\displaystyle (x+5)(x-2)=0.}$ It follows from the zero-product property that either ${\displaystyle x=2}$ or ${\displaystyle x=-5}$ are the solutions, since precisely one of the factors must be equal to zero. All quadratic equations will have two solutions in the complex number system, but need not have any in the real number system. For example, ${\displaystyle x^{2}+1=0}$ has no real number solution since no real number squared equals −1. Sometimes a quadratic equation has a root of multiplicity 2, such as: ${\displaystyle (x+1)^{2}=0.}$ For this equation, −1 is a root of multiplicity 2. This means −1 appears twice, since the equation can be rewritten in factored form as ${\displaystyle [x-(-1)][x-(-1)]=0.}$ #### Complex numbers All quadratic equations have exactly two solutions in complex numbers (but they may be equal to each other), a category that includes real numbers, imaginary numbers, and sums of real and imaginary numbers. Complex numbers first arise in the teaching of quadratic equations and the quadratic formula. For example, the quadratic equation ${\displaystyle x^{2}+x+1=0}$ has solutions ${\displaystyle x={\frac {-1+{\sqrt {-3}}}{2}}\quad \quad {\text{and}}\quad \quad x={\frac {-1-{\sqrt {-3}}}{2}}.}$ Since ${\displaystyle {\sqrt {-3}}}$ is not any real number, both of these solutions for x are complex numbers. ### Exponential and logarithmic equations The graph of the logarithm to base 2 crosses the x axis (horizontal axis) at 1 and passes through the points with coordinates (2, 1), (4, 2), and (8, 3). For example, log2(8) = 3, because 23 = 8. The graph gets arbitrarily close to the y axis, but does not meet or intersect it. An exponential equation is one which has the form ${\displaystyle a^{x}=b}$ for ${\displaystyle a>0}$,[38] which has solution ${\displaystyle X=\log _{a}b={\frac {\ln b}{\ln a}}}$ when ${\displaystyle b>0}$. Elementary algebraic techniques are used to rewrite a given equation in the above way before arriving at the solution. For example, if ${\displaystyle 3\cdot 2^{x-1}+1=10}$ then, by subtracting 1 from both sides of the equation, and then dividing both sides by 3 we obtain ${\displaystyle 2^{x-1}=3}$ whence ${\displaystyle x-1=\log _{2}3}$ or ${\displaystyle x=\log _{2}3+1.}$ A logarithmic equation is an equation of the form ${\displaystyle log_{a}(x)=b}$ for ${\displaystyle a>0}$, which has solution ${\displaystyle X=a^{b}.}$ For example, if ${\displaystyle 4\log _{5}(x-3)-2=6}$ then, by adding 2 to both sides of the equation, followed by dividing both sides by 4, we get ${\displaystyle \log _{5}(x-3)=2}$ whence ${\displaystyle x-3=5^{2}=25}$ from which we obtain ${\displaystyle x=28.}$ ${\displaystyle {\overset {}{\underset {}{{\sqrt[{2}]{x^{3}}}\equiv x^{\frac {3}{2}}}}}}$ Radical equation showing two ways to represent the same expression. The triple bar means the equation is true for all values of x A radical equation is one that includes a radical sign, which includes square roots, ${\displaystyle {\sqrt {x}},}$ cube roots, ${\displaystyle {\sqrt[{3}]{x}}}$, and nth roots, ${\displaystyle {\sqrt[{n}]{x}}}$. Recall that an nth root can be rewritten in exponential format, so that ${\displaystyle {\sqrt[{n}]{x}}}$ is equivalent to ${\displaystyle x^{\frac {1}{n}}}$. Combined with regular exponents (powers), then ${\displaystyle {\sqrt[{2}]{x^{3}}}}$ (the square root of x cubed), can be rewritten as ${\displaystyle x^{\frac {3}{2}}}$.[39] So a common form of a radical equation is ${\displaystyle {\sqrt[{n}]{x^{m}}}=a}$ (equivalent to ${\displaystyle x^{\frac {m}{n}}=a}$) where m and n are integers. It has real solution(s): n is odd n is even and ${\displaystyle a\geq 0}$ n and m are even and ${\displaystyle a<0}$ n is even, m is odd, and ${\displaystyle a<0}$ ${\displaystyle x={\sqrt[{n}]{a^{m}}}}$ equivalently ${\displaystyle x=\left({\sqrt[{n}]{a}}\right)^{m}}$ ${\displaystyle x=\pm {\sqrt[{n}]{a^{m}}}}$ equivalently ${\displaystyle x=\pm \left({\sqrt[{n}]{a}}\right)^{m}}$ ${\displaystyle x=\pm {\sqrt[{n}]{a^{m}}}}$ no real solution For example, if: ${\displaystyle (x+5)^{2/3}=4}$ then {\displaystyle {\begin{aligned}x+5&=\pm ({\sqrt {4}})^{3},\\x+5&=\pm 8,\\x&=-5\pm 8,\end{aligned}}} and thus ${\displaystyle x=3\quad {\text{or}}\quad x=-13}$ ### System of linear equations There are different methods to solve a system of linear equations with two variables. #### Elimination method The solution set for the equations ${\displaystyle x-y=-1}$ and ${\displaystyle 3x+y=9}$ is the single point (2, 3). An example of solving a system of linear equations is by using the elimination method: ${\displaystyle {\begin{cases}4x+2y&=14\\2x-y&=1.\end{cases}}}$ Multiplying the terms in the second equation by 2: ${\displaystyle 4x+2y=14}$ ${\displaystyle 4x-2y=2.}$ Adding the two equations together to get: ${\displaystyle 8x=16}$ which simplifies to ${\displaystyle x=2.}$ Since the fact that ${\displaystyle x=2}$ is known, it is then possible to deduce that ${\displaystyle y=3}$ by either of the original two equations (by using 2 instead of x ) The full solution to this problem is then ${\displaystyle {\begin{cases}x=2\\y=3.\end{cases}}}$ This is not the only way to solve this specific system; y could have been resolved before x. #### Substitution method Another way of solving the same system of linear equations is by substitution. ${\displaystyle {\begin{cases}4x+2y&=14\\2x-y&=1.\end{cases}}}$ An equivalent for y can be deduced by using one of the two equations. Using the second equation: ${\displaystyle 2x-y=1}$ Subtracting ${\displaystyle 2x}$ from each side of the equation: {\displaystyle {\begin{aligned}2x-2x-y&=1-2x\\-y&=1-2x\end{aligned}}} and multiplying by −1: ${\displaystyle y=2x-1.}$ Using this y value in the first equation in the original system: {\displaystyle {\begin{aligned}4x+2(2x-1)&=14\\4x+4x-2&=14\\8x-2&=14\end{aligned}}} Adding 2 on each side of the equation: {\displaystyle {\begin{aligned}8x-2+2&=14+2\\8x&=16\end{aligned}}} which simplifies to ${\displaystyle x=2}$ Using this value in one of the equations, the same solution as in the previous method is obtained. ${\displaystyle {\begin{cases}x=2\\y=3.\end{cases}}}$ This is not the only way to solve this specific system; in this case as well, y could have been solved before x. ### Other types of systems of linear equations #### Inconsistent systems The equations ${\displaystyle 3x+2y=6}$ and ${\displaystyle 3x+2y=12}$ are parallel and cannot intersect, and is unsolvable. Plot of a quadratic equation (red) and a linear equation (blue) that do not intersect, and consequently for which there is no common solution. In the above example, a solution exists. However, there are also systems of equations which do not have any solution. Such a system is called inconsistent. An obvious example is {\displaystyle {\begin{cases}{\begin{aligned}x+y&=1\\0x+0y&=2\,.\end{aligned}}\end{cases}}} As 0≠2, the second equation in the system has no solution. Therefore, the system has no solution. However, not all inconsistent systems are recognized at first sight. As an example, consider the system {\displaystyle {\begin{cases}{\begin{aligned}4x+2y&=12\\-2x-y&=-4\,.\end{aligned}}\end{cases}}} Multiplying by 2 both sides of the second equation, and adding it to the first one results in ${\displaystyle 0x+0y=4\,,}$ which clearly has no solution. #### Undetermined systems There are also systems which have infinitely many solutions, in contrast to a system with a unique solution (meaning, a unique pair of values for x and y) For example: {\displaystyle {\begin{cases}{\begin{aligned}4x+2y&=12\\-2x-y&=-6\end{aligned}}\end{cases}}} Isolating y in the second equation: ${\displaystyle y=-2x+6}$ And using this value in the first equation in the system: {\displaystyle {\begin{aligned}4x+2(-2x+6)=12\\4x-4x+12=12\\12=12\end{aligned}}} The equality is true, but it does not provide a value for x. Indeed, one can easily verify (by just filling in some values of x) that for any x there is a solution as long as ${\displaystyle y=-2x+6}$. There is an infinite number of solutions for this system. #### Over- and underdetermined systems Systems with more variables than the number of linear equations are called underdetermined. Such a system, if it has any solutions, does not have a unique one but rather an infinitude of them. An example of such a system is {\displaystyle {\begin{cases}{\begin{aligned}x+2y&=10\\y-z&=2.\end{aligned}}\end{cases}}} When trying to solve it, one is led to express some variables as functions of the other ones if any solutions exist, but cannot express all solutions numerically because there are an infinite number of them if there are any. A system with a higher number of equations than variables is called overdetermined. If an overdetermined system has any solutions, necessarily some equations are linear combinations of the others. ## References • Leonhard Euler, Elements of Algebra, 1770. 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# CAT Time and Work Very Important Questions PDF [With Solutions] 0 1388 Time and Work for CAT is an important topic in the CAT Quant section. Over the past few years, Time and Work has made a recurrent appearance in the Quant Section of the CAT. You can check out these Time and Work CAT Previous year questions. Practice a good amount of sums on CAT Time and Work questions. In this article, we will look into some important Time and Work Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT Time and Work Questions PDF below, which is completely Free. Question 1:Â Karan and Arjun run a 100-meter race, where Karan beats Arjun 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race? a)Â Karan and Arjun reach the finishing line simultaneously. b)Â Arjun beats Karan by 1 metre c)Â Arjun beats Karan by 11 metres. d)Â Karan beats Arjun by 1 metre. Solution: The speeds of Karan and Arjun are in the ratio 10:9. Let the speeds be 10s and 9s. Time taken by Karan to cover 110 m = 110/10s = 11/s Time taken by Arjun to cover 100 m = 100/9s = 11.11/s Therefore, Karan reaches the finish line before Arjun. From the options, the only possible answer is d). Question 2:Â Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? a)Â 3 b)Â 3.5 c)Â 4 d)Â 4.5 e)Â 5 Solution: Let the distance be D. Time taken by Arun = D/30 Time taken by Barun = D/40 Now, D/40 = D/30 – 2 => 3D = 4D – 240 => D = 240 Therefore time taken by Arun to cover 240 km = 240/30 = 8 hr Time Kiranmala takes to cover 240 km = 240/60 = 4 hr So, Kiranmala has to start 4 hours after Arun. Question 3:Â Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyomâ€™s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up? a)Â 40 b)Â 50 c)Â 60 d)Â 80 Solution: Let the number of steps on the escalator be x. So, by the time Shyama covered 25 steps, the escalator moved ‘x-25’ steps. Hence, the ratio of speeds of Shyama and escalator = 25:(x-25) Similarly, the ratio of speeds of Vyom and escalator = 20:(x-20) But the ratio is 3:2 Ratio of speeds of Shyama and Vyom = 25(x-20)/20(x-25) = 3/2 => 10(x-20) = 12(x-25) => 2x = 100 => x = 50 Question 4:Â On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend to the patient at the hospital. Assume that a total ofÂ 1 min is elapsed for taking the patient into and out of the ambulance? a)Â 4 min b)Â 2.5 min c)Â 1.5 min d)Â The patient died before reaching the hospital Solution: Let the distance between gutter 1 and A be x and between gutter 1 and 2 be y. Hence, x + y + 2y + x = 20 => 2x+3y=20 Also x = 30kmph 5/60 = 2.5km Hence, y = 5km After the ambulance doubles its speed it goes at 60kmph i.e. 1km per min. Hence, time taken for the rest of the journey = 152 + 2.5 = 32.5 Hence, total time = 5 + 32.5 + 1 = 38.5 mins So, the doctor would get 1.5 min to attend to the patient. Question 5:Â It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed? [CAT 2002] a)Â 6.40 pm b)Â 7 pm c)Â 7.20 pm d)Â 8 pm Solution: Let the work done by each technician in one hour be 1 unit. Therefore, total work to be done = 60 units. From 11 AM to 5 PM, work done = 66 = 36 units. Work remaining = 60 – 36 = 24 units. Work done in the next 3 hours = 7 units + 8 units + 9 units = 24 units. Therefore, the work gets done by 8 PM. Question 6:Â Three small pumps and a large pump are filling a tank. Each of the three small pump works at 2/3 the rate of the large pump. If all four pumps work at the same time, they should fill the tank in what fraction of the time that it would have taken the large pump alone? [CAT 2002] a)Â 4/7 b)Â 1/3 c)Â 2/3 d)Â 3/4 Solution: Let the work done by the big pump in one hour be 3 units. Therefore, work done by each of the small pumps in one hour = 2 units. Let the total work to be done in filling the tank be 9 units. Therefore, time taken by the big pump if it operates alone = 9/3 = 3 hours. If all the pumps operate together, the work done in one hour = 3 + 23 = 9 units. Together, all of them can fill the tank in 1 hour. Required ratio = 1/3 Question 7:Â A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is a)Â 25, 130 b)Â 30,150 c)Â 20, 90 d)Â None of these Solution: Time difference, when second time car’s engine failed at a distance of 30 km., is of 9 min. Hence putting this in equation: $\frac{12}{\frac{4v}{5}} – \frac{12}{v} = \frac{9}{60}$ hr. (Because difference of time is considered with extra travelling of 12 km. in second case) We will get $v (velocity) = 20$ km/hr. Now for distance $\frac{d-18}{16} + \frac{18}{20} – \frac{d}{20} = \frac{45}{60}$ hr. (As car is 45 min. late after engine’s faliure in first case) So $d$ = 78 km. Hence none of these will be our answer. Question 8:Â Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously? a)Â 12 hours b)Â 10 hours c)Â 8 hours d)Â 6 hour Solution: As machine B’s efficiency is twice as of A’s, Hence, it will complete its work in 30 hours. And C’s efficiency is putting A and B together i.e. = 20 hours $( (\frac{1}{60} + \frac{1}{30})^{-1})$ Now if all three work together, then it will be completed in x (say) days. $\frac{1}{x} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60}$ or x = 10 hours Question 9:Â Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk? a)Â 1 hour b)Â 50 minutes c)Â 1/2 hour d)Â 55 minutes Solution: Since we know that Neera’s husband drives at a uniform speed to and from his residence. If he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence. If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time. When the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM. So, Neera must have walked for 55 minutes from 5PM. Question 10:Â In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.) a)Â Akshay, $\frac{1}{12}$ mile b)Â Chinmay, $\frac{1}{32}$ mile c)Â Akshay, $\frac{1}{24}$ mile d)Â Chinmay, $\frac{1}{16}$ mile Solution: Akshay can complete 1600 – 128 = 1472 m and Bhairav can completer 1600 m in the same time. Bhairav can complete 100 m and Chinmay can complete 96 m in the same time. => Bhairav can complete 1600 m and Chinmay can complete 1536 m in the same time. => Akshay can complete 1472 m and Chinmay can complete 1536 m in the same time. 1.5 miles => 2400 m Distance travelled by Akshay by the time Chinmay completes 1.5 miles = $\frac{1472}{1536}2400$ = 2300 m => Akshay lost by 100 m, which is $\frac{1}{16}th$ of a mile.
# In this lesson you will dive deep into how the LAD 7 MULTIPLY routine of the previous RsLogix 500 - Sequencing Machine Steps uses integers and their associated bits to sequence through steps. This method is probably used more than it should be in industry but is another great exercise in manipulating 16 bits. Note that by simply multiplying by 2, you shift to the “1” to the next bit. The advantages of this method have diminished in my mind over the years although it used to be my preferred programming method. It does allow you to manipulate 16 bits at a time but from a maintenance point of view I think the Word method is clearer while accomplishing the same thing. The disadvantages are it isn't as clear for maintenance troubleshooting plus if you have over 16 steps it gets more complicated than the other methods.. 1. Download the CougarTurnSignalMultiply.RSS file. This is a simplified version of the RSS file you used in the Sequencing Machine Steps lesson, only including the Multiply sequence programming. 3. Open up the LAD 7 MULTIPLY routine. 4. Turn on I:0/0. It should still be wired to the Green button from the previous lesson. 5. Your screen should be sequencing through rungs 0-5 similarly to the video below. You may have to right click the image below and open and new tab depending on your browser. Below is the explanation of how the integer keeps track of which step in the sequence the PLC is on. We are ignoring the XIC I:0/0 left turn signal in the explanation below for simplicity. The key to understanding this method is understanding that the integer B3:1 is directly linked to the values of B3:1/0 through B3:1/15 since they are in fact the same number. So B3:1 equaling 1 will always mean B3:1/0 is a 1 and B3:1/1-B3:1/15 are all 0. Then understanding that multiplying a binary number by two simply shifts its position by 1. So if B3:1 equals 1, which means B3:1/0 is a 1 and B3:1/1-B3:1/15 are all 0, then multiplying that 1 by 2 makes B3:1 equal 2, which means B3:1/1 is now 1, and B3:1/0 and B3:1/1-B3:1/15 are now all 0. After that, this one sequences exactly the same as the previous RsLogix 500 - Sequencing Machines with Cascading Timers and RsLogix 500 - Sequencing Machines with Counters sequencers. This one is a little less common but you will run into it from time to time and it is an excellent exercise in manipulating bits at the word level. Continue to the next lesson, RsLogix 500 - Sequencing Machines Part 7 - Sequencer Instructions. ## Next Steps Go to the Allen Bradley RsLogix 500 PLC Training PLC Training Getting Started Lesson series to select your next lesson. There are also many other Lesson Series on PLC Programming and Industrial Automation.
Home » Maths » Fraction to Percent Conversion # Fraction to Percent- Conversion Formula, Steps, Table, Examples Fractions and Percent are the two most common mathematical concepts used in daily life. You must have come across these two mathematical entities quite regularly. If you are someone who have studied mathematics, then you must know that you can convert fraction to percent and vice versa. In this article, we will learn about the fraction to percent conversion step by step. We will also check the Fraction to Percent conversion formula for quick calculation. ## Fraction to Percent Fraction and Percent are two parts of the same coin. Before understanding it, let us recall what does a fraction and percent means. In Mathematics, a fraction is described as a component of the entire object. For instance, if a pizza is cut into four equal slices, each slice is denoted by ¼. Fractions make it easier to distribute and compare numbers, leading to quicker calculations. The way fractions are represented is simpler compared to using decimal values. A percent or percentage in math is a value that can be shown as a fraction out of 100. To find the percentage of a number, divide the number by the total amount and then multiply by 100. Therefore, the percentage represents a fraction per hundred. The term “per cent” indicates out of every 100. The symbol “%” is used to represent it. ## Fraction to Percent Conversion The fraction to percent conversion is one of the common types of problems encountered in both school as well as competitive exams. To grasp the idea of fraction and percent, contemplate that in a class of 38 students, 23 are male. Then, what percentage of students are male? The answer should be 23 out of 38. It is written as: 23/28. In simpler terms, the fraction 23/38 is approximately equal to 0.60526315789473684210526315789474, which is roughly 60%. That is how, we can interchange a fcation into a percentage and vice versa. ## Fraction to Percent Formula For faster calculation, candidates can use the direct formula for fraction to percent conversion. To change a fraction to percentage, we convert it to a decimal number and multiply it by 100 followed by the symbol %. The formula to convert fraction m/n to percentage is: Percent Value = m/n * 100 where, m = numerator of the fraction n = denominator of the fraction And students must remember that it is compulsory to add the percentage sing (%) sign after the result value. ## Fraction to Percent Conversion Steps Converting fractions into percentages allows for simple comparison and ranking of two quantities. Fractions, decimals, and percentages are concepts that are all connected to each other. There are two major steps involved in the conversion of a fraction into percent. In order to change a fraction into a percent, we must initially convert it to a decimal. To calculate the percentage value, you need to multiply the provided decimal by 100. The two steps are show below: • Convert the fraction in a decimal by dividing the numerator by its denominator • Multiply the decimal result to get the percent value of that fraction ## How to Convert Fraction to Percent To change a fraction into a percentage, convert the fraction to a decimal, multiply the decimal by 100, and include the percentage symbol. Multiplying the fraction by 100, simplifying it, and adding the percentage sign is an alternative method for converting a fraction to a percentage. Let us take an example to understand this conversion. Suppose we want to convert the fraction 4/5 into percent Step 1: Change the fraction to a decimal. In order to accomplish this, you should divide the top number by the bottom number. On following the step 1, we divide 4 by 5 to get the decimal value of 0.8 Step 2: In order to calculate a percentage, you need to multiply the decimal number that you have by 100. The result is the fraction converted into a percentage. Write the percentage as a number followed by the percentage symbol next to the solution. On following the step 2, we get: 0.8 * 100 = 80 and on adding the percentage symbol, it becomes 80 %. Hence, the percent value for the fraction 4/5 is 80 %. Another pictorial example of the fraction to percent conversion has been given below for more clarity. Alternatively, we can also multiply the numerator by 100 and then simplify the obtained value to get the desired result. For Example: 4/5 * 100 = 400/5 on dividing the 400 by 5, we get 80 and putting the % sign after it makes it 80 %, the desired answer. ## Fraction to Percent Table The table representing the percentage value of different fractions have been provided below for quick consultation. These values are often asked in aptitude test. Make sure that you can derive the following conversions. Fraction Value Percent Value 1 / 2 50 % 1 / 3 33.33 % 2 / 3 66.67 % 1 / 4 25 % 2 / 4 50 % 3 / 4 75 % 1 / 5 20 % 2 / 5 40 % 3 / 5 60 % 4 / 5 80 % 1 / 6 16.67 % 2 / 6 33.33 % 3 / 6 50 % 4 / 6 66.67 % 5 / 6 83.33 % 1 / 7 14.285714 % 2 / 7 28.571429 % 3 / 7 42.857143 % 4 / 7 57.142858 % 5 / 7 71.428571 % 6 / 7 85.714286 % 1 / 8 12.5 % 2 / 8 25 % 3 / 8 37.5 % 4 / 8 50 % 5 / 8 62.5 % 6 / 8 75 % 7 / 8 87.5 % 1 / 9 11.111111 % 2 / 9 22.222222 % 3 / 9 33.333333 % 4 / 9 44.444444 % 5 / 9 55.555556 % 6 / 9 66.666667 % 7 / 9 77.777778 % 8 / 9 88.888889 % 1 / 10 10 % 2 / 10 20 % 3 / 10 30 % 4 / 10 40 % 5 / 10 50 % 6 / 10 60 % 7 / 10 70 % 8 / 10 80 % 9 / 10 90 % ## Fraction to Percent Example Solutions Practice the example solutions provided below for mastering this concept. Example 1: Convert 4 / 16 to percent. Solution: Step 1: Convert the fraction 44/ 16 into decimal Step 2: 4 / 16 =0.25 Step 3: Multiply the decimal by 100: 0.25 × 100 = 25.00 % Therefore, the solution is 25.00 % Example 2: Change 8 / 12 to a percent. Solution: Step 1: Convert the fraction 9 / 12 into decimal Step 2: 9 / 12 = 0.75 Step 3: Multiply the obtained decimal by 100: 0.75 × 100 = 75% Therefore, the solution is 75% Example 3: In a football tournament, team Royal has won 7 games out of 8 games played, while team Beast has won 9 out of 10 games played. Which cricket team has a higher percentage of wins? Solution: Team Royal has won 7 out of 8 games played, i.e., 7 / 8 Step 1: Convert the fraction 7 / 8 into decimal Step 2: 7 / 8 = 0.875 Step 3: Multiply the decimal by 100: 0.875 × 100 = 87.5% Team Beast has won 9 out of 10 games played: 9 / 10 Step 1: Convert the fraction 9 / 10 into decimal Step 2: 9 / 10 = 0.9 Step 3: Multiply the decimal by 100: 0.9 × 100 = 90% Team Royal has 87.5% of winning rate, while team Beast has a 90 % winning rate Therefore, team Beast has a higher percentage of wins with 90 % Sharing is caring! ## FAQs ### What is the percent value of number 1. The value of 1 in percentage is 100%.
The new Gynzy Flash-free You are here : Home > Mathematics > # Fractions Sort ### Multiplying Fractions - Lesson plan - Review different methods for finding the product of two fractions, and apply to finding the product of negative fractions. ### Dividing Fractions & Mixed Numbers - Lesson plan - Dividing integers, mixed numbers and fractions by fractions and mixed numbers. Solve word problems involving mixed numbers and fractions. ### Adding & Subtraction Fractions with Unlike Denominators - Lesson plan - Use fraction models to add and subtract fractions with unlike denominators, and progressing to solve addition and subtraction problems without modeling. ### Graphmaker - Activity - Make, use and fill in a graph. ### Adding & Subtracting Mixed Numbers with Unlike Denominators - Lesson plan - Use fraction models to add and subtract mixed numbers with unlike denominators, and progressing to solving mixed number problems without modeling. ### Fractions, Decimals & Percents - Lesson plan - Write numbers as percents, fractions and decimals. Match equivalent fractions, decimals and percents. ### Algebraic Fractions - Lesson plan - Simplify algebraic fractions. Add, subtract, multiply and divide algebraic fractions. ### Adding & Subracting Fractions and Mixed Numbers - Lesson plan - Using fraction models to add and subtract fractions and mixed numbers with unlike denominators, and progressing to solving fraction problems without modeling. ### Fractions & Decimals - Lesson plan - Write fractions as either terminating or repeating decimals. Convert between fractions and decimals. Identify terminating and repeating decimals. ### Estimating with Percents - Lesson plan - Estimating percents of a number using near fractions. Solve percent problems by estimation. ### Simplifying Fractions - Lesson plan - Simplify fractions. Find the GCF of the numerator and denominator and use this to reduce fractions to their least terms. - Lesson plan - Develop fluency in adding and subtracting non-negative rational numbers (halves, fourths, eighths; thirds, sixths, twelfths; fifths, tenths, hundredths, thousandths; mixed numbers). ### Fractions - Lesson plan - Use 1/2 and 1/4 appropriately. Picture and name fractions. ### Equivalent Fractions - Lesson plan - Use models and points of reference to find equivalent fractions. ### Equivalent Fractions - Lesson plan - Use models, points of reference and equivalent forms to judge the size of fractions. Round fractions to the nearest half. ### Fractions - Lesson plan - Recognize and model equivalent fractions with concrete objects or pictures. Identify fractional parts of a group.
# Solving Quadratics by Factoring and Completing the Square This section covers: Note that factoring the difference of cubes, and more advanced factoring, including Factoring with Exponents can be found here in the Rational Expressions and Functions section, and Solving by Factoring section. Again, there are several ways to solve quadratics, or find the solution (also known as the x-intercepts, roots, zeros, or values) to a quadratic equation. (Remember that we solve quadratic equations most easily by getting everything to one side of the equal sign, which sets the quadratic to 0). In this section, we’ll talk about two other ways: • Factoring and setting factors to 0 (there are many ways to factor, but not everything can be factored) • Completing the square and taking the square root of each side (a way where we don’t have to set the quadratic to 0!) # Factoring Methods Note that there is more information on factoring in the Solving by Factoring section here. So just the way we learned how to multiply binomials (via FOILING), we need to learn how to do the opposite, or factor (or “unfoil”) the resulting trinomials. After we’ve done enough multiplying binomials and factoring trinomials, this will become second nature. We’ll also learn other basic polynomial factoring methods, like taking out the Greatest Common Factors (GCF) of polynomials, and factoring the difference of two squares and factoring perfect square trinomials. Think of factoring as just “pulling apart” things that are multiplied together. It’s the same principle of factoring 35 and getting 5 and 7. When we factor in algebra, we do the same thing, but with variables. NOTE: Remember that when we factor, we want to set each factor with a variable in it to 0, and solve for the variable to get the roots. This is because any factor that becomes 0 makes the whole expression 0. (This is the zero product property: if ab = 0, than a = 0 and/or b = 0). Also remember that when we factor to solve quadratics or any polynomials, we can never just divide by factors (with variables) on both sides to get rid of them. If we do this, we may be missing solutions! Note that not every quadratic can be factored; if it can’t, we’ll have to use one of the other methods. But let’s learn the factoring techniques so we can solve the quadratics that can be factored; it’s a pretty simple way to solve them. And later, when we learn how to solve more generic polynomials in the Graphing and Solving Polynomial Functions section, we’ll know how to factor! ## Taking out the Greatest Common Factor (GCF) So let’s first start with polynomials that need some simple factoring. We always need to pull out the GCF (largest coefficients/variables that go into all the terms) first. (We learned about the GCF with regular numbers in the Multiplying and Dividing Section here.) Remember from the Exponents and Radicals in Algebra section that we add exponents when we multiply with the same base, and subtract exponents when we divide with the same base. Here are some examples of factoring polynomials by taking the GCF out: See how it’s reversing the distributive property? It’s “undistributing”. And it’s always a good idea to multiply back to check your answers! When we factor quadratics, we try to “unfoil” to get two binomials. Again, remember that when factoring trinomials, we always need to take out any GCF’s first! And remember that this method does not always work; certain quadratics are “unfactorable” (prime), and must be solved with another method. Also remember that, with a trinomial, if the discriminant where a, b, and c are from , isn’t a perfect square (like 25 or 49), you can’t factor it. Here are some methods that we use: ### Guess and Check “Guess and Check” is just what it sounds; we have certain rules, but we try combinations to see what will work. Let’s start with an example; let’s “UNFOIL” or factor one of the first examples that we worked with in Quadratics. Here’s how we FOIL’ed, and how we “UNFOIL”: So if we were to solve this trinomial, we would set the factors equal to zero: Note the signs when you “unfoil”; multiply these back to make sure they work: Sometimes it’s easier and more visual if you use an “X” when you begin to factor these; we’ll first show this when there are no numbers (coefficients) before the x2 term. We can put the coefficient of the middle term on the top, the constant term (no variable) on the bottom, and then use the two sides to find 2 numbers multiplied together to equal the bottom, but added together to equal the top. Let’s try this for the last problem above: Let’s try another one that’s a little bit more complicated. Note again that to make the factoring easier, we always want to factor the GCF out of the polynomial before we “unfoil”. Example: Let’s factor . First we take out the GCF (2) to get : We still need to “unfoil” . This is a little more complicated, since we have the 4x2. Now we need two factors of 4 and two factors of 7, so that multiplying the inside terms and multiplying the outside terms and adding them together will give us –11. We can see that this trinomial is factorable, since (the discriminant) is which is a perfect square. When we’re “guessing and checking”, we’re actually “foiling back” to see if we got the right answer! You may want to use your factor trees (that we learned in the Multiplying and Dividing section) to write down all the factors of the first and last coefficients. Here are all the combinations we could have. Note that the factors of 4 are 2 and 2, or 4 and 1. The factors of 7 are just 7 and 1. So the complete factoring is: 2(4x – 7)(x – 1). If we were to solve this trinomial, we would get by setting each factor to 0 and solving for x. (Ignore the factor of 2, since 2 can never be 0. If we had an x on the outside, an additional factor would be 0). Note that if we had , we’d also have to split the y’s at the end, so we would factor to 2(4x – 7y)(x – y). Multiply it all to together to show that it works! (This is one we wouldn’t be asked to solve). ## Special Products of Binomials – Easily Factored There are two special cases where you can use a shortcut to factor the trinomials: Difference of Two Squares and Perfect Square Trinomials. We saw these in the Introduction to Polynomials section in the table here. Here are the two cases and how to factor them: ## Grouping Method, or the “ac” Method The Grouping Method for factoring trinomials has gotten more popular since the FOILing hasn’t been taught as much recently. (The reason it’s not taught as much is because it can be only used with binomial multiplication. I still like FOILing since I believe learning to factor is easier if we learn FOILing first). We will turn the trinomial into a quadratic with four terms, to be able to do the grouping. Then we have to find a pattern of binomials so we can use the distributive property to put them together (like a puzzle!). Let’s look at the same problem as above (with the 2 already factored out, but let’s keep the y’s at the end): Let’s use our “X” again to help us solve by the grouping method. We can put the middle term’s coefficient (–11) on the top part of the “X”, but this time we multiply the first and last coefficients (4 x 7 = 28) and put it on bottom part of an “X”. Remember that the sign of a term comes before it, and pay attention to signs. So since we found that –4 and –7 “work”, we can rewrite the trinomial and separate the middle term, so we can do some grouping and factor with the grouping method: ## Grouping Method with Four Terms Here are some examples of the grouping method when we start out with 4 terms. Notice that the first one is a 4-term quadratic and the second is a cubic polynomial that includes factoring with the difference of squares. Remember the difference of squares factoring is Note that having four terms to factor doesn’t always work with the grouping method above; sometimes we have to look for differences of squares (sometimes combined with the grouping method): ## The Box Method One more method that is getting popular is called the “box” method. This is a modified “ac” method, but we use greatest common factors (GCF) to help us factor. This is sort of the opposite of what we did with the “multiplication box” earlier. We put the first term in the upper left hand corner, the last term in the lower right hand corner, and we divide up the middle term in the remaining two corners so they add up to the middle term, and are factors of the first term times the last term (thus, the “ac” method). You can put the middle terms (upper right and lower left corners) in any order, but make sure the signs are correct so they add up to the middle term. If you have set up the box correctly, the diagonals should multiply to the same product. Then we get the GCFs across the columns and down the rows, using the same sign of the closest box (boxes either on the left or the top). Let’s try this for : Then read across and down to get the factors: (5x + 2)(2x – 3). Foil it back, and we see that we got it correct! If we were to solve this trinomial, we would get by setting each factor to 0 and solving for x. Note: when we do the box method, we have to make sure the x2 has a positive coefficient; otherwise, we have to take the negative out across all three terms, do the factoring, and then put the negative in the front of the factored answer. ## The “ac” Method without Grouping There’s another new method to factor trinomials out there and it’s kind of cool. It’s similar to the “ac” method above, but a little simpler. Again, let’s use : # Completing the Square (Square Root Method) Completing the square is what is says: we take a quadratic in standard form and manipulate it to have a binomial square in it, like . This way we can solve it by isolating the binomial square (getting it on one side) and taking the square root of each side. This is commonly called the square root method. We can also complete the square to find the vertex more easily, since the vertex form is . What we want to do for the square root method is to make a square out of the side with the variable, and move the numbers (constants) to the other side, so we can take the square root of both sides. Then we don’t have to use the quadratic equation, or “unfoil” to solve. Let’s first think about what happens when we square a binomial by looking at: (We first saw these perfect square trinomials in Introduction To Polynomials.) Since we add the product of the middle terms twice, we have twice the product of the first (x) and second (3) terms in the middle (to get 6x). See also how we have the square of the second term (3) at the end (9). So to complete the square of a trinomial that isn’t a perfect square, we need to halve the second term and take the square of it – and then add that number so the square can be complete. Then we have to make sure to add the same thing to the other side. Then we take the square root of each side, remember that we need to include the plus and minus of the right hand side, since by definition, the square root is just the positive. Another way to think of it is the absolute value of the left side equals the right side, so we have to include the plus and minus of the right side. (Note: you may want to review Solving Exponential and Radical Equations to review how to solve square root equations.) Let’s work with first, since it just starts with an (coefficient of of 1): Note that it would have been much easier to factor this quadratic, but, like the quadratic equation, we can use the completing the square method for any quadratic. (We may not always get a real number for the answer(s); we’ll talk about imaginary numbers later). Here’s one that’s a little trickier, because of the radical in the coefficient of x: Here’s one where the coefficient of the isn’t 1. (Remember again that if we can take out any factors across the whole trinomial, do it first and complete the square with the trinomial only.) This gets a little more complicated, but it’s not too bad: ## Completing the Square to get Vertex Form We can use the same technique to put (standard form, or ) into vertex form (). In this case, we want to leave everything on one side, so we have to undo what we’ve added or subtracted by completing the square: Note: There is another way to convert from Standard Form to Vertex Form. We can use a General Vertex Form equation $$y=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}-\frac{{{{b}^{2}}-4ac}}{{4a}}$$. Let’s try this for the example above: a = 2, b = –5, and c = –12. We would get $$y=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}-\frac{{{{b}^{2}}-4a}}{{4a}}=2{{\left( {x+\frac{{-5}}{{2\left( 2 \right)}}} \right)}^{2}}-\frac{{{{{\left( {-5} \right)}}^{2}}-4\left( 2 \right)\left( {-12} \right)}}{{4\left( 2 \right)}}=2{{\left( {x-\frac{5}{4}} \right)}^{2}}\,\,-\,\,\frac{{121}}{8}\,\,$$. Pretty cool! # Obtaining Quadratic Equations from a Graph or Points Sometimes you will be asked to look at a quadratic graph (or given the vertex and a point) and write the equation (in all three forms) for that graph. The easiest way to do this is to find the vertex from the graph (usually it’s obvious!), put the equation in vertex form, and then compute the “a” (coefficient of ) from another point on the graph, such as a root or y-intercept, if given. If you’re given the x-intercepts (roots), you can also put it in factored form, and use another point (not one of the roots) to find the “a” part of the equation. Remember that the “a” in all three forms (standard, factored, and vertex) will be the same. Here are some examples: Here’s another type of problem you might see where you have to write a Quadratic Function given a Parabola’s Axis of Symmetry and two Non-Vertex Points. Note that we have to use a System of Equations: So again, here are different forms of Quadratics and also the methods for finding roots of Quadratics. Note that the “a” (coefficient of the ) is the same for all three forms! Note: Sometimes we’ll have a Quadratic in “almost” vertex form and play around with it to get it in vertex form. For example, if we have $$y=-8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3$$ and want to change it to vertex form, we can take out the coefficient of x and do some algebra: $$y=-8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3=-8{{\left( {\left( {\frac{1}{2}} \right)\left( {x+4} \right)} \right)}^{2}}+3$$$$=-8{{\left( {\frac{1}{2}} \right)}^{2}}{{\left( {x+4} \right)}^{2}}+3=-2{{\left( {x+4} \right)}^{2}}+3$$, so the vertex is (–4, 3). Note: if we are really having a difficult time factoring a quadratic trinomial, we could, as a last resort, use the Quadratic Formula, get the roots, and, if they are rational (integers or fractions), put the quadratic back in factored form. For example, if we used the Quadratic Formula and got roots and –3, our factors would be . Learn these rules, and practice, practice, practice! Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems. If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets. You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy! ## 19 thoughts on “Solving Quadratics by Factoring and Completing the Square” 1. Help! Math is not my forte and my daughter is struggling with competing the square- square root method. Her big question is why does the x term ostensibly disappear after taking half of the coefficient and then squaring it? While she doesn’t understand the reasoning re: the formula that requires that she take half the coef. then square it, she can do it, but the next step feels like voodoo to her (me too!). Can someone please explain in detail what leads to that jump? Thanks in advance and light to whomever can help! We have visited several touted websites w/ videos and my last resort was to put the word “girls” in front of our search on this topic! I am just so happy to have an opportunity to ask the question! • Hi! Thanks so much for your questions and hope this helps. The x term “disappears” because if you were to FOIL (multiply) it back out, it would be there. So for example (x + 1)^2 doesn’t really have an x term, but x^2 + 2x + 1 does. So when you add the squared to the binomial (like (x + 1)^2), the x term is in there – but you just don’t see it. Does that make sense? (I know this is a dated comment, but wanted to put it on the site 🙂 2. The article was nicely and fully explained. Since solving quadratic equations is a main core subject of high school math, studies on solving methods has been numerous and diversified. There are so far 8 existing common solving methods. They are: graphing, completing the square, factoring FOIL Method, quadratic formula, the Bluma Method, the Diagonal Sum Method, the improved factoring AC Method, and recently the New Transforming Method (Yahoo and Google Search) that may be the fastest and simplest solving method. I sincerely suggest that the article’s author keeps tracks of the new methods and explains them to students. • Thank you so much for your comment, and I will certainly investigate these new methods and add to my blog. Thanks again! Lisa 3. Thank you, Lisa. Very glad that you respond to my comments. I sincerely suggest that you read these articles: – Best method to solve quadratic equation (Math Concentration website) – Best methods to solve quadratic inequalities (Math Concentration website) – Solving quadratic equations by the new Transforming Method (Adobe Edu Exchange) – Quadratic equations – Compare the factoring AC Method and the new Transforming Method (Math Concentration) – Quadratic equations – Compare the factoring AC Method and the Diagonal Sum Method (Math Concentration) – Solving quadratic equations by the Diagonal Sum Method (Math Concentration) – Shortcut in solving linear equations (Math Concentration) 4. thank you so much factorizing by completing square method is done to me 5. there is a mistake in the ‘Grouping Method with 4 Terms’ the 25z squared is written as 5z squared…. • Thank you so much for finding the error; I’ve fixed it. Let me know if you find any other mistakes 😉 Lisa 6. There is another way to convert f(x) = ax^2 +bx + c to vertex form f(x) = a(x + p)^2 + q by using the general vertex form: f(x) = a(x + b/2a)^2 – d^2/4a, where d^2 = b^2 – 4ac, and (-b/2a) is the x-coordinate of the parabola axis. By replacing values of a, b, c into the general form, we get the specific vertex form. Example: Convert f(x) = 2x^2 – 5x – 12 to vertex form. We have: a = 2; (b/2a) = -5/4; and d^2 = b^2 – 4ac = 121; and (-d^2/4a) = -121/8. Answer. Vertex form: f(x) = 2(x – 5/4)^2 – 121/8. • Thanks so much for the tip – I’m going to add it to my webpage now. Let me know if you have other examples! lisa 7. Example. Convert y = 2x^2 + 9x + 10 to vertex form. Solution. a = 2; and (-b/2a) = -9/4, and d^2 = (b^2 – 4ac) = 81 – 80 = 1 –>(- d^2/4a) = -1/8 Answer. Vertex form: y = 2(x + 9/4) – 1/8. Note. We also can convert a quadratic equation in intercept form, with radical roots, to vertex form by using the general expression; y = a(x – x1)(x – x2) = a(x – b/2a + d/2a)(x – b/2a – d/2a), with d^2 = b^2 – 4ac, Exp. Convert y = (x – 2/3 + sqr.3/3)(x – 2/3 – sqr.3/3) to vertex form. Solution. a = 9; and (x1 + x2)/2 = -2/3 –> x-coordinate of vertex = -2/3 y-coordinate of vertex: f(-2/3) = 9(sqr.3/3)(-sqr.3/3) =- 9*(3/9) = -3. Answer: y = 9(x + 2/3)^2 – 3. 8. thanks for the articles .is there any techniques to cancel terms,like adding on both sides etc
# Math Insight ### Initial dynamical systems exploration Let's explore some basic ideas about a dynamical system, which is a system where things change over time according to some fixed rule. We will use Geogebra to visualize concepts such as the state space of a dynamical system and the rule for the evolution of the state variables. #### Evolution of the size of an anteater population ##### Introduction Let's use a dynamical system to model the evolution of the size of a population of anteaters. The first step in defining a dynamical system is to determine the state variable(s), i.e., variables whose values indicate the current state of the system. In this case, we'll assume we can base our model on just the total anteater population size (and not worry about other details such as the the locations or ages of the anteaters). This assumption means we can use a single state variable. Let's denote the state variable by $a$ and define $a$ to be the total number of anteaters in the population. It's clear only non-negative values of $a$ make sense, so the state space will be the set of non-negative numbers, which we can represent by a number line. The number $a$ of anteaters will change over time as anteaters are born and die. In other words, we have a dynamical system where the state changes. Let's use $t$ to denote time in years. Then we can use the notation $a(t)$ to indicate the value of the state variable $a$ after $t$ years from the time we start modeling the anteater population. We'll use a continuous-time dynamical system, meaning we will let time vary continuously so that $t$ can be any non-negative real number. With this convention, $a(10.1532)=132$ indicates that the anteater population contains 132 anteaters after 10.1532 years from when we started modeling the population. ##### Exploring the anteater model The second component of a dynamical system, after definition of the state variables, is the rule for the evolution of the state variables. Since a continuous-time dynamical system requires derivatives, and we don't expect knowledge of derivatives at this point, we won't give the equations for the evolution of our anteater model. Instead, we'll use the below Geogebra applet to explore the behavior of the anteater dynamical system. (If you are familiar with derivatives and would like to see the equation behind the model, you can check out the applet page.) In the applet's initial state, time is fixed at $t=0$. The left panel shows the state space depicted as a blue number line. This state space is labeled as population size $x(t)$, but we want to use $a$ for our state variable since we are dealing with a population of anteaters. In the box labeled state variable, change the $x$ to an $a$ (and press Enter). All references to the state variable should now be in terms of $a$. The large blue dot represents the value of the state variable at time $t=0$, i.e., $a(0)$. We call this initial value the initial condition of the dynamical system. We often write the initial condition as \begin{align} a(0) = a_0, \end{align} so we can use $a(0)$ and $a_0$ interchangeably. You can change the initial condition by dragging the blue point with your mouse or by typing a new value in the box labeled by $a_0$ (assuming you changed the state variable to $a$). If you left the time at $t=0$, then $a(t)$ is the same as $a_0$ and changes with it. The size of the population of anteaters is also indicated by the green diamonds in the right panel, where each diamond represents one anteater. Evolution of a bistable population. The evolution of the size $x(t)$ of a population is illustrated in three different ways. In the right panel, each green diamond represents one individual in the population, so the the population size $x(t)$ is given by the number of green diamonds. You can change time $t$ by dragging the red point on the red slider in the left panel or by clicking the triangle in the lower-left corner of one of the panels to start the animation. The state space, or space of possible values of $x(t)$, is represented by the blue line in the left panel. The initial condition $x_0$, or population size at time $t=0$, is shown by the large blue point in the state space. The population size $x(t)$ at time $t$ is shown by the small red point on the state space (visible only if $t > 0$). If you check the “plot versus time” box, then the right panel displays a green curve indicating how the population size changes as a function of time. The height of the blue dot on the curve indicates the initial population size $x_0$ at time $t=0$, and the height of the red dot on the curve indicates $x(t)$ for the time $t$ shown in the left panel. The evolution of the population size shows bistable behavior so that the population size either crashes to zero (i.e., the population dies out) or the population size approaches a larger value (i.e., the population survives). The fate of the population depends on the initial condition $x_0$ (which you can change by typing a value in the box or dragging one of the blue points). You can change the value of the larger population size (and other parameters affecting how the population evolves) by checking the “advanced options” and changing the values of the parameters. You can also change the name of the state variable $x$ to another variable by entering a different letter in the state variable box. Next, let the population of anteaters evolve. Press the play (triangle) button that is in the lower left corner of one of the two applet panels to start the animation. Then, time $t$ increases steadily up to $t=30$ years before reseting back to zero, as indicated by the red slider. The corresponding value of $a(t)$ is shown by the red point on the blue number line representing the state space, and the number of diamonds changes to match $a(t)$. What happens to the anteater population as time begins to increase from $t=0$. Does the population begin to increase or decrease, or does it stay at the same size? As more time passes and $t$ approaches 30 years, do the anteaters survive or does the population size drop toward zero? Does the population continue the trend (increasing or decreasing) it had in the beginning, or does it oscillate up and down? Does it grow faster and faster so that the population size explodes or does the population stabilize toward a steady value? One can determine how the population size evolves by analyzing the movie of $a(t)$ changing with time (represented by either the red point in the left panel or the green diamonds in the right panel). It is easier, though, to analyze a summary plot that captures all values $a(t)$ for the thirty years with $0 \le t \le 30$. To view a plot of population size $a$ versus time $t$, you can click the “plot versus time” checkbox. In the resulting plot (right panel), the state space for $a$ is shown flipped vertically. The blue vertical axis is the same as the blue number line in the left panel for the initial time $t=0$. Hence, you can also change the initial condition $a_0$ by moving the blue point in the right panel up and down. The values of the population size for later times is shown by the green curve. When the animation is playing, a red dot representing $a(t)$ moves steadily to the right along the green curve, in sync with the red point in the left panel. The moving thin blue line represents the state space at time $t$. With the plot versus time, you can quickly verify your analysis from the movie of $a(t)$. This plot will also help you analyze the dependence of the population behavior on initial condition $a_0$. How does the behavior of the population (and its ultimate fate) depend on the initial population size? As you vary the initial population size, how many different long term outcomes are possible? Can you find initial conditions where the population dies out, explodes to arbitrarily large values, approaches a steady value, or oscillates up and down? (Not all of these options are possible for this model.) For each of these possible outcomes, what is the range of initial conditions that leads to that outcome? #### The decay of lead in the bloodstream ##### Introduction As lead is a strong poison, small amounts of lead in the body can cause lead poisoning, leading to serious health problems. The body does eliminate lead, such as through urine, but the elimination is slow. Elimination from the bones can take decades, but we'll model how the lead level in the bloodstream decays, which is a faster process. We can use a simple dynamical system model where the state variable $p$ is the concentration of lead in the blood, measured in μg/dl (micrograms per deciliter). If no further sources of lead are introduced, then lead will be slowly eliminated from the body. Let's imagine that the amount of lead decreases by 11% each week. Although the lead level changes continuously over time, let's just model weekly snapshots of the lead concentration. (You could imagine that a doctor is testing a patient's lead concentrations each week, and we'll work with just these weekly numbers.) We'll use a discrete-time dynamical system, where the time interval is one week. Let $t$ measure time in weeks, and we'll allow $t$ to take only non-negative integer values. We'll start at $t=0$. The value $t=1$ will represent one week later, $t=2$ will represent two weeks later, etc. We'll use $p_t$ to denote the lead level at time $t$. We use the notation convention as a subscript for $t$, using $p_t$, because we have a discrete-time dynamical system and $t$ is always an integer. (If we were using a continuous time model, then we would have used $p(t)$ instead, as we did for the anteater population size $a(t)$.) We'll denote the lead level at time $t=0$ (the initial condition) by $p_0$, the lead level after one week by $p_1$, the lead level after two weeks by $p_2$, etc. Since the lead level decreases by 11% each week, we can write a simple formula for the dynamical rule: the lead level after a week is 89% of the previous lead level. Therefore, the rule for the first week is $p_1=0.89 p_0$, the rule for the second week is $p_2=0.89p_1$, the rule for the third week is $p_3=0.89p_2$, etc. We can summarize all these steps by saying that if the lead level in week $t$ is $p_t$, then the lead level the next week (in week $t+1$) is 0.89 times $p_t$, i.e., \begin{align} p_{t+1} = 0.89 p_t. \label{leaddecay} \end{align} ##### Obtaining the applet for the lead decay model To explore the lead decay, we'll use Geogebra again. This time, though, we'll interact with the Geogebra applet in a different way. For the anteater model, we just manipulated the applet embedded in the web page. For the lead model, you'll need more control over the applet, so you'll run Geogebra on your own computer and manipulate the Geogebra construction directly. When you open the file within Geogebra, it should look like the above applet. ##### Setting up the applet for the lead decay model The next step is to use the applet to calculate the evolution of our lead concentration dynamical system of equation \eqref{leaddecay} and create a plot of the results. The following video demonstrates how to use the integration of the Geogebra spreadsheet and Graphics window to create an applet for our model. You can either watch the video or read the rest of the text of this subsection, as they both explain the same procedure. Using applet to explore lead level decay. Let's focus on the spreadsheet, which has three columns. Column A is for the time points and column B is for the corresponding value of the state variable $p_t$. (The first column is labeled in cell B1 as $x_t$, but you can change it to $p_t$ if you like: double click cell B1 and change x_t to p_t.) Column C creates the points $(t,p_t)$, which are displayed in the left panel. (You can change the label in cell C1 from $(t,x_t)$ to $(t,p_t)$ if you like.) Geogebra automatically creates points in the graphics window if you type an ordered pair like (3,2) into a spreadsheet cell. You can experiment by typing an ordered pair into any cell of the spreadsheet. (Select the cell with your mouse and press Delete to get rid of the point.) We'll use this feature to create a plot of how the lead concentration decreases over time. Our goal is to use this applet to plot the solution to the dynamical system of equation \eqref{leaddecay}. We'll calculate the evolution using column B and use column C to plot the result. The initial condition $p_0$ is in cell B2. Let's imagine that we started with a high level of lead in the blood, say 64 μg/dl, so change the initial condition by typing 64 into cell B2. (The points may disappear in the Graphics window because they are beyond the scale of the vertical axis. You can change the scale of the vertical axis by holding shift and dragging up or down on the axis with your mouse.) To program the correct dynamical rule of equation \eqref{leaddecay}, we need to change the formula in cell B3. By default, it just adds one to the initial condition in B2; we'll make it multiply B2 by 0.89. Right-click cell B3, click Object Properties. In the Basic tab, change the Definition to be “0.89B2” and press Enter. With that change, the applet should calculate the value of $$p_1 = 0.89p_0 = 0.89(64) = 56.96$$ in cell B3 and plot the point $(1,56.96)$ in the graphics window. The formula in cell B3 now implements to our rule that the lead level should decay to 89% its previous level after one week. As a result, the lead level should be down to 56.96 μg/dl in week 1. To evolve the discrete dynamical system, one just needs to keep iterating the rule “multiply by 0.89” to march time forward week by week. We can use the spreadsheet to quickly do this calculation for us. Simply highlight the last row in the table (cells A3, B3, and C3), press Ctrl+C to copy them, click cell A4, and press Ctrl+V to paste. This copies the formulas into cells A4, B4, and C4, so that B4 contains the value \begin{align} p_2 = 0.89 p_1 = 0.89(56.96) \approx 50.69, \end{align*} and C4 contains the point $(2,p_2)=(2,50.69)$, which is also plotted in the Graphics window. You can continue this process by pasting the formula into A5. Or, you can paste into many rows at once. However, unlike other spreadsheets, you have to highlight more than just a range of cells in column. Instead, you need to highlight a rectangle of cells in columns A-C and as many rows as you like. Then, press Ctrl+V to paste the formulas into all those rows, and calculate the lead concentration $p_t$ for many time points $t$. The result should be a list of the values of $p_t$ in column $B$ and sequence of points of concentration versus time point in the Graphics window. Geogebra labels all the points with their spreadsheet cell: C2, C3, C4, etc. We don't want to show those labels, as they are irrelevant for the dynamical system model. To hide the labels, select the points in the Graphics window by dragging a rectangle around them with the mouse, right click a point, and click Show Label to turn off the labels. To make the plot understandable, we should label it to make the content clear. The horizontal axis represents $t$, which is in weeks. Label it “time t (weeks)” by clicking the Graphics Window, clicking the Insert Text Tool, clicking where you want the text, then entering the text in the dialog box that appears. The vertical axis represents the lead concentration $p_t$; label it “lead level p_t (μg/dl),” where you can get the μ from clicking the Symbols button and looking in the Basic section. This long label for the vertical axis should be rotated, but it's complicated to get rotated text, Greek symbols, and superscripts together in Geogebra. Instead, you can just press Enter between words to space the words out vertically. Geogebra makes it easy to change the symbols or colors used in the plot. Just right click an object, select Object Properties, and take a look around. ##### Exploring the lead decay model Since the applet is doing all the calculations, you are free to look at the big picture to understand the dynamics of the elimination of lead from the bloodstream. In particular, you can determine how fast the lead is eliminated as predicted from the model. About how long does it take for the lead concentration to drop down to half its original concentration? One quarter? One eighth? How does the time required to drop to these levels depend on the initial lead level $p_0$? For example, does it seem like it will take longer for the lead level to drop to half of $p_0$ if you increase $p_0$ above 64 μg/dl or if you decrease $p_0$ below 64 μg/dl?
SAT / ACT Prep Online Guides and Tips Variables, exponents, and more variables, whoo! ACT operations questions will involve all of these (and so much more!). So if you ever wondered what to do with or how to solve some of those extra long and clunky algebra problems (“What is the equivalent to \${2/3}a^2b - (18b - 6c) +\$ …” you get the picture), then this is the guide for you. This will be your complete guide to ACT operations questions--what they’ll look like on the test, how to perform operations with multiple variables and exponents, and what kinds of methods and strategies you’ll need to get them done as fast and as accurately as possible. You'll see these types of questions at least three times on any given ACT, so let's take a look. What Are Operations? There are four basic mathematical operations--adding, subtracting, multiplying, and dividing. The end goal for any particular algebra problem may be different, depending on the question, but the operations and the methods to solve them will be the same. For example, when solving a single variable equation or a system of equations (coming soon!), your ultimate objective is to solve for a missing variable. However, when solving an ACT operations problem, you must use your knowledge of mathematical operations to identify an equivalent expression (NOT solve for a missing variable). This means that the answer to these types of problems will always include a variable or multiple variables, since we are not actually finding the value of the variable. Let’s look at two examples, side-by-side. This is a single variable equation. Your objective is to find \$x\$. This is an ACT operations problem. You must find an equivalent expression after performing a mathematical operation on a polynomial. (We will go through exactly how to solve this problem shortly) Let's break down each component of an operations problem, step-by-step. (Also, bonus french braid lesson!) Operation Question How-To’s Let us look at how to identify operations questions when you see them and how to solve for your answer. How to Identify an Operations Problem As we said before, the end goal of an operations problem is not to solve for a missing variable. Because of this, you can identify an operations problem by looking at your answer choices. If the question involves variables (instead of integers) in the given equation AND in the answer choices, then it is likely you are dealing with an operations problem. This means that if the problem asks you to identify an “equivalent” expression or the “simplified form” of an expression, then it is highly likely that you are dealing with an operations problem. How to Solve an Operations Problem In order to solve these types of questions, you have two options: you can either solve your problems by using algebra, or by using the strategy of plugging in numbers. Let’s begin by looking at how algebraic operations work. First, you must understand how to add, multiply, subtract, and divide terms with variables and exponents. (Before we go through how to do this, be sure to brush up on your understanding of exponents and integers.) So let us look at the rules of how to manipulate terms with variables and exponents. When adding or subtracting terms with variables (and/or exponents), you can only add or subtract terms that have the exact same variable. This rule includes variables with exponents--only terms with variables raised to the same power may be added together (or subtracted). For example, \$x\$ and \$x^2\$ CANNOT be combined into one term (i.e. \$2x^2\$ or \$x^3\$). It can only be written as \$x + x^2\$. To add terms with variables and/or exponents, simply add the numbers before the variable (the coefficients) just as you would add any numbers without variables, and keep the variables intact. (Note: if there is no coefficient in front of the variable, it is worth 1. \$x\$ is the same thing as \$1x\$.) Again, if one term has an additional variable or is raised to a different power, the two terms cannot be added together. Yes \$x + 4x = 5x\$ \$10xy - 2xy = 8xy\$ No \$6x + 5y\$ \$xy - 2x - y\$ \$x + x^2 + x^3\$ These expressions all have terms with different variables (or variables to different powers) and so CANNOT be combined into one term. How they are written above is as simplified as they can ever get. Multiplication and Division When multiplying terms with variables, you may multiply any variable term with another. The variables do not have to match in order for you to multiply the terms--the variables instead are combined, or taken to an additional exponent if the variables are the same, after multiplying. (For more on multiplying numbers with exponents, check out the section on exponents in our guide to advanced integers) \$x * y = xy\$ \$ab * c = abc\$ \$z * z = z^2\$ The variables in front of the terms (the coefficients) are also multiplied with one another as usual. This new coefficient will then be attached to the combined variables. \$2x * 3y = 6xy\$ \$3ab * c = 3abc\$ Just as when we multiplying variable terms, we must take each component separately when we divide them. This means that the coefficients will be reduced/divided with regard to one another (just as with regular division), as will the variables. (Note: again, if your variables involve exponents, now might be a good time to brush up on your rules of dividing with exponents.) \${8xy}/{2x} = 4y\$ \${5a^2b^3}/{15a^2b^2} = b/3\$ \${30y + 45}/5 = 6y + 9\$ When working on operations problems, first take each component separately, before you put them together. Typical Operation Questions Though there are several ways an operations question may be presented to you on the ACT, the principles behind each problem are essentially the same--you must manipulate terms with variables by performing one (or more) of the four mathematical operations on them. Most of the operations problems you’ll see on the ACT will ask you to perform a mathematical operation (subtraction, addition, multiplication, or division) on a term or expression with variables and then ask you to identify the “equivalent” expression in the answer choices. More rarely, the question may ask you to manipulate an expression in order to present your equation “in terms of” another variable (e.g. “which of the following expressions shows the equation in terms of \$x\$?”). Now let’s look at the different kinds of operations problems in action. Here, we have our problem from earlier, but now we know how to go about solving it using algebra. We also have a second method for solving the question (for those of you are uninterested in or unwilling to use algebra), and that is to use the strategy of plugging in numbers. We’ll look at each method in turn. Solving Method 1--Algebra operations Knowing what we know about algebraic operations, we can multiply our terms. First, we must multiply our coefficients: \$2 * 3 = 6\$ This will be the coefficient in front of our new term, so we can eliminate answer choices F and J. Next, let us multiply our individual variables. \$x^4 * x^5\$ \$x^[4 + 5]\$ \$x^9\$ And, finally, our last variable. \$y * y^8\$ \$y^[1 + 8]\$ \$y^9\$ Now, combine each piece of our term to find our final answer: \$6{x^9}y^9\$ Our final answer is H, \$6{x^9}y^9\$ Solving Method 2--Plugging in our own numbers Alternatively, we can find our answer by plugging in our own numbers (remember--any time the question uses variables, we can plug in our own numbers). Let us say that \$x = 2\$ and \$y = 3\$ (Why those numbers? Why not! Any numbers will do--except for 1 or 0, which is explained in our PIN guide--but since we are working with exponents, smaller numbers will give us more manageable results.) So let us look at our first term and convert it into an integer using the numbers we selected to replace our variables. \$2{x^4}y\$ \$2(2^4)(3)\$ \$2(16)(3)\$ \$96\$ Now, let us do the same to our second term. \$3{x^5}{y^8}\$ \$3(2^5)(3^8)\$ \$3(32)(6,561)\$ \$629,856\$ And finally, we must multiply our terms together. \$(2{x^4}y)(3{x^5}{y^8})\$ \$(96)(629,856)\$ \$60,466,176\$ Now, we need to find the answer in our answer choices that matches our result. We must plug in our same values for \$x\$ and \$y\$ as we did here and then see which answer choice gives us the same result. If you are familiar with the process of using PIN, you know that our best option is usually to start with the middle answer choice. So let us test answer choice H to start. \$6{x^9}y^9\$ \$6(2^9)(3^9)\$ \$6(512)(19,683)\$ \$60,466,176\$ Success! We have found our correct answer on the first try! (Note: if our first option had not worked, we would have seen whether it was too low or too high and then picked our next answer choice to try, accordingly.) Our final answer is again H, \$6{x^9}y^9\$ Now let us look at our second type of problem. This question requires us to translate the problem first into an equation. Then, we must manipulate that equation until we have isolated a different variable than the original. Again, we have two methods with which to solve this question: algebra or PIN. Let us look at both. Solving Method 1--Algebra First, let us begin by translating our equation into an algebraic one. We are told that the product of \$c\$ and 3 is equal to \$b\$. A “product” means we must multiply \$c\$ and 3 and so our equation looks like this: \$3c = b\$ Now we are asked to find the sum of \$c\$ and 3. This means we must isolate \$c\$ so that we can add them together. So let us first isolate \$c\$ by using our knowledge of algebraic operations. \$3c = b\$ \$c = b/3\$ Now, we can sum \$c\$ and 3 by replacing our \$c\$ with \$b/3\$. \$c + 3\$ \${b/3} + 3\$ Solving Method 2--Plugging in numbers Alternatively, we can use our technique of plugging in numbers. Because our question deals with variables, we can choose our own numbers (so long as they follow the rules of our given information.) We are told that the product of \$c\$ and 3 is equal to \$b\$. So let us assign a value to \$c\$ and use this information to find the value of \$b\$. So let us say that \$c = 4\$. (Why 4? Why not!) If \$c = 4\$, then the product of \$c\$ and 3 is: \$3c = b\$ \$3(4) = b\$ \$b = 12\$ So, when \$c\$ equals 4, \$b\$ equals 12. Now we must find the sum of \$c\$ and 3. \$3 + c\$ \$3 + (4)\$ \$7\$ Now that we have found our sum, we must identify the answer choice that gives us this sum. All of our answer choices are presented to us in terms of \$b\$, so we will use our found value of 12 to replace \$b\$ for each. \$3(b + 3)\$ We can tell just by looking at it that this will be far larger than 7, but we can always test this out. \$3(12 + 3)\$ \$3(15)\$ \$45\$ We can eliminate answer choice C. Just by glancing, we can see that answer choices A and B will also be larger than 12, which means we can eliminate them as well. Let us try answer choice D. \${b + 3}/3\$ \${12 + 3}/3\$ \$15/3\$ \$5\$ Answer choice D did not match our sum, which means we can eliminate it as well. By process of elimination, we are left with answer choice E, but let us test it to be sure. \${b/3} + 3\$ \${12/3} + 3\$ \$4 + 3\$ \$7\$ Success! We have found the answer choice that matches the sum we found. Our final answer is, once again, E, \${b/3} + 3\$. As you can see, the answer to your operations questions will always be in variables and the problem will always require you to interpret and manipulate expressions with variables, but there are always multiple options for how to solve these types of problems. You've got the power to decide how you would like to solve and manipulate your operations problems. Magic! Strategies for Solving Operations Questions Now that we’ve seen the types of operations questions you’ll see on the ACT, let’s review our solving strategies. If you ever feel concerned that you may be going down the wrong path while manipulating your operations problems, or if you simply want to double-check your answer, it's never a bad idea to use the strategy of plugging in numbers Although it can take a little longer plug in your own numbers for your variables, you'll never have to fear misremembering how to manipulate your exponents, your variables, or your equations as a whole. Once you're able to use real numbers for your variables, the math will be a piece of cake. 2) Focus on one aspect of the term at a time It can become all too easy to lose yourself when working with multiple variables at once, especially when it comes to multiplication and division. The test-makers know this and will provide bait answers for any number of common mistakes. In order to keep all your components organized, focus on just one piece of each expression at a time. First, look at the coefficients, then look at the variables. This will help keep all your moving pieces in order and lessen the odds of mix-ups and mistakes. Operations problems can sometimes mess with your head, not because they are inherently difficult, but because the ACT is a marathon and your brain can get tired and confused (and lazy). This, combined with the fact that all the answer choices generally look quite similar, with only small differences--a minus sign instead of a plus sign, one coefficient difference, etc.--can lead you to select the wrong answer, even when you know what the correct one should be. To avoid this kind of careless error (the worst kind of error!), eliminate your answer choices as you go through your problem. Know that the coefficient for your \$y\$ value must be 3? Immediately cross out any answer choices that give you anything other than \$3y\$. It may seem inefficient to solve problems this way, but it will keep your answers much more clear. 4) Keep careful track of your negatives Not only can it be difficult to keep track of multiple variables, but it's even easier to mix-up the proper negative and positive signs. Many students make careless errors with their negative signs and the ACT test-makers are all too aware of this. They will provide all manner of bait answers for anyone who misplaces even a single negative sign, so be very careful. For a problem like this, we are being asked to subtract the entire expression, \$4a + 6b - 5c\$, from the entire expression, \$a + 2b + 3c\$. This means that the negative sign will be negating every term in the expression \$4a + 6b - 5c\$. So we must put a negative sign in front of each term. \$4a\$ becomes \$-4a\$ \$6b\$ becomes \$-6b\$ \$-5c\$ becomes \$- -5c\$ or \$+5c\$. Now let us put these pieces together with the first expression. \$a - 4a = -3a\$ \$2b - 6b = -4b\$ \$3c + 5c = 8c\$ Our final expression will be: \$-3a - 4b + 8c\$ Our final answer is E\$-3a - 4b + 8c\$. [Note: many (many!) students put a negative sign only in front of the first term in the parenthesis, which in this case the \$4a\$. If you had done this, you would have gotten: \$a - 4a = -3a\$ \$2b + 6b = 8b\$ \$3c - 5c = -2c\$. This would have given you answer choice C, \$-3a + 8b - 2c\$. Again the test-makers know this is a common error and there will always be a bait answer to tempt anyone who makes this kind of mistake.] Operations in the "real world." Hyuk, yuk, yuk. Now that we’ve gone through the tips and tricks of operations questions, it’s time to put your knowledge to the test with more real ACT math problems. 1) 2) 3) 4) 1) As always, we can solve this question using algebra or using PIN. Let us look at both ways. Method 1--Algebra First, we must distribute out our terms. Only afterwards will we subtract them. Let us take each half of our expression by itself. \$2(4x + 7)\$ \$8x + 14\$ \$ -3(2x - 4)\$ \$-6x + 12\$ (Note: keep careful track of your negatives here, especially in the second half of our expression.) Now, we can put the two together. \$8x + 14 - 6x + 12\$ \$2x + 26\$ We cannot go any further, as we have combined all our like terms. Our final answer is H, \$2x + 26\$ Method 2--PIN As an alternative to algebra, we can always use plugging in numbers. So let us assign our own value to \$x\$, which we will call 3. (Why 3? Why not!) This means that we will replace any \$x\$ in our given equation with a 3. \$2(4x + 7) - 3(2x - 4)\$ \$2(4(3) + 7) - 3(2(3) - 4)\$ 2(12 + 7) - 3(6 - 4)\$ \$2(19) - 3(2)\$ \$38 - 6\$ \$32\$ Now, let us find the answer choice that matches with our found answer of 32, once we replace the \$x\$ with 3. \$2x + 26\$ \$2(3) +26\$ \$6 + 26\$ \$32\$ Success! We found our answer on the first try. But remember--when using PIN, always check your other answer options to make sure there are not repeat correct answers. We can see straightaway that answer choices F and G will be too small, since answer choice H was a match. So let us try answer choice J. \$3x + 10\$ \$3(3) + 10\$ \$9 + 10\$ \$19\$ This answer choice is too small and we can see just by looking that answer choice K will be too small as well (since they only differ by 1). This means we are safe with our answer choice H, as no others produced a match. Our final answer is H, \$2x + 26\$. As we saw from earlier in the guide and from the example problem above, we can always use algebra or PIN for our operations problems. Knowing that, we will only go through one method each for the rest of our answer explanations. 2) For this problem, let us do our solve using algebra (again, we could also use PIN, but for the sake of brevity, we are only choosing one method for each problem). We are given the equation: \${1/2}y^2(6x + 2y + 12x - 2y)\$ Now, let us first make life easier by combining the like terms in the parenthesis. \$(6x + 2y + 12x - 2y)\$ \$(6x + 12x + 2y - 2y)\$ \$(18x)\$ The \$y\$ terms cancel one another out, so we are left with only \$18x\$ in the parenthesis. Now, we must multiply our \$18x\$ by \${1/2}y^2\$. As always, when multiplying, we must multiply first the coefficients and then combine them with the combined variables. So: \${1/2}y^2 * 18x\$ \$(1/2) * 18 = 9\$ \$y^2 * x = y^2x\$ Put the two together and we have: \$9y^2x\$ So our final answer is A, \$9xy^2\$ 3) Because we used algebra last time, let us try our hand at solving this question using PIN. Because we are using our own numbers, we don’t have to worry about whether or not we are matching up the right terms, or if we are combining them incorrectly; we can bypass all the mess and use numbers instead. We have one variable, \$t\$, so let us say that \$t = 2\$. (Why 2? As always, why not!) \$t^2 - 59t + 54 - 82t^2 + 60t\$ \$(2)^2 - 59(2) + 54 - 82(2)^2 + 60(t)\$ \$4 - 118 + 54 - 328 + 120\$ \$-268\$ Now, we must find the answer choice that matches our found answer of 102, once we replace \$t\$ with 2. Let us start in the middle, with answer choice H. \$-81t^4 + t^2 + 54\$ \$-81(2)^4 + (2)^2 + 54\$ \$-81(16) + 4 + 54\$ \$-1296 + 58\$ \$-1238\$ We can see just by looking that answer choice G will be too small as well (\$-26 * 16 = -416\$), and answer choice F will be too large (-26 * 4 = -104). So let us try answer choice J. \$-81t^2 + t + 54\$ \$-81(2)^2 + 2 + 54\$ \$-81(4) + 56\$ \$-324 + 56\$ \$-268\$ Success! And we can also see that the only difference between answer choices J and K are the coefficient in front of \$t^2\$ (-81 vs. -82), so we know that answer K would produce an incorrect and smaller number than answer choice J. Our final answer is J, \$-81t^2 + t + 54\$ 4) Because we used PIN last time, let us use algebra for this problem. Because we do not have like terms in the parenthesis, we must distribute out our expression using multiplication. \$-8x^3(7x^6 - 3x^5)\$ \$-8x^3(7x^6) - -8x^3(3x^5)\$ And take each piece separately. \$-8x^3(7x^6)\$ => \$-8 * 7 = -56\$ and \$x^3 * x^6 = x^[3 + 6] = x^9\$ (for more on this, look to the section on exponents in our advanced integers guide). So, combined, we have: \$-56x^9\$ And the other half of our expression will be the same. \$- -8x^3(3x^5)\$ \$8x^3(3x^5)\$ => \$8 * 3 = 24\$ and \$x^3 * x^5 = x^[3 + 5] = x^8\$ So, combined, we have: \$24x^8\$ Now our equation looks like this: \$-56x^9 + 24x^8\$ Our final answer is A, \$-56x^9 + 24x^8\$ (Take care! The only difference between answer choice A and B is the negative sign. If you weren’t careful with your double negatives, you may have fallen for this bait answer.) Ten thousand gold stars for solving your operations problems! The Take Aways Though operations problems are easy to get wrong if you’re going too quickly through the test (or trying to solve them in your head), the basic elements are the same as any problem with variables--combine like terms, keep your work organized, and use PIN if you feel overwhelmed (or simply want to double-check your answer). You have a multitude of options for solving ACT algebra questions, so don’t be afraid to use them. What’s Next? Still in the mood for math? Well we've got you covered! First, take a gander at exactly what's tested on the ACT math section in order to get a feel for your strong and weak points. Next, dive right into our ACT math guides for any topic you feel you haven't quite mastered (or just any topic you want to refresh). From circles to ratios, slopes to polygons, we've got your back. Running out of time on the ACT math section? Check out our guide on how to help maximize your avaialable time in order to get your best score possible. Nervous about test day? Ease your mind by taking a look at what to do the night before and the day of the test. Trying for a perfect score? Look no further than our guide to getting a perfect 36 on the ACT math, written by a perfect-scorer. Want to improve your ACT score by 4 points? Check out our best-in-class online ACT prep program. We guarantee your money back if you don't improve your ACT score by 4 points or more. Our program is entirely online, and it customizes what you study to your strengths and weaknesses. If you liked this Math lesson, you'll love our program. Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step program to follow so you'll never be confused about what to study next. Check out our 5-day free trial: Courtney Montgomery Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time. You should definitely follow us on social media. You'll get updates on our latest articles right on your feed. Follow us on all 3 of our social networks:
## Equation of a circle Consider a fixed complex number $z_{0}$ and let $z$ be any complex number which moves in such a way that its distance from $z_{0}$ is always equal to r. This implies $z$ would lie on a circle whose centre is $z_{0}$ and radius r. And, its equation would be $\|z-z_{0}\|=r$ or $\|z-z_{0}\|^{2}=r^{2}$ or $(z-z_{0})(\overline{z}-\overline{z_{0}})=r^{2}$, or $z\overline{z}-z \overline{z_{0}}-\overline{z}z_{0}-r^{2}=0$ Let $-a=z_{0}$ and $z_{0}\overline{z_{0}}-r^{2}=b$. Then, $z\overline{z}+a\overline{z}+\overline{a}z+b=0$ It represents the general equation of a circle in the complex plane. Now, let us consider a circle described on a line segment AB $(A(z_{1}), B(z_{2}))$ as its diameter. Let $P(z)$ as its diameter. Let $P(z)$ be any point on the circle. As the angle in the semicircle is $\pi/2$, so $\angle {APB}=\pi/2$ $\Longrightarrow (\frac{z_{1}-z}{z_{2}-z})=\pm \pi/2$ $\Longrightarrow \frac{z-z_{1}}{z-z_{2}}$ is purely imaginary. $\frac{z-z_{1}}{z-z_{2}}+\frac{\overline{z}-\overline{z_{1}}}{\overline{z}-\overline{z_{2}}}=0$ $\Longrightarrow (z-z_{1})(\overline{z}-\overline{z_{2}})+(z-z_{2})(\overline{z}-\overline{z_{1}})=0$ Condition for four points to be concyclic: Let ABCD be a cyclic quadrilateral such that $A(z_{1})$, $B(z_{2})$, $C(z_{3})$ and $D(z_{4})$ lie on a circle. (Remember the following basic property of concyclic quadrilaterals: opposite angles are supplementary). The above property means the following: $\arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})+\arg (\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$ $\Longrightarrow \arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$ $(\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})$ is purely real. Thus, points $A(z_{1})$, $B(z_{2})$, $C(z_{3})$, $D(z_{4})$ (taken in order) would be concyclic if the above condition is satisfied. More later, Nalin Pithwa This site uses Akismet to reduce spam. Learn how your comment data is processed.
## Binary Search Binary Search method works only for sorted list. This method of searching is more efficient as compared to linier search method. Because we does not have to compare the search element with all the elements in the list until the element is found. The binary search method of searching an element works as follow: Suppose we want to search an element X from the list of N element. Determine the Lower and Upper limit of the list by assigning Lower index of the list to LOW and Upper Index of the list to HIGH. Now calculate the MIDDLE position using following equation: MIDDLE = (LOW + HIGH)/2 After finding the MIDDLE index from the list we compare the value of the MIDDLE element with X. If the value of the middle element is greater then X then it will exist in the lower interval of the list. So we change the value of HIGH index by MIDDLE – 1. If the value of the middle element is smaller then X then it will exist in the upper interval of the list. So we change the value of LOW index by MIDDLE + 1 This process is repeated until entire list is searched or the element is found. Consider following example: LOW MIDDLE HIGH 0 1 2 3 4 5 6 7 8 9 23 45 67 89 123 189 210 235 345 545 Suppose we want to find the element X=189 in the above list. First, LOW = 0, HIGH = 9 MIDDLE = (LOW + HIGH)/2 = (0+9)/2 = 4.5 = 4 Here the value of MIDDLE element is 123 which is less then 189 so the element is in the upper interval of the list. Now change the value of LOW = MIDDLE + 1 = 4 +1 =5. LOW MIDDLE HIGH 5 6 7 8 9 189 210 235 345 545 Now, LOW = 5, HIGH = 9 Middle=(LOW+HIGH)/2 = (5+9)/2 = 14/2 = 7 Here the value of MIDDLE element is 235 which is greater then 189. So the element is in the lower interval of the list. Now change the value of HIGH = MIDDLE - 1 = 7 - 1 =6. Now, LOW = 5, HIGH = 6 Middle=(LOW+HIGH)/2 = (5+6)/2 = 11/2 = 5.5 = 5 Here the value of MIDDLE element is 189 which is equal to 189. So the element is found in the list.
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # What is the Prime Factorization Of 139? • Prime factors of 139: 1 * 139 ## Is 139 A Prime Number? • Yes the number 139 is a prime number. • It's a prime because one hundred and thirty-nine has no positive divisors other than 1 and itself. ## How To Calculate Prime Number Factorization • How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself. • It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself. • Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors. ## What is a prime number? Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. ## What is Prime Number Factorization? • In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
Enable contrast version # Tutor profile: Vaidehi T. Inactive Vaidehi T. Mathematics Expert at an Engineering college Tutor Satisfaction Guarantee ## Questions ### Subject:Trigonometry TutorMe Question: If \$\$cos{x} = 4\$\$, then find the value of \$\$ cos{2x}.\$\$ Inactive Vaidehi T. As we know that, \$\$ cos{2x} = 2cos^2{x} - 1.\$\$ Plug the value of \$\$cos{x}\$\$, \$\$\therefore cos{2x} = 2(4)^2 - 1\$\$ \$\$= 32 - 1\$\$ \$\$= 31.\$\$ ### Subject:Algebra TutorMe Question: Find the equation of the line that passes through two points \$\$(2, 1),\, (3,4)\$\$. Inactive Vaidehi T. When we are given two points and we have to find the equation of the line passing through those points, we can use two point formula. Let's say \$\$A(x_1, y_1)\$\$ and \$\$B(x_2, y_2)\$\$ are two points then the formula is defined as \$\$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1} (x - x_1) \$\$. For the given data, we can take \$\$x_1 = 2,\, x_2 = 3,\, y_1 = 1,\, y_2 = 4.\$\$ Plugging all these values in the formula to find the equation of a line: \$\$y - 1 = \dfrac{4 - 1}{2 - 1} (x - 2) \$\$ \$\$\therefore y - 1 = 3(x-2)\$\$ \$\$\therefore y - 1 = 3x - 6\$\$ \$\$ \therefore y = 3x - 5.\$\$ ### Subject:Differential Equations TutorMe Question: Solve the following first order differential equation: \$\$\dfrac{dy}{dx} = x^2 y^2.\$\$ Inactive Vaidehi T. In order to solve this first order differential equation, we can separate variables like this \$\$ \dfrac{dy}{y^2} = x^2\, dx.\$\$ Now, integrating both the sides, we get \$\$\dfrac{y^{-2+1}}{-2+1} = \dfrac{x^{2+1}}{2+1} + C\$\$ \$\$ \therefore \dfrac{-1}{y} = \dfrac{x^3}{3} + C\$\$ \$\$ \therefore \dfrac{x^3}{3} + \dfrac{1}{y} + C = 0.\$\$ ## Contact tutor Send a message explaining your needs and Vaidehi will reply soon. Contact Vaidehi ## Request lesson Ready now? Request a lesson. Start Lesson ## FAQs What is a lesson? A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard. How do I begin a lesson? If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson. Who are TutorMe tutors? Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you. BEST IN CLASS SINCE 2015 TutorMe homepage Made in California by Zovio © 2013 - 2021 TutorMe, LLC High Contrast Mode On Off
# What are the rules for percentages? ## What are the rules for percentages? If you have to turn a percentage into a decimal, just divide by 100. For example, 25% = 25/100 = 0.25. To change a decimal into a percentage, multiply by 100. So 0.3 = 0.3 × 100 =30% . ## How do I calculate percentage of a total? The following formula is one of the most common strategies to determine the percentage of something:Determine the whole or total amount of what you want to find a percentage for. Divide the number that you wish to determine the percentage for. Multiply the value from step two by 100. ## How do I figure out the percentage of a number? 1. How to calculate percentage of a number. Use the percentage formula: P% * X = YConvert the problem to an equation using the percentage formula: P% * X = Y.P is 10%, X is 150, so the equation is 10% * 150 = Y.Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10. ## How do you teach percentages of a number? If you set up a proportion to find the percent of a number, you’ll always end up dividing by 100 in the end. Another way to get the same answer is to divide by 100 first and then multiply the numbers together. 24/100 = . 24 and (. ## How do you calculate percentages quickly? Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. ## How do you do percentages mentally? For example:To find 5%, find 10% and divide it in two.To find 15%, find 10%, then add 5%.To find 20%, find 10% and double it.To find 25%, find 50% and then halve it.To find 60%, find 50% and add 10%.To find 75%, find 50% and add 25%.and so on. ## How do you find percentages without a calculator? If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72. ## How do you solve percent problems tricks? Learn the important concepts and tricks to solve questions based on percentages.Percentages: Concepts & Tricks.For example: 13 = 1300/100 = 1300%For example: 16(2/3) % of 300 = (50/3) × (300/100) = 50.A. Fraction to Percent: Multiply the fraction by 100 to convert it into a percent. ## What is effective percentage? The effective annual interest rate is the real return on a savings account or any interest-paying investment when the effects of compounding over time are taken into account. It is also called the effective interest rate, the effective rate, or the annual equivalent rate. ## How do you write a short percentage? It is often denoted using the percent sign, “%”, although the abbreviations “pct.”, “pct” and sometimes “pc” are also used. A percentage is a dimensionless number (pure number); it has no unit of measurement. ## How do I get a percentage shortcut? Percentages: Formulas, Tricks and ShortcutsPercent implies “for every hundred”. To calculate p % of y. To find what percentage of x is y: y/x × 100.To calculate percentage change in value. Percentage point change = Difference of two percentage figures.Increase N by S % = N( 1+ S/100 )Decrease N by S % = N (1 – S/100) ## How do you find the percentage of a question? Solve simple percent problemsFinding 100% of a number: Remember that 100% means the whole thing, so 100% of any number is simply the number itself: Finding 50% of a number: Remember that 50% means half, so to find 50% of a number, just divide it by 2: ## How can I calculate average? The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count. ## What is the formula to calculate average percentage? To find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage. So, 0.27 multiplied by 100 equals 27 or 27%.
Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math ``` Date: 05/06/99 at 07:57:30 From: Jayant Subject: Permutations and Combinations Ten sticks of lengths 1,2,3,....,10 are given. Four are selected to so formed. Also find in how many of these quadrilaterals a circle can be inscribed. ``` ``` Date: 05/07/99 at 11:19:20 From: Doctor Anthony Subject: Re: Permutations and Combinations The condition for a quadrilateral ABCD to have an inscribed circle is that AB + CD = AD + BC You can prove this easily by considering that the tangent to the circle from point A along AB is equal to the length of the tangent along AD, similarly the tangents from B are equal along BA and BC, and so on with tangents from C along CB and CD and tangents from D along DC and DA. Adding up these equal lengths leads to the result quoted above. So the four sticks must satisfy the condition that the sum of one pair is equal to the sum of the other pair So 1+10 = 2+9 = 3+8 = 4+7 = 5+6 This gives: C(5,2) = 10 quadrilaterals Similarly 2+10 = 3+9 = 4+8 = 5+7 giving C(4,2) = 6 quadrilaterals " 3+10 = 4+9 = 5+8 = 6+7 " C(4,2) = 6 " " 4+10 = 5+9 = 6+8 " C(3,2) = 3 " " 5+10 = 6+9 = 7+8 " C(3,2) = 3 " " 6+10 = 7+9 " C(2,2) = 1 " " 7+10 = 8+9 " C(2,2) = 1 " " 1+9 = 2+8 = 3+7 = 4+6 " C(4,2) = 6 " " 1+8 = 2+7 = 3+6 = 4+5 " C(4,2) = 6 " " 1+7 = 2+6 = 3+5 " C(3,2) = 3 " " 1+6 = 2+5 = 3+4 " C(3,2) = 3 " " 1+5 = 2+4 " C(2,2) = 1 " " 1+4 = 2+3 " C(2,2) = 1 " -------------------------- Since the order must stay as opposite pairs, we could keep one pair fixed and swap over the other pair giving double this answer, but if we can view the quadrilateral from either side, we would divide by 2 again, so we stick at 50 quadrilaterals which can have an inscribed circle. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ ``` Associated Topics: High School Conic Sections/Circles High School Geometry High School Permutations and Combinations High School Triangles and Other Polygons Search the Dr. Math Library: Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words Submit your own question to Dr. Math Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search
# How do you find the "m" and "b" of any linear equation? Mar 19, 2018 $m$ is the slope, while $b$ is the y-intercept. #### Explanation: Any linear equation has the form of $y = m x + b$ • $m$ is the slope of the equation • $b$ is the y-intercept The slope of the line, $m$, is found by $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ where $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are the coordinates of any two points in the line. The y-intercept, $b$, is found by plugging in $x = 0$ into the equation, which results in $y = b$, and therefore is the y-intercept. In some cases, if the equation is already arranged for you nicely, like $y = 3 x + 5$, we can easily find the y-intercept for this line, which is $5$. Other times, the equation might not be arranged nicely, with cases such as $\frac{1}{2} x + 3 y = 5$, in which we solve for the y-intercept: $\frac{1}{2} x + 3 y = 4$ $3 y = 4 - \frac{1}{2} x$ $y = \frac{- \frac{1}{2} x + 4}{3}$ $y = - \frac{1}{6} x + \frac{4}{3}$ So, the y-intercept of this line is $\frac{4}{3}$.
## Sunday, August 24, 2014 ### What Makes the Golden Ratio so Golden? Now that I’ve concluded my Golden Ratio-themed comic serial, I wanted to get into a little what the Golden Ratio actually is. Just saying that it’s some number that’s approximately 1.618, doesn’t quite do it justice. What’s so special about that number? And what makes that a ratio? Second point, first. A ratio is a comparison of two numbers, so it can cause confusion if only a single number is written, even if “to 1” is implied. What makes it so special is what is being compared. Look at these two images, for example. They are basically the same thing, only the notation for the lengths are different. Suppose we wanted to find a ratio of the length of the square to the length of the rectangle, and we wanted that ratio to equal the ratio of the length of the smaller rectangle to that of the bigger one, what value of x would accomplish that? It terms out that we could set up the problem in either of these two ways (and many more, besides!). On the left, the square has a length of 1, the smaller rectangle has a width of x, so the bigger rectangle has a total length of x + 1. One the right, the square still has a length of 1, but the bigger rectangle has a total length of x, so the smaller rectangle has a length of x – 1. Setting up the proportions and cross-multiplying we find: In either case, we get the same quadratic equation: x2 – x – 1 = 0. That’s not something easily factorable, so we have to use Ye Olde Quadratic Formula, after which we discover that the roots are If we take the positive value, we get x = 1.618033988749894848204586... But what about the negative value? If we subtract root 5 from 1 and divide by 2, we get x = -0.618033988749894848204586... Look at the decimal portion. They are the same! Keep in mind, that we’re dealing with irrational numbers here, which aren’t supposed to conform to patterns, but this one is just brilliant. And, yes, there is a reason for it. Take a look at that second proportion, above, on the right, with the removable 1 in the denominator. This is saying that one less than the number is the reciprocal of that number! If you divide 1 by 1.618033988749894848204586..., you will get 0.618033988749894848204586..., the same as if you just subtracted one. This works for the negative value as well: If you divide 1 by -0.618033988749894848204586..., you will get -1.618033988749894848204586..., the same as if you just subtracted one. One last point, and a hat tip to William Ricker for mentioning it, how else can we write the reciprocal of x, 1/x? What exponent gives us the reciprocal of x? An exponent of -1. That means that this equation, this proportion, can be written as: Absolutely brilliant. And quite Golden, if I say so myself.
# what is the meaning of associative property Nội dung bài viết Associative Property In arithmetic, the associative property is a property of some main arithmetic operations, which supplies the identical end result even after rearranging the parentheses of any expression. Allow us to study the associative property with a couple of solved examples. You're reading: what is the meaning of associative property 1. What Is Associative Property? 2. Associative Property of Addition 3. Associative Property of Multiplication 4. Verification of Associative Property 5. FAQs on Associative Property In any given expression containing two or extra numbers together with an associative operator, the order of operations doesn’t change the ultimate end result so long as we maintain the sequence of the operands the identical. That is legitimate even after altering the place of the parentheses current within the expression. In different phrases, we will add/multiply the numbers in an equation no matter the grouping of these numbers. ### Associative Property Definition Two main arithmetic operations + and × on any given set M known as associative if it satisfies the given associative regulation that’s (p ∗ q) ∗ r = p ∗ (q ∗ r) for all p, q, r in M. Right here, ∗ may be both changed by an addition image or multiplication image. This property known as the associative property. So, the associative property exists in solely addition and multiplication operations. The associative property of addition and multiplication is given as: Allow us to talk about intimately the associative property of addition and multiplication with examples. You might want to know: what is a serial port used for Suppose we have now three numbers: a, b, and c. We are going to present the associative property of addition as: Associative Property System for Addition: The sum of three or extra numbers stays the identical no matter the best way numbers are grouped. (A + B) + C = A + (B + C) Instance: (1 + 7) + 3 = 1 + (7 + 3) = 11. We are saying that addition is associative for the given set of three numbers. Suppose we have now three numbers: a, b, and c. We are going to present the associative property of multiplication as: Associative Property System for Multiplication: The product of three or extra numbers stays the identical no matter the best way numbers are grouped. (A × B) × C = A × (B × C) Allow us to think about an associative property of multiplication instance For instance, (1 × 7) × 3 = 1 × (7 × 3) = 21. Right here we discover that multiplication is associative for the given set of three numbers. Allow us to attempt to justify how and why the associative property is just legitimate for addition and multiplication operations. We are going to apply the associative regulation individually on the 4 fundamental operations. For Addition: The final associative property regulation for addition is expressed as (A + B) + C = A + (B + C). Allow us to attempt to repair some numbers within the components to confirm the identical. For instance, (1 + 4) + 2 = 1 + (4 + 2) = 7. We are saying that addition is associative for the given set of numbers. For Subtraction: The final associative property components is expressed as (A – B) – C ≠ A – (B – C). Allow us to attempt to repair some numbers within the components to confirm the identical. For instance, (1 – 4) – 2 ≠ 1 – (4 – 2) i.e., -5 ≠ -1. We are saying that subtraction shouldn’t be associative for the given set of numbers. For Multiplication: For any set of three numbers (A, B, and C) associative property for multiplication is given as (A × B) × C = A × (B × C). For instance, (1 × 4) × 2 = 1 × (4 × 2) = 8. Right here we discover that multiplication is associative for the given set of numbers. For Division: For any three numbers (A, B, and C) associative property for division is given as A, B, and C, (A ÷ B) ÷ C ≠ A ÷ (B ÷ C). For instance, (9 ÷ 3) ÷ 2 ≠ 9 ÷ (3 ÷ 2) = 3/2 ≠ 6. You can find that expressions on each side will not be equal. So division shouldn’t be associative for the given three numbers. ☛ Associated Articles Take a look at these attention-grabbing articles associated to associative property regulation for in-depth understanding. • Associative Property of Addition Worksheets • Distributive Property of Multiplication • Commutative Property Allow us to check out a couple of examples to higher perceive the associative property. You might want to know: what is form 2848 used for
T4.1-2 # T4.1-2 - <?xml version="1.0"... This preview shows pages 1–5. Sign up to view the full content. (1) = n i 0 i = n(n+1)/2, n 0 for all n N, = n i 0 i = n(n+1)/2 Let P(n): = n i 0 i = n(n+1)/2 We may restate (1) as 2200 nP(n), Universe= N How can we prove such statement? RULE OF INFERENCE . P(0) Premise 2200 n(P(n) -> P(n+1)) Premise ————————— 2200 nP(n) Conclusion If we can prove the two premises (somehow), then we are allowed to say we have proved the conclusion. This rule of inference has the name mathematical induction . 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Intuitively, if we prove P for the value 0, then the truth of the second premise says that P is true for the value 1. But if it is true for 1, the second premise says that it is true for 2, and so ad infinitum. In practice P(0) is proved directly. This step is the basis. The second premise is proved by using universal instantiation, and then using the proof method for implications, which is to assume the antecedent and prove the consequent. When we combine these we get a statement like this: Assume P(n) is true for an arbitrary n. Based on this assumption , show that P(n+1) is true. 2 Example: P(n): = n i 0 i = n(n+1)/2 P(0) is true. (Basis) because = 0 0 i i=0(0+1)/2=0 Assume the truth of P(n). Based on the assumption, prove P(n+1): + = 1 0 n i i = (n+1)(n+2)/2. Proof: + = 1 0 n i i = = n i 0 i + (n+1) = n(n+1)/2 + (n+1) = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2 QED 3 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document To prove 2200 nP(n): 1. Identify the propositional function P(n)! 2. Prove P(0) 3. Assume P(n) for arbitrary n 4. Based on that assumption, prove P(n+1) E.g. Prove n < 2 n (Same as 2200 n(n<2 n )) 1. P(n): n < 2 n 2. P(0): 0<2 0 =1 is True 3. Assume n < 2 n 4. Prove: n+1 < 2 n+1 Proof: n < 2 n Induction hypothesis 1 2 n Fact about natural nos. n+1 < 2 This is the end of the preview. Sign up to access the rest of the document. ## T4.1-2 - <?xml version="1.0"... This preview shows document pages 1 - 5. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# Intro to the imaginary numbers CCSS Math: HSN.CN.A.1 Learn about the imaginary unit i, about the imaginary numbers, and about square roots of negative numbers. In your study of mathematics, you may have noticed that some quadratic equations do not have any real number solutions. For example, try as you may, you will never be able to find a real number solution to the equation $x^2=-1$. This is because it is impossible to square a real number and get a negative number! However, a solution to the equation $x^2=-1$ does exist in a new number system called the complex number system. ## The imaginary unit The backbone of this new number system is the imaginary unit, or the number $i$. The following is true of the number $i$: • $i=\sqrt{-1}$ • $i^2=-1$ The second property shows us that the number $i$ is indeed a solution to the equation $x^2=-1$. The previously unsolvable equation is now solvable with the addition of the imaginary unit! ## Pure imaginary numbers The number $i$ is by no means alone! By taking multiples of this imaginary unit, we can create infinitely many more pure imaginary numbers. For example, $3i$, $i\sqrt{5}$, and $-12i$ are all examples of pure imaginary numbers, or numbers of the form $bi$, where $b$ is a nonzero real number. Taking the squares of these numbers sheds some light on how they relate to the real numbers. Let's investigate this by squaring the number $3i$. The properties of integer exponents remain the same, so we can square $3i$ just as we'd imagine. \begin{aligned}(3i)^2&=3^2i^2\\ \\ &=9{i^2}\\\\ \end{aligned} Using the fact that $i^2=-1$, we can simplify this further as shown. \begin{aligned}\phantom{(3i)^2} &=9\goldD{i^2}\\\\ &=9(\goldD{-1})\\\\ &=-9 \end{aligned} The fact that $(3i)^2=-9$ means that $3i$ is a square root of $-9$. What is $(4i)^2$? \begin{aligned}(4i)^2&=4^2i^2\\ \\ &=16\goldD{i^2}\\\\ &=16(\goldD{-1})&\small{\gray{\text{(Because {i^2=-1})}}}\\\\ &=-16 \end{aligned} Which of the following is a square root of $-16$? The fact that $(4i)^2=-16$ means that $4i$ is a square root of $-16$. The number $-4$ is a square root of positive $16$ since $(-4)^2=16$. In this way, we can see that pure imaginary numbers are the square roots of negative numbers! ## Simplifying pure imaginary numbers The table below shows examples of pure imaginary numbers in both unsimplified and in simplified form. Unsimplified formSimplified form $\sqrt{-9}$$3i$ $\sqrt{-5}$$i\sqrt{5}$ $-\sqrt{-144}$$-12i$ But just how do we simplify these pure imaginary numbers? Let's take a closer look at the first example and see if we can think through the simplification. Original equivalenceThought process \begin{aligned}\sqrt{-9} = 3i \end{aligned}The square root of $-9$ is an imaginary number. The square root of $9$ is $3$, so the square root of negative $9$ is $\textit 3$ imaginary units, or $3i$. The following property explains the above "thought process" in mathematical terms. For $a>0$, $\Large\sqrt{-a}=i\sqrt{a}$ If we put this together with what we already know about simplifying radicals, we can simplify all pure imaginary numbers. Let's look at an example. ### Example Simplify $\sqrt{-18}$. ### Solution First, let's notice that $\sqrt{-18}$ is an imaginary number, since it is the square root of a negative number. So, we can start by rewriting $\sqrt{-18}$ as $i\sqrt{18}$. Next we can simplify $\sqrt{18}$ using what we already know about simplifying radicals. The work is shown below. \begin{aligned}\sqrt{-18}&=i\sqrt{18}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\cdot\sqrt{9\cdot 2}&&\small{\gray{\text{9 is a perfect square factor of 18}}}\\\\ &=i\sqrt{9}\cdot\sqrt{2}&&\small{\gray{\sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \text{ when } a, b\geq0}} \\\\ &=i\cdot 3\cdot \sqrt2&&\small{\gray{\sqrt{9}=3}}\\\\ &=3i\sqrt{2}&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned} So it follows that $\sqrt{-18}=3i\sqrt{2}$. Sure. Watch this video to see Sal simplify $\sqrt{-52}$. ## Let's practice some problems ### Problem 1 Simplify $\sqrt{-25}$. \begin{aligned}\sqrt{-25}&=i\sqrt{25}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\cdot 5&&\small{\gray{\sqrt{25}=5}}\\\\ &=5i&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned} So it follows that $\sqrt{-25}=5i$. ### Problem 2 Simplify $\sqrt{-10}$. \begin{aligned}\sqrt{-10}&=i\sqrt{10}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ \end{aligned} Since there are no perfect square factors of $10$, this is as far as the simplification goes. So it follows that $\sqrt{-10}=i\sqrt{10}$ or $\sqrt{10}i$. ### Problem 3 Simplify $\sqrt{-24}$. \begin{aligned}\sqrt{-24}&=i\sqrt{24}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\sqrt{4\cdot 6}&&\small{\gray{\text{4 is a perfect square factor of 24}}}\\\\ &=i\sqrt{4}\cdot\sqrt{6}&&\small{\gray{\sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \text{ when } a, b\geq0}} \\\\ &=i\cdot 2\cdot \sqrt6&&\small{\gray{\sqrt{4}=2}}\\\\ &=2i\sqrt{6}&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned} So it follows that $\sqrt{-24}=2i\sqrt{6}$ or $2\sqrt{6}i$. ## Why do we have imaginary numbers anyway? The answer is simple. The imaginary unit $i$ allows us to find solutions to many equations that do not have real number solutions. This may seem weird, but it is actually very common for equations to be unsolvable in one number system but solvable in another, more general number system. Here are some examples with which you might be more familiar. • With only the counting numbers, we can't solve $x+8=1$; we need the integers for this! • With only the integers, we can't solve $3x-1=0$; we need the rational numbers for this! • With only the rational numbers, we can't solve $x^2=2$. Enter the irrational numbers and the real number system! And so, with only the real numbers, we can't solve $x^2=-1$. We need the imaginary numbers for this! As you continue to study mathematics, you will begin to see the importance of these numbers.
# How to Calculate Displacement in a Physics Problem Displacement is the distance between an object’s initial position and its final position and is usually measured or defined along a straight line. Since this is a calculation that measures distance, the standard unit is the meter (m). ## How to find displacement In physics, you find displacement by calculating the distance between an object’s initial position and its final position. In physics terms, you often see displacement referred to as the variable s. The official displacement formula is as follows: s = sf – si • s = displacement • si = initial position • sf = final position ## Calculating displacement example Say, for example, that you have a fine new golf ball that’s prone to rolling around. This particular golf ball likes to roll around on top of a large measuring stick and you want to know how to calculate displacement when the ball moves. You place the golf ball at the 0 position on the measuring stick, as shown in the below figure, diagram A. The golf ball rolls over to a new point, 3 meters to the right, as you see in the figure, diagram B. The golf ball has moved, so displacement has taken place. In this case, the displacement is just 3 meters to the right. Its initial position was 0 meters, and its final position is at +3 meters. The displacement is 3 meters. Scientists, being who they are, like to go into even more detail. You often see the term si, which describes initial position, (the i stands for initial). And you may see the term sf used to describe final position. In these terms, moving from diagram A to diagram B in the figure, si is at the 0-meter mark and sf is at +3 meters. The displacement, s, equals the final position minus the initial position: Displacements don’t have to be positive; they can be zero or negative as well. If the positive direction is to the right, then a negative displacement means that the object has moved to the left. In diagram C, the restless golf ball has moved to a new location, which is measured as –4 meters on the measuring stick. The displacement is given by the difference between the initial and final position. If you want to know the displacement of the ball from its position in diagram B, take the initial position of the ball to be si = 3 meters; then the displacement is given by When working on physics problems, you can choose to place the origin of your position-measuring system wherever is convenient. The measurement of the position of an object depends on where you choose to place your origin; however, displacement from an initial position si to a final position sf does not depend on the position of the origin because the displacement depends only on the difference between the positions, not the positions themselves.
# Class 8 Maths MCQ – Applications of Linear Equation (Create and Solve the Equations) This set of Class 8 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Linear Equation (Create and Solve the Equations)”. 1. The digits of a two-digit number differ by 4. If the digits are interchanged, and the resulting number is added to the original number, we get 152. What can be the original number? a) 95 b) 40 c) 73 d) 59 Explanation: Let us take the two digit number such that the digit in the units place is x. The digit in tens place differs by 4 ∴ the digit in tens place is (x + 4). ∴the two digit number obtained = [10 * (x + 4)] + [x] ∴the two digit number obtained = 10x + 40 + x ∴the two digit number obtained = 11x + 40 When the digits are interchanged we obtain the number = [10 * x] + [(x + 4)] The new number obtained = 11x + 4 As we know the original number and the new number add up to 152. ∴ [11x + 40] + [11x + 4] = 152 ∴ 22x + 44 = 152 ∴ 22x = 110 ∴ x = 5 The original number = [10 * ( x + 4 )] + [x] The original number = 11x + 40 The original number = 11 * (5) + 40 The original number = 55 + 40 The original number = 95. 2. Jon is thrice as old as Kavya. Five years ago his age was two times Kavya’s age. Find their present age. a) Kavya’s age = 15 years; Jon’s age = 5 years b) Kavya’s age = 5 years; Jon’s age = 15 years c) Kavya’s age = 5 years; Jon’s age = 5 years d) Kavya’s age = 15 years; Jon’s age = 15 years Explanation: Let us consider Kavya’s age as x years. Then Jon’s age would be 3x years. Kavya’s age five years ago was (x – 5) years. Jon’s age five years ago was (3x – 5) years. It is given that Jon’s age five years ago was two times Kavya’s age. Thus, (3x – 5) = 2 * (x – 5) Or 3x -5 = 2x – 10 Or x = 5 ∴ Jon’s age = 3x ∴ Jon’s age = 3 * 5 ∴ Jon’s age = 15 years ∴ Kavya’s age = x ∴ Kavya’s age = 5 years. 3. Arya takes a number adds $$\frac{13}{3}$$ to it and then divides it by 3. At the end of all operations he gets 12. What would be the original number? a) $$\frac{23}{3}$$ b) $$\frac{3}{23}$$ c) 23 d) 3 Explanation: Let the number chosen by Arya be x. The first operation carried out by Arya was adding $$\frac{13}{3}$$ to the number. Hence, the equation formed is x + $$\frac{13}{3}$$ = 12 The second operation is to divide throughout by 3. Hence, the equation is modified to $$\frac{x}{3} + \frac{13}{9}$$ = 4 $$\frac{x}{3} + \frac{13}{9}$$ = 4 ∴ $$\frac{3x}{9} + \frac{13}{9} = \frac{36}{9}$$ ∴ 3x + 13 = 36 ∴ 3x = 23 ∴ x = $$\frac{23}{3}$$. 4. When a number is subtracted from 484, we get 459. The number subtracted is square of? a) 5 b) 4 c) 3 d) 2 Explanation: Let the number subtracted be x. 484 – x = 459 ∴ 484 – 459 = x ∴ x = 25 ∴ the subtracted number is 25 and 25 is square of number 5. 5. Raj buys books worth rupees four hundred, he has coins of denomination two-rupees. How many coins does he need to pay the bill? a) 200 b) 100 c) 400 d) 150 Explanation: Let the number of coins required be x. The amount to be paid is rupees 400. (2 * x) = 400 ∴ 2x = 400 ∴ x = 200 Raj needs 200 coins each of two rupees in order to pay the bill amount. Note: Join free Sanfoundry classes at Telegram or Youtube 6. Form an equation for all multiples of 12. a) 3x b) 12x c) 4x d) 3x Explanation: If a set is formed consisting all the multiples of 12 it would be like [12,24,36,48,60,…..] The set of multiples of 12 is formed by multiplying the set of natural numbers with 12. Let the set of natural number be represented by x ∴ the general equation of multiples of 12 = 12x 7. Sita wants to buy books of five hundred-rupees and she has 12 fifty-rupees notes. How many notes will she have after the payment? a) 1 b) 2 c) 3 d) 4 Explanation: If Sita wants to pay five hundred-rupees in fifty-rupees notes then she has to give the shopkeeper 10 notes each of fifty-rupees. After giving 10 notes, she will be left with 2 notes. Hence, the correct answer to this question is 2. 8. If Ram’s present age is 3 years and Shyam is twice Ram’s present age. What will be Shyam’s age after 10 years? a) 16 b) 17 c) 18 d) 19 Explanation: Let Ram’s present age be x years. ∴ Shyam’s present age will be 2x years. After 10 years Shyam’s age would be, (2x + 10) years. ∴ Shyam’s age after 10 years = 2 * 3 + 10 ∴ Shyam’s age after 10 years = 6 + 10 ∴ Shyam’s age after 10 years = 16 years. 9. If the perimeter of a regular hexagon is 192 m then find the Length of each side of the regular hexagon. a) 32 cm b) 32 m c) 23 m d) 23 cm Explanation: Perimeter of a regular hexagon = 6 * (side) ∴ 192 = 6 * (side) ∴ side = 32 m. 10. If the perimeter of a scalene triangle is 23 cm, with side 1 with Length 12 cm and side 2 with Length 3 cm. Find the Length of third side. a) 23 cm b) 12 cm c) 2 cm d) 8 cm Explanation: Perimeter of a scalene triangle = Length of side 1 + Length of side 2 + Length of side 3 ∴ 23 = 12 + 3 + Length of side 3 ∴ 23 = 15 + Length of side 3 ∴ Length of side 3 = 8 cm. Sanfoundry Global Education & Learning Series – Mathematics – Class 8. To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
# A projectile is shot from the ground at a velocity of 8 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land? Jul 21, 2017 $t = 1.58$ $\text{s}$ #### Explanation: We're asked to find the time $t$ of an object's flight, given its initial velocity. To do this, we recognize that when the object lands (assuming it launches and lands at the same height), its height $\Delta y$ will be $0$; we can use the equation $\Delta y = {v}_{0} \sin {\alpha}_{0} t - \frac{1}{2} g {t}^{2}$ to find this time. For this equation, • $\Delta y$ is the change in height ($0$) • ${v}_{0}$ is the initial speed ($8$ $\text{m/s}$) • ${\alpha}_{0}$ is the launch angle ($\frac{7 \pi}{12}$) • $g$ is the acceleration due to gravity near earth's surface ($9.81$ ${\text{m/s}}^{2}$) • $t$ is the time (what we're trying to find) Plugging in known quantities: $0 = \left(8 \textcolor{w h i t e}{l} {\text{m/s")sin((7pi)/12)t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$ $\frac{1}{2} \left(9.81 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (7.73color(white)(l)"m/s}\right) t$ (4.905color(white)(l)"m/s"^2)t = 7.73color(white)(l)"m/s" t = (7.73color(white)(l)"m/s")/(4.905color(white)(l)"m/s"^2) = color(red)(1.58 color(red)("s"
02. Vector Arithmetic and Recycling Vectors can use simple arithmetic expressions (+, -, , /) to perform basic operations. Let's first look at addition, then discuss a caveat of vector arithmetics. You can add or subtract the corresponding elements of two or more vectors of the same length together. ``````> c(1,2,3) + c(99,98,97) [1] 100 100 100 > c(1,2,3) + c(4,5,6) [1] 5 7 9 > c(1,2,3) - c(1,1,1) [1] 0 1 2`````` But what would happen if all the vectors weren't of the same length? Instead of erroring out, R performs recycling. Recycling Recycling occurs when vector arithmetic is performed on multiple vectors of different sizes. R takes the shorter vector and repeats them until it becomes long enough to match the longer one. ``````> c(1,2,3,4,5,6) + c(1,3) [1] 2 4 3 7 6 9`````` As you can see, the `c(1,3)` vector repeated itself to form `c(1,3,1,3,1,3)` so that it could successfully match the previous term. If the shorter vector is not a vector of the longer one, then a warning message appears, but the operation still takes place. ``````> c(1,2,3,4,5) + c(1,3) [1] 2 5 4 7 6 Warning message: In c(1, 2, 3, 4, 5) + c(1, 3) : longer object length is not a multiple of shorter object length`````` Multiplication and Division Multiplying or dividing vectors is similar to addition and subtraction in that each corresponding element matches up and a product is formed. When the sizes differ, recycling occurs. ``````> c(1,2,3) c(0,3,6) [1] 0 6 18 > c(1,3,5) * c(2,4) [1] 2 12 10``` Warning message: In c(1, 3, 5) * c(2, 4) : longer object length is not a multiple of shorter object length``` Operators are mere functions One small detail to notice is that these common arithmetic expressions are actually functions. Thus, they can be with a similar function notation. ``````> "*"(5,6) [1] 30`````` Modulo We can also perform the modulo operator, which outputs the remainder after division of two numbers. ``````> c(55,54,53) %% c(3) [1] 1 0 2`````` You can also apply linear algebra on your vectors in R. To calculate the cross product, use `crossprod()`: ``````> crossprod(1:3, 4:6) [,1] [1,] 32`````` You'll notice that the return type isn't a new vector, but instead a one-dimensional matrix. We'll look at matrices in the next lesson. Become a Bioinformatics Whiz! Introduction to Bioinformatics Vol. 2 This is Volume 2 of Bioinformatics Algorithms: An Active Learning Approach. This book presents students with a light-hearted and analogy-filled companion to the author's acclaimed course on Coursera. Each chapter begins with an interesting biological question that further evolves into more and more efficiently solutions of solving it. \$ Check price (5+ reviews) Learn to be a Pythonista! Python Playground Python Playground is a collection of fun programming projects that will inspire you to new heights. You'll manipulate images, build simulations, and interact with hardware using Arduino & Raspberry Pi. With each project, you'll get familiarized with leveraging external libraries for specialized tasks, breaking problems into smaller, solvable pieces, and translating algorithms into code. \$ Check price (14+ reviews)
English # Find if a Number is Divisible by 20 Given here is the tutorial explaining the shortcut method to determine if the given number is divisible by 20. This tutorial explains the Divisibility Rule for 20. Using this rule you can determine if the number is divisible by 20 without performing the division. ## Divisibility Rule for 20 ##### Steps to be Followed: Check whether the last two digits are even and divisible by 10. Alternatively, check whether the last two digits are divisible by 20. If both the conditions are satisfied, then the original number is also divisible by 20. ##### Example Find if the number 2880 is divisible by 20 ##### Solution : Applying the divisibility rule for 20, ###### Step 1: Here, last two digit 80 is even and the last digit is also 0 Hence, we know that the given number 2880 is divisible by 10 ###### Step 2: The last two digits of the number 2880 is 80 80 is divisible by 20 (i.e) 20 x 4 = 80 Therefore, 2880 is also divisible by 20 ###### Step 3: The above given number is divisible by both 10 and 20. Hence, the original number (i.e) 2880 is also divisible by 20.
# What rule of inference is used in each of these arguments? “Rules of inferences” is an important topic in Discrete Mathematics. We can use a rule (s) of inferences to prove if an argument is valid or not. We can also use rules of inference to produce valid arguments. In mathematics, an argument is a sequence of statements. We have a valid argument if and only if is impossible for all the premises to be true, and the conclusion to be false. We can decide whether an argument is valid or not by using rules of inference. Let’s see the solution to an exercise. What rule of inference is used in each of these arguments? ## a) Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major. Let p=”Alice is a mathematics major” and q=”Alice is a computer science major” Rewriting the argument, we have: ``````p ------ ∴p∨q `````` We can also affirm that this is a valid argument because of the rule of Addition. ## b) Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major. Following a similar reasoning to the previous exercise: p = “Jerry is a mathematics major”, q = “Jerry is a computer science major” ``````p^q ------ ∴p `````` ## c) If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed. p=”it is rainy”, q=”the pool will be closed” ``````p p->q ------ ∴q`````` ## d) If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today. p=”it snows today”, q=”the university will close” ``````p->q ¬q ------- ∴¬q `````` ## e) If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn. p=”I go swimming”, q=”I will stay in the sun too long”, r=”I will sunburn” ``````p->q q->r -------- ∴q->r ``````
Study Guide # Differential Equations - Slope Fields ## Slope Fields If you build it, they will come. Consider the first-order differential equation We can't solve this d.e. by thinking backwards, but we can still think about how a solution to this d.e. would behave. We know means the slope of y with respect to x. This means any solution to this differential equation is a function y whose slope at any point (x, y) on the function is equal to x + y. We're going to go back to Leibniz Notation. ### Sample Problem Suppose we have a solution y to the d.e. that goes through the point (2, 3): The slope of y at the point (2, 3) must be Although we don't know what exactly the function y looks like, we do know that at the point (2, 3) the slope of the function (and therefore the slope of its tangent line) is 5: If we have some other solution y to the d.e. that goes through the point (4,-1), the slope of y at the point (4,-1) must be This means the tangent line to y at (4,-1) has slope 3: ### Sample Problem Suppose f is a solution to the d.e. '(x) = 4xy. If f passes through the point (3, 10) then the derivative of f at that point is 4(3)(10) = 120. A graph with lots of little tangent lines, like the one we just drew, is a called a slope field or a vector field. Slope fields are useful for visualizing the solutions to a given differential equation. If you know how the slope of the function behaves, you can see what the overall function looks like. Because we're just drawing little lines, drawing slope fields is a bit boring and tedious. If we draw the slope field for the d.e. and include more points, we get something like this: We're sure you can imagine how awful it would be to need to figure out all those slopes in this picture by hand. Thankfully, this is one place where getting computers to do your work is usually encouraged. Here's an online tool for drawing slope fields: You will be asked to match slope fields with their differential equations, or to match differential equations with their slope fields. Our stroll through the slope fields above gave some examples of things you can look for. Here are some questions you can ask yourself when trying to match slope fields and differential equations (you can ask these questions when looking at either a d.e. or a slope field): • Where is the slope positive? Negative? Zero? • What's the slope when x = 0? When y = 0? • What's the slope when y = x? When y = -x? • Does the slope depend on both x and y? Just x? Just y? If the slope only depends on y, then all lines at the same height y will have the same slope: • If the slope only depends on x, then all lines at the same x-position will have the same slope: • As x approaches ∞ how does the slope change? How about as x approaches -∞? As x approaches 0? • What happens as y approaches ∞, -∞, or 0? • ### Slope Fields and Solutions If we have a slope field for a d.e. and a point in that slope field, we can sketch the solution that goes through that point. The little bits of tangent lines are like arrows telling the function which way to go. If we follow the arrows, we get the solution. ### Sample Problem Here's the slope field for the d.e. and a point: Draw the solution that passes through the point. Answer. The little tangent lines trace out the shape of the solution: When we use slope fields to sketch solutions to differential equations, the solutions won't necessarily be functions. ### Sample Problem Here's the slope field for the d.e. and a point: This isn't a function because it fails the vertical line test. This is cool, because even if we can't get a single-variable function that's a solution to a differential equation, we can still see the shape of the solution on a graph. Slope fields are the visual equivalent of IVPs for first-order differential equations. A slope field is the visual equivalent of a differential equation, and a point is the visual equivalent of an initial condition. We can draw the solution that goes through a point, given a slope field and a point because there's a Uniqueness Theorem that says such a problem has exactly one answer. Here's the fancy math language, written as bad poetry: Given a first-order differential equation and an initial condition (so long as the derivative is continuous) there is exactly one solution that satisfies both the differential equation and the initial condition. Here's the translation for slope fields: Given a slope field and a point (so long as the derivative is continuous) there is exactly one solution that goes through that point and has the shape specified by the slope field. Since it's very unlikely that you'll ever be given such a problem where the derivative isn't continuous, you can probably get away with thinking that an IVP involving a first-order differential equation has exactly one solution. • ### Equilibrium Solutions An equilibrium solution is a solution to a d.e. whose derivative is zero everywhere. On a graph an equilibrium solution looks like a horizontal line. Given a slope field, we can find equilibrium solutions by finding everywhere a horizontal line fits into the slope field. Equilibrium solutions come in two flavors: stable and unstable. These terms are easiest to understand by looking at slope fields. A stable equilibrium solution is one that other solutions are trying to get to. If we pick a point a little bit off the equilibrium in either direction, the solution that goes through that point tries to snuggle up to the equilibrium solution. An unstable equilibrium solution is one that the other solutions are trying to get away from. If we pick a point a little bit off the equilibrium, the solution that goes through that point is trying to run away from the equilibrium solution. If the solutions are trying to get away on one side and snuggle up on the other side, the equilibrium is still unstable. If we're given a differential equation instead of a slope field, we can determine whether each equilibrium solution is stable or unstable by using the differential equation to sketch a very rough slope field. ## Tired of ads? Join today and never see them again.
Polynomials Questions We provide polynomials practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. We focus on polynomials skills mastery so, below you will get all questions that are also asking in the competition exam beside that classroom. List of polynomials Questions Question NoQuestionsClass 168. If x= 6+ then the value of * + 4 is (1) 1448 (3) 1444 (2) 1442 (4) 1446 9 2Find the product of ( (boldsymbol{m}-mathbf{1})(boldsymbol{m}- ) 2)( (m-3) ) 9 360. If x2 = y + z, y2 = 2 + x and z = x + y. then the value of 1.11 1+x 1+ y 1+ z is (1)-1 (2) 1 (3) 2 (4) O 9 4Find the Quotient and the Remainder when the first polynomial is divided by the second. ( left(x^{4}-2 x^{3}+2 x^{2}+x+4right) ) by ( left(x^{2}+x+right. ) 1) A. Quotient ( =x^{2}+3 x+4 ), Remainder ( =0 ) B. Quotient= ( x^{2}-3 x-4 ), Remainder ( =0 ) c. Quotient ( =-x^{2}-3 x+4 ), Remainder ( =0 ) D. Quotient ( =x^{2}-3 x+4 ), Remainder ( =0 ) 10 561. If 7x = 14, then the value 2x of x- 3 is : 8x (1) 9 (3) 27 (2) 8 (4) 11 9 6If ( frac{boldsymbol{a}}{boldsymbol{b}}=frac{boldsymbol{c}}{boldsymbol{d}}=frac{boldsymbol{e}}{boldsymbol{f}} ) and ( frac{2 a^{4} b^{2}+3 a^{2} c^{2}-5 e^{4} f}{2 b^{6}+3 b^{2} d^{2}-5 f^{5}}=left(frac{a}{b}right)^{n} ) then the value if ( n ) is ( A ) B. 2 ( c cdot 3 ) ( D ) 10 7Find the cube of 499 860. If (x – 1) and (x + 3) are the fac- tors of x2 + kx + k, then (1) k, = -2, k, = -3 (2) k, = 2, k = -3 (3) k, = 2, k, = 3 (4) k, = -2, k, = 3 9 9When ( x^{3}-2 x^{2}+a x-b ) is divided by ( x^{2}-2 x-3, ) the remainder is ( x-6 . ) The values of ( a ) and ( b ) are respectively: ( A cdot-2 ) and -6 B. 2 and -6 c. -2 and 6 D. 2 and 6 10 1064. If m+ = 4, find the value of (m – 2)2 + (m-22 (1) -2 (3) 2 (2) 0 (4) 4 9 11Which polynomial represents this plot? A. cubic D. linear 9 12The remainder in the division of ( 14 x^{2}- ) ( 53 x+45 ) by ( 7 x-9 ) is A . -3 B. ( 7 x ) c. 90 D. 10 13If ( left(x^{2}+4 x-21right) ) is divided by ( x+7 ) then the quotient is ( mathbf{A} cdot x+3 ) B. ( x-3 ) c. ( x^{2}-2 ) D. ( x-4 ) 10 14Factorize: ( 4 a^{2} b-9 b^{3} ) ( mathbf{A} cdot b(2 a+3 b)(2 a+3 b) ) B. ( b(2 a+3 b)(a-3 b) ) ( mathbf{c} cdot b(a+3 b)(2 a-3 b) ) D. ( b(2 a+3 b)(2 a-3 b) ) 9 15If ( f(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b ) is a polynomial such that when it is divide by ( (x-1) ) and ( (x+1) ) the remainders are 5 and 19 respectively the remainder when ( f(x) ) is divisible by ( (x-2) ) is ( A cdot 7 ) B. 8 c. 9 D. 10 9 16Find the value of ( b ) for which the polynomial ( 2 x^{3}-9 x^{2}-x-b ) is divisible by ( 2 x+3 ) 10 17( frac{x^{2}+9 x+14}{x+7}= ) ( mathbf{A} cdot x+2 ) B. ( x+7 ) ( c cdot 2 ) D. 10 18Write the degree of each of the following polynomials: ( frac{1}{2} y^{7}-12 y^{6}+48 y^{5}-10 ) 10 19Which of the following is a constant polynomial? ( mathbf{A} cdot p(x)=7+3 x ) B ( . p(x)=7 ) C ( . p(x)=7 x+7 ) D. ( p(x)=4 x+3 ) 10 20What is the degree of the given monomial ( boldsymbol{x} boldsymbol{y}^{2} boldsymbol{z}^{2} ? ) ( A cdot 3 ) B. 4 ( c .5 ) D. 6 10 21Which of the following is a cubic polynomial? ( mathbf{A} cdot p(x)=x^{2}-16 ) в. ( p(x)=x-16 ) C ( cdot p(x)=x^{3}-27 ) D ( cdot p(x)=27^{3} ) 10 22Write the polynomial in standard form and also write down their degree. ( left(frac{5}{6} z-frac{3}{4} z^{2}-frac{2}{3} z^{3}+1right) )10 23If ( a+b+c=8 ) and ( a b+b c+c a=20 ) find the value of ( a^{3}+b^{3}+c^{3}-3 a b c ) 9 24If ( a=frac{1}{3-2 sqrt{2}}, b=frac{1}{3+2 sqrt{2}}, ) then the value of ( a^{3}+b^{3} ) is A ( cdot 194 ) в. 196 ( c cdot 198 ) D. 200 9 25The degree of the equation, given by ( (x+2)(x-1)=(x+1)(x+3), ) is ( A cdot 2 ) B. 3 ( c . ) D. 9 26Degree of polynomial 5 is ( A cdot 1 ) B. 2 ( c cdot 0 ) D. Not defined 9 27Which one of the following is a A ( cdot x^{2}+3 ) B. ( x^{3}+x^{2}+4 ) ( mathbf{c} cdot 2 x^{4}+4 x^{3}+3 x^{2}+6 ) D. None of the above 9 28Divide ( left(y^{3}-3 y^{2}+5 y-1right) div(y-1) ) 10 2956. If 2x + 3y = 13 and xy = 6 then the value of 8x + 27yº will be (1) 799 (2) 797 (3) 795 (4) 793 9 30Find the quotient the and remainder of the following division: ( left(5 x^{3}-8 x^{2}+5 x-7right) div(x-1) ) 10 31( f(x+2) ) is a factor of ( left(x^{4}-x^{2}-aright) ) then find ( a ) 10 32Solve: ( frac{m^{2}-3 m-108}{m+9}=0, ) then ( m=? ) 10 33Which of the following is NOT a quadratic polynomial? ( mathbf{A} cdot p(x)=16-4 x ) в. ( p(x)=13-x ) C ( cdot p(x)=12 x^{3}-x ) D. All of the above 9 3461. If x + y + z = 1, xy + yz + 2x = -1, xyz = -1, then xy + y + z is (1)-2 (2)-1 (3) O (4) 1 9 35If ( boldsymbol{x} neq-mathbf{5}, ) then the expression ( frac{mathbf{3} boldsymbol{x}}{boldsymbol{x}+mathbf{5}} div ) ( frac{6}{4 x+20} ) can be simplified to A ( .2 x ) в. ( frac{x}{2} ) c. ( frac{9 x}{2} ) D. ( 2 x+4 ) 10 36Find the missing terms such that the given polynomial become a perfect square trinomial: [ -12 x+9 ] 10 37Simplify: ( (7 m-8 n)^{2}+(7 m+8 n)^{2} ) A ( cdot 198 m^{2}+28 n^{2} ) B. ( 98 m^{2}+128 n^{2} ) ( mathbf{c} .98 m+128 n ) D. ( 98 m^{2}-128 n^{2} ) 9 38What should be added to ( x^{5}-1 ) to be completely divisible by ( x^{2}+3 x-1 ? ) 10 39Show that ( (x-2) ) is a factor of ( x^{3}- ) ( 3 x^{2}-10 x+24 ) 10 4068. If x2 + y2 + 2 + 2 = 2(y-2), then value of x + y + z is equal to (1) 0 (2) 1 (4) 3 de (3) 2 9 41If ( m-frac{1}{m}=5, ) then find ( m^{2}+frac{1}{m^{2}} ) B. ( sqrt{27} ) c. ( 25 sqrt{29} ) D. ( 25 sqrt{27} ) 9 4258. If ab + bc + ca = 0, then the val- ue of +- a? – bc b? – ac c2 – ab * (1) 2 (2) -1 (3) O (4) 1 9 43Degree of the polynomials ( frac{x^{23}+x^{14}-x^{16}}{x^{2}} ) is ( A cdot 2 ) B . 23 c. 14 D. 21 9 44If the polynomial ( boldsymbol{f}(boldsymbol{x})= ) ( left(6 x^{4}+8 x^{2}+17 x^{2}+21 x+7right) ) is divided by another polynomial ( g(x)= ) ( 3 x^{2}+4 x+1 ) the remainder is ( (a x+b) ) Find ( a ) and ( b ) 10 45If ( p(t)=t^{3}-1, ) find the values of ( boldsymbol{p}(mathbf{1}), boldsymbol{p}(-mathbf{2}) ) 10 46The remainder obtained when ( t^{6}+ ) ( 3 t^{2}+10 ) is divided by ( t^{3}+1 ) is A ( cdot t^{2}-11 ) B . ( 3 t^{2}+11 ) ( mathbf{c} cdot t^{3}-1 ) D. ( 1-t^{3} ) 10 47f ( a=3, b=-3, ) find the value of ( (a-2)^{2}+(b-2)^{2} ) 9 48Substituting ( x=-3 ) in ( x^{2}-5 x+4 ) A . -2 B . 28 ( c cdot 2 ) D. -1 10 49Classify the following polynomial based on their degrees: ( boldsymbol{y}^{2}-boldsymbol{4} ) 10 50If a zero of ( p(x)=x^{2}+3 x+g ) is ( 2, ) then value of ( g ) is A . -10 B. 10 ( c .5 ) D. -5 10 51Factorize ( 3 a^{5}-108 a^{3} ) A ( cdot 3 a^{3}(a-6)(2 a-6) ) В ( cdot 2 a^{3}(a+6)(7 a-6) ) c. ( 2 a^{3}(a-6)(3 a+6) ) D. ( 3 a^{3}(a+6)(a-6) ) 9 52Verify whether the following are zeros of the polynomial indicated against them: ( boldsymbol{g}(boldsymbol{x})=mathbf{5} boldsymbol{x}^{2}+mathbf{7} boldsymbol{x}, boldsymbol{x}=mathbf{0},-frac{mathbf{7}}{mathbf{5}} ) A. True B. False 10 5355. If x = 1.75, y = 0.5, then find the value of 4×2 + 4xy + y2. (1) 15.75 (2) 16.00 (3) 16.25 (4) 16.75 9 54Perform division ( left(y^{2}+7 y+10right) div 6(y+5) ) 10 557. The age of a man is same as his wife’s age with the digits reversed. Then sum of their ages is 99 years and the man is 9 years older than his wife. The age of man and his wife is (a) 50 years (b) 45 years (c) 54 years (d) 44 years 10 56If the polynomial ( left(x^{3}-3 x^{2}+a x+18right) ) is divided by ( (x-4), ) the reminder is 58 Find the value of a. 9 57Carry out the following divisions ( -54 l^{4} m^{3} n^{2} ) by ( 9 l^{2} m^{2} n^{2} )10 5825. The The real number k for which the equation, 2×3 + 3x +k=0 has two distinct real roots in [0, 1] [JEEM 2013] (a) lies between 1 and 2 (6) lies between 2 and 3 © lies between-1 and 0 (d) does not exist. 10 59Prove if ( cot theta+frac{1}{cot theta}=2 ) then ( cot ^{2} theta+ ) ( frac{1}{cot ^{2} theta}=2 ) 9 60When ( p(x)=x^{3}+a x^{2}+2 x+a ) is divided by ( (x+a), ) the remainder is ( mathbf{A} cdot mathbf{0} ) B. ( a ) ( c .-a ) D. ( 2 a ) 9 61Divide ( 4 x^{2} y^{2}(6 x-24) div 4 x y(x-4) ) ( mathbf{A} cdot 6 x y ) B. ( 4 x y ) c. ( x-4 ) D. ( x y(x-4) ) 10 62If ( left(x^{3 / 2}-x y^{1 / 2}+x^{1 / 2} y-y^{3 / 2}right) ) is divided by ( left(x^{1 / 2}-y^{1 / 2}right), ) the quotient is: ( mathbf{A} cdot x+y ) в. ( x-y ) C ( cdot x^{1 / 2}-y^{1 / 2} ) D. ( x^{2}-y^{2} ) 10 6361. If x + y2 + z = xy + y2 + zx, lx # O). then the value of 4x + 2y – 32 is 2x (2) 1 (1) o 9 64Say true or false: The zeros of the polynomial ( x^{2}-14 x+ ) 49 are equal to 7 A. True B. False 10 65Write each of the following polynomials in the standard form. Also, write their degree: ( left(x^{3}-1right)left(x^{3}-4right) ) 9 66If ( x-y=7 ) and ( x^{3}-y^{3}=133 ; ) find: the value of ( x y ) A . 12 B. – – ( c .-10 ) D. 5 9 6751. For real a, b, c if a + b + c a+c ab + bc + ca, the value of – (1) 3 (3) 2 (2) (40 9 68Divide: ( a^{4}+4 b^{4} ) by ( a^{2}+2 a b+b^{2} )10 69Simplify: ( (sqrt{2} x-2 y)^{2} )9 7067. If 5x 2×2 + 5x +1 then the value of (x+2x is of X +- (1) 15 (3) 20 (2) 10 (4) 5 9 71Divide ( left(frac{boldsymbol{y}}{boldsymbol{6}}+frac{boldsymbol{2} boldsymbol{y}}{boldsymbol{3}}right) divleft(boldsymbol{y}+frac{boldsymbol{2} boldsymbol{y}-mathbf{1}}{boldsymbol{3}}right) )10 72Factorise the following: ( a^{6}-b^{6} ) ( mathbf{A} cdot(a+b)(a-b)left(a^{2}+b^{2}+a bright)left(a^{2}-b^{2}+a bright) ) B ( cdot(a+b)(a-b)left(a^{2}+b^{2}+a bright)left(a^{2}+b^{2}-a bright) ) ( mathbf{c} cdot(a+b)(a-b)left(a^{2}-b^{2}-a bright)left(a^{2}+b^{2}-a bright) ) D. None of these 9 7313. If a2 + b2 + c2=1, then ab + bc + ca lies in the interval (1984 – 2 Marks @ 15,2] (b) (-1,2] @ 1-1 () [-1,] 9 74The degree of the polynomial ( x^{2}- ) ( 5 x^{4}+frac{3}{4} x^{7}-73 x+5 ) is A. 7 B. ( frac{3}{4} ) ( c cdot 4 ) D. -73 9 75( 10 UUIU 3. D . Ratio of my present age to age twenty years ago is (a) 3:2 (b) 2:1 (c) 3:1 (d) 1:2 10 76Choose the correct options: This question has multiple correct options A. Remainder obtained on dividing ( p(x)=x^{3}+1 ) by ( (x+ ) 1) is 0 B. ( x=1 ) and ( x=2 ) are zeroes of polynomial ( P(x)= ) ( 5 x^{5}-20 x^{4}+5 x^{3}+50 x^{2}-20 x-40 ) C ( cdot 6 x^{7}-5 x^{4}+2 x+3 ) is a polynomial of degree 7 D. ( x+1 ) and ( 2 x-3 ) are factors of ( 2 x^{3}-9 x^{2}+x+12 ) 9 77If ( x^{3}-frac{1}{x^{3}}=14, ) then ( x-frac{1}{x}= ) ( A cdot 2 ) B. 4 ( c .5 ) D. 9 78Divide and write the quotient and remainder. (a) ( left(y^{2}+10 y+24right) div(y+4) ) (b) ( left(p^{2}+7 p-5right) div(p+3) ) 10 79If ( a^{2}+10 b^{2}+5 c^{2}+6 a b+2 b c-16 c+ ) ( mathbf{1 6}=mathbf{0} ) then the possible value of ( boldsymbol{a}- ) ( b+c= ) A . 5 B. -3 ( c cdot-2 ) D. 10 9 80( R_{1} ) and ( R_{2} ) are the reminders when the polynomial ( a x^{3}+3 x^{2}-3 ) and ( 2 x^{3}- ) ( 5 x+2 a ) are divided by ( (x-4) ) respectively. If ( 2 R_{1}-R_{2}=0, ) then find the value of a A ( cdot a=frac{1}{6} ) в. ( _{a=frac{1}{3}} ) c. ( _{a=frac{1}{7}} ) D. ( a=frac{1}{5} ) 9 81Find if polynomial ( x^{4}-3 x^{3}+7 x^{2}- ) ( 8 x+12 ) is exactly divisible by ( x^{2}- ) ( 2 x+2 ) 10 82The expansion of ( (2 x-3 y)^{2} ) is: A ( cdot 2 x^{2}+3 y^{2}+6 x y ) B. ( 4 x^{2}+9 y^{2}-12 x y ) c. ( 2 x^{2}+3 y^{2}-6 x y ) D. ( 4 x^{2}+9 y^{2}+12 x y ) 9 83The quotient and remainder when ( 3 x^{4}+6 x^{3}-6 x^{2}+2 x-7 ) is divided by ( x-3 ) are A. Quotient: ( 3 x^{3}+15 x^{2}+39 x+119 ) and Remainder: 350 B. Quotient: ( 3 x^{3}+10 x^{2}+39 x+119 ) and Remainder: 35 C . Quotient: ( 3 x^{3}+15 x^{2}+39 x+119 ) and Remainder: 50 D. Quotient: ( 3 x^{3}+15 x^{2}+119 ) and Remainder: 350 9 84If ( frac{x^{a^{2}}}{x^{b^{2}}}=x^{16}, x>1, ) and ( a+b=2, ) what is the value of ( a-b ? ) ( A cdot 8 ) B. 14 ( c cdot 16 ) D. 18 9 85( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{4}+boldsymbol{2} boldsymbol{x}^{3}-boldsymbol{2} boldsymbol{x}^{2}+boldsymbol{x}-mathbf{1} ) and ( boldsymbol{q}(boldsymbol{x})=boldsymbol{x}^{2}+boldsymbol{2} boldsymbol{x}-boldsymbol{3} ) Then ( p(x) ) is divisible by ( q(x), ) if we A. Add ( (x-2) ) B. Add ( (x-3) ) c. Add ( (2-x) ) D. Add ( (3-x) ) 10 86Solve: ( left(5 p^{2}-25 p+20right) div(p-1) )9 87Divide the polynomial ( p(x) ) by the polynomial ( g(x) ) and find the quotient and remainder. ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{3}-boldsymbol{3} boldsymbol{x}^{2}+mathbf{5} boldsymbol{x}-boldsymbol{3} ) ( g(x)=x^{2}-2 ) ( mathbf{A} cdot q(x)=x-3, r(x)=7 x+9 ) в. ( q(x)=x+3, r(x)=-7 x-9 ) C ( . q(x)=x+3, r(x)=7 x-9 ) D. ( q(x)=x-3, r(x)=7 x-9 ) 10 88If ( x^{2}+frac{1}{x^{2}}=7 ) find the value of ( x^{3}+frac{1}{x^{3}} )9 89Find the remainder when we divide ( boldsymbol{x}^{7} boldsymbol{y}-boldsymbol{x} boldsymbol{y}^{7} ) by ( (boldsymbol{x}+boldsymbol{y})left(boldsymbol{x}^{2}-boldsymbol{x} boldsymbol{y}+boldsymbol{y}^{2}right) ) 10 90Find the zeroes of the quadratic polynomial ( x^{2}+14 x+48 ) and verify them 9 91Factorise the polynomial: ( a x^{2}+b x^{2}+ ) ( a y^{2}+b y^{2} ) 9 92A cubic polynomial is a polynomial of degree A . B. ( c cdot 3 ) ( D ) 9 93Simplify ( (7 a-5 b)left(49 a^{2}+35 a b+right. ) ( left.25 b^{2}right) ) 9 94( (3-sqrt{7})(3+sqrt{7})= ) ( mathbf{A} cdot mathbf{4} ) B . 2 ( c cdot 6 ) D. 8 9 95Divide as directed ( 5(2 x+1)(3 x+5) div ) ( (2 x+1) ) 10 96Simplify: ( left(a^{2}-b^{2}right)^{2} )9 97( p(x)=25 ) is a polynomial A. linear c. constant D. cubic 9 98Divide the polynomial ( p(x) ) by the polynomial ( p(g) ) and find the quotient an in each of the following: ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{3}-boldsymbol{3} boldsymbol{x}^{2}+boldsymbol{5} boldsymbol{x}-boldsymbol{3}, boldsymbol{g}(boldsymbol{x})= ) ( x^{2}-2 ) ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{4}-boldsymbol{3} boldsymbol{x}^{2}+boldsymbol{4} boldsymbol{x}+mathbf{5}, boldsymbol{g}(boldsymbol{x})= ) ( boldsymbol{x}^{2}+mathbf{1}-boldsymbol{x} ) ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{4}-boldsymbol{5} boldsymbol{x}+boldsymbol{6}, boldsymbol{g}(boldsymbol{x})=boldsymbol{2}-boldsymbol{x}^{2} ) 10 99( (x)^{n}+(a)^{n} ) is completely divisible by ( x+a, ) then ( n ) can be A . 7028 в. 861 c. 26 D. 782 9 100( boldsymbol{f}(boldsymbol{x})=mathbf{3} boldsymbol{x}^{5}+mathbf{1} mathbf{1} boldsymbol{x}^{4}+mathbf{9} mathbf{0} boldsymbol{x}^{2}-mathbf{1} mathbf{9} boldsymbol{x}+mathbf{5} mathbf{3} ) is divided by ( x+5, ) then the remainder is: A. 100 B . -100 ( c cdot-102 ) D. 102 9 101Factorize: ( a^{2}-(2 a+3 b)^{2} ) ( mathbf{A} cdot-3(a+b)(a+3 b) ) B ( cdot 3(a+b)(a+3 b) ) ( mathbf{c} cdot-3(a-b)(a-3 b) ) D. ( 3(a+b)(a-3 b) ) 9 102Find the degree of the given algebraic expression ( boldsymbol{a} boldsymbol{x}^{2}+boldsymbol{b} boldsymbol{x}+boldsymbol{c} ) ( mathbf{A} cdot mathbf{0} ) B. ( c cdot 2 ) D. 3 9 1031 -b))2 + ab = p (a + b)2, then the value of p is (assume that a # -b) (2) 1 (3) = (4) 9 104If ( a^{2}=b^{2}+b c ) then ( b+c ) always equals A ( cdot frac{a^{2}}{b} ) в. ( frac{1}{2} b c ) ( c cdot 1 ) D. ( b c ) 9 105If ( a+frac{1}{a}=2, ) then find ( a^{4}+frac{1}{a^{4}} )9 106Divide the polynomial ( p(x) ) by the polynomial ( g(x) ) and find the quotient and remainder in each of the following: (i) ( p(x)=x^{3}-3 x^{2}+5 x-3, g(x)= ) ( x^{2}-2 ) (ii) ( p(x)=x^{4}-3 x^{2}+4 x+5, g(x)= ) ( boldsymbol{x}^{2}+mathbf{1}-boldsymbol{x} ) (iii) ( p(x)=x^{4}-5 x+6, g(x)=2-x^{2} ) 10 107Divide the polynomial ( 2 x^{4}-4 x^{3}- ) ( 3 x-1 ) by ( (x-1) ) and verify the remainder with zero of the divisor. 10 108Verify whether the following are zeroes of the polynomial, indicated against them. ( boldsymbol{p}(boldsymbol{x})=boldsymbol{l} boldsymbol{x}+boldsymbol{m}, boldsymbol{x}=-frac{boldsymbol{m}}{boldsymbol{l}} )10 109Degree of the polynomial ( boldsymbol{p}(boldsymbol{x})=-10 ) is A . -10 B. 10 ( c cdot 0 ) D. 10 110Factorise the following: ( (5 x-6 y)^{3}+ ) ( (7 z-5 x)^{3}+(6 y-7 z)^{3} ) A. ( 3(5 x-6 y)(7 z-5 x)(6 y-7 z) ) в. ( (x-6 y)(7 z-x)(y-7 z) ) c. ( 3(x-6 y)(z+5 x)(8 y-z) ) D. ( (x-y)(7 z+5 x)(6 y-7 z) ) 9 11166. If a – b – – 3abc = 0, then (1) a = b = c (2) a + b + c = 0 (3) a + c = b (4) a = b + c 9 112Arrange ( boldsymbol{x}^{8}+boldsymbol{x}+boldsymbol{x}^{12}-boldsymbol{3} boldsymbol{x}^{7}+boldsymbol{x}^{9}+1 ) in descending powers of ( x ) 10 113Using the reals ( a_{n} ;(n=1,2, dots, 5), ) if ( l, m, n in{1,2,3,4,5} m<n ) A ( cdotleft(sum a_{n}right)^{2}=sumleft(a_{l}^{2}right)+2left(sum a_{m} a_{n}right) ) B . ( 0=sumleft(a_{l}^{2}right)-sumleft(a_{m} a_{n}right) ) C ( cdotleft(sum a_{n}right)^{2}=sumleft(a_{l}^{2}right)-2left(sum a_{m} a_{n}right) ) D. ( left(sum a_{n}right)^{2}=sumleft(a_{l}^{2}right)+sumleft(a_{m} a_{n}right) ) 9 114The degree of the term ( x^{3} y^{2} z^{2} ) is: ( A cdot 3 ) B. 2 c. 12 D. 9 115Using identity ( (a+b)^{2}=left(a^{2}+2 a b+right. ) ( b^{2} ) ) Evaluate (i) ( (609)^{2}left(text { ii) }(725)^{2}right. ) 9 116Solve: ( left(y^{2}+10 y+24right) div(y+4) ) 10 11751. The value of [(0.87)2+(0.13)2 + (0.87) x (0.26)] 2013 (1) O (3) 1 (2) 2013 (4) -1 9 118Divide ( 81 x^{3}left(50 x^{2}-98right) ) by ( 27 x^{2}(5 x+ ) 7) 10 119If ( a^{2}+b^{2}+c^{2}=250 ) and ( a b+b c+ ) ( c a=3 ), then find ( a+b+c ) 9 120If ( a neq 0 ) and ( a-frac{1}{a}=4, ) find: ( a^{4}+frac{1}{a^{4}} ) A .92 в. 112 ( c .322 ) D. 122 9 121Zeroes of polynomial ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{2}-boldsymbol{3} boldsymbol{x}+ ) 2 are This question has multiple correct options ( A cdot 3 ) B. ( c cdot 4 ) D. 2 9 122Obtain all the zeroes of ( 3 x^{4}+6 x^{3}- ) ( 2 x^{2}-10 x-5, ) if of its zeroes are ( sqrt{frac{5}{3}} ) ( mathfrak{Q}-sqrt{frac{mathbf{5}}{mathbf{3}}} )10 123Find: ( (5 m+7 n)^{2} )9 124Perform the division ( left(a^{4}-a^{3}+a^{2}-right. ) ( a+1) divleft(a^{3}-2right) ) 10 125Write the polynomial in standard form and also write down their degree. ( left(p^{2}+2right)left(p^{2}+7right) )10 126State true or false: If ( a+2 b=5 ; ) then ( a^{3}+8 b^{3}+30 a b=125 ) A. True B. False 9 127f ( 2 x+y=14 ) and ( x y=6, ) Find the value of ( 4 x^{2}+y^{2} ) 9 128Find the number of zeroes of the quadratic ( x^{2}+7 x+10 ) and verify the relationship between the zeroes and the the co-efficients. 10 129What must be added to ( f(x)=4 x^{4}+ ) ( 2 x^{3}+2 x^{2}+x-1 ) so that the resulting polynomial is divisible by ( g(x)=x^{2}+ ) ( 2 x-3 ) A. ( -61 x+65 ) в. ( 2 x-15 ) c. ( -15 x+2 ) D. None of these 10 130ff ( x+2 y+3 z=0 ) and ( x^{3}+4 y^{3}+ ) ( mathbf{9} z^{3}=18 x y z ; ) evaluate: ( frac{(x+2 y)^{2}}{x y}+frac{(2 y+3 z)^{2}}{y z}+ ) ( frac{(3 z+x)^{2}}{z x} ) A . 18 B. 23 c. 16 D. 11 9 131If the product of two numbers is 21 and their difference is ( 4, ) then the ratio of the sum of their cubes to the difference of their cubes is ( mathbf{A} cdot 185: 165 ) B . 165: 158 c. 185: 158 D. 158: 145 9 132Find the roots of the following equation ( 2 y^{2}+frac{15}{y^{2}}=12, ) then A ( cdot quad y=pm sqrt{frac{6+sqrt{6}}{4}}, y=pm sqrt{frac{6-sqrt{6}}{2}} ) в. ( quad y=pm sqrt{frac{6+sqrt{6}}{2}}, y=pm sqrt{frac{6-sqrt{5}}{2}} ) c. ( y=pm sqrt{frac{6+sqrt{6}}{2}}, y=pm sqrt{frac{6-sqrt{6}}{2}} ) D. ( y=pm sqrt{frac{6+sqrt{5}}{2}}, y=pm sqrt{frac{6-sqrt{6}}{2}} ) 10 133For ( frac{boldsymbol{x}^{mathbf{3}}+mathbf{2} boldsymbol{x}+mathbf{1}}{mathbf{5}}-frac{mathbf{7}}{mathbf{2}} boldsymbol{x}^{2}-boldsymbol{x}^{mathbf{6}}, ) write i) the degree of the polynomial ii) the coefficient of ( x^{3} ) iii) the coefficient of ( x^{6} ) iv) the constant term 10 134Which of the following is INCORRECT? ( mathbf{A} cdot p(x)=5 x+5, ) degree ( =1 ) B ( cdot p(x)=4 x^{4}+4, ) degree ( =4 ) ( mathbf{C} cdot p(x)=x^{8}, ) degree ( =8 ) D ( . p(x)=9, ) degree ( =9 ) 9 135Simplify the following. ( b^{4} div b^{5} ) 10 136If ( boldsymbol{x}=mathbf{5}-mathbf{2} sqrt{mathbf{6}}, ) find the value of ( boldsymbol{x}+frac{mathbf{1}}{boldsymbol{x}} )9 137Tuu 5 -1 4. Assertion : x +11=1 is a linear equation. Reason: In a linear equation power ofx cannot be negati 10 138Expand: ( (x y+8)(x y-8) )9 139Given that ( a(a+b)=36 ) and ( b(a+ ) ( b)=64, ) where ( a ) and ( b ) are positive, ( (a-b) ) equals: A . 2.8 B. 3.2 ( c .-2.8 ) D. – 2.5 9 140Find the polynomials whose zeroes are three times the zeroes of ( 2 x^{2}-3 x+1 ) 10 141If ( a, b, c ) are the roots of the equation ( boldsymbol{x}^{3}+boldsymbol{p} boldsymbol{x}^{2}+boldsymbol{q} boldsymbol{x}+boldsymbol{r}=boldsymbol{0}, ) form the equation whose roots are ( a-frac{1}{b c}, b-frac{1}{c a}, c-frac{1}{a b} ) 10 142Prove the following identities: ( boldsymbol{a b c}left(sum boldsymbol{a}right)^{3}-left(sum boldsymbol{b c}right)^{3}=boldsymbol{a b c} sum boldsymbol{a}^{3}- ) ( sum b^{3} c^{2}=left(a^{2}-b cright)left(b^{2}-c aright)left(c^{2}-a bright) ) 9 143( f(alpha, beta, gamma ) are the zeros of the polynomial ( boldsymbol{x}^{3}+boldsymbol{p} boldsymbol{x}^{2}+boldsymbol{q} boldsymbol{x}+boldsymbol{2} ) such that ( boldsymbol{alpha} boldsymbol{beta}+mathbf{1}= ) ( 0, ) find the value of ( 2 p+q+5 ) 10 144Which type of polynomial is ( (2+x) ? ) A. Linear Polynomial c. Cubic Polynomial D. None of above 9 145Q 1 ) If the polynomial ( a z^{3}+4 z^{2}+3 z- ) 4 and ( z^{3}-4 z+a ) leave the same remainder when divided by ( z-3, ) find the value of ( a ) ( Q ) 2) If both ( x-2 ) and ( x-frac{1}{2} ) are the factors of ( p x^{2}+5 x+r, ) show that ( p=r ) Q 3) Without actually division prove that ( 2 x^{4}-5 x^{3}+2 x^{2}-x+2 ) is divisible by ( x^{2}-3 x+2 ) 9 146Perform the following divisions and give the remainders. ( left(x^{2}+15 x+56right) div(x+8) ) 10 147Carry out the following division. ( 28 x^{4} div 56 x )10 148The integral part of ( (sqrt{2}+1)^{2} ) is: ( A cdot 2 ) B. 3 ( c cdot 4 ) D. 9 149Total number of pencils required are ( operatorname{given} operatorname{by} 4 x^{4}+2 x^{3}-2 x^{2}+62 x-66 . ) If each box contains ( x^{2}+2 x-3 ) pencils, then find the number of boxes to be purchased. 10 150Find all zeroes of polynomial ( 3 x^{4}+ ) ( 6 x^{3}-2 x^{2}-10 x-5 ) if two zeroes ( operatorname{are} sqrt{5 / 3} ) and ( -sqrt{5 / 3} ) 10 151The degree of the differential equation ( left(frac{d^{2} y}{d x^{2}}right)^{2}+left(frac{d y}{d x}right)^{2}=x sin left(frac{d^{2} y}{d x^{2}}right) ) ( A cdot 1 ) B. 2 ( c .3 ) D. None of these 10 152If ( x-frac{1}{x}=5, ) then ( x^{3}-frac{1}{x^{3}} ) equals A . 125 в. 130 c. 135 D. 140 9 153Evaluate each of the following using suitable identities: ( (99)^{3} ) 9 154State whether the statement is True or False. The square of ( (x+3 y) ) is equal to ( x^{2}+ ) ( 6 x y+9 y^{2} ) A. True B. False 9 155f ( 3 x+y+z=0 ) show that ( 27 x^{3}+ ) ( boldsymbol{y}^{3}+boldsymbol{z}^{3}=mathbf{9} boldsymbol{x} boldsymbol{y} boldsymbol{z} ) 9 156The value of ( frac{(0.31)^{3}-(0.21)^{3}}{0.0961+0.0651+0.0441} ) is ( mathbf{A} cdot mathbf{0} ) B. ( 0 . ) c. 0.2 D. 0.04 9 157Find the value of ( (a+b)^{2}-(a-b)^{2} ) ( mathbf{A} cdot a b ) в. ( 2 a ) ( c cdot 3 a b ) D. ( 4 a ) 9 158When ( 4 g^{3}-3 g^{2}+g+k ) is divided by ( g-2, ) the remainder is ( 27 . ) Find the value of ( k ) ( mathbf{A} cdot mathbf{3} ) B. 5 c. 8 D. 12 E . 10 9 159Solve: ( frac{x^{2}-(y-z)^{2}}{(x+z)^{2}-y^{2}}+frac{y^{2}-(x-z)^{2}}{(x+y)^{2}-z^{2}}+ ) ( frac{z^{2}-(x-y)^{2}}{(y+z)^{2}-x^{2}}= ) ( A ) B. ( c cdot 1 ) ( D ) 9 160( left(2 m^{2}-3 m+10right) div(m-5) )10 161If the polynomials ( 2 x^{3}+m x^{2}+3 x-5 ) and ( x^{3}+x^{2}-4 x+m ) leaves the same remainder when divided by ( x-2, ) then the value of ( m ) is A ( cdot-frac{3}{13} ) B. ( -frac{13}{3} ) ( c cdot frac{3}{13} ) D. ( frac{13}{3} ) 9 162When a positive integer ( y ) is divided by ( 47, ) the remainder is ( 11 . ) Therefore, when ( y^{2} ) is divided by ( 47, ) the remainder will be ( A cdot 7 ) B. 17 c. 27 D. 37 9 163Using remainder theorem, find the remainder when ( 2 x^{3}-3 x^{2}+4 x-5 ) is divided by ( boldsymbol{x}+mathbf{3} ) ( mathbf{A} cdot 204 ) B . -136 ( c .-98 ) D. 42 9 164Simplify: ( (8 a-5 b)^{2} ) 9 165Write the degree of the following polynomials. (i) ( boldsymbol{p}+boldsymbol{p}^{mathbf{3}}+boldsymbol{p}^{boldsymbol{7}} ) (ii) ( a+a^{3}-a^{0} ) (iii) ( boldsymbol{m}+boldsymbol{a}^{4} boldsymbol{m}+boldsymbol{a}^{5} boldsymbol{m}^{3}-boldsymbol{m}^{2}-boldsymbol{a}^{4} boldsymbol{m}^{boldsymbol{7}} ) 10 166If ( p-frac{1}{p}=4, ) find the value of ( p^{4}+frac{1}{p^{4}} ) A . 16 B. 18 c. 324 D. 322 9 167Use remainder theorem to find remainder when ( p(x) ) is divided by ( q(x) ) in the following questions: ( p(x)=x^{4}+ ) ( boldsymbol{x}^{3}+boldsymbol{x}^{2}-mathbf{5} boldsymbol{x}+mathbf{1}, boldsymbol{q}(boldsymbol{x})=boldsymbol{x}+mathbf{1} ) 9 168Evaluate: ( frac{xleft(8 x^{2}-32right)}{8 x(x-4)} ) ( mathbf{A} cdot x+2 ) B. ( x+4 ) c. ( frac{x^{2}-4}{x-4} ) D. ( x^{2}-4 ) 10 169( frac{6 a b-b^{2}+12 a c-2 b c}{b+2 c} )10 17051. 1fx–2, then the value of x’ (1) 15 (3) 14 (2) 2 (4) 11 9 171Find all zeroes of ( 2 x^{4}-3 x^{3}-3 x^{2}+ ) ( 6 x-2 ) if 2 zeroes ( sqrt{2} ) and ( -sqrt{2} ) 10 17269. If x + y = 15, then (x – 10)3 + (y-5) is (1) 25 (2) 125 (3) 625 (4) O 10 173Divide ( frac{x^{3}+x+1}{x^{2}-1} ) 10 174Simplify : ¡) ( frac{-14 x^{8} y^{5}+21 x^{10} y-28 x^{7} y^{6}}{7 x^{7} y^{8}} ) ii) ( frac{15 a^{4} x^{8}-30 a^{7} x^{5}-45 a^{6} x^{6}}{20 a^{14} x^{5}} ) iii) ( frac{-60 x^{4} a^{5}-75 x^{3} a^{6}+8 x^{5} a^{4}}{-20 x^{8} a^{4}} ) 10 175Shikhaa has Piggy bank. It is full of one-rupee and fifty- paise coins. It contains 3 times as many fifty paise coins as one rupee coins. The total amount of the money in the bank is 35. How many coins of each kind are there in the bank? (a) 14 (b) 16 (c) 48 (d) 42 10 176ff ( a+b-12=0 ) and ( a b=27 ), then find ( boldsymbol{a}^{boldsymbol{3}}+boldsymbol{b}^{boldsymbol{3}} ) 9 177( frac{sqrt{a^{2}-b^{2}}+a}{sqrt{a^{2}+b^{2}}+b} div frac{sqrt{a^{2}+b^{2}}-b}{a-sqrt{a^{2}-b^{2}}} )10 178Factorise the following ( : 27(a-b)^{3}+ ) ( (2 a-b)^{3}+(4 b-5 a)^{3} ) A ( cdot(a-b)(a-b)(4 b-8 a) ) В. ( 9(a-b)(2 a-b)(4 b-5 a) ) c. ( 9(a-b)(a-2 b)(4 b-a) ) D. ( 94-b)(2 a-b)(b-a) ) 9 17961. If x is a rational number and (x + 1)3 – (x – 13 (x + 1)2 –(x – 1)2 sum of numerator and denomi- nator of x is -101(1) 3 (2) 400 (3) 5 (4) 7100 9 180Find the values of polynomial ( 3 x^{3}- ) ( 4 x^{2}+7 x-5 ) when ( x=3 ) & ( x=-3 ) A. 61,-143 В. -61,142 c. 61,142 D. -61,-142 10 181What is the type of polynomial ( 11= ) ( -4 x^{2}-x^{3} ? ) A. Cubic c. Linear D. None of these 10 182If ( x=sqrt{6}+sqrt{5}, ) then ( x^{2}+frac{1}{x^{2}}-2=? ) A ( cdot 2 sqrt{6} ) B. ( 2 sqrt{5} ) ( c cdot 24 ) D. 20 9 183( (x+y-z)^{2}= ) A ( cdot x^{2}+y^{2}-z^{2}+2 x y+2 x z-2 y z ) B ( cdot x^{2}+y^{2}+z^{2}+2 x y-2 x z-2 y z ) C ( cdot x^{2}+y^{2}-z^{2}-2 x y+2 x z+2 y z ) D. None of the above 9 184If the polynomial ( 6 x^{4}+8 x^{3}-5 x^{2}+ ) ( a x+b ) is exactly divisible by the polynomial ( 2 x^{2}-5, ) then find the product of the values of ( a ) and ( b ) 10 185What is the degree of the following polynomial expression: ( frac{4}{3} x^{7}-3 x^{5}+2 x^{3}+1 ) A. 7 B. 4 ( c cdot 5 ) D. 10 186A polynomial of 4 is called a c. cubic polynomial D. none of these 9 187( frac{boldsymbol{x}-boldsymbol{y}}{sqrt{boldsymbol{x}}+sqrt{boldsymbol{y}}}=ldots ldots ) ( mathbf{A} cdot sqrt{x-y} ) B. ( sqrt{x}+sqrt{y} ) c. ( -(sqrt{x}+sqrt{y}) ) D. ( sqrt{x}-sqrt{y} ) 9 188Factorize ( (5 x-3 y)^{3}+(3 y-8 z)^{3}+ ) ( (8 z-5 x)^{3} ) 9 189What is the degree of the polynomial ( 2 a^{2}+4 b^{8} ? ) ( A cdot 2 ) B. 10 c. 8 D. 10 1902 2. Divide 34 into two parts in such a way that of one part is equal to of the other. (a) (c) 10 14 (b) 24 (d) 20 10 191The polynomial ( 4 x^{2}+2 x-2 ) is a A. Linear polynomial c. Cubic polynomial D. constant polynomial 10 192The zeros of the polynomial ( n^{3}+9 n^{2}+ ) ( 23 n+15 ) are ( a-d, a ) and ( a+d . ) What is the value of ‘a’? A . 4 B. -3 ( c .6 ) D. 3 9 A. has the highest power equal to 2 B. has the highest power equal to 1 C. has the highest power equal to 3 D. None of the above. 9 194State whether the statement is True or False. The cube of ( left(2 x+frac{1}{x}right) ) is equal to ( 8 x^{3}+ ) ( 12 x+frac{6}{x}+frac{1}{x^{3}} ) A. True B. False 9 195Write the polynomial in standard form and also write down their degree. ( 4 p+15 p^{6}-p^{5}+4 p^{2}+3 ) 9 196Simplify: ( left(left(3 x^{2}-2 a xright)+3 a^{2}right)^{3} ) 9 197Which of the following should be added to ( 9 x^{3}+6 x^{2}+x+2 ) so that the sum is divisible by ( (3 x+1) ? ) A . -4 B. – ( c cdot-2 ) D. – 9 198If ( boldsymbol{x}-frac{mathbf{1}}{boldsymbol{x}}=mathbf{9}, ) the value of ( boldsymbol{x}^{2}+frac{mathbf{1}}{boldsymbol{x}^{2}} ) is A. 83 B. 79 c. 11 D. 9 199Simplify the following into their lowest form: ( frac{6 x^{2}+9 x}{3 x^{2}-12 x} ) A ( cdot frac{2 x+3}{x-4} ) в. ( frac{2 x-3}{x-4} ) c. ( frac{2 x+3}{x+4} ) D. None of these 10 200Which of the following represents a linear polynomial? A ( cdot p(x)=3 x+4 ) В ( cdot p(x)=3 x^{3}+4 ) C ( cdot p(x)=4 x^{3}+3 ) D ( cdot p(x)=2 x^{2} ) 9 201Consider the polynomial ( frac{x^{3}+2 x+1}{5}-frac{7}{2} x^{2}-x^{6} ) Write the degree of the above polynomial ( mathbf{A} cdot mathbf{6} ) B. 3 c. 1 D. 9 202By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial: ( boldsymbol{x}^{4}+mathbf{1} ; boldsymbol{x}-mathbf{1} ) A ( cdot x^{3}+x^{2}+x+1,3 ) B. ( x^{3}+x^{2}+x+1,1 ) c. ( x^{3}+x^{2}+x+1,5 ) D. ( x^{3}+x^{2}+x+1,2 ) 10 203Find the remainder when the polynomial ( 4 y^{3}-3 y^{2}-5 y+1 ) is divided by ( 2 y+3 ) 10 204Find the roots of equation ( x^{2}-3 x- ) ( mathbf{1 0}=mathbf{0} ) 10 205Find the quotient and remainder when ( x^{5}-5 x^{4}+9 x^{3}-6 x^{2}-16 x+13 ) is divided by ( x^{2}-3 x+2 ) 10 206Divide the following and write your answer in lowest terms: ( frac{3 x^{2}-x-4}{9 x^{2}-16} div ) ( frac{4 x^{2}-4}{3 x^{2}-2 x-1} ) A ( frac{3 x+1}{4(3 x+4)} ) В. ( frac{3 x-1}{4(3 x+4)} ) c. ( frac{3 x+1}{4(3 x-4)} ) D. None of these 10 207( 25 x^{2}-left(x^{2}-36right)^{2}= ) A ( cdot(x-4)(x+4)(x+9)(x-9) ) B ( cdot(x-4)(4+x)(x+9)(9-x) ) ( mathbf{c} cdot(x+4)(x+4)(x-9)(x-9) ) D ( cdot(x-4)(4-x)(x+9)(9+x) ) 9 20855.- =? (0.87)4 -(0.13) 0.87% 0.87 +0.13×0.13 (1) 1 (2) 0.87 (3) 0.13 (4) 0.74 9 209A number was divided successively in order by 4,5 and 6 The remainders were respectively 2,3 and 4 Then find out the number9 210Choose the correct answer from the alternatives given If the expression ( 2 x^{2}+14 x-15 ) is divided by ( (x-4) ). then the remainder is A . 65 B. 0 ( c cdot 73 ) D. 45 10 211Which of the following is NOT a A ( cdot p(x)=x^{2}+4 x-16 ) ( mathbf{B} cdot p(y)=y^{2}+8 y-10 ) ( mathbf{c} cdot p(x)=73 x-84 ) ( mathbf{D} cdot p(y)=2 y^{2}-x^{2} ) 9 212Find the degree of given polynomial: ( 4 x^{3}-1 ) 9 213Divide ( 4left(2 x^{2}+5 x+3right) ) by ( 2(2 x+3) )10 214f ( p(x)=x^{42}-2 k ) is divided by ( (x+1) ) the remainder is ( 9, ) what is the value of k? 9 215Find the remainder when ( p(x)=2 x^{2}- ) ( 5 x-1 ) is divided by ( x-3 ) A . B. ( c cdot 2 ) D. 3 9 216State True Or False, if the following expression is a quadratic polynomial: ( boldsymbol{x}^{2}+boldsymbol{x} ) A . True B. False 9 217Multiply: ( left(m^{2}-5right) timesleft(m^{3}+2 m-2right) ) 10 218Solve ( left(4 x^{4}-5 x^{3}-7 x+1right) div(4 x-1) )10 219The remainder of ( frac{(5 m+1)(5 m+3)(5 m+4)}{5} ) is ( mathbf{A} cdot mathbf{1} ) B. 2 ( c cdot 3 ) ( D ) 9 220Find the degree of following polynomial ( 4 x-sqrt{5} ) A ( cdot frac{1}{2} ) B. ( c cdot 2 ) D. 9 221The polynomial ( 3 x-2 ) is a. A. Linear polynomial c. Cubic polynomial D. constant polynomial 10 222Divide ( p(x)=x^{3}-3 x^{2}+5 x- ) ( mathbf{3} ) by ( boldsymbol{g}(boldsymbol{x})=boldsymbol{x}^{2}-mathbf{2} ) 10 223In the following case, use the remainder theorem and find the remainder when ( boldsymbol{p}(boldsymbol{x}) ) is divided by ( boldsymbol{g}(boldsymbol{x}) cdot boldsymbol{p}(boldsymbol{x})=boldsymbol{4} boldsymbol{x}^{3}- ) ( 12 x^{2}+14 x-3 g(x)=2 x-1 ) 9 224Find and correct errors of the following mathematical expressions: ( frac{3 x}{3 x+2}=frac{1}{2} ) 10 225Factors of ( a^{2}+4 a+4 ) are: A ( cdot(a+2)^{2} ) в. ( (a+1)^{2} ) c. ( (a-2)^{2} ) D. ( (a-1)^{2} ) 9 226Find the remainder obtained on dividing ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{3}+mathbf{1} ) by ( boldsymbol{x}+mathbf{1} ) 10 227Prove the following identities: [ begin{array}{l} frac{boldsymbol{a}^{3}(boldsymbol{b}+boldsymbol{c})}{(boldsymbol{a}-boldsymbol{b})(boldsymbol{a}-boldsymbol{c})}+frac{boldsymbol{b}^{3}(boldsymbol{c}+boldsymbol{a})}{(boldsymbol{b}-boldsymbol{c})(boldsymbol{b}-boldsymbol{a})}+ \ frac{boldsymbol{a}^{3}(boldsymbol{a}+boldsymbol{b})}{(boldsymbol{c}-boldsymbol{a})(boldsymbol{c}-boldsymbol{b})}=boldsymbol{b} boldsymbol{c}+boldsymbol{c} boldsymbol{a}+boldsymbol{a} boldsymbol{b} end{array} ] 9 228When the polynomial ( a^{3}+2 a^{2}- ) ( 5 a x-7 ) is divided by ( a+1, ) the remainder is ( R_{1} . ) If ( R_{1}=14 ), find the value of ( boldsymbol{x} ) ( mathbf{A} cdot mathbf{1} ) B . 2 ( c .3 ) D. 9 229If ( boldsymbol{a}+boldsymbol{b}=mathbf{7} boldsymbol{a} boldsymbol{n} boldsymbol{d} boldsymbol{a} boldsymbol{b}=boldsymbol{6}, boldsymbol{f} boldsymbol{i} boldsymbol{n} boldsymbol{d} boldsymbol{a}^{2}-boldsymbol{b}^{2} ) ( mathbf{A} cdot pm 35 ) B. ±12 ( c .pm 22 ) ( mathrm{D} cdot pm 39 ) 9 23068. If the sum of and its recipro- cal is 1 and a # O, b# 0, then the value of a + b3 is (1) 2 (2) – 1 0 (3) (4) 1 9 231Find the value of ( 52^{2} ) using standard identity. A . 2604 в. 2704 c. 2804 D. 2904 9 23257. If 3 (a + b + c) = (a + b + c)2 and a, b, care non-zero real num- bers, then (1) a + b = c (2) a + c = b (3) b + c = a (4) a = b = c 9 233Find the zeroes of the polynomial in each of the following: ( h(y)=2 y ) ( mathbf{A} cdot mathbf{0} ) B. ( 2 y ) c. ( y ) D. – 10 234If a polynomial ( a z^{3}+4 z^{2}+2 z-4 ) and ( z^{2}+4 z+a ) leave the same remainder divide by ( (z-3) . ) Find the value of a. A . 1 в. ( frac{-17}{26} ) c. -1 D. 17 9 23563. If a = 2.234, b= 3.121 and c = -5.355, then the value of a+b3 + c3-3 abc is (1)-1 (2) O (3) 1 (4) 2Y 9 236Find the value of ( 99 times 101 ) using standard identity A . 9999 B. 9989 ( mathrm{c} .9979 ) D. 1009 9 237If ( x+frac{1}{x}=3, ) then ( x^{4}+frac{1}{x^{4}}= ) A. 81 B . 23 c. 25 D. 47 9 238Find the value of ( left(frac{a}{b-1}right)+ ) ( left(frac{a}{b-1}right)^{2}+left(frac{a}{b-1}right)^{3} ) if ( a+2 b=2 ) A . -6 B. 8 c. 10 D. -12 E . 14 10 239Find the degree of the polynomial ( left(x^{2}+right. ) ( mathbf{9})left(mathbf{5}-boldsymbol{x}^{2}right) ) 9 240Factorise: ( (boldsymbol{a}+mathbf{2 b}-mathbf{3 c})^{mathbf{3}}-boldsymbol{a}^{mathbf{3}}-mathbf{8 b}^{mathbf{3}}+ ) ( mathbf{2 7 c}^{mathbf{3}} ) 9 241If ( x^{2}+frac{1}{x^{2}}=83 . ) Find the value of ( x^{3}- ) ( frac{1}{m^{3}} ) 9 242Perform the division. ( (10 x-25) div 5 ) 10 243Which of the following is a rational function? A ( cdot frac{1}{3} sqrt{4 x^{3}+4 x+7} ) в. ( frac{3 x^{3}-7 x+1}{x-2}, x neq 2 ) c. ( frac{3 x^{5}+5 x^{3}+2 x+7}{x^{3 / 2}}, x>0 ) D. ( frac{sqrt{1+x}}{2+5 x}, x neq-2 / 5 ) 9 244Expand ( left[boldsymbol{x}-frac{boldsymbol{2}}{boldsymbol{3}} boldsymbol{y}right]^{boldsymbol{3}} )9 245What should be added to ( 1+2 x-3 x^{2} ) to get ( x^{2}-x-1 ? ) 9 246For polynomial ( boldsymbol{P}(boldsymbol{x})=mathbf{6} boldsymbol{x}^{mathbf{3}}+mathbf{2} mathbf{9} boldsymbol{x}^{mathbf{2}}+ ) ( mathbf{4 4} boldsymbol{x}+mathbf{2 1}, ) find ( mathbf{P}(-mathbf{2}) ) 9 247The value of the expression ( frac{left(x^{2}-y^{2}right)^{3}+left(y^{2}-z^{2}right)^{3}+left(z^{2}-x^{2}right)^{3}}{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}} ) is A ( cdotleft(x^{2}-y^{2}right)left(y^{2}-z^{2}right)left(z^{2}-x^{2}right) ) B. ( 3(x-y)(y-z)(z-x) ) C. ( (x+y)(y+z)(z+x) ) D. ( 3(z+y)(y+z)(z+x) ) 9 248f ( p=5+2 sqrt{6} ) and ( q=frac{1}{p}, ) find ( p^{2}+q^{2} ) is : A .49 B. 98 ( c .100 ) D. None of these 9 249Find the Quotient and the Remainder when the first polynomial is divided by the second. ( left(6 x^{2}-31 x+47right) ) by ( (2 x-5) ) A. Quotient ( =3 x-8, ) Remainder ( =7 ) B. Quotient = 3x + 8, Remainder = 7 c. Quotient ( =-3 x-8, ) Remainder ( =7 ) D. Quotient ( =-3 x+8, ) Remainder ( =7 ) 10 250The factors of ( 1-p^{3} ) are A ( cdot(1-p)left(1+p+p^{2}right) ) B ( cdot(1+p)left(1-p-p^{2}right) ) C . ( (1+p)left(1+p^{2}right) ) D. ( (1+p)left(1-p^{2}right) ) 9 251If the quotient obtained on dividing ( x^{4}+10 x^{3}+35 x^{2}+50 x+29 ) by ( (x+ ) 4) is ( x^{3}-a x^{2}+b x+6 ) then find ( a ) and b. Also find the remainder 9 252If ( boldsymbol{y}=mathbf{3}, ) then ( boldsymbol{y}^{3}left(boldsymbol{y}^{3}-boldsymbol{y}right)= ) A. 300 в. 459 ( c cdot 648 ) D. 999 E . 1099 10 253Evaluate the following: ( (7 x-2 y)^{2} ) ( (3 x+7 y)^{2} ) 9 254What is the degree of the polynomial ( boldsymbol{p}(boldsymbol{x})=mathbf{5} boldsymbol{x}^{3}-boldsymbol{8} boldsymbol{x}^{2}+boldsymbol{4} boldsymbol{x} ? ) ( A cdot 3 ) B . 2 ( c cdot 1 ) D. 9 255( f(alpha, beta, gamma ) are the zeroes of the cubic polynomial ( x^{3}+4 x+2, ) then find the value of: ( frac{1}{alpha+beta}+frac{1}{beta+gamma}+frac{1}{gamma+alpha} ) 10 256If the sum and difference of two numbers are 20 and 8 respectively then the difference of their squares is : A ( cdot 12 ) B. 28 ( c cdot 160 ) D. 180 9 257( mu^{2}+frac{1}{mu^{2}}=79 ) find ( mu+frac{1}{mu} ) 9 258The difference of the degrees of the polynomials ( 3 x^{2} y^{3}+5 x y^{7}-x^{6} ) and ( 3 x^{5}-4 x^{3}+2 ) is ( A cdot 2 ) B. 3 ( c ) D. None 10 259( operatorname{Let} fleft(x+frac{1}{x}right)=x^{2}+frac{1}{x^{2}}(x neq 0), ) then ( boldsymbol{f}(boldsymbol{x}) ) equals: A ( cdot x^{2}-2 ) B . ( x^{2}-1 ) ( mathbf{c} cdot x^{2} ) D. None of these 9 260Write a polynomial of degree 5 using variable ( x )10 261Find the product ( (3+sqrt{2})(3-sqrt{2}) )9 26258. If a2 + b2+ c2-ab-bc- ca = 0 then (1) a = b c (2) a = b = c (3) a +b= c (4) a= buc 9 263What is the degree of the given monomial ( -11 y^{2} z^{2} ? ) ( mathbf{A} cdot mathbf{0} ) B . 2 ( c cdot 4 ) D. 10 264Using remainder theorem, find the reminder when ( x^{3}-a x^{2}+2 x-a ) is divided by ( boldsymbol{x}-boldsymbol{a} ) ( A cdot a ) B. ( a+2 ) ( mathbf{c} cdot a+1 ) D. ( a-2 ) 9 265( a^{12}-1 ) can be factorised as: ( mathbf{A} cdot(a-1)(a-2)(a-3)(a-4) ) B ( cdot(a-1)left(a^{2}+a+1right)(a+1)left(a^{2}+a+1right) ) C ( cdotleft(a^{2}+a+1right)left(a^{2}-a+1right) ) D ( cdot(a-1)left(a^{2}+a+1right)(a+1)left(a^{2}-a+1right)left[left(a^{2}+1right)left(a^{4}-right.right. ) ( left.a^{2}+1right) ) 9 266The value of ( 7 x-42 ) is A. ( 7(x-6) ) B. ( 7(x+6) ) ( c cdot-7(x-6) ) D. ( -7(x+6) ) 9 267Verify whether ( x=3 ) is a zero of the polynomial ( boldsymbol{x}^{2}+mathbf{2} boldsymbol{x}-mathbf{1 5} ) 10 268Two number differ by 5. If their product is ( 336, ) then the sum of the two numbers is : A . 21 B . 28 ( c .37 ) D. 51 9 269Evaluate ( (4 a+3 b)^{2}-(4 a-3 b)^{2}+ ) ( 48 a b ) A . ( 76 a b ) B. ( 96 a b ) ( c .46 a b ) D. ( 106 a b ) 9 270Write the degree of the following polynomial : ( 5 x^{2} y z^{3}+x y^{4} z^{2} ) ( A cdot 3 ) B. 2 ( c cdot 7 ) D. 10 271Simplify: ( mathbf{2 0}(boldsymbol{y}+mathbf{4})left(boldsymbol{y}^{2}+mathbf{5} boldsymbol{y}+mathbf{3}right) div mathbf{5}(boldsymbol{y}+mathbf{4}) ) A ( cdot 5left(y^{2}+5 y+3right) ) B ( cdot 4left(y^{2}-5 y+3right) ) ( mathbf{c} cdot 4left(y^{2}+5 y-3right) ) D. ( 4left(y^{2}+5 y+3right) ) 10 272Find the product of ( (a-3)(a-5)(a-7) )9 273If ( boldsymbol{x}=mathbf{2}, boldsymbol{x}=mathbf{0} ) are roots of the polynomials ( f(x)=2 x^{3}-5 x^{2}+a x+ ) ( b, ) then find the values of ( a ) and ( b ) A ( cdot a=2, b=0 ) В. ( a=7, b=0 ) c. ( a=3, b=0 ) D. ( a=1, b=0 ) 10 274Find the degree of the given algebraic expression ( 2 y^{2} z+10 y z ) ( mathbf{A} cdot mathbf{1} ) B . 2 ( c .3 ) D. 10 9 275What is the quotient when ( left(x^{3}+8right) ) is divided by ( left(x^{2}-2 x+4right) ? ) ( mathbf{A} cdot x-2 ) B. ( x+2 ) c. ( x+1 ) D. ( x-1 ) 10 276Assertion : Vx+9x = 3 is not a linear equation. Reason: Linear equation involves only linear polynomials. 10 277Find the remainder when ( 10 x-4 x^{2}-3 ) is divided by ( x+2 ) using remainder theorem. A . -60 B. 39 ( c .-39 ) D. None of these 9 278The degree of ( 3 x^{2} y^{4} z^{6} ) is ( A cdot 2 ) B. 12 ( c cdot 4 ) D. 6 10 279Factorize: ( (x-y)^{3}-8 x^{3} ) ( mathbf{A} cdot-2(x+y)left(7 x^{2}-4 x y+y^{2}right) ) B . ( -(x-y)left(x^{2}-4 x y+6 y^{2}right) ) ( mathbf{c} cdot-(x+2 y)left(x^{2}-4 x y+y^{2}right) ) ( mathbf{D} cdot-(x+y)left(7 x^{2}-4 x y+y^{2}right) ) 9 280If two of the zeros of equation ( 2 x^{4}- ) ( 3 x^{3}-3 x^{2}+6 x-2, ) are ( sqrt{2} ) and ( -sqrt{2} ) and other two are ( ^{prime} a^{prime} ) and ( ^{prime} b^{prime}, ) then ( a b= ) 0. ( P ), then the value of ( p= ) 10 281Find the degree of the following polynomial ( boldsymbol{x}^{2}-mathbf{9} boldsymbol{x}+mathbf{2 0} ) 10 282If ( a=2 overline{3}+2 overline{3}, ) then A. ( a^{3}-6 a-6=0 ) В . ( a^{3}-6 a+6=0 ) c. ( a^{3}+6 a-6=0 ) D. ( a^{3}+6 a+6=0 ) 9 283Find the value of ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{2}-boldsymbol{x}+mathbf{1} ) at ( boldsymbol{x}=mathbf{2} ) ( A cdot 3 ) B. – – ( c ) D. – 10 284Perform the division: ( 2 x^{2}+2 x+11 ) by ( boldsymbol{x}+mathbf{3} ) 10 285Using the identity ( (boldsymbol{a}-boldsymbol{b})^{2}=boldsymbol{a}^{2}- ) ( 2 a b+b^{2} ) compute ( (5 a-4 b)^{2} ) 9 286( boldsymbol{p}(boldsymbol{y})=mathbf{5} boldsymbol{y}^{3}-boldsymbol{2} boldsymbol{y}^{2}+boldsymbol{y}+mathbf{1 0} ) is a polynomial in ( y ) of degree ( A cdot 0 ) B. ( c cdot 2 ) D. 3 9 287Divide ( (10 x-25) div(2 x-5) )10 288Factorise: ( boldsymbol{p}^{3}(boldsymbol{q}-boldsymbol{r})^{3}+boldsymbol{q}^{3}(boldsymbol{r}-boldsymbol{p})^{3}+boldsymbol{r}^{3}(boldsymbol{p}-boldsymbol{q})^{3} ) 9 289What is ( frac{x^{2}-3 x+2}{x^{2}-5 x+6} div frac{x^{2}-5 x+4}{x^{2}-7 x+12} ) equal to A. ( frac{x+3}{x-3} ) B. c. ( frac{x+1}{x-1} ) D. 10 290Divide ( boldsymbol{a}+boldsymbol{b} ) by ( boldsymbol{a}^{1 / 3}+boldsymbol{b}^{1 / 3} )10 291If ( a^{3}-3 a^{2} b+3 a b^{2}-b^{3} ) is divided by ( (a-b), ) then the remainder is A ( cdot a^{2}-a b+b^{2} ) B ( cdot a^{2}+a b+b^{2} ) ( c ) D. 10 292Solve: ( 8 l^{3}-36 l^{2} m+54 l m^{2}-27 m^{3} ) 10 293Divide the following and write your answer in lowest terms: ( frac{x}{x+1} div ) ( frac{x^{2}}{x^{2}-1} ) A ( cdot frac{x-1}{x} ) в. ( frac{x+1}{x} ) c. ( frac{x-1}{x^{2}} ) D. None of these 10 294Using appropriate identity, factorise the following: ( (1) 49 a^{2}+70 a b+25 b^{2} ) ( (2) 9 a^{2}-30 a b+25 b^{2} ) 9 295Factorise and then divide the given algebraic expressions. 1. ( left(12 r^{2}+8 r^{2}-4 r^{2}right) b yleft(-4 r^{2}right) ) 2. ( left(frac{4}{9} x^{2}-49 z^{2}right) b y(2 x+21 z) ) ( 3 cdotleft(5 a^{2} b^{2}+15 a^{2} b^{2}-20 a^{4} bright) b y 25 a b^{2} ) 4. ( left(49 a^{2}-56 aright) b y(21 a-24) ) 10 296Factorize: ( 4 a^{2}-12 a+9-49 b^{2} ) A ( cdot(2 a+7 b-3)(8 a-7 b-3) ) в. ( (2 a-7 b-3)(8 a-7 b+3) ) c. ( (2 a-7 b-3)(2 a-7 b+3) ) D. ( (2 a+7 b-3)(2 a-7 b-3) ) 9 297Find the degree of the polynomial given below: ( boldsymbol{x}^{5}-boldsymbol{x}^{4}+mathbf{3} ) 9 298Find the degree of the following polynomial ( 2 x+4+6 x^{2} )10 299Decide using factor theorem, whether ( (x-2) ) is a factor of ( x^{3}-4 x^{2}-4 ) 9 300Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm. ( x^{3}- ) ( mathbf{3} boldsymbol{x}+mathbf{1}, boldsymbol{x}^{mathbf{5}}-mathbf{4} boldsymbol{x}^{mathbf{3}}+boldsymbol{x}^{mathbf{2}}+mathbf{3} boldsymbol{x}+mathbf{1} ) A. Yes B. No c. Ambiguous D. Data insufficient 10 301If ( a+b+2 c=0, ) then the value of ( a^{3}+ ) ( b^{3}+8 c^{3} ) is equal to A . ( 3 a b c ) B. ( 4 a b c ) ( c cdot a b c ) D. ( 6 a b c ) 9 302( left(5 x^{2}-9 x+3right),left(10 x^{4}+17 x^{3}-right. ) ( left.62 x^{2}+30 x-3right) ) 10 303Expand ( (k+4)^{3} )10 304The polynomial ( boldsymbol{p}(boldsymbol{x})=boldsymbol{a} boldsymbol{x}^{3}+boldsymbol{4} boldsymbol{x}^{2}+ ) ( 3 x-4 ) and ( q(x)=x^{3}-4 x+a ) leave same remainder when divided by ( (x-3) . ) Find ( a ) and hence find the remainder when ( p(x) ) is divided by ( (x-2) ) 9 305Which of the following is a cubic polynomial? A. ( x^{3}+3 x^{2}-4 x+3 ) B. ( x^{2}+4 x-7 ) ( c cdot 3 x^{2}+4 ) D. ( 3left(x^{2}+x+1right) ) 10 306If ( a^{2}+b^{2}=34 ) and ( a b=12 ; ) find ( 7(a-b)^{2}-2(a+b)^{2} ) A . 48 B. -46 c. -45 D. 46 9 307What is the degree of the given monomial ( -x y^{2} ? ) ( A cdot 2 ) B. 3 ( c cdot 4 ) D. none 9 308If ( a^{2}+b^{2}=34 ) and ( a b=12 ; ) find : ( 7(a-b)^{2}-2(a+b)^{2} ) A. 186 B . 46 c. -46 D. -186 9 309Use the identity ( (x+a)(x+b)=x^{2}+ ) ( (a+b) x+a b ) to find the following product. ( left(2 a^{2}+9right)left(2 a^{2}+5right) ) 9 310Evaluate using expansion of ( (a+b)^{2} ) or ( (a-b)^{2}: ) ( (20.7)^{2} ) is 428.49 If true then enter 1 and if false then enter 0 9 311Write the degree of the following polynomial. ( 12-x+4 x^{3} )10 312( 1+5 x ) is a quadratic polynomial A. True B. False 9 313Verify ( t=1 ) is a zero of the polynomial ( 2 t^{3}-3 t^{2}+7 t-6 ) 10 314Assertion Degree of a zero polynomial is not defined. Reason Degree of a non-zero constant polynomial is 0 A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct 10 315For a polynomial, dividend is ( x^{4}+4 x- ) ( 2 x^{2}+x^{3}-10, ) quotient is ( x^{2}+3 x- ) ( 3 x^{2}+4 x+12 ) and remainder is 14 then divisor is equal to A. ( x^{2}+2 ) B . ( x^{2}-2 ) c. ( x+2 ) D. None of these 10 316Evaluate ( frac{(-40+80 a)}{8 a} )10 317UCHUU U UU ULULLAH 3. a 8 6x -2 Assertion : 4x + = is a linear equation. Reason: Solution of the equation is-2. 10 318Which of the following is NOT a ( mathbf{A} cdot p(x)=p x^{2}+q x+r ) B ( cdot p(x)=a x^{2}+b x+c ) ( mathbf{c} cdot p(x)=x cdot x^{2}+x cdot x+x ) D ( cdot p(x)=m x^{2}+n x+l ) 10 319Expand ( (boldsymbol{p}-boldsymbol{2} boldsymbol{q}+boldsymbol{r})^{2} ) ( mathbf{A} cdot p^{2}+4 q^{2}+r^{2}-4 p q-4 q r+2 r p ) B ( cdot p^{2}-4 q^{2}+r^{2}-4 p q-4 q r+2 r p ) C ( cdot p^{2}+4 q^{2}-r^{2}-4 p q-4 q r+2 r p ) D. None of these 9 320Simplify ( -3 x y z div z^{2} )10 321Evaluate ( left(x^{2}-8 x+12right) div(x-6) ) A. ( x-2 ) B. ( x+2 ) c. ( x ) D. None 10 322Factorize ( 2 x^{2}+4 x+2=0 ) 9 323Find the cube of ( 3 a-2 b ) B . ( a^{3}-54 a^{2} b+6 a b^{2}-b^{3} ) ( mathbf{c} cdot 27 a^{3}-54 a^{2} b+6 a b^{2}-8 b^{3} ) D. ( a^{3}-54 a^{2} b-36 a b^{2}-8 b^{3} ) 9 324If ( x^{2}-(a+b) x+a b=0, ) then the value of ( (x-a)^{2}+(x-b)^{2} ) is ( mathbf{A} cdot a^{2}+b^{2} ) B. ( (a+b)^{2} ) c. ( (a-b)^{2} ) D. ( a^{2}-b^{2} ) 9 325Find the product ( :(x-3)(x+3)left(x^{2}+right. ) ( mathbf{9} ) 9 326If ( a^{2}+b^{2}=29 ) and ( a b=10, ) then find ( a-b ) A . 10 B. 3 ( c cdot 9 ) D. 19 9 327The degree of a polynomial ( 2 x^{5}-5 x^{3}- ) ( 10 x+9 ) is A . 5 B. 3 c. 1 D. 9 328f ( x+a ) is one of the factors of ( p(x)= ) ( 2 x^{2}+2 a x+5 x+10, ) then find ( a ) 9 329The degree of the polynomial ( 2 x-1 ) is A. 0 B. ( frac{1}{2} ) ( c cdot-1 ) D. 10 330( x^{2}+frac{1}{x^{2}}=6 ) Find (i) ( x^{3}-frac{1}{x^{3}} ) (i) ( x^{6}+frac{1}{x^{6}} ) 9 331If the degree of ( 12 x^{3} y^{8} z^{n} ) is ( 14, ) then ( boldsymbol{n}= ) 9 332( 4 r^{3} ) is a quadratic polynomial A. True B. False 9 333Let ( Q(x) ) denotes the quotient which results from the division of the polynomial ( x^{5}+3 x^{4}-x^{3}+8 x^{2}-x+ ) ( 8 mathrm{by} x^{2}+1 ) The sum of the square of the coefficient of ( Q(x) ) is A . 36 B. 37 c. 38 D. 39 10 33454. ( + ) simplifies to9 3354. Let a>0, b>0 and c>0. Then the roots of the equation ax2 + bx+c=0 (1979) (a) are real and negative (b) have negative real parts (c) both (a) and (b) (d) none of these 10 33661. If x + y = z, then the expression x + y – 2+ 3xyz will be equal to : (1) O (2) 3xyz (3) -3xyz (4) z 9 337If ( x^{3}+6 x^{2}+4 x+k ) is exactly divisible by ( x+2, ) then ( k ) is equal to A . – 6 B. – – ( c cdot-8 ) D. -10 9 338Classify the following as linear, quadratic and cubic polynomials ( x-1 ) 10 339If ( boldsymbol{x}+frac{mathbf{1}}{boldsymbol{x}}=mathbf{7}, ) find ( boldsymbol{x}^{mathbf{3}}+frac{mathbf{1}}{boldsymbol{x}^{mathbf{3}}} )9 340Write the degree of each of the following polynomials: (i) ( 5 x^{3}+4 x^{2}+7 x ) (ii) ( 4-y^{2} ) (iii) ( 5 t-sqrt{7} ) (iv) 3 10 341Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer (i) ( 4 x^{2}-3 x+7 ) (ii) ( y^{2}+sqrt{2} ) (iii) ( 3 sqrt{t}+t sqrt{2} ) (iv) ( boldsymbol{y}+frac{mathbf{2}}{boldsymbol{y}} ) (v) ( x^{10}+y^{3}+t^{50} ) 10 34264. If x = 2 – 21/3 + 22/3, then the value of x-6×2 + 18x + 18 is (1) 22 (2) 33 (3) 40 (4) 45 (1) 22 (2) 23 9 343Show that: ( (a-b)(a+b)+(b-c)(b+c)+(c- ) ( boldsymbol{a})(boldsymbol{c}+boldsymbol{a})=mathbf{0} ) 9 344Test whether ( x^{5}+5 x^{3}+3 x ) is divisible by ( (x-1) ) 9 345Find the degree of the following polynomial ( x^{3}+2 x^{2}-5 x-6 )9 346Shew that ( left(a^{2}+b^{2}+c^{2}-b c-c a-right. ) ( boldsymbol{a b})left(boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{z}^{2}-boldsymbol{z}^{2}-boldsymbol{y} boldsymbol{z}-boldsymbol{z} boldsymbol{x}-boldsymbol{x} boldsymbol{y}right) ) may be put into the form ( A^{2}+B^{2}+ ) ( C^{2}-B C-C A-A B ) 9 347Find the degree of the given polynomials. ( x^{0} ) 9 348Find the quotient the and remainder of the following division: ( left(2 x^{2}-3 x-14right) div(x+2) ) 10 349If ( f(x)=a x^{2}+b x+c ) is divided by ( (b x+c), ) then the remainder is: This question has multiple correct options A ( cdot frac{a c^{2}}{b^{2}} ) B. ( frac{a c^{2}}{b^{2}}+2 c ) c. ( fleft(-frac{c}{b}right) ) D. ( frac{a c^{2}+2 b^{2} c}{b^{2}} ) 9 350Use remainder theorem to find remainder when ( p(x) ) is divided by ( q(x) ) in the following questions: ( boldsymbol{p}(boldsymbol{x})=mathbf{2} boldsymbol{x}^{2}-mathbf{5} boldsymbol{x}+mathbf{7}, boldsymbol{q}(boldsymbol{x})=boldsymbol{x}-mathbf{1} ) 9 351Find the cube of 1089 352If the quotient on dividing, ( 8 x^{4}-2 x^{2}+ ) ( 6 x-7 ) by ( 2 x+1 ) is ( 4 x^{3}+p x^{2}-q x+3 ) then find ( boldsymbol{q}-boldsymbol{p} ) 10 353Divide the polynomial ( p(x) ) by the polynomial ( g(x) ) and find the quotient and remainder in each of the following. ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{3}-boldsymbol{3} boldsymbol{x}^{2}+mathbf{5} boldsymbol{x}-boldsymbol{3} quad boldsymbol{g}(boldsymbol{x})= ) ( x^{2}-2 ) 10 354Without finding the cubes, factorise the following: ( (x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3} ) 9 355If ( boldsymbol{x}+boldsymbol{y}+boldsymbol{z}=mathbf{0} ) then what is the value of ( frac{x^{2}}{y z}+frac{y^{2}}{z x}+frac{z^{2}}{x y} ? ) A ( cdot(x y z)^{2} ) B. ( x^{2}+y^{2}+z^{2} ) c. 9 D. 3 9 356Following polynomars write the degree of each. 1). ( 6 x^{2}+x+1 ) 2). ( y^{9}-3 y^{7}+frac{3}{2} y^{2}+4 ) 3). ( 2 x+1 ) 4). ( 5 t-sqrt{11} ) 9 357Factorise the expression: ( (x-2 y+ ) ( 3 z)^{2} ) ( mathbf{A} cdot x^{2}+y^{2}+9 z^{2}-4 x y+12 y z+6 z x ) B . ( x^{2}+y^{2}+9 z^{2}-4 y-12 y z+6 z x ) ( mathbf{c} cdot x^{2}+4 y^{2}+9 z^{2}-4 x y-12 y z+5 z x ) D. ( x^{2}+4 y^{2}+9 z^{2}-4 x y-12 y z+6 z x ) 9 358Factorise : ( (2 a+b)^{3}-(a+3 b)^{3} ) A ( cdot(a+b)left(7 a^{2}+13 a b+13 b^{2}right) ) B . ( (a-2 b)left(7 a^{2}+17 a b+13 b^{2}right) ) C ( cdot(a-b)left(7 a^{2}+13 a b+17 b^{2}right) ) D. ( (2 a-b)left(7 a^{2}+17 a b+17 b^{2}right) ) 9 359If ( boldsymbol{x}^{mathbf{3}}-boldsymbol{a} boldsymbol{x}^{mathbf{2}}+boldsymbol{b} boldsymbol{x}- ) ( boldsymbol{6} ) is exactly divisible by ( boldsymbol{x}^{2}- ) ( mathbf{5 x + 6 . t h e n} frac{a}{b} ) is A. an integer B. an irrational number ( c cdot frac{6}{11} ) D. 9 360Assertion :=x+4= -x is a linear equation. Reason: Four-fifth of a number is more than three fourth of the number by 4 is a statement of linear equation. 10 361Simplify: ( (boldsymbol{x}+boldsymbol{y})^{2}+(boldsymbol{x}-boldsymbol{y})^{2} )9 362If ( sqrt{boldsymbol{x}}-sqrt{mathbf{1 2}}=sqrt{mathbf{4}}-sqrt{boldsymbol{x}}, ) then find ( boldsymbol{x} ) A. ( 1+sqrt{3} ) B. ( 2+2 sqrt{3} ) c. ( 4+sqrt{3} ) D. ( 4+2 sqrt{3} ) 9 363The degree of constant polynomial is A . B. 2 c. 0 ( D ) 10 36480. If x + = 1, then the value of then the value of is x²+x+2 (1-x) (1) 1 (3) 2 (2) -1 (4) -2 9 365On dividing the polynomial ( 9 x^{4}- ) ( 4 x^{2}+5 ) by another polynomial ( 3 x^{2}+ ) ( x-1, ) the remainder comes out to be ( a x-b, ) Find ( ^{prime} a^{prime} ) and ( ^{prime} b^{prime} ) 9 366( left(frac{4}{3} x-frac{3}{4} yright)^{3} ) is equal to A ( cdot frac{64}{27} x^{3}+frac{27}{64} y^{3}+4 x^{2} y+frac{9}{4} x y^{2} ) B. ( frac{64}{27} x^{3}+frac{27}{64} y^{3}-4 x^{2} y-frac{9}{4} x y^{2} ) C ( frac{64}{27} x^{3}-frac{27}{64} y^{3}-4 x^{2} y+frac{9}{4} x y^{2} ) D. ( frac{64}{27} x^{3}-frac{27}{64} y^{3}+4 x^{2} y-frac{9}{4} x y^{2} ) 9 367Evaluate the following (using identities): ( (11)^{3} ) 9 368OLOT 57. If 2x + 3y = and xy = then the value of 8×3 + 27yº is (1) 583 (2) 583 (3) 187 (4) 671 9 369Evaluate: ( left(y^{3}-216right) div(y-6) ) 10 370Find the quotient and remainder on dividing ( p(x) ) by ( g(x) ) in the following case, without actual division. ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{3}+boldsymbol{4} boldsymbol{x}^{2}-boldsymbol{6} boldsymbol{x}+boldsymbol{2} ; boldsymbol{g}(boldsymbol{x})=boldsymbol{x}- ) 3 10 371Factors of ( (boldsymbol{a}+boldsymbol{b})^{3}-(boldsymbol{a}-boldsymbol{b})^{3} ) are A ( cdot 2 a bleft(3 a^{2}+b^{2}right) ) B . ( a bleft(3 a^{2}+b^{2}right) ) c. ( 2 bleft(3 a^{2}+b^{2}right) ) D. ( 3 a^{2}+b^{20} ) 9 372Calculate ( left(frac{mathbf{3} boldsymbol{x}}{boldsymbol{x}+mathbf{5}}right) divleft(frac{mathbf{6}}{mathbf{4} boldsymbol{x}+mathbf{2 0}}right) ) given that ( boldsymbol{x} neq-mathbf{5} ) ( mathbf{A} cdot 2 x ) в. ( frac{x}{2} ) c. ( frac{9 x}{2} ) D. ( 2 x+4 ) 10 373Use suitable identities to find the product of ( (x+5)(x+2) ) 9 374Use Remainder theorem to factorize the following polynomial. ( 2 x^{3}+3 x^{2}-9 x- ) ( mathbf{1 0} ) 9 375Evaluate ( left(frac{7}{8} x+frac{4}{5} yright)^{2} ) A. ( frac{49}{64} x^{2}+frac{6}{25} y^{2}+frac{7}{5} x y ) B ( cdot frac{18}{77} x^{2}+frac{16}{5} y^{2}+frac{1}{5} x y ) c. ( frac{13}{22} x^{2}+frac{16}{25} y^{2}+frac{1}{5} x y ) D. ( frac{49}{64} x^{2}+frac{16}{25} y^{2}+frac{7}{5} x y ) 9 376Read the following statements (a) ( x^{2}-5 x+sqrt{2} ) is a polynomial in ( x ) (b) ( 4 x^{2}-3 sqrt{x}+7 ) is not a polynomial in ( x ) (c) ( frac{x^{2}+2 x+5}{x+3}(x neq-3) ) is a rational expression. (d) ( frac{x^{3}-5 sqrt{x}-1}{x^{2}+x+4} ) is not a rational expression. Correct options is – A . acd B. abc ( c cdot a b d ) D. abcd 9 377Factorise the expressions and divide them as directed.5pq ( left(p^{2}-q^{2}right) div ) ( mathbf{2} p(boldsymbol{p}+boldsymbol{q}) ) A ( cdot frac{5}{2} q(p-q) ) в. ( frac{3}{2} p q ) c. ( frac{5}{2} q(p+q) ) D ( cdot frac{2}{5} q(p q) ) 10 378What must be added to ( x^{3}-3 x^{2}- ) ( 12 x+19 ) so that the result is exactly divisible by ( x^{2}+x-6 ? ) A ( .2 x-5 ) B. ( 2 x+5 ) c. ( -2 x-5 ) D. ( x+5 ) 9 379Say true or false: The degree of the sum of two polynomials each of degree 5 is always ( mathbf{5} ) A . True B. False 9 380Find the volume of the cuboid with dimensions ( (x-1),(x-2) ) and ( (x- ) ( mathbf{3}) ) 9 381Determine whether the following polynomial has ( (x+1) ) as a factor. ( boldsymbol{x}^{4}-boldsymbol{x}^{3}+boldsymbol{x}^{2}-boldsymbol{x}+mathbf{1} ) 9 382If ( x ) is real, then find the solution of ( sqrt{x+1}+sqrt{x-1}=1 ) A ( cdot frac{5}{4} ) B. ( frac{4}{5} ) ( c cdot frac{3}{5} ) D. No such ( x ) exists 9 383Divide the first expression by the second. Write the quotient and the remainder. ( a^{2}-b^{2} ; a-b ) A. Quotient ( =b ), Remainder ( =0 ) B. Quotient ( =a+b ), Remainder ( =0 ) c. Quotient ( =a b ), Remainder ( =a ) D. Quotient ( =a-b ), Remainder ( =a ) 10 384Perform the division: ( 6 x^{3}-23 x+x^{2}+ ) 12 by ( 2 x-3 ) 10 385Solve: ( x^{4}+frac{1}{x^{4}} ) 9 386On dividing the polynomial ( 3 x^{3}+ ) ( 4 x^{2}+5 x-13 ) by a polynomial ( g(x), ) the quotient and the remainder were ( (3 x+ ) 10) and ( (16 x-43) ) respectively. Find ( boldsymbol{g}(boldsymbol{x}) ) 10 387Find the remainder when ( p(x)= ) ( -3 x^{3}-4 x^{2}+10 x-7 ) is divided by ( boldsymbol{x}-mathbf{2} ) A . -26 в. -27 c. 26 D. 27 9 388The zero polynomial is the identity of the additive group of polynomials. A. multiplicative c. multiplicative inverse 9 389Find the degree of the following polynomial ( x^{3}+17 x-21-x^{2} )9 390Expand ( (4 x-3 y)^{3} ) ( mathbf{A} cdot 64 x^{3}-144 x^{2} y+108 x y^{2}-27 y^{3} ) B. ( 64 x^{3}+144 x^{2} y+108 x y^{2}-27 y^{3} ) ( mathbf{c} cdot 64 x^{3}-144 x^{2} y-108 x y^{2}-27 y^{3} ) D. None of these 9 39156. If (x + 1) and (x-2) be the fac- tors of x + (a + 1)x2 – (b-2)x-6, then the values of a and b will be (1) 2 and 8 (2) 1 and 7 (3) 5 and 3 (4) 3 and 7 9 392Identify the cubic polynomial. A ( cdot x^{5}+y^{3}-x^{3}+x^{4} ) В. ( x^{3}+y^{3}-x^{2} ) c. ( 2 x^{3}+8 y^{3}-9 x^{5} ) D. ( -7 x^{7}+x^{3} ) 10 393What is the degree of the polynomial ( (x+1)left(x^{2}-x-x^{4}+1right) ? ) 9 394Divide : ( 15 x^{3} y^{3} ) by ( 3 x y^{2} ) 10 39562. If (3-2)-3 x # o, then the value of x? + (2)23 (1) 25 (3) 2 (427 9 396ff ( x=2 ) and ( y=-8 ) find the value of ( (x-y)^{3} ) 10 397Evaluate: ( (a+b)(a-b)left(a^{2}+b^{2}right) ) is ( a^{4}-b^{4} ) If true then enter 1 and if false then enter 0 9 398Expand using suitable identities ( (-2 a+5 b-3 c)^{2} ) 9 399Which type of polynomial is ( 5 t-sqrt{7} ? ) A. Linear Polynomial c. cubic Polynomial D. None of above 10 400When ( x^{3}-6 x^{2}+12 x-4 ) is divided by ( x-2, ) the remainder is A .4 B. 0 ( c .5 ) D. 6 9 401Zero of the zero polynomial is ( mathbf{A} cdot mathbf{0} ) B. 1 C . Any real number D. Not defined 10 402Determine whether the following polynomial has ( (x+1) ) as a factor. ( x^{3}-x^{2}-(3-sqrt{3}) x+sqrt{3} )9 403State whether the statement is True or False. Evaluate: ( (a+b c)(a-b c)left(a^{2}+b^{2} c^{2}right) ) is equal to ( a^{4}-b^{4} c^{4} ) A. True B. False 9 404( mathbf{f} boldsymbol{p}(boldsymbol{x})=boldsymbol{a} boldsymbol{x}^{3}+boldsymbol{3} boldsymbol{x}-mathbf{1} boldsymbol{3} ) and ( boldsymbol{q}(boldsymbol{x})= ) ( 2 x^{3}-5 x+1 ) are divided by ( x+2 ) remainder is same in each case. find the value of ( a ) 9 405The term, that should be added to (4×2 + 8xd so that resulting ex- pression be a perfect square, is (1) 2 (2) 4 (3) 2x (4) 1 9 406Let ( boldsymbol{f}(boldsymbol{x}) ) be polynomial in ( boldsymbol{x} ) of degree not less than 1 and ( ^{prime} a^{prime} ) be a real number. If ( f(x) ) is divided by ( (x-a), ) then the remainder is ( boldsymbol{f}(boldsymbol{a}) ). If ( (boldsymbol{x}-boldsymbol{a}) ) is a factor of ( f(x), ) then ( f(a)=0 . ) Find the remainder of ( x^{4}+x^{3}-x^{2}+2 x+3 ) when divided by ( boldsymbol{x}-mathbf{3} ) A. 108 B. 98 c. 165 D. 170 9 407Write the degree of the following polynomial: ( mathbf{7} p^{2} boldsymbol{q}^{3} boldsymbol{t}-mathbf{1} mathbf{1} boldsymbol{p}^{4} boldsymbol{t}+mathbf{2} boldsymbol{p}^{8} ) ( A cdot 3 ) B. 5 ( c cdot 7 ) ( D ) 10 408( 20 a^{2}-45= ) ( mathbf{A} cdot 5(3-2 a)(3+2 a) ) B. ( 5(2 a-3)(2 a-3) ) ( mathbf{c} cdot 3(5+2 a)(5-2 a) ) D. ( 3(2 a+5)(2 a-5) ) 9 409( frac{a^{2}-b^{2}-2 b c-c^{2}}{a^{2}+b^{2}+2 a b-c^{2}} ) is equivalent to A ( cdot frac{a+b+c}{a-b+c} ) в. ( frac{a-b-c}{a+b-c} ) c. ( frac{a-b-c}{a-b+c} ) D. ( frac{a-b+c}{a+b+c} ) 10 410The degree of a polynomial ( x^{3}-27 ) is ( mathbf{A} cdot mathbf{3} ) B. c. 26 D. 27 10 411Solve: ( frac{5 a^{3}-4 a^{2}+3 a+18}{a^{2}-2 a+3} )10 412Divide the first expression by the second. Write the quotient and the remainder. ( x^{2}-frac{1}{4 x^{2}} ; x-frac{1}{2 x} ) A ( cdot ) Quotient ( =x+frac{2}{2 x}, ) Remainder ( =1 ) B. Quotient ( =2 x+frac{1}{2 x} ), Remainder = 0 c. quotient ( =x-frac{1}{2 x}, ) Remainder ( = ) D. Quotient ( =x+frac{1}{2 x}, ) Remainder ( =0 ) 10 413The remainder obtained when ( t^{6}+ ) ( 3 t^{2}+10 ) is divided by ( t^{3}+1 ) is: A ( cdot t^{2}-11 ) B . ( t^{3}-1 ) ( c cdot 3 t^{2}+11 ) D. none of these 9 41467. The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is : (1) 10 (2) 46 (3) 70 (4) 90 9 415( frac{0.86 times 0.86 times 0.86+0.14 times 0.14 times 0.1}{0.86 times 0.86-0.86-0.14+0.14 times 0.1} ) is equal to A . 1 B. 0 c. 2 D. 10 9 416Polynomials of degrees 1,2 and 3 are called polynomials respectively. A. cubic, linear, quadratic B. linear, quadratic, cubic c. quadratic, linear, cubic D. none of the above 10 417The value of ( (a+b)^{2}-2(a-b)^{2}+ ) ( (a-b)(a+b) ) is A ( cdot 4 a b-b^{2} ) B . ( 2 a b-b^{2} ) ( c cdot 3 a b-b^{2} ) D. ( 6 a b-2 b^{2} ) 9 41859. If a = 11 and b = 9, then the la? +62 + ab value of 23 – 63 is (1) (2) 2 (3) 2 (4) 20 9 419Classify the following as a constant, linear, quadratic and cubic polynomials ( sqrt{2} x-1 ) A. Linear c. cubic D. None of these 9 420Given a function ( f ) such that ( f(4)=5 ) then which of the following is/are true for ( boldsymbol{f} ? ) A ( . f(x) neq x+1 ) в. ( f(x) neq 2 x-3 ) c. ( f(x) neq 3 x-2 ) D. ( f(x) neq 4 x-11 ) 10 421Find the value of ( 1.05 times 0.95 ) using standard identity A . 0.9985 B. 0.9975 c. 0.9875 D. 0.9995 9 422Expand the following using identities ( (x+7)(y+5) )9 423Perform the division: ( x^{4}-16 ) by ( x-2 )10 424Find the degree of the expression ( left[x+left(x^{3}-1right)^{frac{1}{2}}right]^{5}+left[x-left(x^{3}-1right)^{frac{1}{2}}right]^{5} )9 425The polynomial having atmost 3 zero A. constant polynomial B. linear polynomial D. cubic polynomial 10 426If ( a-b-2=0 ) and ( a^{3}-b^{3}-6 a b=k ) then find the value of ( k ) 9 427The greatest index of a variable in the polynomial ( 5 x^{2}+3 x+1 ) is ( mathbf{A} cdot mathbf{1} ) B . 2 ( c .3 ) D. 4 9 428Factorise ( : 8 x^{3}-27 y^{3}-2 x+3 y ) A ( cdot(2 x+3 y)left(4 x^{2}-x y+9 y^{2}+1right) ) В ( cdot(2 x+y)left(4 x^{2}-6 x y-9 y^{2}right) ) C ( cdot(x-3 y)left(4 x^{2}-3 x y+9 y^{2}right) ) D ( cdot(2 x-3 y)left(4 x^{2}+6 x y+9 y^{2}-1right) ) 9 429Factorize: ( 4 x y-x^{2}-4 y^{2}+z^{2} ) ( mathbf{A} cdot(z+x-2 y)(z-x+2 y) ) B . ( (z-x-2 y)(z-x+2 y) ) ( mathbf{c} cdot(z+x-2 y)(z-x-2 y) ) D. ( (z+x-2 y)(z-x+y) ) 9 430( (5 x+2 y+3 z)^{2} ) ( mathbf{A} cdot 25 x^{2}+4 y^{2}+9 z^{2}+20 x y+12 y z+30 z x ) B ( cdot 25 x^{2}-4 y^{2}+9 z^{2}+20 x y+12 y z+30 z x ) ( mathbf{c} cdot 25 x^{2}+4 y^{2}-9 z^{2}+20 x y+12 y z+30 z x ) D. None of these 9 431Workout the following divisions ( mathbf{3 6}(boldsymbol{x}+mathbf{4})left(boldsymbol{x}^{2}+mathbf{7} boldsymbol{x}+mathbf{1 0}right) div mathbf{9}(boldsymbol{x}+mathbf{4}) ) ( mathbf{A} cdotleft(x^{2}+7 x+10right) ) B. ( 4left(x^{2}+7 x+6right) ) c. ( 4left(x^{2}+7 x+10right) ) D. None 10 432If ( alpha ) and ( beta ) are the zeroes of a polynomial ( x^{2}-4 sqrt{3} x+3, ) then find the value of ( boldsymbol{alpha}+boldsymbol{beta}-boldsymbol{alpha} boldsymbol{beta} ) 10 433What must be added to ( x^{3}-3 x^{2}- ) ( 12 x+19, ) so that the result is exactly divisible by ( x^{2}+x-6 ? ) A ( .2 x+5 ) в. ( 4 x-4 ) c. ( 2 x+3 ) D. ( -2 x-5 ) 9 434Divide and write the quotient and the remainder. ( left(p^{2}+7 p-5right) div(p+3) ) 10 435Find the zeroes of polynomial ( boldsymbol{p}(boldsymbol{x})= ) ( 6 x^{2}-3 ) 10 436How to divide the equation in algebraic expression9 43771. I x)2, then the value of (1) 212 (3) 25 (2) 2 (4) 27 9 438Write down the degree of the following polynomial: ( 7 x^{3}-5 x^{3} y^{2}+4 y^{4}-9 ) ( A cdot 3 ) B. 4 ( c cdot 5 ) ( D ) 9 439Evaluate : ( frac{6 x^{2}-x-2}{3 x-2} ) 10 440Classify the following as linear, quadratic and cubic polynomials: ( boldsymbol{x}^{boldsymbol{3}}-boldsymbol{4} ) A. Cubic B. Linear D. None of the above 9 441Write each of the following polynomials in the standard form. Also, write their degree: ( a^{2}+4+5 a^{6} ) 9 442If ( 3 x-frac{1}{3 x}=5, ) then find ( 81 x^{4}+frac{1}{81 x^{4}} )9 443Simplify: ( left[3 a^{2}-2 a^{2}+9 a^{2}-left{6 a^{2}-right.right. ) ( left.left(-2 a^{2}+3 a^{2}right)right} ) 9 444boty 26. For the equation 3×2 + px +3 = 0,p>0, if one of the root is square of the other, then p is equal to (2000) (a) 1/3 (b) 1 (C) 3 (d) 2/3 10 445State whether the following expressions are polynomials in one variable or not. Give reasons for your answer. ( sqrt[3]{t}+2 t )10 446Find ( y^{2}+frac{1}{y^{2}} ) and ( y^{4}+frac{1}{y^{4}} ) if ( y+frac{1}{y}=9 )9 447The variable in the quadratic polynomial ( t^{2}+4 t+5 ) is ( mathbf{A} cdot mathbf{1} ) B. 4 ( c cdot t ) D. 5 10 448Solve ( :left[left(4 x^{4}-3 x^{3}-2 x^{2}+4 x-3right)right. ) divide by ( (x-1) ) 10 449Degree of the polynomial ( left(a^{2}+1right)(a+2)left(a^{3}+3right) ) is ( A cdot 3 ) B. 6 ( c cdot 2 ) ( D ) 10 450Evaluate: ( 96 a b c(3 a-12)(5 b-30) div ) ( mathbf{1 4 4}(boldsymbol{a}-mathbf{4})(boldsymbol{b}-mathbf{6}) ) ( mathbf{A} cdot 10 a b c ) B. 2abc ( c cdot 2 a b ) D. ( 2 b c ) 10 451( boldsymbol{p}(boldsymbol{x})=left(boldsymbol{x}^{2}-mathbf{1 0} boldsymbol{x}-mathbf{2 4}right), ) when divided by ( x+2 ) and ( x neq-2 ) gives the quotient Q. Find ( Q ) A . ( x-22 ) B. ( x-12 ) c. ( x+12 ) D. ( x+22 ) 10 452The value of a polynomial A. changes with the change in variable B. doesn”t change with the change in variable C. many or may not change with the change in variable D. all of the above 10 453According to the remainder theorem when we divide a polynomial ( f(x) ) by ( (x-c), ) the remainder equals A ( . f(c) ) в. ( f(-c) ) c. 0 D. None of the above 9 454Evaluate using expansion of ( (a+b)^{2} ) or ( (a-b)^{2}: ) ( (9.4)^{2} ) A . 88.36 B. 88.46 c. 89.16 D. 89.56 9 455( left(x^{2}+3 x+1right)=(x-2)^{2} ) is an equation of degree A. three B. one c. four D. two 9 456Find the reduced form of the expression ( frac{20 u^{3} v^{2}-15 u^{2} v}{10 u^{4} v+30 u^{3} v^{3}} ) A ( cdot frac{5 u v}{40 u^{7} v^{4}} ) в. ( frac{2 v-1}{u+2 u v^{2}} ) c. ( frac{4 u v-3}{2 u^{2}+6 u v^{2}} ) D. ( frac{2 u v-3 u v^{2}}{u^{2}+6} ) 10 457Find the degree of the polynomial: 5 ( boldsymbol{x}^{2} ) 10 458Simplify: ( (3 a-5 b)^{3}-(3 a+5 b)^{3} )9 45969. If x + y + 5+= 4, then the value of x + y2 is (1) 2 (2) 4 (3) 8 (4) 16 9 460Factorise the expression and divide them as directed. ( 12 x yleft(9 x^{2}-16 y^{2}right) div 4 x y(3 x+4 y) ) 10 461The polynomial ( a x^{3}+b x^{2}+x-6 ) has ( (x+2) ) as a factor and leaves a remainder 4 when divided by ( (x-2) ) Find ( a ) and ( b ) This question has multiple correct options ( mathbf{A} cdot a=0 ) в. ( b=2 ) ( mathbf{c} cdot a=2 ) ( mathbf{D} cdot b=0 ) 9 462The product of two numbers is 120 and the sum of their squares is ( 289 . ) The sum of the numbers is : A . 20 B. 23 ( c cdot 16 ) D. None of these 9 463Simplify: ( (2 x+3 y)^{2} ) 9 464If ( p^{2}-6 p+7 ) is divided by ( (p-1) ) the remainder will be A. positive B. zero c. negative D. none of these 9 465( boldsymbol{f}(boldsymbol{x})=mathbf{2} boldsymbol{x}^{3}-mathbf{5} boldsymbol{x}^{2}+boldsymbol{a} boldsymbol{x}+boldsymbol{a} ) Given that ( (x+2) ) is a factor of ( f(x) ) find the value of the constant ( a ) A . -16 B. 32 ( c .-36 ) D. 42 10 466If ( a x^{3}+b x^{2}+c x+d ) is divided by ( x ) 2, then the remainder is equal A ( . d ) B. ( a-b+c-d ) c. ( 8 a+4 b+2 c+d ) D. ( -8 a+4 b-2 c+d ) 9 467A quadratic polynomial has at the most zero(es). A . zero B. one c. three D. two 10 468Evaluate: ( (a-3 b)^{2}-4(a-3 b)-21 ) 9 469If ( x^{3}+a x^{2}+b x+6 ) divided by ( x-2 ) as factor then remainder becomes ( 0, ) and leaves remainder 3 when divided by ( x-3 ) find the values of a and ( b ) 9 470The sum of two numbers is 9 and their product is 20. Find the sum of their cubes A . 189 в. 130 c. 76 D. 39 9 9 471If ( frac{a}{b}+frac{b}{a}=1, ) then the value of ( a^{3}+b^{3} ) is- A . B. ( a ) ( c cdot b ) D. 9 472Show that ( x+1 ) and ( 2 x-3 ) are factors of ( 2 x^{3}-9 x^{2}+x+12 ) 10 473 23. Ifa and B(a<B) are the roots of the equation x2 + bx+c=0. where c<0<b, then (20005) (a) 0<a<B (b) a<0<B<l al (c) a<B<0 (d) a<0<al<B 0<a<e a<<\$<la 10 474Simplify: ( (3 m+5 n)^{2}-(2 n)^{2} )9 475Simplify: Find ( boldsymbol{x}(boldsymbol{x}+mathbf{1})(boldsymbol{x}+mathbf{2})(boldsymbol{x}+boldsymbol{3}) div boldsymbol{x}(boldsymbol{x}+mathbf{1}) ) A ( cdot(x+2)(x+3) ) B. ( x+2 ) c. ( x+3 ) D. None of these 10 476Remainder when ( p(x)=x^{4}-5 x+6 ) is divided by ( g(x)=2-x^{2} ) is ( -m x+2 m ) Find ( boldsymbol{m} ) 10 47758. If a, b, c are real and al + b2 + 2 = 2 (a – b -c) – 3, then the value of 2a – 3b + 4c is (1) -1 (2) O (3) 1 (4) 2 9 478If ( n ) is an integer,what is the remainder when ( 5 x^{2 n+1}-10 x^{2 n}+3 x^{2 n-1}+5 ) is divided by ( x+1 ? ) A . B. 2 ( c cdot 4 ) ( D cdot-8 ) ( E cdot-13 ) 9 479Determine the factors of ( 216 u^{3}+1 ) A ( cdot(6 u-1)left(36 u^{2}-6 u+1right) ) B . ( (6 u+1)left(36 u^{2}-6 u+1right) ) c. ( (6 u+1)left(6 u^{2}-6 u+1right) ) D. ( (u+1)left(6 u^{2}-6 u+1right) ) 9 480f ( a=x(y-z), b=y(z-x) ) and ( c= ) ( z(x-y) . ) What is the value of ( frac{x y z}{a b c}left(frac{a^{3}}{x^{3}}+frac{b^{3}}{y^{3}}+frac{c^{3}}{z^{3}}right) ? ) 9 48152. 9×2 +25-30xcan be expressed as the square of (1) -3x-5 (2) 3x + 5 (3) 3x – 5 (4) 3×2 – 25 9 482Say true or false: For polynomials ( p(x) ) and any non-zero polynomial ( g(x), ) there are polynomials ( boldsymbol{q}(boldsymbol{x}) ) and ( boldsymbol{r}(boldsymbol{x}) ) such that ( boldsymbol{p}(boldsymbol{x})= ) ( boldsymbol{g}(boldsymbol{x}) boldsymbol{q}(boldsymbol{x})+boldsymbol{r}(boldsymbol{x}), ) where ( boldsymbol{r}(boldsymbol{x})=mathbf{0} ) or ( operatorname{degree} r(x)<operatorname{degree} g(x) . ) This statement is correctly explains the remainder theorem. A . True B. False 9 483( frac{x^{-1}}{x^{-1}+y^{-1}}+frac{x^{-1}}{x^{-1}-y^{-1}} ) is equal to A. ( frac{2 y^{2}}{y^{2}-x^{2}} ) в. ( frac{2 x^{2}}{y^{2}-x^{2}} ) c. ( frac{2 y^{2}}{y^{2}+x^{2}} ) D. none 10 484Factorise: ( 7 y^{3}+12 z^{3} )9 48566. If a + b2 + 4c2 = 2 (a + b – 2c) – 3 and a, b, c are real, then the value of (a + b + c) is (1) 3 (2) 3- (3) 2 (4) 2- 9 48658. If p= 102 then the value of p (p2 – 6p + 12) is (1) 1000008 (2) 10000008 (3) 999992 (4) 9999992 9 487Evaluate using expansion of ( (a+b)^{2} ) or ( (a-b)^{2}: ) ( (45)^{2} ) 9 488Which of the following is not a constant polynomial? A ( cdot p(x)=3^{3} ) B ( cdot p(x)=2^{3} ) ( mathbf{c} cdot p(x)=x^{3} ) D ( cdot p(x)=4^{3} ) 9 489The value of ( frac{(mathbf{1 1 9})^{2}+(mathbf{1 1 9})(mathbf{1 1 1})+(mathbf{1 1 1})^{mathbf{2}}}{(mathbf{1 1 9})^{mathbf{3}}-(mathbf{1 1 1})^{mathbf{3}}} ) is ( A ) B. ( frac{1}{8} ) ( c cdot 230 ) D. ( frac{1}{23} ) 9 490Give examples of polynomials ( boldsymbol{p}(boldsymbol{x}), boldsymbol{g}(boldsymbol{x}), boldsymbol{q}(boldsymbol{x}) ) and ( boldsymbol{r}(boldsymbol{x}), ) which satisfy the division algorithm and (i) ( operatorname{deg} p(x)=operatorname{deg} q(x) ) (ii) ( operatorname{deg} boldsymbol{q}(boldsymbol{x})= ) ( operatorname{deg} r(x) ) (iii) ( operatorname{deg} r(x)=0 ) 10 491The quotient when ( left(2 x^{4}-3 x^{3}-x^{2}+4 x-2right) ) divided by is: 10 49262. The value of 204 x 197 is (1) 40218 (2) 40188 (3) 40212 (4) 39812 9 493If ( x^{2}+y^{2}+10=(2 sqrt{2 x}+4 sqrt{2} y) ) then the value of ( (x+y) ) is A ( .4 sqrt{2} ) B. ( 3 sqrt{2} ) ( c cdot 6 sqrt{2} ) D. ( 9 sqrt{2} ) 9 494If ( sqrt{2 x-1}-sqrt{2 x+1}+4=0, ) then ( 128 x ) is equal to A . 120 в. 260 c. 165 D. 200 10 495f ( x=3, ) then the value of ( 20 x^{7}+x^{5} 3 )9 496Classify the following polynomials as monomials, binomials and trinomials: ( 3 x^{2}, 3 x+2, x^{2}-4 x+2, x^{5}-7, x^{2}+ ) ( mathbf{3} boldsymbol{x} boldsymbol{y}+boldsymbol{y}^{2}, boldsymbol{s}^{2}+boldsymbol{3} boldsymbol{s} boldsymbol{t}-boldsymbol{2} boldsymbol{t}^{2}, boldsymbol{x} boldsymbol{y}+boldsymbol{y} boldsymbol{z}+ ) ( z x, a^{2} b+b^{2} c, 2 l+2 m ) 10 497Find the cubic polynomial with three different variables. A ( cdot x^{3}+y^{3}+2^{3}-5^{3} ) B ( cdot a^{3}+b^{3}-c^{2}+1 ) c. ( x^{3}+x^{2}+x ) D. ( x y^{2}+x y^{3}-x y+6^{3} ) 9 498( f(x)=x^{3}-3 x^{2}+2 x ) then find the value of ( p(x) ) at ( x=2 ) 9 499Expand ( frac{1}{x^{4}-5 x^{3}+7 x^{2}+x-8} ) in descending powers of ( x ) to four terms, and find remainder. 10 500Identify zero polynomial among the following. A . 0 B. ( x ) ( mathbf{c} cdot x^{2} ) D. None of the above 9 501Find the degree of the expression ( [x+ ) ( left.left(x^{3}-1right)^{frac{1}{2}}right]^{5}+left[x-left(x^{3}-1right)^{frac{1}{2}}right]^{5} )10 502Expand: ( left(a^{2}+4 b^{2}right)(a+2 b)(a-2 b) )9 503The polynomials ( left(2 x^{3}-5 x^{2}+x+aright) ) and ( left(a x^{3}+2 x^{3}-3right) ) when divided by ( (x-2) ) leave the remainders ( R_{1} ) and ( R_{2} ) respectively. Find the value of ( ^{prime} a^{prime} ) in the following case, if ( boldsymbol{R}_{1}-boldsymbol{2} boldsymbol{R}_{2}=mathbf{0} ) 9 504( left{x^{2}+10 x+25right} div(x+5) )10 505When ( left(x^{3}-2 x+p x-qright) ) is divided by ( left(x^{2}-2 x-3right), ) the remainder is ( (x-6) ) What are the values of ( p, q ) respectively. A . -2,-6 в. 2,-6 c. -4,12 D. 2,6 9 506The degree of the polynomial ( x^{2}- ) ( 5 x^{4}+frac{3}{4} x^{7}-73 x+5 ) is A. 7 B. ( frac{3}{4} ) ( c cdot 4 ) D. -73 10 507If ( a+b=5 ) and ( a^{2}+b^{2}=13, ) find ab9 508What is the remainder, when ( left(4 x^{3}-3 x^{2}+2 x-1right) ) is divided by ( (x+2) ? ) A . – 49 B. 55 ( c cdot-30 ) D. 37 10 509( (x-2) ) is a factor of the expression ( x^{3}+ ) ( a x^{2}+b x+6 . ) when this expression is divided by ( (x-3), ) it leaves the remainder ( 3 . ) find the values of a and ( b ) 9 510State True or False. ( mathbf{2} a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a= ) ( left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}right] ) A. True B. False 9 511Choose the correct answer from the alternatives given. If ( x-frac{1}{x}=3 ) then find the value of ( x^{3}+frac{1}{x^{3}} ) A. ( 10 sqrt{13} ) (3) 5 B. ( 100 sqrt{3} ) c. ( 13 sqrt{10} ) D. ( 130 sqrt{10} ) 9 512( frac{x^{3}+x^{2}+2 x-12}{x-3} )10 513If ( A(x) ) and ( B(x) ) be two polynomials and ( boldsymbol{f}(boldsymbol{x})=boldsymbol{A}left(boldsymbol{x}^{3}right)+boldsymbol{x} boldsymbol{B}left(boldsymbol{x}^{3}right) . ) If ( boldsymbol{f}(boldsymbol{x}) ) is divisible by ( x^{2}+x+1 ) then show that it is divisible by ( x-1 ) also. 10 514Write the quadratic polynomial with zeros -2 and ( frac{1}{3} )10 515Find the last digit of ( 1^{5}+2^{5}+ldots+99^{5} )9 516Find the cube of ( :left(2 a+frac{1}{2 a}right) ; a neq 0 ) A ( cdot 8 a^{3}+6 a+frac{3}{2 a}+frac{1}{8 a^{3}} ) в. ( 8 a^{3}+3 a+frac{3}{a}+frac{1}{8 a^{3}} ) c. ( 8 a^{3}+3 a+frac{6}{a}+frac{1}{8 a^{3}} ) D. ( 8 a^{3}+6 a+frac{6}{a}+frac{1}{8 a^{3}} ) 9 517Find the remainder when ( x^{3}+p x^{2}+ ) ( boldsymbol{q} boldsymbol{x}+boldsymbol{r} ) is divided by ( boldsymbol{x}^{2}+boldsymbol{p} boldsymbol{x}+boldsymbol{q} ) 10 518The least positive value of ( x ) satisfying ( frac{sin ^{2} 2 x+4 sin ^{4} x-4 sin ^{2} x cos ^{2} x}{4-sin ^{2} 2 x-4 sin ^{2} x}=frac{1}{9} ) is A ( cdot frac{pi}{3} ) B. ( frac{pi}{6} ) c. ( frac{2 pi}{3} ) D. ( frac{5 pi}{6} ) 10 519Find the degree of : a) ( x^{3}-3 x^{3} y^{6}+8 y^{3} ) b) ( 6 x^{4}-9 x^{3} y^{2}-6 y^{6} ) 9 520( left(x^{2}-9 x-10right) div(x+1)= ) A . ( x-10 ) B. ( x+10 ) ( mathbf{c} cdot x-8 ) D. ( x+8 ) 10 521Which of the following is NOT a constant polynomial? A ( cdot p(y)=y^{circ} ) В . ( p(x)=x^{o} ) c. ( p(y)=frac{y}{y} ) D. ( p(x)=y x ) 9 522Evaluate the following using suitable identities ( (999)^{3} ) 9 523Perform the division ( frac{left(x^{2}+1right)left(x^{2}+2right)}{left(x^{2}+3right)left(x^{2}+4right)} )10 524Find a quadratic polynomial each with the give numbers as the sum and product of its zeros respectively. ( sqrt{mathbf{2}}, frac{mathbf{1}}{mathbf{3}} )10 525Solve by using suitable identity: ( a^{4}- ) ( b^{4}+2 b^{2}-1 ) 9 526f ( p=2-a ) then prove that ( a^{3}+6 a p+ ) ( boldsymbol{p}^{3}-boldsymbol{8}=boldsymbol{0} ) 9 527On dividing ( 3 x^{3}+x^{2}+2 x+5 ) by a polynomial ( g(x), ) the quotient and remainder are ( (3 x-5) ) and ( (9 x+10) ) respectively. Find ( g(x) ) 10 528If ( boldsymbol{x}-boldsymbol{y}=-boldsymbol{6} ) and ( boldsymbol{x} boldsymbol{y}=mathbf{4}, ) find the value of ( boldsymbol{x}^{mathbf{3}}-boldsymbol{y}^{mathbf{3}} ) ( mathbf{A} cdot-288 ) в. 288 ( c cdot-28 ) D. None of these 9 529Factorize : ( (2 a+1)^{3}+(a-1)^{3} ) 9 530Twenty years ago, my age was one-third of what it is now 1. My present age is (a) 66 years (b) 30 years (c) 33 years (d) 36 years 10 531Factorise: ( 27 a^{2}-75 b^{2} ) ( mathbf{A} cdot(3 a+5 b)(3 a-5 b) ) B. ( 3(3 a+5 b)(3 a+5 b) ) ( mathbf{c} cdot(3 a+5 b)(a-b) ) D. ( 3(3 a+5 b)(3 a-5 b) ) 9 532There are ( x^{4}+57 x+15 ) pens to be distributed in a class of ( x^{2}+4 x+2 ) students. Each student should get the minimum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed ( (boldsymbol{x} epsilon boldsymbol{N}) ) в. ( 9 x-15 ) c. ( 9 x-20 ) D. ( 9 x+13 ) 10 53385. If x y 1 then the value of x f- z 1 1 +-+- y z is (1) 9 (3) 4 (2) 3 (4) 6 10 534Find the degree of the given algebraic expression ( 3 x-15 ) ( mathbf{A} cdot mathbf{1} ) B. 3 ( c cdot 2 ) D. -15 10 5353. Ifx,y and z are real and different and (1979) u=x2 + 4y2 +9z2 – 6yz – 3zx – 2xy, then u is always. (a) non negative (b) zero c) non positive (d) none of these 9 536( boldsymbol{p}(boldsymbol{x})=mathbf{7} boldsymbol{x}^{2}-boldsymbol{9} ) is a polynomial B. linear c. constant D. cubic 10 537Solve ( (boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} )9 538Find the product of ( (sqrt{2}+sqrt{3})(sqrt{2}+sqrt{5})(sqrt{2}+sqrt{7}) )9 539( operatorname{Let} p(x)=a x^{2}+b x+c ) be a quadratic polynomial. It can have at most A. One zero B. Two zeros c. Three zeros D. None of these 10 540If ( alpha ) and ( beta ) are the zeroes of the polynomial ( 2 y^{2}+7 y+5, ) write the values of ( boldsymbol{alpha}+boldsymbol{beta}+boldsymbol{alpha} boldsymbol{beta} ) 10 541( x^{3}-15 x^{2}+59 x-45=0 ) solve for ( x )10 542State the following statement is True or False The zero of the polynomial ( x^{3}-27 ) is 3 A. True B. False 9 543If ( a-b=7, a b=8, ) find ( a^{2}+b^{2} )9 544What are the solution(s) to the system of equations ( boldsymbol{y}=boldsymbol{x}^{2}-mathbf{9} ) and ( boldsymbol{y}-boldsymbol{3}= ) ( x ? ) A. -3,0 and 4,7 в. -3,0 c. 4,7 D. 4,-3 10 545Divide ( p(x) ) by ( g(x) ) in the following case and verify division algorithm ( boldsymbol{p}(boldsymbol{x})=boldsymbol{x}^{2}+boldsymbol{4} boldsymbol{x}+boldsymbol{4} ; boldsymbol{g}(boldsymbol{x})=boldsymbol{x}+boldsymbol{2} ) 10 546Simplify: ( (boldsymbol{p}-boldsymbol{q})^{2}+mathbf{4} boldsymbol{p} boldsymbol{q} ) A ( cdot p^{2}-q^{2} ) в. ( (p+q)^{2} ) c. ( (2 p-q)^{2} ) D. ( (2 p-2 q)^{2} ) 9 547The value of ( boldsymbol{p}(boldsymbol{x})=boldsymbol{3} boldsymbol{x}-boldsymbol{2}-boldsymbol{x}^{2} ) at ( boldsymbol{x}= ) 3 ( A cdot 0 ) B. – 1 ( c cdot-2 ) D. – 3 10 548Write whether the following statements are True or False. Justify your answer The degree of ( x^{2}+2 x y+y^{2} ) is 2 A . True B. False 10 549Simplify: ( left(boldsymbol{m}^{2}+boldsymbol{2} boldsymbol{n}^{2}right)^{2}-boldsymbol{4} boldsymbol{m}^{2} boldsymbol{n}^{2} )9 550Simplify: ( (boldsymbol{a}+boldsymbol{b})^{boldsymbol{3}} )9 551Find the polynomial which represents the perimeter of the rectangle whose length is 2 more than its breadth.10 552If 1 is zero of the polynomial ( p(x)= ) ( a x^{2}-3(a-1) x-1 ) then the value of ‘a’ is A . B. – ( c cdot-2 ) D. 10 553If the quotient on dividing ( 2 x^{4}+x^{3}- ) ( 14 x^{2}-19 x+6 ) by ( 2 x+1 ) is ( x^{3}+ ) ( a x^{2}-b x-6 ) Find the values of a and ( b ), also the remainder 10 5547. (1980) If(x2 + px + 1) is a factor of (ax3 + bx+c), then (a) a² + c = – ab (b) al-c2 = – ab (c) a2-c2 = ab (d) none of these 10 555Which of the following does NOT represent a zero polynomial? ( mathbf{A} cdot p(x)=0 ) B ( cdot p(x)=0 . x^{0} ) ( mathbf{C} cdot p(x)=x^{0} ) D ( cdot p(x)=x^{0}-1 ) 9 556Find the degree of following polynomial ( -frac{3}{2} ) A . B. 2 c. 0 D. Cannot be determined 9 557Use synthetic division to find the quotient ( Q ) and remainder ( R ) when dividing ( f(x)=x^{2}+2 x+3 ) by ( x-i ) ( i=sqrt{(}-1) ) the imaginary unit) 9 558(0 JU JUU My age twenty years ago is (a) 40 years (b) 15 years (c) 22 years (d) 10 years D 10 559Which type of polynomial is ( 3 x^{3} ? ) A. Cubic Polynomial B. Linear Polynomial c. square Polynomial D. None of These 10 560Evaluate ( (2 x+3 y+5 z)^{2} )9 561Factorise the following: ( 2 a^{3}-54 b^{3} ) ( mathbf{A} cdot 2(a b+3 b)left(a^{2}+3 b+9 a b^{2}right) ) B ( cdot 2(a+2 b)left(a^{3}+3 a b-9 b^{3}right) ) C ( cdot(a-3 b)left(a^{2}+3 b+9 a b^{2}right) ) D ( cdot 2(a-3 b)left(a^{2}+3 a b+9 b^{2}right) ) 9 562Simplify: (i) ( left(a^{2}-b^{2}right)^{2} ) (ii) ( (2 x+5)^{2}-(2 x-5)^{2} ) (iii) ( (7 m-8 n)^{2}+(7 m+8 n)^{2} ) ( (i v)(4 m+5 n)^{2}+(4 n+5 m)^{2} ) ( (v)(2.5 p-1.5 q)^{2}-(1.5 p-2.5 q)^{2} ) ( (v i)(a b+b c)^{2}-2 a b^{2} c ) ( (v i i)left(m^{2}-n^{2} mright)^{2}+2 m^{3} n^{2} ) 9 563The remainder obtained when the polynomial ( boldsymbol{x}^{4}-mathbf{3} boldsymbol{x}^{mathbf{3}}+mathbf{9} boldsymbol{x}^{2}-mathbf{2 7} boldsymbol{x}+mathbf{8 1} ) is divided by ( (x-3) ) is: ( mathbf{A} cdot mathbf{0} ) B. 3 c. 81 D. 27 9 564The remainder when ( 4 a^{3}-12 a^{2}+ ) ( 14 a-3 ) is divided by ( 2 a-1, ) is A ( cdot frac{2}{3} ) в. ( frac{5}{3} ) ( c cdot frac{6}{7} ) D. ( frac{3}{2} ) 10 565f ( p(x)=x+3, ) then ( p(3)+p(-3), ) is equal to ( A cdot 3 ) B . 2 ( c cdot 0 ) D. 6 10 566( left(3 a-frac{1}{a}right)^{3} )9 567The polynomial ( left(a x^{2}+b x+cright)left(a x^{2}-d x-cright), a c neq 0 ) has? 10 5685. The sum of two numbers is 45 and their ratio is 7: 8. Find the numbers. (a) 21 (b) 24 (c) 25 (d) 20 10 569The number of cubic polynomials ( boldsymbol{P}(boldsymbol{x}) ) satisfying ( P(1)=2, P(2)=4, P(3)=6 ) ( P(4)=8 ) is ( mathbf{A} cdot mathbf{0} ) B. c. more than one but finitely manyt D. infinitely many 10 570Simplify: ( -45 p^{3} div 9 p^{2} ) 10 571If the sum of squares of the zeroes of the polynomials ( 6 x^{2}+x+k ) is ( frac{25}{36} ) find the value of k? 10 572If -1 is a zero of the polynomial ( p(x)= ) ( a x^{3}-x^{2}+x+4, ) find the value of ( a )10 573Divide: ( x^{7}-y^{7} ) by ( x-y ) 10 574If ( a+2 b=9 ) and ( a b=7 . ) Find the value of ( a^{2}+4 b^{2} ) 9 57557. If xy (x + y) = 1, then the val- ue of -x- yº is : (1) O (3) 3 (2) 1 (4)-2 9 576If ( a x^{n-1}+b x^{n-2}+c x^{n-3} ) is a cubic polynomial where ( n in N, ) then find the value of ( boldsymbol{n} ) 9 577Factorize: ( a^{4}-343 a ) ( mathbf{A} cdot a(a-7)left(a^{3}+a+36right) ) B ( cdot a(a-7)left(a^{2}+7 a+49right) ) ( mathbf{c} cdot a(4 a-7)left(a^{2}-a+49right) ) D. ( a(4 a-7)left(a^{3}+7 a-36right) ) 9 578Divide ( 3 x^{4}-5 x^{3} y+6 x^{2} y^{2}-3 x y^{3}+ ) ( boldsymbol{y}^{4} ) by ( boldsymbol{x}^{2}-boldsymbol{x} boldsymbol{y}+boldsymbol{y}^{2} ) What should be subtracted from the quotient to make it a perfect square? A ( cdot 2 x^{2} ) В. ( x^{2} ) ( c cdot y^{2} ) D. -xy 10 57966. If x-1=1, then the value of (1) (2) (4) O 10 580If ( x+frac{1}{x}=5, ) then find the value of ( x^{3}+frac{1}{x^{3}} ) A . 110 B. 115 ( c cdot 105 ) D. 100 9 581Divide: ( left(6 a^{5}+8 a^{4}+8 a^{3}+2 a^{2}+right. ) ( 26 a+35) ) by ( left(2 a^{2}+3 a+5right) ) Answer: ( 3 a^{3}-3 a^{2}+a+7 ) A . True B. False 10 582Give the zeros of polynomial and list their multiplicities: ( boldsymbol{P}(boldsymbol{x})=(boldsymbol{x}+mathbf{2})(boldsymbol{x}-mathbf{1}) ) 10 583( 26 z^{3}left(32 z^{2}-18right) div 13 z^{2}(4 z-3) ) A ( cdot z(4 z+3) ) в. ( (4 z+3) ) c. ( 4 z(4 z+3) ) D. None 10 584State if True or False Check whether the polynomial ( p(y)= ) ( 2 y^{3}+y^{2}+4 y-15 ) is a multiple of ( (2 y-3) ) A . B. can not say ( c ) D. can not be determine 9 585State whether True or False. Divide: ( x^{6}-8 ) by ( x^{2}-2, ) then the answer is ( x^{4}+2 x^{2}+4 ) A. True B. False 10 586State whether the statement is True or False. The cube of ( left(x-frac{1}{2}right) ) is equal to ( x^{3}- ) ( frac{3 x^{2}}{2}+frac{3 x}{4}-frac{1}{8} ) A . True B. False 9 587Find the zero of the polynomial given below: ( boldsymbol{p}(boldsymbol{x})=boldsymbol{9} boldsymbol{x}-boldsymbol{3} ) A ( cdot frac{6}{7} ) B. ( frac{1}{2} ) c. ( frac{1}{3} ) D. ( frac{7}{3} ) 10 588The value of ( frac{(1.5)^{2}+(4.7)^{3}+(3.8)^{3}-3 times 1 .}{(1.5)^{2}+(4.7)^{2}+(3.8)^{2}-1.5 times 4.7-4} ) ( A cdot 8 ) B. 9 c. 10 ( D ) 9 589Using the identity ( (boldsymbol{a}-boldsymbol{b})^{2}=boldsymbol{a}^{2}- ) ( 2 a b+b^{2} ) compute ( (x-6)^{2} ) 9 590Solve: ( 4 x^{4}-5 x^{3}-7 x+1 div 4 x-1 ) 10 591Work out the following divisions: ( (10 x-25) div(2 x-5) ) 10 592Illustration 2.13 Plot a graph for the equation y = x2 – 4x. CO3 10 593Factorize ( 25 a^{2}-9 b^{2} ) A ( cdot(5 a+b)(5 a-3 b) ) B. ( (a+3 b)(5 a-3 b) ) c. ( (5 a+3 b)(5 a-3 b) ) D. ( (5 a-3 b)(5 a+3 b) ) 9 594Find the remainder when ( 4 x^{3}-3 x^{2}+ ) ( 4 x-2 ) divided by ( (text { i) } x-1 ) (ii) ( x-2 ) 10 595If ( x+frac{1}{x}=7 ) the value of ( x^{4}+frac{1}{x^{4}} ) is A .2401 в. 2023 c. 2209 D. 2207 9 596If the polynomials ( left(2 x^{3}+a x^{2}+3 x-5right) ) and ( left(x^{3}+x^{2}-4 x-aright) ) leave the same remainder when divided by ( (x-1), ) find the value of ( a ) 9 597Is ( x^{8}+a^{8} ) divisible by ( x+a ? )10 598Using remainder theorem, find the remainder when ( 3 x^{2}+x+7 ) is divided by ( boldsymbol{x}+mathbf{2} ) A . 17 B. -23 c. 13 D. -29 9 599The polynomial ( p(x)=x^{4}-2 x^{3}+ ) ( 3 x^{2}-a x+3 a-7 ) when divided by ( x+ ) 1 leaves the remainder 19 Find the value of ( a ). Also find the remainder when ( p(x) ) is divided by ( x+ ) 2 A ( . a=5 ; 62 ) В. ( a=4 ; 62 ) c. ( a=5 ; 60 ) D. ( a=4 ; 60 ) 9 600ff ( x^{4}+2 x^{3}-3 x^{2}+x-1 ) is divided by ( x-2 . ) then the remainder is A . 12 B. 14 c. 16 D. 21 10 601Degree of the polynomial ( left(x^{3}-2right)left(x^{2}+right. ) 11) is ( mathbf{A} cdot mathbf{0} ) B. 5 ( c .3 ) D. 10 602If ( x=frac{1}{5-x} ) and ( x neq 5, ) find the value of ( x^{3}+frac{1}{x^{3}} ) A . 80 в. 240 c. 110 D. 530 9 603Divide the polynomial ( 39 y^{3}left(50 y^{2}-98right) ) by ( 36 y^{2}(5 y+7) ) 10 604Expand using suitable identities ( (2 a-3 b)^{3} ) 9 605Divide ( (24 x-42) ) by ( (4 x-7) ) A ( .4 x-7 ) B. 3 ( c cdot 6 ) D. 7 10 606If ( P(x), q(x) ) and ( r(x) ) are three polynomials of degree 2 , then prove that [ left\|begin{array}{lll} boldsymbol{p}(boldsymbol{x}) & boldsymbol{q}(boldsymbol{x}) & boldsymbol{r}(boldsymbol{x}) \ boldsymbol{p}^{prime}(boldsymbol{x}) & boldsymbol{q}^{prime}(boldsymbol{x}) & boldsymbol{r}^{prime}(boldsymbol{x}) \ boldsymbol{p}^{prime prime}(boldsymbol{x}) & boldsymbol{q}^{prime prime}(boldsymbol{x}) & boldsymbol{r}^{prime prime}(boldsymbol{x}) end{array}right\| text { is independent } ] of ( X ) 10 607Which type of polynomial, the given expression ( 5 x^{2}+x-7 ) is? 9 608Find a quadratic polynomial each with the give numbers as the sum and product of its zeros respectively. 4,1 10 609State whether the statement is True or False. The square of ( left(5-x+frac{2}{x}right) ) is equal to ( 21+x^{2}+frac{4}{x^{2}}-10 x+frac{10}{x} ) A. True B. False 9 610Write the degree of each polynomial given below: ( boldsymbol{x} boldsymbol{y}+boldsymbol{y} boldsymbol{z}-boldsymbol{z} boldsymbol{x}^{3} ) 10 611Perform the division: ( x^{3}-5 x^{2}+8 x-4 ) by ( boldsymbol{x}-mathbf{2} ) 10 612Simplify: ( frac{mathbf{6 x}^{2}+mathbf{9 x}}{mathbf{3 x}^{2}-mathbf{1 2 x}} )10 613Factorize: ( 9 a^{2}+frac{1}{9 a^{2}}-2-12 a+frac{4}{3 a} ) A ( cdotleft(3 a-frac{1}{3 a}right)left(3 a-frac{1}{3 a}-4right) ) B. ( left(a-frac{1}{3 a}right)left(7 a-frac{1}{3 a}-1right) ) c. ( left(3 a-frac{1}{3 a}right)left(2 a-frac{1}{3 a}-4right) ) D. ( left(3 a-frac{1}{3 a}right)left(4 a-frac{1}{3 a}-1right) ) 9 614Which of the following is a polynomial with only one zero? ( mathbf{A} cdot p(x)=2 x^{2}-3 x+4 ) B ( cdot p(x)=x^{2}-2 x+1 ) C ( . p(x)=2 x+3 ) D ( . p(x)=5 ) 10 615Write the polynomial in standard form and also write down their degree. ( left(x^{2}-frac{2}{3}right)left(x^{2}+frac{4}{3}right) )10 61666. If p, q, r are all real numbers, then (p-q) + (q-1)3 + (r-p)3 is equal to (1) (p -q (q-) (r-p) (2) 3(p-q) (9-) (r-p) (3) O (4) 1 9 617Solve: ( x^{3}-(x+1)^{2}=2001 ) ( A cdot 13 ) B. 16 c. 10 D. 21 10 618Carry out the following divisions ( 11 x y^{2} z^{3} ) by ( 55 x y z )10 619Evaluate the following using suitable identity: ( (10 x-1)^{2} ) ( (x-8 y)^{2} ) 9 620Simplify : ( left(frac{3}{2} x-0.45 yright)^{2} ) A. ( 2.25 x^{2}-1.35 x y+0.2015 y^{2} ) B. ( 2.15 x^{2}-1.35 x y+0.2025 y^{2} ) c. ( 2.25 x^{2}-1.35 x y+0.2025 y^{2} ) D. ( 2.25 x^{2}-1.25 x y+0.2025 y^{2} ) 9 621Write the degree of the following polynomial: ( 5 y+sqrt{2} ) 9 622Degree of the polynomial ( 4 x^{4}+0 x^{3}+ ) ( mathbf{0} boldsymbol{x}^{boldsymbol{5}}+mathbf{5} boldsymbol{x}+mathbf{7} ) is A . 4 B. 5 ( c cdot 3 ) D. 9 623Solution: ( 10 a^{2}(0.1 a-0.5 b) ) 9 624( (2 x+3 y)^{2}-16 z^{2} )9 625The simplified form of the expression given below is 🙁 y^{4}-x^{4}-y^{3} ) ( frac{x(x+y) x}{y^{2}-x y+x^{2}} ) A . 1 B. ( c cdot-1 ) D. 2 10 626Simplify: i) ( left(a^{2}-b^{2}right)^{2} ) ii) ( (2 x+5)^{2}-(2 x-5)^{2} ) 9 62759. If a and b are two odd positive integers, by which of the follow- ing integers is (a – b) always divisible? (2) 6 (3) 8 (4) 12 (1) 3 9 628Use the identity and expand the following ( (2 x+y)^{2} )9 629What is meant by division algorithm give example? 10 630If ( x+frac{1}{x}=sqrt{3}, ) find the value of ( x^{3}+ ) ( frac{1}{x^{3}} ) 9 631The product of two zeroes of the polynomial ( p(x)=x^{3}-3 x^{2}-6 x+8 ) is ( (-2) . ) Find all the zeroes of the polynomial 10 632Solve: ( left(y^{2}+10 y+24right) div(y+4) ) 10 633Find the degree of the following polynomial ( boldsymbol{x}^{9}-boldsymbol{x}^{4}+boldsymbol{x}^{12}+boldsymbol{x}-boldsymbol{2} ) A ( cdot 12 ) B. ( c cdot 4 ) D. 10 634Factorise the following: ( y^{6}+32 y^{3}-64 ) A ( cdot(y+2)left(y^{4}-2 y^{3}+8 y^{3}+8 y+16right) ) B . ( left(y^{2}+2 y-4right)left(y^{4}-2 y^{3}+8 y^{2}+8 y+16right) ) C ( cdotleft(y^{2}+2 y-4right)left(y^{4}-2 y^{3}+8 y^{2}right) ) D. ( (y+2)left(y^{4}-2 y^{3}+8 y^{2}+8 yright) ) 9 635What is the degree of the following polynomial expression: ( boldsymbol{u}^{frac{-1}{2}}+boldsymbol{3} boldsymbol{u}+boldsymbol{2} ) ( mathbf{A} cdot mathbf{1} ) B. 0 ( c cdot frac{-1}{2} ) D. Not Defined 9 63665. If x2 + y2 – 4x – 4y + 8 = 0, then the value of x-y is (1) 4 (2) 4 (3) O (4) 8 9 637Find the quotient and remainder when ( p(x)=x^{3}-3 x^{2}+5 x-3 ) is divided by ( g(x)=x^{2}-2 ) 10 638Find the remainder when ( p(x)=x^{2}+ ) ( 3 x+4 ) is divided by ( x+1 ) ( mathbf{A} cdot mathbf{0} ) B. ( c cdot 2 ) D. 3 9 639Using remainder theorem, find the remainder when ( 4 x^{3}+5 x-10 ) is divided by ( boldsymbol{x}-mathbf{3} ) A . 197 в. 113 ( c cdot-1 ) D. 0 9 640Find ( 302 times 308 )9 641Divide the following polynomial ( p(x) p(x) ) by the polynomial ( S(x) 5(x) ) ( boldsymbol{P}(boldsymbol{x})=frac{2}{3} boldsymbol{x}^{2}+mathbf{5} boldsymbol{x}+boldsymbol{6}, boldsymbol{S}(boldsymbol{x})=boldsymbol{x}+boldsymbol{6} ) 10 642Find the remainder if ( x-1 ) is divided by ( 5 x^{3}-2 x^{2}+3 x-22 ) 9 643The factors of ( boldsymbol{x}^{mathbf{4}}+boldsymbol{y}^{mathbf{4}}+boldsymbol{x}^{mathbf{2}} boldsymbol{y}^{mathbf{2}} ) are ( mathbf{A} cdotleft(x^{2}+y^{2}right)left(x^{2}+y^{2}-x yright) ) ( mathbf{B} cdotleft(x^{2}+y^{2}right)left(x^{2}-y^{2}right) ) C ( cdotleft(x^{2}+y^{2}+x yright)left(x^{2}+y^{2}-x yright) ) D. Factorization is not possible 9 644Find the value of ( (2-sqrt{1-x})^{6}+(2+ ) ( sqrt{1-x})^{6} ) 9 645Work out the following divisions. ( 96 a b c(3 a-12)(5 b+30) div 144(a- ) 4)( (b+6) ) ( mathbf{A} cdot 96 a b c ) B. ( 10 a b c ) c. ( 144 a b c(b-12) ) D. ( 24(a-b+c) ) 10 646Using the Remainder and Factor Theorem, factorise the following polynomial: ( x^{3}+10 x^{2}-37 x+26 )10 647If ( a^{2}+b^{2}+c^{2}=35 ) and ( a b+b c+ ) ( c a=23 ; ) find ( a+b+c ) ( A cdot pm 7 ) B. ±2 ( c .pm 9 ) ( mathrm{D} cdot pm 14 ) 9 648Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder: ( boldsymbol{y}^{4}+boldsymbol{y}^{2}, boldsymbol{y}^{2}-boldsymbol{2} ) 10 649If the zeros of the polynomial ( boldsymbol{f}(boldsymbol{x})= ) ( x^{3}-3 x^{2}+x+1 ) are ( a-b, a, a+b, ) find ( a ) and b. 10 650Using long division method, divide the polynomial ( 4 p^{3}-4 p^{2}+6 p-frac{5}{2} ) by ( 2 p-1 ) A ( cdot 2 p^{2}-p-frac{5}{2} ) В ( cdot 2 p^{2}+p+frac{5}{2} ) c. ( _{2 p^{2}-p+frac{5}{2}} ) D. None of the above 10 651Divide ( boldsymbol{x}(boldsymbol{x}+mathbf{1})(boldsymbol{x}+mathbf{2})(boldsymbol{x}+boldsymbol{3}) ) by ( (x+3)(x+2) ) A. ( x(x+3) ) в. ( x(x+2) ) c. ( (x+1) ) D. ( x(x+1) ) 10 652If ( a x^{3}+b x^{2}+c ) is divided by ( (x-3) ) then the remainder is: A. ( -27 a+9 b+c ) B . ( -27 a-9 b ) c. ( 27 a+9 b+c ) D. ( 27 a+9 b ) E ( .-27 a+9 b ) 9 653Factorise the expression and divide them as directed. ( 5 p qleft(p^{2}-q^{2}right) div 2 p(p+q) ) 10 654If ( a+b=1 a n d a^{2}+b^{3}+3 a b=k, ) then the value of k is ( mathbf{A} cdot mathbf{1} ) B. 3 ( c .5 ) D. 9 655Solve ( left(15 y^{4}+10 y^{3}-3 y^{2}right) div 5 y^{2} ) 10 656Evaluate ( 8.5 times 9.5 ) using suitable standard identity. 9 657ff both ( (x-2) ) and ( left(x-frac{1}{2}right) ) are factors of ( p x^{2}+5 x+r, ) show that ( p=r )9 658Find the cube of 239 659Which of the following is not a linear polynomial? A ( . p(y)=8 y+6 ) В. ( p(x)=8 y+2 x ) c. ( p(x)=4+frac{5 x}{x}+8 x^{circ} ) D. ( p(x)=4+5 x ) 9 660Expand ( left(2 y-frac{3}{y}right)^{3} ) A ( cdot 8 y^{3}-36 y+frac{54}{y}-frac{27}{y^{3}} ) в. ( 8 y^{3}+36 y+frac{54}{y}-frac{27}{y^{3}} ) c. ( 8 y^{3}-36 y-frac{54}{y}-frac{27}{y^{3}} ) D. None of these 9 661Find the degrees of the following polynomial ( 3-4 a b+5 b^{3}+2 a b^{2} ) 9 662Prove the following result by using suitable identities. ( (x-y)^{2}+(y-z)^{2}+(z-x)^{2}= ) ( 2left(x^{2}+y^{2}+z^{2}-x y-y z-z xright) ) 9 663Which of the following is a factor of the polynomial ( -2 x^{2}+7 x-6 ? ) A. ( -2 x-3 ) B. ( 2 x+2 ) c. ( x-6 ) D. ( 2 x-2 ) E . ( -2 x+3 ) 10 664Find the product. ( (3+sqrt{2})(2+sqrt{3})(3-sqrt{2})(2-sqrt{3}) )9 6653+ x + V3 -X=2 then x is 66. V3+r 13 + x – 13- equal to 9 666Simplify: ( a^{6}-b^{6} ) 9 667Divide ( 4 x^{3}+3 x^{2}-2 x+8 ) by ( x-2 )10 668Write the polynomial in standard form and also write down their degree. ( 4 z^{3}-3 z^{5}+2 z^{4}+z+1 ) 10 669Find the remainder when ( x^{4}+x^{3}- ) ( 2 x^{2}+x+1 ) is divided by ( x-1 ) ( A cdot 2 ) B. – ( c cdot-2 ) D. 9 67067. If a+b= 10 and a + b = 58, then a + b3 will be equal to (1) 340 (2) 540 (3) 270 (4) 370 9 671Use the identity ( (a+b)(a-b)=a^{2}- ) ( b^{2} ) to find the product of ( left(frac{2 x}{3}+1right)left(frac{2 x}{3}-right. ) ( mathbf{1} ) 9 672Find the degree of the polynomial: ( mathbf{7} boldsymbol{x}^{mathbf{3}}+mathbf{2} boldsymbol{x}^{mathbf{2}}+boldsymbol{x} ) 10 673If ( b ) is zero of the polynomial ( p(x)= ) ( a x^{2}-3(a-1) x-1, ) then find the value of ( boldsymbol{a} ) 9 674Divide ( x^{4}-y^{4} ) by ( x-y )10 675Factorize ( 27 m^{3}-216 n^{3} )9 676Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial (i) ( t^{2}-3,2 t^{4}+3 t^{3}-2 t^{2}-9 t-12 ) (ii) ( x^{2}+3 x+1,3 x^{4}+5 x^{3}-7 x^{2}+ ) ( 2 x+2 ) (iii) ( x^{3}-3 x+1, x^{5}-4 x^{3}+x^{2}+3 x+ ) ( mathbf{1} ) 10 677Find the remainder when ( p(x)=x^{3}- ) ( 4 x^{2}+3 x+1 ) is divided by ( (x-1) ) 9 678Classify the following polynomial based on their degrees: ( boldsymbol{y}+mathbf{3} ) 10 679Two angles in a triangle are in the ratio 4:5. If the sum of these angles is equal to the third angle, then the angles are (a) 180° (b) 40° (c) 50° (d) 90° . Ten 10 680Divide ( left(2 a^{2}-13 a b+15 b^{2}right) b y(a-5 b) ) 10 681Factorise the expressions and divide them as directed. ( 12 x yleft(9 x^{2}-16 y^{2}right) div 4 x y(3 x+4 y) ) A. ( 3(3 x-4 y) ) B. ( 3(3 x+4 y) ) c. ( 4(3 x-4 y) ) D. ( 3(4 x-3 y) ) 10 68252. If x + y + z = 6 and xy + y2 + 2x = 10 then the value of x3 + y + 22 – 3.xyz is : (1) 36 (2) 48 (3) 42 (4) 40 9 68357. If xy (x + y) = 1, then the val- ue of – +3,3--y is : (1) o (3) 3 (2) 1 (4) -2 9 68466. If a = 2-361, b = 3.263 and c = 5.624, then the value of a + b3 – + 3abc is (1) 35-621 (2) (3) 19.277 (4) 1 9 68570. If x 3 then the value of x1 + x2 + x + 1 is (1) (2) 1 (3) 2 (4) 3 9 68655. For what value of a, (x + a) is a factor of polynomial f(x)= x + ax? – 2x + a +6? (1) 2 (2) 3 (3) – 2 (4) – 3 9 687If ( a+frac{1}{a}=4 ) and ( a neq 0, ) find ( a^{2}+frac{1}{a^{2}} ) A . 14 B. 12 c. 17 D. 19 9 688Convert into factors ( a^{2}+4 a+4-b^{2} )9 689Evaluate ( (10 x-25) div(2 x-5) ) ( mathbf{A} cdot mathbf{5} ) B. 4 ( c cdot 3 ) D. 10 690Write the constant term of each of the following algebraic expressions: ( a^{3}- ) ( 3 a^{2}+7 a+5 ) 9 691If ( l^{2}+m^{2}+n^{2}=5, ) then ( (l m+m n+ ) ( (n) ) is ( mathbf{A} cdot geq(-5 / 2) ) B ( cdot geq(-1) ) ( c cdot leq 5 ) ( D . leq 3 ) 9 692Find ( alpha ) in order that ( x^{3}-7 x+5 ) may be a factor of ( x^{5}-2 x^{4}-4 x^{2}+19 x^{2}- ) ( 31 x+12+alpha ) 10 693Find the value of ‘a’ ( operatorname{in} 4 x^{2}+a x+ ) ( mathbf{9}=(mathbf{2} boldsymbol{x}+mathbf{3})^{2} ) A ( . a=34 ) B. ( a=11 ) c. ( a=12 ) D. ( a=33 ) 9 694If ( x=frac{4}{3} ) is a zeroes of the polynomial ( p(x)=6 x^{3}-11 x^{2}+k x-20, ) find the value of ( k ) 10 695Divide ( 4 sqrt{2} y^{3} ) by ( 3 sqrt{2} y^{2} ) The answer is ( frac{m y}{3} ) Then ( m= ) 10 696If ( a-frac{1}{a}=5 ; ) find ( a^{2}+frac{1}{a^{2}}-3 a+frac{3}{a} ) A . 10 B. 14 ( c cdot 6 ) D. 12 9 697Which type of polynomial is ( 45 y^{2} ? ) A. Linear Polynomial c. Cubic Polynomial 9 698Find the missing terms such that the given polynomial become a perfect square trinomial: ( 81 x^{2}+dots-1 ) 10 699Factorise: ( mathbf{2} sqrt{mathbf{2}} boldsymbol{a}^{mathbf{3}}+mathbf{1 6} sqrt{mathbf{2}} boldsymbol{b}^{mathbf{3}}+boldsymbol{c}^{mathbf{3}}-mathbf{1 2 a b} boldsymbol{c} ) 9 700If ( (n+1)^{3}-(n)^{3}=n+1, ) then which of the following can be the value of ( n ? ) ( mathbf{A} cdot mathbf{0} ) B . 2 c. -2 D. Cannot be determined 9 701Find the square of: 607 A. 368549 B. 368449 ( c .368349 ) D. 368249 9 702Zero of the polynomial ( p(x)=2-5 x ) is10 703Use synthetic division to find the quotient ( Q ) and remainder ( R ) when dividing ( boldsymbol{f}(boldsymbol{x})=boldsymbol{x}^{5}+boldsymbol{x}^{4}+boldsymbol{x}^{3}+boldsymbol{2} boldsymbol{x}-boldsymbol{5} ) by ( boldsymbol{x}+boldsymbol{i} ) ( (i=sqrt{(}-1) text { imaginary unit }) ) 9 704Verify ( f(x)=2 x^{3}+11 x^{2}-7 x-6 ) is the factor of ( (x-1) ) using factor theorem. 10 705Find the remainder by using remainder theorem when polynomial ( x^{3}-3 x^{2}+ ) ( x+1 ) is divided by ( x-1 ) ( A cdot 8 ) B. ( c cdot-2 ) D. -7 9 706State whether the statement is True or False. Expanding ( (a-b+c)^{2} ) we get ( a^{2}+ ) ( b^{2}+c^{2}-2 a b-2 b c+2 c a ) A. True B. False 9 707Find the value of ( 87^{2}-13^{2} ) A . 7300 B. 7350 ( c .7400 ) D. 7450 9 708Divide ( 6 x^{4}+13 x^{3}+39 x^{2}+37 x+45 ) by ( 3 x^{2}+2 x+9 ) 10 709Degree of a constant term of a polynomial is A . B. 0 ( c cdot 2 ) D. Not defined 9 7106. Two numbers are such that the ratio between them is 3:5. If each is increased by 10, the ratio between the new numbers so formed is 5:7. Find the original numbers. (a) 12 (b) 20 (c) 25 (d) 15 10 711Using identity find the value of ( (4.7)^{2} )9 712Divide ( left(36 x^{2}-4right) ) by ( (6 x-2) ) A. ( 3 x-1 ) B. ( 3 x+1 ) c. ( 6 x-2 ) D. ( 6 x+2 ) 10 713Find the remainder when ( p(x)=x^{3}+ ) ( 3 x^{2}+3 x+1, ) is divided by ( x ) ( A ) B. ( c cdot-1 ) D. 9 714State true or false: ( frac{3}{4 x+3}=frac{1}{4} ) A. True B. False 10 715( boldsymbol{p}(boldsymbol{x})=mathbf{2} boldsymbol{x}^{3}-mathbf{3} boldsymbol{x}^{2}-mathbf{5} ) is a polynomial A . linear B. cubic D. constant 10 716State whether True or False. Divide: ( x^{2}+3 x-54 ) by ( x-6, ) then the answer is ( x+9 ) A. True B. False 10 717Expand ( (sqrt{10} x-sqrt{5} y)^{2} ) using appropriate identity 9 718When ( 5 x^{13}+3 x^{10}-k ) is divided by ( x+ ) 1, the remainder is 20. The value of k is A . – 22 B . -12 ( c cdot 8 ) D. 28 ( E cdot 14 ) 9 719Find the degree of each of the polynomials given below ( 5 t-sqrt{3} )10 720Divide and write the quotient and remainder: ( left(4 x^{4}-5 x^{3}+0 x^{2}-7 x+1right) div(4 x-1) ) 10 721Find whether ( x-1 ) is a factor of ( 2 x^{2}- ) ( mathbf{5} boldsymbol{x}+mathbf{3} ) 10 722What number should be added to ( x^{3}- ) ( 9 x^{2}-2 x+3 ) so that the remainder may be 5 when divided by ( (x-2) ) 9 723Find the product of ( left(frac{1}{2} m-frac{1}{3}right)left(frac{1}{2} m+right. ) ( left.frac{1}{3}right)left(frac{1}{4} m^{2}+frac{1}{9}right) ) 9 724Degree of which of the following polynomial is zero? ( A cdot x ) B . 15 c. ( y ) D. ( _{x+frac{1}{x}} ) 9 725In each of the following two polynomials find the value of ( a, ) if ( x+a ) is a factor. ( x^{4}-a^{2} x^{2}+3 x-a ) 10 726. x² + 8 60. Simplify: x4 + 4×2 +16 x + 2 021- x+ 2 x² + 2 3 + 2×2 + 4 (3) x² + 2 x3 + 3×2 +8 1 9 Hope you will like above questions on polynomials and follow us on social network to get more knowledge with us. If you have any question or answer on above polynomials questions, comments us in comment box.
## 17Calculus Precalculus - Use Elimination to Solve Systems of Linear Equations ##### 17Calculus Solve by Elimination This is the best technique to solve systems of equations since it works all the time, you can control the numbers to avoid fractions until near the end of the solution and it will prepare you for the fourth technique using matrices. So it is important to learn this technique well. Here is a good video explaining this technique while working an example, AND he explains why this technique works. ### Khan Academy - Solving Systems of Equations by Elimination [12min-43secs] Okay, this technique is pretty straight-forward, so we are going to let you jump right into the practice problems. Practice Unless otherwise instructed, solve these linear systems using elimination. $$2x+3y=4$$ $$-2x+7y=16$$ Problem Statement $$2x+3y=4$$ $$-2x+7y=16$$ Solution ### PatrickJMT - 1719 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$x-3y=6$$ $$4x-3y=10$$ Problem Statement $$x-3y=6$$ $$4x-3y=10$$ Solution ### PatrickJMT - 1720 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$4x-2y=16$$ $$5x+2y=11$$ Problem Statement $$4x-2y=16$$ $$5x+2y=11$$ Solution ### MIP4U - 1721 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$4x+3y=8$$ $$x-3y=7$$ Problem Statement $$4x+3y=8$$ $$x-3y=7$$ Solution ### MIP4U - 1724 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$3x+5y=4$$ $$-2x+3y=10$$ Problem Statement $$3x+5y=4$$ $$-2x+3y=10$$ Solution ### MIP4U - 1725 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$2x-3y=-1$$ $$-4x+6y=5$$ Problem Statement $$2x-3y=-1$$ $$-4x+6y=5$$ Solution ### MIP4U - 1726 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$5x+2y=4$$ $$5x+2y=-2$$ Problem Statement $$5x+2y=4$$ $$5x+2y=-2$$ Solution ### MIP4U - 1727 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$3x+y=-10; 7x+5y=-18$$ Problem Statement $$3x+y=-10; 7x+5y=-18$$ Solution ### MIP4U - 1728 video solution video by MIP4U Log in to rate this practice problem and to see it's current rating. $$2x=-6y+8$$ $$3x-5y=2$$ Problem Statement $$2x=-6y+8$$ $$3x-5y=2$$ $$(13/7,5/7)$$ Problem Statement $$2x=-6y+8$$ $$3x-5y=2$$ Solution The instructor in this video runs out of time and does not finish the problem. He gets $$y=5/7$$ and then stops. Using his work, here is how to get the final answer. Substituting $$y=5/7$$ into the first original equation, we have $$\begin{array}{rcl} 2x & = & -6y+8 \\ 2x & = & -6(5/7)+8 \\ 2x & = & -30/7+56/7 \\ 2x & = & 26/7 \\ x & = & 13/7 \end{array}$$ $$(13/7,5/7)$$ Log in to rate this practice problem and to see it's current rating. $$4x-4y+8z=20; 8x+4y-4z=4; 12x-8y-12z=-40$$ Problem Statement $$4x-4y+8z=20; 8x+4y-4z=4; 12x-8y-12z=-40$$ Solution ### PatrickJMT - 1722 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. $$2x-y+z=3; 5x+2y-3z=1; 2x+y-z=2$$ Problem Statement $$2x-y+z=3; 5x+2y-3z=1; 2x+y-z=2$$ Solution ### PatrickJMT - 1723 video solution video by PatrickJMT Log in to rate this practice problem and to see it's current rating. When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications. DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.
# The Ultimate Guide: Exploring the Fascinating Math Behind 42 Divided by 10 ## Understanding the Concept: 42 Divided by 10 ### The Basics of Division When it comes to understanding the concept of dividing numbers, it is important to start with the basics. Division is an arithmetic operation that allows us to split a given number into equal parts. In the case of 42 divided by 10, we are essentially dividing 42 into 10 equal groups or parts. Key Point: Division is the inverse operation of multiplication. It can be thought of as the process of finding out how many times one number can be evenly divided by another. You may also be interested in: Convert 36 ml to oz: An Easy Guide to Converting Milliliters to Ounces ### The Quotient and Remainder In division, we often come across two important terms: quotient and remainder. The quotient is the whole number part that results from the division, while the remainder is the leftover amount that cannot be evenly divided. In the case of 42 divided by 10, the quotient would be 4, meaning that 42 can be divided into 10 equal groups with 4 in each group. The remainder would be 2, as there are 2 left over after dividing 42 evenly into 10 groups. Key Point: The quotient tells us how many groups we have and the remainder tells us what is left after dividing the number evenly into those groups. It is important to consider both the quotient and remainder when interpreting the result of a division. ### Decimal and Fractional Division In some cases, division may result in a decimal or fractional answer. When dividing 42 by 10, the answer is a decimal 4.2. This means that 42 can be evenly split into 10 groups, with each group containing 4.2 units. Alternatively, we can express the result as a fraction, which would be 21/5. This fraction shows that 42 can be divided into 10 groups, with each group containing 4 whole units and a remainder of 2 units. Key Point: Division can produce decimal or fractional results when the number being divided cannot be evenly split into the desired number of groups. These types of results are helpful when dealing with measurements or when a more precise answer is needed. Understanding the concept of 42 divided by 10 gives us a foundation for further exploring division and its applications. Whether it’s dividing numbers in math problems, calculating averages or ratios, or even splitting resources among a group, division is a fundamental operation that helps us make sense of quantities and distributions. Remembering the basics of division, the role of quotient and remainder, and the potential for decimal or fractional answers will help us confidently tackle division-related challenges in various contexts. You may also be interested in: 660 ml to oz: Ultimate Conversion Guide and Tips for Easy Measurements ## Simple Steps to Divide 42 by 10 ### Step 1: Understand the Basics of Division Before we delve into the steps to divide 42 by 10, let’s quickly refresh our understanding of division. Division is a mathematical operation used to distribute a quantity equally into smaller parts. The dividend is the number being divided, in this case, 42, and the divisor is the number by which we divide, which is 10. The quotient is the result or the answer to the division problem. Step 2: Long Division Method The most common method used to divide large numbers is the long division method. This method allows us to break down the division problem into smaller, more manageable steps. To divide 42 by 10 using long division, we start by dividing the first digit of the dividend (4) by the divisor (10). Since 4 is less than 10, we add a decimal point to the quotient and bring down the next digit (2) to continue the division. Step 3: Continue the Division Now that we have 2 as our new dividend, we divide it by 10. The quotient is 0.2. However, we still have a remainder of 2. To include the remainder in the final answer, we can express it as a fraction, making the complete division result 4.2. Step 4: Verify and Simplify To ensure the accuracy of our division, we should verify our answer. We can multiply the quotient (4.2) by the divisor (10) and check if it equals the dividend (42). In this case, 4.2 multiplied by 10 does indeed equal 42, confirming that our division was done correctly. Remember, practice is key when it comes to mastering division, and using these simple steps can help you divide numbers accurately. ## Explore the Calculation: 42 ÷ 10 Calculating the result of 42 ÷ 10 is a fundamental math problem that can be easily solved. Dividing 42 by 10 will give you the answer of 4.2. It is important to note that this is a decimal number. The division symbol, ÷, represents the operation of dividing one number by another. Dividend and Divisor: In this calculation, 42 is the dividend and 10 is the divisor. The dividend is the number that is being divided, while the divisor is the number by which the dividend is divided. Understanding the role of both the dividend and divisor is essential to correctly solve a division problem. Quotient: The result of a division problem is called the quotient. In this case, the quotient is 4.2. The quotient represents the answer or the number of times the divisor can be evenly divided into the dividend. Decimal Places: When dividing a number with decimals, like 42 ÷ 10, it is important to consider the decimal places in the quotient. In this example, the quotient has one decimal place, which indicates that the result is not a whole number. In summary, dividing 42 by 10 results in a quotient of 4.2. Understanding the concept of the dividend, divisor, and quotient is crucial for correctly solving division problems. Remember to consider the decimal places when dealing with division involving decimal numbers. ## The Importance of 42 Divided by 10 in Mathematics You may also be interested in: Converting Distance: Discover How to Easily Convert 6.5 km to Miles ### Understanding the Value of the Division Operation In the realm of mathematics, division is a fundamental operation used to distribute a quantity into parts or groups of equal size. The division of numbers plays a vital role in solving equations, calculating proportions, and determining ratios. One particular division that holds significance is the division of 42 by 10. The Importance of Ratios and Proportions: Mathematical concepts such as ratios and proportions heavily rely on division. When dividing 42 by 10, we obtain the decimal value of 4.2. This result can be interpreted as a ratio of 4.2:1 or 4.2 parts to 1 whole. Ratios and proportions are prevalent in various fields such as finance, cooking, and engineering. Understanding the division of 42 by 10 provides a foundation for comprehending and applying these concepts. Application in Real-World Scenarios: The division of 42 by 10 has tangible applications in everyday life. For instance, in financial planning, this division can be used to determine the interest rate on a loan or the percentage increase or decrease in a stock’s value. It also proves invaluable in converting units of measurement. By dividing 42 centimeters by 10, we can ascertain the length in decimeters. These practical applications highlight the importance of mastering division skills, including the operation of dividing 42 by 10. In conclusion, the division of 42 by 10 holds significance in mathematics due to its role in understanding ratios, proportions, and practical applications in various fields. By grasping the fundamentals of division, including this specific calculation, individuals can gain a greater understanding of mathematical concepts and apply them in real-world scenarios. ## Common Mistakes to Avoid When Dividing 42 by 10 ### 1. Ignoring the Order of Operations When dividing numbers, it’s crucial to follow the order of operations, also known as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). One common mistake is to neglect this rule and perform multiplication or division before addition or subtraction. In the case of dividing 42 by 10, you should perform the division after any multiplication or addition that occurs within the equation to get the correct result. ### 2. Forgetting to Account for Remainders Another mistake that can be made when dividing numbers is failing to consider remainders. Dividing 42 by 10 will give you a quotient of 4, but there will be a remainder of 2. Ignoring the remainder or rounding it off can lead to inaccurate results. It’s essential to take the remainder into account, especially when dealing with values that require precise calculations or further analysis. ### 3. Using the Wrong Mathematical Operator The choice of the mathematical operator is vital when dividing numbers. Using the wrong operator, such as multiplication instead of division, can lead to incorrect results. To divide 42 by 10 in HTML, use the forward slash (/) as the division operator between the numbers. Mistakenly using an asterisk (*) for multiplication would yield an erroneous outcome. It’s important to be aware of these common mistakes when dividing numbers like 42 by 10. By following the order of operations, considering remainders, and using the correct mathematical operator, you can ensure accurate results in your mathematical calculations. Remembering these principles will help avoid errors in HTML coding when dividing numbers and contribute to a smoother web development process.
Problem of the Week Updated at Aug 1, 2022 12:33 PM This week we have another equation problem: How can we solve the equation $${(\frac{n}{5}-3)}^{2}-3=1$$? Let's start! ${(\frac{n}{5}-3)}^{2}-3=1$ 1 Add $$3$$ to both sides.${(\frac{n}{5}-3)}^{2}=1+3$2 Simplify $$1+3$$ to $$4$$.${(\frac{n}{5}-3)}^{2}=4$3 Take the square root of both sides.$\frac{n}{5}-3=\pm \sqrt{4}$4 Since $$2\times 2=4$$, the square root of $$4$$ is $$2$$.$\frac{n}{5}-3=\pm 2$5 Break down the problem into these 2 equations.$\frac{n}{5}-3=2$$\frac{n}{5}-3=-2$6 Solve the 1st equation: $$\frac{n}{5}-3=2$$.1 Add $$3$$ to both sides.$\frac{n}{5}=2+3$2 Simplify $$2+3$$ to $$5$$.$\frac{n}{5}=5$3 Multiply both sides by $$5$$.$n=5\times 5$4 Simplify $$5\times 5$$ to $$25$$.$n=25$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$n=25$7 Solve the 2nd equation: $$\frac{n}{5}-3=-2$$.1 Add $$3$$ to both sides.$\frac{n}{5}=-2+3$2 Simplify $$-2+3$$ to $$1$$.$\frac{n}{5}=1$3 Multiply both sides by $$5$$.$n=1\times 5$4 Simplify $$1\times 5$$ to $$5$$.$n=5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$n=5$8 Collect all solutions.$n=25,5$Donen=25,5
Vectors and Maps Can your students read a map? Vectors are multidimensional numbers used for making maps. Vectors are used by surveyors, archeologists, cartographers, and of course, physicists. Below are simple activities that will work for introducing vectors in elementary school as part of map making and reading exercises. Objective: Students will learn how to write and read vectors by doing activities which use vectors. Reformatted: January 2010 Part 1: Distance-Distance Vectors Distance-Distance vectors represent how far over and how far forward. They are properly written like: However for ease they are frequently written as [2, 3]. Activity 1: Classroom Vectors Make 2 sets of signs with numbers starting with 1 going as high as you need. Place the signs on two walls every foot starting from a corner numbering every foot across the room. Have students determine what the classroom vector is for their seat. For example if they see the number 13 straight in front of them and the number 8 directly to the side the vector for their seat is [13, 8]. Once they have learned to identify the vector for their location have them make maps of the room by using graph paper and numbering the grid on the graph paper to match the room. This activity may be extended to three dimensional vectors by labeling the feet up from the floor. For example if a student's head vector written as [ 6, 8, 4], the he is standing 6 feet over, 8 feet forward, and his head is 4 feet above the floor. Related pages at this site: Activity 2: Chess board Vectors Label two edges of a chess board with the numbers 1 through 8. Have students place pieces on the board and identify the vector to the location of the piece. Be sure they count over-up not up-over. Have them discuss the difference between [ 5, 2] and [2, 5]. Activity 3: Applied Vectors and maps. Most street maps substitute one of the dimensions as letters rather than numbers. Students may replace the letters on the map with number and give the map vector for various locations. On a map of the world the equator and Greenwich mean time are the standard lines for marking vectors. Notice that locations to the west or south require the use of negative numbers in the vectors. Center city Philadelphia is laid out such that vectors may be used to assign addresses. The East-West starting point is the River with street going 1 to 23 going west. City Hall is the starting point moving north and south with address going up by 100 for each block moved north. Archeologists use vectors to map their sites. Many pictures of digs show a set of strings stretched across the dig to the the vector to objects. Adding distance-distance vectors is very easy. Students may discover the method using the following reasoning: The first vector represents a starting point, the second vector represents how far someone moved and the sum represents where they ended up. This may be demonstrated by moving pieces on the chessboard or having students walk out the steps in the classroom. Here are student started at a point 3 steps over and 5 steps forward. She then moved over 4 more steps and up 6 more steps to stop at 7 steps over, 11 forward. Part 2: distance angle vectors Distance-angle vectors represent how far and at what angle. They may be written as 3/45 -that's 3 followed by the angle symbol followed by 45. Activity 6: Applied Vectors for discussion Surveyors use distance angle vectors when they plot out land. The surveyor holding the striped pole is marking an exact point. The device that the other surveyor looks through calculates the angle and distance to the striped pole. Look at the north pole on a globe. The line radiating from it represent angle and the circles may be used to calculate distance (Notice how this is not true at the equator where the line represent distance-distance vectors.) Pilots use maps that show the distance and angle from various airports.
Open in App Not now # Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.2 • Last Updated : 12 Mar, 2021 ### (iv) sin x = 3/5, x in quadrant I Solution: (i) cot x = 12/5, x in quadrant III As we knew that tan x and cot x are positive in third quadrant and sin x, cos x, sec x, cosec x are negative. By using the formulas, tan x = 1/cot x = 5/12 cosec x = = -13/5 sin x = 1/cosec x =- 5/13 cos x = = -12/13 sec x = 1/cos x = – 13/12 Hence, the values of the other five trigonometric functions are: sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12 (ii) cos x = -1/2, x in quadrant II As we knew that sin x and cosec x are positive in second quadrant and tan x, cot x, cos x, sec x are negative. By using the formulas, we get sin x = = -2 Hence, the values of the other five trigonometric functions are: sin x = âˆš3/2, tan x = -âˆš3, cosec x = 2/âˆš3, cot x = -1/âˆš3, sec x = -2 (iii) tan x = 3/4, x in quadrant III As we knew that tan x and cot x are positive in third quadrant and sin x, cos x, sec x, cosec x are negative. By using the formulas, sin x = Hence, the values of the other five trigonometric functions are: sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3 (iv) sin x = 3/5, x in quadrant I As we knew that, all trigonometric ratios are positive in first quadrant. So, by using the formulas, tan x = Hence, the values of the other five trigonometric functions are: cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3 ### Question 2. If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x. Solution: Given: Sin x = and x lies in the second quadrant. We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative. So, by using the formulas, we get Cos x = tan x = sin x/cos x sec x = 1/cos x Sec x + tan x = ((-13/5) +(-12/5)) = (-13 – 12)/5 = -25/5 = -5 Hence, the value of Sec x + tan x = -5 ### Question 3. If sin x = 3/5, tan y = 1/2, and Ï€â€‹/2 < x < Ï€ < y < 3Ï€â€‹/2 find the value of 8 tan x -âˆš5 sec y. Solution: Given, sin x = 3/5, tan y = 1/2, and Ï€â€‹/2 < x< Ï€< y< 3Ï€â€‹/2 Here, x is in second quadrant and y is in third quadrant. So, cos x and tan x are negative in second quadrant and sec y is negative in third quadrant. So, by using the formula, we get cos x = tan x = sin x/ cos x cos x = We know that sec y = 8tan x â€“ âˆš5 sec y = 8 Ã— (-3)/(4) â€“ âˆš5 Ã— (-âˆš5/2) = -6 + (5/2) = (-12 + 5)/2 = -7/2 8tan x â€“ âˆš5 sec y = -7/2 Hence, the value of 8 tan x â€“ âˆš5 sec y = -7/2 ### Question 4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x. Solution: Given, sin x + cos x = 0 and x lies in fourth quadrant. sin x = -cos x sin x/cos x = -1 So, tan x = -1 (since, tan x = sin x/cos x) cos x and sec x are positive in fourth quadrant and all other ratios are negative. So, by using the formulas, sec x = cos x = 1/sec x sin x = sec x = Hence, the value of sin x = -1/âˆš2 and cos x = 1/âˆš2 ### Question 5. If cos x = -3/5 and Ï€ < x < 3Ï€/2 find the values of other five trigonometric functions and hence evaluate Solution: Given, cos x = -3/5 and Ï€ <x < 3Ï€/2 tan x and cot x are positive in the third quadrant and all other rations are negative. Now, by using the formulas, we get sin x = â€“ tan x = sin x/cos x cot x = 1/tan x sec x = 1/cos x cosec x = 1/sin x sin x = tan x = cot x = sec x = cosec x = Now we evaluate: My Personal Notes arrow_drop_up Related Articles
July 21, 2024 Learn how to solve linear equations with this comprehensive guide. We cover step-by-step instructions, common mistakes, real-world applications, and different strategies, including substitution and graphing methods. Discover advanced applications, including matrix algebra and solving systems of differential equations. ## I. Introduction Linear equations are one of the fundamental concepts in algebra that students encounter in middle school or high school. It involves a basic equation, where the variable is raised to the power of one and consists of linear terms. Solving linear equations is an essential skill for anyone who wants to pursue higher math, sciences or engineering. This article provides a comprehensive guide on solving linear equations. We will cover all the necessary steps, common mistakes to avoid, and different approaches to solving linear equations. Furthermore, we will delve into the real-world applications of linear equations and advanced applications relevant to university-level math. ## II. Step-by-Step Guide to Solving Linear Equations Before we dive into the step-by-step guide let’s first define what we are trying to solve for, the linear equation. A linear equation takes on the general form of ax+b=c, where x is the variable we are solving for, a, b, and c are constants. We may have multiple variables, but the focus would remain on obtaining the solution for the given variable. The following are the steps to solving linear equations: 1. Isolate the variable on one side of the equation. 2. Use inverse operations to solve for the variable. 3. Verify the solution by checking the answer. The first step is to place all variable terms on one side of the equation, leaving constant terms on the other side. The second step is to perform inverse operations, which means undoing any arithmetic operation (addition, subtraction, multiplication, or division) to isolate the variable. Lastly, we assume the value obtained as the solution (x) and check our answers by substituting x back into the original equation. Here’s an example of how to solve a linear equation: 2x+5=15 First, isolate the variable(2x) and move the constant to a separate side by subtracting 5 from both ends. 2x+5-5=15-5 2x=10 Next, divide by 2 on both sides to get the value of x. 2x/2 = 10/2 x=5 Lastly, check the answer by plugging x=5 back into the original equation: 2(5)+5=15 The result satisfies the original equation; hence, x=5 is the correct solution. ## III. Common Mistakes to Avoid When Solving Linear Equations Common mistakes when solving linear equations include errors in simplifying or adding/subtracting values, distributing values inconsistently, or even changing the wrong terms. It is vital always to remember the order of operations (PEMDAS) and avoid rushing through the problem to ensure no details are missed in the calculation. Double-checking the final answer before submission is another useful tip to detect any potential errors. ## IV. Solving Linear Equations Using Graphs Another method to solve linear equations is by graphing them. Graphs are useful when dealing with complex expressions since they provide a visual representation of the problem. To graph a linear equation, we must know either the slope-intercept or point-slope form of the equation given. Here is an example of graphing equation 3x+2y=6 using the slope-intercept form: 1. First, we need to rearrange the equation and solve for y (the slope-intercept form) 2. 2y=-3x+6 3. y=-3/2x+3 The slope-intercept form is y=mx+b , where m is equal to the slope, and b is the y-intercept. From the given equation, we know that the slope (m) is -3/2, and the y-intercept (b) is 3. We can now draw a line on our x-y graph using the slope (-3/2) and y-intercept (3) as a reference point. Next, we find the solution of the equation by locating the point (x,y) where the line intercepts the x-axis. ## V. Real-World Applications of Linear Equations Linear equations have essential practical applications in various fields, such as ‘cost of good sold’ analysis in accounting, calculating depreciation of assets, and calculating rates of change in sciences. One example of a real-world application of linear equations in everyday life is discounts given by retailers to attract customers. For instance, a clothing store provides 25% off on all products. A customer comes into the shop and buys a \$200 jacket. The amount of discount she receives is the product price multiplied by the discount percentage, i.e., Discount = Price x (Discount Percentage/100) Discount= \$200 x (25/100) Discount= \$50 Therefore, the final cost of the jacket the customer pays is \$200 – \$50 = \$150. ## VI. Different Strategies for Solving Linear Equations There are different strategies to solve linear equations, including substitution, elimination, matrix, and graphical methods. The substitution method involves isolating one variable in a linear equation and substituting the result into the other equation to obtain an equation in one variable. The elimination method requires adding or subtracting equations to remove one variable simultaneously. Matrix methods involve Gauss-Jordan elimination and finding the inverse of coefficients of variables. Graphical methods involve plotting linear equations on an x-y plane and determining the point of intersection as a solution. One example of applying substitution is solving a system of equations. Say, for instance, we have two linear equations: 2x + 3y = 7 3x + 2y = 8 From the first equation place 2x=7-3y and substitute in the second equation. 3(7-3y/2) + 2y = 8 21 – 9y/2 + 2y = 8 21-9y+4y=16 21-5y=16 -5y=-5 y=1 Substitute 1 in the first equation 2x+3(1)=7 2x=4 x=2 Thus the solution to the system of equations is (2,1). ## VII. Advanced Applications of Linear Equations Advanced applications of linear equations include solving systems of differential equations in physics, linear programming in economics, and machine learning in computer science. Matrices are essential to solving systems of linear equations in matrix algebra. Example: Consider the following system of equations: x + y + z = 6 2y + 5z = -4 2x + 5y – z = 27 We can create a matrix system for this with one matrix containing all the coefficients and a separate one having the values we are setting equal to: [1 1 1 [6 0 2 5 -4 2 5 -1] =27] Next, we could solve for the identity matrix by using the Row Echelon form: [1 1 1 6 [1 1 1 6 0 2 5 -4 –> 0 2 5 -4/2 =-2 0 3 -3 15] 0 1 1 5 Finally, we would use back-substitution to solve for the variables: x=1, y=2, z=3. ## VIII. Conclusion Solving linear equations is a fundamental concept that we use in many practical areas of life. It can be challenging and intimidating for many students, but with perseverance, anyone can master it. In this article, we have covered all the necessary details, including the step-by-step guide, common mistakes, different strategies, and advanced applications of linear equations. We encourage readers to practice and apply these concepts to excel in their math and science courses.
# Sum of all substrings of a string representing a number \| Set 2 (Constant Extra Space) Last Updated : 12 Jul, 2022 Given a string representing a number, we need to get the sum of all possible sub strings of this string. Examples : ```Input : s = "6759" Output : 8421 sum = 6 + 7 + 5 + 9 + 67 + 75 + 59 + 675 + 759 + 6759 = 8421 Input : s = "16" Output : 23 sum = 1 + 6 + 16 = 23``` We have discussed a solution in below post. Sum of all substrings of a string representing a number \| Set 1 The solution is based on a different approach which does not use any extra space. This problem can be viewed as follows. Let number be s = “6759” ``` 1 10 100 1000 6 1 1 1 1 7 2 2 2 5 3 3 9 4``` The above table indicates that, when all the substrings are converted further to the ones, tens, hundreds etc.. form, each index of the string will have some fixed occurrence. The 1st index will have 1 occurrence each of ones, tens etc..The 2nd will have 2, 3rd will have 3 and so on. One more point is that the occurrence of the last element will only be restricted to ones. Last 2nd element will be restricted to ones and tens. last 3rd will be up to a hundred and so on. From the above points lets find out the sum. ```sum = 6(11 + 110 + 1100 + 11000) + 7(21 + 210 + 2100) + 5(31 + 310) + 9(41) = 61(1111) + 72(111) + 53(11) + 94(1) = 6666 + 1554 + 165 + 36 = 8421``` Now, to handle the multiplication we will be having a multiplying factor which starts from 1. It’s clear from the example that the multiplying factor(in reverse) is 1, 11, 111, … and so on. So the multiplication will be based on three factors. number, its index, and a multiplying factor. Implementation: ## C++ `// C++ program to print sum of all substring of` `// a number represented as a string` `#include ` `using` `namespace` `std;` `// Returns sum of all substring of num` `int` `sumOfSubstrings(string num)` `{` ` ``long` `long` `int` `sum = 0; ``// Initialize result` ` ``// Here traversing the array in reverse` ` ``// order.Initializing loop from last` ` ``// element.` ` ``// mf is multiplying factor.` ` ``long` `long` `int` `mf = 1;` ` ``for` `(``int` `i=num.size()-1; i>=0; i--)` ` ``{` ` ``// Each time sum is added to its previous` ` ``// sum. Multiplying the three factors as` ` ``// explained above.` ` ``// s[i]-'0' is done to convert char to int.` ` ``sum += (num[i]-``'0'``)(i+1)mf;` ` ``// Making new multiplying factor as` ` ``// explained above.` ` ``mf = mf10 + 1;` ` ``}` ` ``return` `sum;` `}` `// Driver code to test above methods` `int` `main()` `{` ` ``string num = ``"6759"``;` ` ``cout << sumOfSubstrings(num) << endl;` ` ``return` `0;` `}` ## Java `// Java program to print sum of all substring of` `// a number represented as a string` `import` `java.util.Arrays;` `public` `class` `GFG {` ` ` ` ``// Returns sum of all substring of num` ` ``public` `static` `long` `sumOfSubstrings(String num)` ` ``{` ` ``long` `sum = ``0``; ``// Initialize result` ` ` ` ``// Here traversing the array in reverse` ` ``// order.Initializing loop from last` ` ``// element.` ` ``// mf is multiplying factor.` ` ``long` `mf = ``1``;` ` ``for` `(``int` `i = num.length() - ``1``; i >= ``0``; i --)` ` ``{` ` ``// Each time sum is added to its previous` ` ``// sum. Multiplying the three factors as` ` ``// explained above.` ` ``// s[i]-'0' is done to convert char to int.` ` ``sum += (num.charAt(i) - ``'0'``) (i + ``1``) * mf;` ` ` ` ``// Making new multiplying factor as` ` ``// explained above.` ` ``mf = mf * ``10` `+ ``1``;` ` ``}` ` ` ` ``return` `sum;` ` ``}` ` ` ` ` ` ``// Driver code to test above methods` ` ``public` `static` `void` `main(String[] args) ` ` ``{` ` ``String num = ``"6759"``;` ` ` ` ``System.out.println(sumOfSubstrings(num));` ` ` ` ``}` `}` `// This code is contributed by Arnav Kr. Mandal.` ## Python3 `# Python3 program to print sum of all substring of` `# a number represented as a string` `# Returns sum of all substring of num` `def` `sumOfSubstrings(num):` ` ``sum` `=` `0` `# Initialize result` ` ``# Here traversing the array in reverse` ` ``# order.Initializing loop from last` ` ``# element.` ` ``# mf is multiplying factor.` ` ``mf ``=` `1` ` ``for` `i ``in` `range``(``len``(num) ``-` `1``, ``-``1``, ``-``1``):` ` ``# Each time sum is added to its previous` ` ``# sum. Multiplying the three factors as` ` ``# explained above.` ` ``# int(s[i]) is done to convert char to int.` ` ``sum` `=` `sum` `+` `(``int``(num[i])) ``` `(i ``+` `1``) ``` `mf` ` ``# Making new multiplying factor as` ` ``# explained above.` ` ``mf ``=` `mf ``` `10` `+` `1` ` ``return` `sum` `# Driver code to test above methods` `if` `__name__``=``=``'__main__'``:` ` ``num ``=` `"6759"` ` ``print``(sumOfSubstrings(num))` `# This code is contributed by` `# Sanjit_Prasad` ## C# `// C# program to print sum of all substring of` `// a number represented as a string` `using` `System;` ` ` `public` `class` `GFG {` ` ` ` ``// Returns sum of all substring of num` ` ``public` `static` `long` `sumOfSubstrings(``string` `num)` ` ``{` ` ` ` ``long` `sum = 0; ``// Initialize result` ` ` ` ``// Here traversing the array in reverse` ` ``// order.Initializing loop from last` ` ``// element.` ` ``// mf is multiplying factor.` ` ``long` `mf = 1;` ` ` ` ``for` `(``int` `i = num.Length - 1; i >= 0; i --)` ` ``{` ` ` ` ``// Each time sum is added to its previous` ` ``// sum. Multiplying the three factors as` ` ``// explained above.` ` ``// s[i]-'0' is done to convert char to int.` ` ``sum += (num[i] - ``'0'``) (i + 1) * mf;` ` ` ` ``// Making new multiplying factor as` ` ``// explained above.` ` ``mf = mf * 10 + 1;` ` ``}` ` ` ` ``return` `sum;` ` ``}` ` ` ` ` ` ``// Driver code to test above methods` ` ``public` `static` `void` `Main() ` ` ``{` ` ``string` `num = ``"6759"``;` ` ` ` ``Console.WriteLine(sumOfSubstrings(num));` ` ` ` ``}` `}` `// This code is contributed by Sam007.` ## PHP `= 0; ``\$i``--)` ` ``{` ` ``// Each time sum is added to ` ` ``// its previous sum. Multiplying ` ` ``// the three factors as explained above.` ` ``// s[i]-'0' is done to convert char to int.` ` ``\$sum` `+= (``\$num``[``\$i``] - ``'0'``) * (``\$i` `+ 1) * ``\$mf``;` ` ``// Making new multiplying ` ` ``// factor as explained above.` ` ``\$mf` `= ``\$mf` `* 10 + 1;` ` ``}` ` ``return` `\$sum``;` `}` `// Driver Code` `\$num` `= ``"6759"``;` `echo` `sumOfSubstrings(``\$num``), ``"\n"``;` `// This code is contributed by m_kit` `?>` ## Javascript `` Output `8421` Time Complexity: O(n) Auxiliary Space: O(1) Previous Next
# polynomial addition in data structure ## Introduction A fundamental mathematical operation, polynomial addition has numerous uses in various disciplines, particularly computer science and data structures. In this thorough investigation, we explore the details of polynomial addition in the context of data structures. Abstractly, polynomials do not merely exist; they also develop algorithms which make it possible to address real-world situations. Linked lists and other similarities can be used to represent them in terms of data structures. It is an approach for combining two or more polynomials in which terms are sequenced in an organized way. It is needed for many applications, including scientific computing, computer graphics, signal processing, and cryptography. ### Understanding Data Structures' Polynomial Representation Understanding how polynomials are stored in data structures is crucial before diving into polynomial addition. One typical method is to record the coefficients of each phrase and their related exponents in arrays or linked lists. The polynomial 3x2+2x+5 can be represented as an array [3,2,5] or a linked list with nodes representing each term. Consider two polynomials: A(x)=4x3+3x2+2x+7 B(x)=2x2+5x+1 A and B's array representations would be [4,3,2,7] and [0,2,5,1], respectively. The coefficients of comparable terms must be added to add these polynomials. The algorithm for adding polynomials is comparable to adding numbers. It entails going through each term in the polynomials iteratively, adding coefficients with the same exponent, and properly addressing missing terms in one of the polynomials. Let's use the sample polynomials A and B from before to demonstrate the algorithm: • Create an array C that is initially zero-filled and will be used to hold the outcome. • Repeat each word from A and B in a single iteration. • Corresponding terms' coefficients are added, and the result is stored in the array C. • Copy the remaining terms to C if one polynomial has more terms than the other. • The sum polynomial is represented by the resulting array C. When A and B are subjected to this method, the sum polynomial results: C(x)=4x3+5x2+7x+8 ### Considerations and improvements for efficiency Any method must be efficient, and polynomial addition is no different. The need to optimize the algorithm increases while working with large polynomials. Memory utilization can be considerably decreased by methods like sparse polynomial representation, where only non-zero components are kept. Switching from arrays to data structures like linked lists can simplify dynamic memory management and allow for the effective addition of polynomials of different degrees. Code Output: Time and Space Complexities The presented polynomial addition algorithm has an O(n) time complexity, where n is the highest degree of the polynomials. This linear time complexity results because each term in both polynomials is only visited once during addition. Therefore, regarding the long pole, algorithms continue sequentially, covering all the parameters involved in the expansion. The space complexity of this algorithm is O(n), where n is a maximum degree of polynomials. The initial loading step involves storing coefficients of the sum polynomial into the result array using the long polynomial. Accordingly, inverse proportionality exists between the maximum degree of polynomial addition and logarithmic polynomial subtraction. ## Conclusion The topic of polynomial addition in the context of data structures has both academic and practical importance. A crucial talent for computer scientists and mathematicians, the manipulation and operation of polynomials are vital in many computing fields. The representation of polynomials in data structures, the technique for adding polynomials, and concerns for efficiency optimization have all been covered in this investigation. A thorough knowledge of polynomial functions within data structures is vital for individuals involved in algorithmic design and execution as technology develops and computational problems become more complex. In essence, polynomial addition is more than just a mathematical operation; it links computer science applications and abstract mathematical ideas, illuminating their mutually beneficial relationship.
# Question Video: Finding the Limit of a Function from Its Graph If the Limit Exists Determine the limit as π₯ βΆ 2 of the function represented by the graph. 01:43 ### Video Transcript Determine the limit as π₯ tends to two of the function represented by the graph. Now, if we look at the graph, we can see that the function is called π of π₯. And weβve being asked to find the limit of π of π₯ as π₯ tends to two. In other words, this is the limit as π₯ tends to two of π of π₯. This can also be described as the value π of π₯ approaches as π₯ tends to two. So we need to find this value of π of π₯ using our graph. We can consider what π of π₯ is doing around the value of π₯ is equal to two. Weβll need to consider π of π₯ on both the left and right of two. Letβs consider π of π₯ on the right of π₯ is equal to two. We can see that as π₯ gets closer and closer to two, π of π₯ is decreasing. And it is decreasing towards this point here, which has a value of three. So we can say that as π₯ tends to two from the right, the value of π of π₯ approaches three. Letβs now consider what happens on the left of π₯ is equal to two. We can again see that as π₯ gets closer and closer to two, the value of π of π₯ is decreasing. And from the graph, we can see that it is decreasing towards the same value that π of π₯ is approaching from the right as π₯ approaches two. And thatβs a value of three. So now we can say that as π₯ approaches two from the left, π of π₯ approaches three. Since π of π₯ approaches the same value from the left and the right of two, we can therefore conclude that the value that π of π₯ approaches as π₯ tends to two is three. And so, we reach our solution, which is that the limit as π₯ approaches two of π of π₯ is equal to three.
# Henry needs to rent a moving truck. Suppose Company A charges a rate of $50 per day and Company B charges$70 fee plus $30 per day. For what number of days is the cost the same? ##### 1 Answer Jul 12, 2018 $3 \frac{1}{2} \mathrm{da} y s$#### Explanation: To solve this set Company A equal to Company B Let $n = \mathrm{da} y s$The cost for Company A is $50n The cost for Company B is $70 +$30n Set the equation $50n =$70 + $30n $50 n = 70 + 30 n$Use additive inverse to combine the variable terms on one side of the equation. $50 n - 30 n = 70 \cancel{+ 30 n} \cancel{- 30 n}$$20 n = 70$Use multiplicative inverse to isolate the variable. $\frac{\cancel{20} n}{\cancel{20}} = \frac{70}{20}$$n = \frac{7}{2}$or $3 \frac{1}{2}\$
# Multiples of 88 Created by: Team Maths - Examples.com, Last Updated: May 30, 2024 ## Multiples of 88 Multiples of 88 are numbers that can be expressed as ( 88 x n ), where ( n ) is an integer. These multiples follow a pattern, increasing by 88 each time (e.g., 88, 176, 264, 352, 440). Multiples of 88 are crucial in mathematics, especially in algebraic concepts, squares, square roots, and fractions. They help in understanding the properties of numbers and performing arithmetic operations efficiently. Recognizing these multiples is essential for grasping complex mathematical ideas and solving algebraic equations. Multiples serve as building blocks in number theory, aiding in exploring patterns, relationships, and the behavior of numbers within mathematical frameworks. Understanding multiples of 88 enhances mathematical proficiency and problem-solving skills. ## What are Multiples of 88? Multiples of 88 are numbers that can be expressed as 88×n, where n is an integer. These numbers are always even and include values like 88, 176, 264, 352, and so on. Prime Factorization of 88 : 2 x 2 x 2 x 11 First 10 Multiples of 88 are 88, 176, 264, 352, 440, 528, 616, 704, 792, 880 First 50 Multiples of 88 are 88, 176, 264, 352, 440, 528, 616, 704, 792, 880, 968, 1056, 1144, 1232, 1320, 1408, 1496, 1584, 1672, 1760, 1848, 1936, 2024, 2112, 2200, 2288, 2376, 2464, 2552, 2640, 2728, 2816, 2904, 2992, 3080, 3168, 3256, 3344, 3432, 3520, 3608, 3696, 3784, 3872, 3960, 4048, 4136, 4224, 4312, 4400 Table of 88 ## Important Notes • Even Numbers : All multiples of 88 are even numbers, meaning they end in 0, 2, 4, 6, or 8. • Divisibility : A number is a multiple of 88 if it can be divided by 88 with no remainder. • Factors : Multiples of 88 have 88 as one of their factors. • Infinite Sequence : There are infinitely many multiples of 88, extending indefinitely as 88, 176, 264, 352, 440, and so on. • Arithmetic Pattern : The difference between consecutive multiples of 88 is always 88. ## Examples on Multiples of 88 ### Simple Multiples 176: 88 × 2 = 176 264: 88 × 3 = 264 352: 88 × 4 = 352 ### Larger Multiples 440: 88 × 5 = 440 880: 88 × 10 = 880 1760: 88 × 20 = 1760 ## Real-Life Examples Time: 5280 seconds in 88 minutes is a multiple of 88 because 88 × 60 = 5280. Money: \$880 (880 dollars) is a multiple of 88 because 88 × 10 = 880. Measurements: 3168 inches in 88 yards is a multiple of 88 because 88 × 36 = 3168. ## Practical Examples of Multiples of 88 1. Time: There are 5280 seconds in 88 minutes, making it a multiple of 88 since 88 × 60 = 5280. 2. Money: If you save \$88 each week, after 10 weeks, you will have saved \$880, which is a multiple of 88. 3. Measurements: A fabric roll of 3168 inches (88 yards) is a multiple of 88 because 88 × 36 = 3168. 4. Event Planning: For a conference with 88 participants, if each table seats 8, you’ll need 11 tables, since 88 is a multiple of 8. 5. Travel: If a train travels 88 miles per hour, in 4 hours, it will cover 352 miles, which is a multiple of 88. ## Practical Applications 1. Time Management: If you have a task that takes 88 minutes to complete, after completing it 5 times, you will have spent 440 minutes (88 × 5). 2. Financial Savings: By saving \$88 every month, in 12 months, you will have saved \$1056 (88 × 12). 3. Material Usage: If you buy fabric by the yard and need 88 yards for a project, you will be purchasing 3168 inches (88 × 36). 4. Seating Arrangement: For a banquet with 88 guests, if each table seats 8 people, you will need 11 tables (88 ÷ 8). 5. Travel Distance: If a vehicle travels at 88 miles per hour, in 7 hours, it will cover 616 miles (88 × 7). ## Is 440 a multiple of 88? Yes, 440 is a multiple of 88 because 88 × 5 = 440. ## Can multiples of 88 be odd numbers? No, multiples of 88 cannot be odd numbers as 88 is an even number. ## What is the 25th multiple of 88? The 25th multiple of 88 is 2200 (88 × 25). ## How many multiples of 88 are there? There are infinitely many multiples of 88. ## Is 352 a multiple of 88? Yes, 352 is a multiple of 88 because 88 × 4 = 352. ## Can multiples of 88 be negative? Yes, multiples of 88 can be negative if you multiply 88 by a negative integer (e.g., 88 × -1 = -88). ## What is the 50th multiple of 88? The 50th multiple of 88 is 4400 (88 × 50). ## Are multiples of 88 divisible by 4? Yes, multiples of 88 are divisible by 4 because 88 itself is divisible by 4. ## Is 792 a multiple of 88? Yes, 792 is a multiple of 88 because 88 × 9 = 792. ## What is the 100th multiple of 88? The 100th multiple of 88 is 8800 (88 × 100). ## How can multiples of 88 be used in real life? Multiples of 88 can be used in various applications, such as time management, financial savings, material measurements, event planning, and calculating travel distances. ## AI Generator Text prompt 10 Examples of Public speaking 20 Examples of Gas lighting
Courses Courses for Kids Free study material Offline Centres More Store # Subtraction of Complex Numbers Reviewed by: Last updated date: 09th Aug 2024 Total views: 176.7k Views today: 2.76k ## Introduction to the Subtraction of Complex Numbers In everyday life, addition and subtraction are the most used mathematical operations. We use it on a daily basis. If we go to the supermarket for shopping, we need an addition or subtraction of money to make the payment. We use addition or subtraction when we have some chocolates, and someone gives us more chocolates or if someone takes some chocolates from us. Confused about complex numbers? Don't worry; all your doubts will be cleared after reading this article. Because in this article today, we are going to learn about what complex numbers are, what their properties are when it comes to subtraction, and what the steps are to solve complex numbers. So now, let's learn about the subtraction of complex numbers. ## What are Complex Numbers? Complex numbers are also numbers, but they are different from normal numbers in many ways. They are made with the help of two numbers that are combined. And in the new combined number, the first number is the real number, and another one is the imaginary number. Like all complex numbers can be expressed in the form a + bi, here, a is the real number, and bi is the imaginary number. Remember that Complex Numbers are denoted by the letter 'C'. Complex numbers also use all mathematical operations: addition, subtraction, multiplication, and division. But today, we will learn about the subtraction of complex numbers, which is as easy as solving a normal number. And the Subtraction of complex numbers requires a formula. Do you know the formula for the subtraction of Complex Numbers? So the formula is (a+ib) - (c + id) = (a-c) +i(b-d). Complex Number ## Properties of Subtracting Complex Numbers 1. Closure property - In the closure property of Subtracting Complex Numbers, the difference between complex numbers is also a complex number. 2. Commutative property- Subtraction of complex numbers is not commutative. 3. Associative property- Subtraction of complex numbers is not associative. Remember that this general form of a complex number is different from the polar form. Similarly, the addition and subtraction of complex numbers in the polar form are different. ## Steps to Subtract Complex Numbers Subtraction of two complex numbers is easy if it's done in the right and systematic way. So let's see the steps of subtraction of complex numbers. $\mathrm{z}_1=\mathrm{a}+\mathrm{ib}, \mathrm{z}_2=\mathrm{c}+\mathrm{id}$ Subtraction $\mathrm{z}_1-\mathrm{z}_2+(a-c)-(c+i d)$ $\mathrm{z}_1-\mathrm{z}_2=(a-c)+(b-d) i$ where $a, b, c, d$ are real number and $i$ is imaginary number Step 1: Disperse the negative Step 2: Now combine the real and imaginary complex numbers in a single group. Step 3: Now, you need to combine and simplify similar terms. Step 4: The answer will be there. Subtraction of two complex numbers example ## Subtraction of Complex Numbers Examples For the subtraction of two complex numbers, we need to Subtract the real number from the real and the imaginary numbers from the imaginary. Let's learn it through examples: Q 1. Subtract $(4-3 i)$ from $(7+5 i)$ Ans. $=7+5 i-(4-3 i)$ $=7+5 i-4+3 i$ $=3+8 i$ Q 2. Subtract ( 8 - 15i ) from (15 - 34i ) Ans. $=15-34 i-(8-15 i)$ $=15-34 i-8+15 i$ $=7-19 i$ Q 3. Subtract $(2+2 i)$ from $(3+5 i)$ Ans. $=3+5 i-(2+2 i)$ $=1+3 i$ ## Points to remember for the Subtraction of Complex Numbers • Subtracting a complex number is like subtracting two binomials; we just need to combine the like terms. • The subtraction of two complex numbers does not hold in the commutative law. • All real numbers are complex numbers, but all complex numbers need not be real numbers. ## Practice Questions We know that maths is not learned just with reading. To understand it, we need to solve the question. So that's why we have provided a practice sheet for you. With the help of this practice sheet, you can better understand the subtraction of two complex numbers. Below are the addition and subtraction of complex numbers worksheets for your practice. Use the subtraction of complex numbers formula mentioned in this article to solve these problems. Q1. $(16+5 i)+(8-3 i)$ Ans. $24+2 i$ Q2. $(5+7 i)-2 i$ Ans. $5+5 i$ Q3. $(4+3 i)-(4-3 i)$ Ans. 6i ## Summary In this article, we learned about the subtraction of complex numbers. With the help of this article, we learned what complex number is, how to subtract complex numbers, and what the properties of Subtracting Complex Numbers are. We learn the steps we need to follow while subtracting a complex number. We also checked an example of subtraction of complex numbers, and at the end, there is a complex number worksheet with the help of which we can easily clear our concept of subtraction of complex numbers. ## FAQs on Subtraction of Complex Numbers 1. Is zero a complex number? We know that real numbers are part of complex numbers, and zero is a real number. That is why we can say that zero is a complex number. 2. Are the complex numbers closed under subtraction? Yes, the complex numbers are closed in Subtraction. 3. Is 50i a complex number? Yes, it's a complex number because it has an imaginary part.
# A Statistical Background ## A.1 Basic statistical terms Note that all the following statistical terms apply only to numerical variables, except the distribution which can exist for both numerical and categorical variables. ### A.1.1 Mean The mean is the most commonly reported measure of center. It is commonly called the average though this term can be a little ambiguous. The mean is the sum of all of the data elements divided by how many elements there are. If we have $$n$$ data points, the mean is given by: $Mean = \frac{x_1 + x_2 + \cdots + x_n}{n}$ ### A.1.2 Median The median is calculated by first sorting a variable’s data from smallest to largest. After sorting the data, the middle element in the list is the median. If the middle falls between two values, then the median is the mean of those two middle values. ### A.1.3 Standard deviation and variance We will next discuss the standard deviation ($$sd$$) of a variable. The formula can be a little intimidating at first but it is important to remember that it is essentially a measure of how far we expect a given data value will be from its mean: $sd = \sqrt{\frac{(x_1 - Mean)^2 + (x_2 - Mean)^2 + \cdots + (x_n - Mean)^2}{n - 1}}$ The variance of a variable is merely the standard deviation squared. $variance = sd^2 = \frac{(x_1 - Mean)^2 + (x_2 - Mean)^2 + \cdots + (x_n - Mean)^2}{n - 1}$ ### A.1.4 Five-number summary The five-number summary consists of five summary statistics: the minimum, the first quantile AKA 25th percentile, the second quantile AKA median or 50th percentile, the third quantile AKA 75th, and the maximum. The five-number summary of a variable is used when constructing boxplots, as seen in Section 2.7. The quantiles are calculated as • first quantile ($$Q_1$$): the median of the first half of the sorted data • third quantile ($$Q_3$$): the median of the second half of the sorted data The interquartile range (IQR) is defined as $$Q_3 - Q_1$$ and is a measure of how spread out the middle 50% of values are. The IQR corresponds to the length of the box in a boxplot. The median and the IQR are not influenced by the presence of outliers in the ways that the mean and standard deviation are. They are, thus, recommended for skewed datasets. We say in this case that the median and IQR are more robust to outliers. ### A.1.5 Distribution The distribution of a variable shows how frequently different values of a variable occur. Looking at the visualization of a distribution can show where the values are centered, show how the values vary, and give some information about where a typical value might fall. It can also alert you to the presence of outliers. Recall from Chapter 2 that we can visualize the distribution of a numerical variable using binning in a histogram and that we can visualize the distribution of a categorical variable using a barplot. ### A.1.6 Outliers Outliers correspond to values in the dataset that fall far outside the range of “ordinary” values. In the context of a boxplot, by default they correspond to values below $$Q_1 - (1.5 \cdot IQR)$$ or above $$Q_3 + (1.5 \cdot IQR)$$. ## A.2 Normal distribution Let’s next discuss one particular kind of distribution: normal distributions. Such bell-shaped distributions are defined by two values: (1) the mean $$\mu$$ (“mu”) which locates the center of the distribution and (2) the standard deviation $$\sigma$$ (“sigma”) which determines the variation of the distribution. In Figure A.1, we plot three normal distributions where: 1. The solid normal curve has mean $$\mu = 5$$ & standard deviation $$\sigma = 2$$. 2. The dotted normal curve has mean $$\mu = 5$$ & standard deviation $$\sigma = 5$$. 3. The dashed normal curve has mean $$\mu = 15$$ & standard deviation $$\sigma = 2$$. Notice how the solid and dotted line normal curves have the same center due to their common mean $$\mu$$ = 5. However, the dotted line normal curve is wider due to its larger standard deviation of $$\sigma$$ = 5. On the other hand, the solid and dashed line normal curves have the same variation due to their common standard deviation $$\sigma$$ = 2. However, they are centered at different locations. When the mean $$\mu$$ = 0 and the standard deviation $$\sigma$$ = 1, the normal distribution has a special name. It’s called the standard normal distribution or the $$z$$-curve. Furthermore, if a variable follows a normal curve, there are three rules of thumb we can use: 1. 68% of values will lie within $$\pm$$ 1 standard deviation of the mean. 2. 95% of values will lie within $$\pm$$ 1.96 $$\approx$$ 2 standard deviations of the mean. 3. 99.7% of values will lie within $$\pm$$ 3 standard deviations of the mean. Let’s illustrate this on a standard normal curve in Figure A.2. The dashed lines are at -3, -1.96, -1, 0, 1, 1.96, and 3. These 7 lines cut up the x-axis into 8 segments. The areas under the normal curve for each of the 8 segments are marked and add up to 100%. For example: 1. The middle two segments represent the interval -1 to 1. The shaded area above this interval represents 34% + 34% = 68% of the area under the curve. In other words, 68% of values. 2. The middle four segments represent the interval -1.96 to 1.96. The shaded area above this interval represents 13.5% + 34% + 34% + 13.5% = 95% of the area under the curve. In other words, 95% of values. 3. The middle six segments represent the interval -3 to 3. The shaded area above this interval represents 2.35% + 13.5% + 34% + 34% + 13.5% + 2.35% = 99.7% of the area under the curve. In other words, 99.7% of values. Learning check Say you have a normal distribution with mean $$\mu = 6$$ and standard deviation $$\sigma = 3$$. (LCA.1) What proportion of the area under the normal curve is less than 3? Greater than 12? Between 0 and 12? (LCA.2) What is the 2.5th percentile of the area under the normal curve? The 97.5th percentile? The 100th percentile? ## A.3 log10 transformations At its simplest, log10 transformations return base 10 logarithms. For example, since $$1000 = 10^3$$, running log10(1000) returns 3 in R. To undo a log10 transformation, we raise 10 to this value. For example, to undo the previous log10 transformation and return the original value of 1000, we raise 10 to the power of 3 by running 10^(3) = 1000 in R. Log transformations allow us to focus on changes in orders of magnitude. In other words, they allow us to focus on multiplicative changes instead of additive ones. Let’s illustrate this idea in Table A.1 with examples of prices of consumer goods in 2019 US dollars. TABLE A.1: TABLE A.2: log10 transformed prices, orders of magnitude, and examples Price log10(Price) Order of magnitude Examples $1 0 Singles Cups of coffee$10 1 Tens Books $100 2 Hundreds Mobile phones$1,000 3 Thousands High definition TVs $10,000 4 Tens of thousands Cars$100,000 5 Hundreds of thousands Luxury cars and houses $1,000,000 6 Millions Luxury houses Let’s make some remarks about log10 transformations based on Table A.1: 1. When purchasing a cup of coffee, we tend to think of prices ranging in single dollars, such as$2 or $3. However, when purchasing a mobile phone, we don’t tend to think of their prices in units of single dollars such as$313 or $727. Instead, we tend to think of their prices in units of hundreds of dollars like$300 or $700. Thus, cups of coffee and mobile phones are of different orders of magnitude in price. 2. Let’s say we want to know the log10 transformed value of$76. This would be hard to compute exactly without a calculator. However, since $76 is between$10 and $100 and since log10(10) = 1 and log10(100) = 2, we know log10(76) will be between 1 and 2. In fact, log10(76) is 1.880814. 3. log10 transformations are monotonic, meaning they preserve orders. So if Price A is lower than Price B, then log10(Price A) will also be lower than log10(Price B). 4. Most importantly, increments of one in log10-scale correspond to relative multiplicative changes in the original scale and not absolute additive changes. For example, increasing a log10(Price) from 3 to 4 corresponds to a multiplicative increase by a factor of 10:$100 to \$1000.
# How do you find x in the equation -2 - 6x = 19? Mar 13, 2018 $x = - \frac{7}{2}$ #### Explanation: $- 2 - 6 x = 19$ Add $2$ on each side of the equation to get rid of the $- 2$ on the left-hand side (whatever you do to one side of the equation, you must do to the other side) $- 2 + 2 - 6 x = 19 + 2$ $0 - 6 x = 19 + 2$ $- 6 x = 21$ To solve for $x$, we must get $x$ by itself. So, divide both sides of the equation by $- 6$ $\frac{- 6 x}{-} 6 = \frac{21}{-} 6$ $x = - \frac{21}{6}$ Simplify as much as possible, in this case, $3$ is common in the denominator and numerator $x = - \frac{7}{2}$
GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 11 Dec 2018, 14:35 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History ## Events & Promotions ###### Events & Promotions in December PrevNext SuMoTuWeThFrSa 2526272829301 2345678 9101112131415 16171819202122 23242526272829 303112345 Open Detailed Calendar • ### Free GMAT Prep Hour December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST. • ### The winning strategy for 700+ on the GMAT December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL. # Math Revolution Approach (PS) Author Message Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 05 Oct 2018, 00:38 [Math Revolution GMAT math practice question] Tom’s Mom is five times as old as him. In 6 years, Tom’s Mom will be three times as old as him. How old is Tom now? A. 3 B. 4 C. 5 D. 6 E. 7 => We can set up this question using the IVY approach. “Is” can be converted to “=” and “times” can be converted to the multiplication sign “x”. Let m and t be Tom’s Mom’s age and Tom’s age now, respectively. Then m = 5t since Tom’s mom is five times as old as Tom. We also have m + 6 = 3(t + 6) since Tom’s mom will be three times as old as Tom in 6 years. Plugging m = 5t into the second equation yields 5t + 6 = 3t + 18 or 2t = 12. So, t = 6. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 07 Oct 2018, 17:38 [Math Revolution GMAT math practice question] Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If lines AB and BC are perpendicular to each other, p=? A. -2 B. -1 C. 0 D. 1 E. 2 => The slope of the line segment joining the two points (x1,y1) and (x2,y2) is (y1-y2)/(x1-x2). The product of the slopes of perpendicular lines is -1. The slope of the line AB is (4-2)/(1-2) = -2. Since the two lines AB and BC are perpendicular each other, the slope of BC, which is (2-1)/(2-p), must be 1/2. So, (2-1)/(2-p)=1/2 1/(2-p)=1/2. Thus, 2 – p = 2, and p = 0. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 07 Oct 2018, 17:39 [Math Revolution GMAT math practice question] \|\|2-3\|-\|2-5\|\|=? A. 1 B. 2 C. 3 D. 4 E. 5 => \|\|2-3\| - \|2-5\|\| = \|\|-1\| - \|-3\|\| = \| 1 – 3 \| = \| -2 \| = 2. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 10 Oct 2018, 00:00 [Math Revolution GMAT math practice question] What is the greatest prime factor of 1+2+3+….+36? A. 2 B. 3 C. 9 D. 31 E. 37 => Since 1 + 2 + 3 + … + n = n(n+1)/2, we have 1 + 2 + 3 + … + 36 = (3637)/2 = 1837 = 23237. Thus, 37 is the greatest prime factor of 23^237. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 11 Oct 2018, 00:38 [Math Revolution GMAT math practice question] The points (1,6) and (7,-2) are two vertices of one diagonal of a square in the xy-plane. What is the area of the square? A. 16 B. 25 C. 32 D. 36 E. 50 => The length of the diagonal of the square is √{ (7-1)^2 + (6-(-2))^2 } = √100 = 10. The side-length of the square is 10/ √2 = 5 √2 since the side-length of a square is equal to the length of its diagonal divided by √2. Thus, the area of the square is (5 √2)^2 = 50. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 12 Oct 2018, 07:37 [Math Revolution GMAT math practice question] If 2 numbers are selected from the first 8 prime numbers, what is the probability that the sum of the 2 numbers selected is an even number? A. 1/2 B. 1/3 C. 2/3 D. 1/4 E. 3/4 => In order for the sum to be even, both primes selected must be odd. As 2 is the only even prime number, the number of selections with an even sum is equal to the number of ways to select 2 numbers from these 7 odd prime numbers, or 7C2. The total number of selections of 2 prime numbers from the first 8 prime numbers is 8C2. Therefore, the probability that the sum of the two numbers selected is even is 7C2 / 8C2 = {(76)/(12)}/{(87)/(12)} = 6/8 = 3/4. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 14 Oct 2018, 17:50 [Math Revolution GMAT math practice question] If x+(1/x)=4, what is the value of x^2+(1/x^2)? A. 4 B. 10 C. 12 D. 14 E. 16 => Since (a+b)^2=a^2+b^2+2ab, a^2+b^2=(a+b)^2-2ab, and x^2 + 1/x^2 = ( x + 1/x )^2 – 2x(1/x) = 4^2 – 2 = 16 -2 = 14. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution Approach (PS) [#permalink] ### Show Tags 14 Oct 2018, 17:51 [Math Revolution GMAT math practice question] What is the solution of the equation (x-1)/(1-x) = 0? A. -1 B. 0 C. 1 D. 0 or 1 E. The equation has no solution. => In order for a fraction to be zero, its numerator must be 0. This occurs when x = 1. However, if x = 1, then the denominator of this fraction is 0. Thus, 0 is an erroneous solution, and the equation has no solution. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only \$99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Re: Math Revolution Approach (PS) &nbs [#permalink] 14 Oct 2018, 17:51 Go to page Previous 1 ... 18 19 20 21 22 23 24 25 26 27 28 29 [ 568 posts ] Display posts from previous: Sort by # Math Revolution Approach (PS) Moderator: souvonik2k Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
# Difference Between Mean And Median tl;dr The main difference between mean and median is that the mean is calculated by summing all the values in a dataset and dividing the total by the number of values, while the median is calculated by sorting the values in ascending or descending order and finding the middle value, and the mean is more sensitive to outliers than the median. Mean and median are the two basic statistical measures that are commonly used to understand a dataset. Though both are measures of central tendency, they are different in terms of how they calculate an average. Mean: The mean, also known as the average, is simply the sum of all the numbers in a dataset divided by the total number of values. It is the most commonly used measure of central tendency and is used to represent the typical value of a dataset. It is important to note that the mean is greatly influenced by extreme values, also known as outliers, and therefore may not be the most accurate representation of the dataset. For example, if a dataset contains the values 3, 5, 6, 8, and 12, their sum is 34. Dividing 34 by the total number of values, which is 5, gives us the mean of 6.8. Median: The median is the middle value of a dataset once it has been sorted in ascending or descending order. It is the value separating the dataset into two halves; half of the values are above the median, and half are below it. The median is not affected by the outliers present in a dataset, as it only depends on the order of the values. For example, the median of the dataset {2, 5, 8, 10, 12} is 8. If the dataset contains an even number of values, the median can be found by calculating the mean of the two middle values. For example, the median of the dataset {2, 5, 7, 8, 10, 12} is (7+8)/2 = 7.5. Differences between mean and median: The main differences between mean and median are their calculations and their sensitivity to the outliers. 1. Calculation: The mean is calculated by summing all the values in a dataset and dividing the total by the number of values. The median is calculated by sorting the values in ascending or descending order and finding the middle value. 2. Sensitivity to outliers: The mean is greatly influenced by outliers, as they can significantly affect the sum of the values. This means that if a dataset has a few extreme values, the mean may not be a good representation of the dataset. On the other hand, the median is not affected by outliers, as it only depends on the order of the values. This makes it a more robust measure of central tendency. For example, imagine a dataset containing the following annual salaries: \$40,000, \$50,000, \$60,000, \$70,000, \$80,000, \$90,000, \$100,000, and \$1,100,000. The mean salary would be (\$40,000+\$50,000+\$60,000+\$70,000+\$80,000+\$90,000+\$100,000+\$1,100,000)/9 = \$205,556. This means that the mean salary is being greatly inflated by the outlier value of \$1,100,000. However, the median salary of this dataset would be \$80,000, which represents the typical salary in this dataset. When to use mean and median: The choice between mean and median depends on the dataset and what you are trying to analyze. If the dataset contains extreme values or outliers, it is better to use the median, as it is less susceptible to their influence. On the other hand, if the dataset does not have any significant outliers, the mean can be used as an accurate representation of the dataset. The mean is more suitable for datasets that have a symmetrical distribution, while the median is more suitable for datasets with skewed distributions. A symmetrical distribution has equal frequencies of values on either side of the median, while a skewed distribution has a longer tail on one side than the other. Conclusion: In summary, mean and median are two measures of central tendency commonly used in statistics to represent a dataset. While mean is the sum of all values divided by the total number of values, median is the middle value found by sorting the dataset in ascending or descending order. The main difference between them is their sensitivity to outliers; mean is greatly affected by outliers, while median is not. The choice between mean and median depends on the dataset and what you are trying to analyze.
# 4.3 Proving Triangles are Congruent: SSS and SAS – PART 2 - PowerPoint PPT Presentation 4.3 Proving Triangles are Congruent: SSS and SAS – PART 2 1 / 14 4.3 Proving Triangles are Congruent: SSS and SAS – PART 2 ## 4.3 Proving Triangles are Congruent: SSS and SAS – PART 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. 4.3 Proving Triangles are Congruent: SSS and SAS – PART 2 2. Congruent Triangles in a Coordinate Plane AC FH ABFG Use the SSS Congruence Postulate to show that ABCFGH. SOLUTION AC = 3 and FH= 3 AB = 5 and FG= 5 3. Congruent Triangles in a Coordinate Plane d = (x2 – x1 )2+ (y2 – y1 )2 d = (x2 – x1 )2+ (y2 – y1 )2 BC = (–4 – (–7))2+ (5– 0)2 GH = (6 – 1)2+ (5– 2)2 = 32+ 52 = 52+ 32 = 34 = 34 Use the distance formula to find lengths BC and GH. 4. Congruent Triangles in a Coordinate Plane BCGH BC = 34 and GH= 34 All three pairs of corresponding sides are congruent, ABCFGH by the SSS Congruence Postulate. 5. Congruent Triangles in a Coordinate Plane MN DE PMFE Use the SSS Congruence Postulate to show that NMPDEF. SOLUTION MN = 4 and DE= 4 PM = 5 and FE= 5 6. Congruent Triangles in a Coordinate Plane d = (x2 – x1 )2+ (y2 – y1 )2 d = (x2 – x1 )2+ (y2 – y1 )2 PN = (–1 – (– 5))2+ (6– 1)2 FD = (2 – 6)2+ (6– 1)2 = 42+ 52 = (-4)2+ 52 = 41 = 41 Use the distance formula to find lengths PN and FD. 7. Congruent Triangles in a Coordinate Plane PNFD PN = 41 and FD= 41 All three pairs of corresponding sides are congruent, NMPDEF by the SSS Congruence Postulate. 8. SAS  postulate SSS  postulate 9. T C S G The vertex of the included angle is the point in common. SSS  postulate SAS  postulate 10. SSS  postulate Not enough info 11. SSS  postulate SAS  postulate 12. Not Enough Info SAS  postulate 13. SSS  postulate Not Enough Info 14. SAS  postulate SAS  postulate
# Number Series: Fundamentals When it comes to number series questions, a series of numbers is given with a number missing. This missing number can be anywhere in the sequence. There can also be questions where one of the numbers in the series is wrong. One is supposed to find a consistent logic in the given series and answer accordingly. In this article, we will go through number series questions. I have also added my solutions to these questions. Through these questions, we will look at various methods of solving the same question (wherever possible) and understand what’s best in the exam situation. For these questions, I have not provided options so that you develop the ability to crack these questions without any support/guesswork using options. Q.1 – 7, 12, 19, ?, 39 Solution: Take the difference between consecutive terms which is 5, 7, and so on. Which means that either it is odd number difference or prime number difference. If it is odd, we will get 19 + 9 = 28, and then 28 + 11 = 39. Satisfies. It won’t be prime as 19 + 11 will give us 30 and then the next difference will become 39 – 30 = 9. Hence, the logic here is odd number difference and the next term will be 19 + 9 = 28. Another logic which a lot of people won’t be able to crack is the n^2 – something logic. Here, 7 = 3^2 – 2 12 = 4^2 – 4 19 = 5^2 – 6 missing term 39 = 7^2 – 10. Missing term will be 6^2 – 8 = 28. Lengthy approach for sure and this typically won’t strike during the test. Q.2 – 0, 6, 24, 60, 120, 210, ? Solution: Most of the times, you will encounter some common numbers where it is important to remember the logic. For example, 210 which is nothing but 6^3 – 6. That gives away the logic of this question that it is n^3 – n. So the sequence is: 1^3 – 1, 2^3 – 2, 3^3 – 3, 4^4 – 4, 5^5 – 5, 6^3 – 6, 7^3 – 7 = 336 will be the answer. Any alternate logic? yes! 6 = 23 24 = 46 = 4(3+3) 60 = 610 = 6(6+4) 120 = 815 = 8(10+5) 210 = 1021 = 10(15+6) Next number will be 12(21+7) = 1228 = 336. Slightly lengthy compared to the first approach but gives the solution. Any other method? Let’s try taking the difference between consecutive terms. 6 – 18 – 36 – 60 – 90 61, 63, 66, 610, 615. The next term will be 621 = 126. 210 + 126 = 336! Q.3 – 4, 6, 12, 14, 28, 30, ? Solution: Simple. 4 + 2 = 6; 62 = 12; 12 + 2 = 14; 142 = 28; 28 + 2 = 30. The next term will be 302 = 60 Q.4 – 1, 3, 3, 6, 7, 9, ?, 12, 21. Solution: Here, once you try the first difference between terms, you will get 2, 0, 3, 1, 2, and so on which doesn’t make much sense. In such cases, it is advisable to check for alternate terms. We will find two sequences here. 1, 3, 7, ?, 21 and 3, 6, 9, 12. As we need to tackle the first one with the missing number, the second sequence doesn’t really matter. 1, 3, 7, ?, 21 will give us term differences as 2, 4, 6, 8 and hence, the term will be 7 + 6 = 13. Q.5 – 1, 2, 5, 26, ? Solution: It is easier to figure out 5 = 5^2 + 1 or 55 + 1 than starting from 1-2. If we work the logic backwards, we will get it as: 1^2 + 1 = 2; 2^2 + 1 = 5; 5^2 + 1 = 26; and the next term will be 26^2 + 1 = 677. Q.6 – 96, 6, 48, 18, 24, 54, ? Solution: In this question, one can always start from 96/16 = 6 and so on but when I observe 6, 18, 54 the logic starts materializing. These are two alternate sequences. Hence, 96 -> 48 -> 24. The next will be half of this = 12. Taking difference as the first attempt at cracking the logic is fair, but since it is ridiculously common, you won’t find it in a lot of moderate to difficult level questions. Q.7 – 1, 10, 66, 469, ? Solution: When the difference between two numbers increases substantially, the logic is generally related to multiplications and powers. So, I will start with 66 and 469. 469 = 767 which gives me a starting point that it is somehow related to 66. So 667 = 462 + 7 = 469. Working backwards, this is what I can establish: 15 + 5 = 10 106 + 6 = 66 667 + 7 = 469 4698 + 8 = 3760 Q.8 – 2, 17, 101, 362, ? Solution: Where did 362 come from? Oh, I know! It is close to 361 which is 19^2. Can the logic be squares + or – something? Absolutely. 1^2 + 1 = 2 4^2 + 1 = 17 10^2 + 1 = 101 19^2 + 1 = 362 What’s the logic behind 1-4-10-19? The difference is increasing by 3-6-9. So the next term will be 19 + 12 = 31. 31^2 + 1 will be 961 + 1 = 962 Q.9 – 15, 31, 95, 239, ?, 895, 1471 Solution: If I take difference in consecutive terms, we will get 16 – 64 – 144 – x – y – 576 and these are squares of 4, 8, 12 and the ones in between should be 16^2 = 256 and 20^ = 400 to give us 24^2 after 895. So 239 + 256 = 495 will be the answer. One might start with 15, 152 + 1 = 31, 313 + 2 = 95, but that doesn’t help. If you started with this logic, you should stop here and think of alternate method. Q.10 – 1, 2, 8, 33, 148, ? Solution: If we take the difference, we get 1, 6, 25, 115 which I don’t think is taking us anywhere. As the difference is increasing significantly after 33, the logic can be into something + or – something. 11 + 1 = 2 22 + 4 = 8 83 + 9 = 33 334 + 16 = 148. The next term will be 148*5 + 25 = 740 + 25 = 765. Hope you got a fair idea of how to approach these questions. To master number series questions, you need a strong command over tables, squares, cubes, exponents, and a knack of identifying pattern by splitting numbers. Typically, if you apply 5/6 initial approaches (difference, alternate series, difference of difference, into something + something, and powers), you will make some breakthrough. However, it is always possible that the logic is completely arbitrary. Spend a minute or so. If you still don’t get it, make an intelligent guess and move on. If there are any queries, please leave them in the comments below and allow me a few days to get back. Share this with your friends and co-aspirants. Happy prepping! 🙂 1. Hello sir!
### 数学代写\|数论作业代写number theory代考\|MAST90136 statistics-lab™ 为您的留学生涯保驾护航在代写数论number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 数学代写\|数论作业代写number theory代考\|Divisibility Definition: Suppose $b$ is an integer and $a$ is a non-zero integer. We say that $a$ divides $b$ if there is an integer $q$ so that $b=a q$. If there are such integers, we denote the fact that $a$ divides $b$ by using the notation $a \mid b$. Be aware that the notation $a \mid b$ is a sentence with the verb being “divides.” Contrast this with the notation $\frac{a}{b}$, which is an element of the rational numbers $\mathbb{Q}$ (see Appendix $B$, not a sentence. Example 1.1. Clearly $2 \mid 8$ since $8=2(4) ; 36 \mid 108$ since $108=$ $36(3) ; 3 \mid(-36)$ since $-36=3(-12)$; and for any integer $m, 3 \mid(15 m+$ 3) since $15 m+3=3(5 m+1)$. On the other hand, 3 does not divide 13 as there is no integer $q$ with $13=3 q$. In the following lemma, we provide a few basic properties involving the divisibility of integers. (Note: A “lemma” is a “helping theorem,” i.e., an often easily proved result which is then used to establish bigger results.) Lemma 1.1. Let $a, b, c, d$ be integers with $a>0$ and $d>0$. (i) If $a \mid b$ and $a \mid c$, then $a \mid(b+c)$; (ii) If $a \mid b$ and $a \mid c$, then $a \mid(b-c)$; (iii) If $a \mid b$ and $a \mid c$, then $a \mid(m b+n c)$ for any integers $m$ and $n$; (iv) If $d \mid a$ and $a \mid b$, then $d \mid b$. Proof. To prove Part (i), we may assume that $b=a q$ and $c=a s$ where $q$ and $s$ are integers. Then $b+c=a q+a s=a(q+s)$ so that $a$ divides $b+c$ (since $q+s$ is an integer). The proof of Part (ii) is similar and hence omitted. Proofs of the remaining parts are left to the reader in Supplementary Problem 1.14. ## 数学代写\|数论作业代写number theory代考\|The Division Algorithm What if a positive integer $a$ does not divide an integer $b$ ? Here is where the seemingly simple but very important Division Algorithm comes into play when doing computations in $\mathbb{Z}$. Theorem 1.2. (Division Algorithm) Let $a$ and $b$ be integers with $a>0$. Then there are integers $q$ and $r$ with $0 \leq r<a$ so that $b=a q+r$. This is simply a formal statement of the long division process. The integer $b$ is often called the dividend, $a$ the divisor, $q$ the quotient, and $r$ the remainder. The key is that the remainder $r$ must be non-negative and must be less than the divisor $a$. It is clear that $a \mid b$ if and only if $r=0$. Example 1.2. Given $b=436$ and $a=17$, we can compute by long division that $436=17(25)+11$. Note that, as required, the remainder 11 is greater than or equal to 0 and is less than the divisor 17. Given $b=-67$ and $a=12$, we get $-67=12(-6)+5$, so in this case the quotient $-6$ is negative, but again the remainder 5 must be non-negative and below the divisor 12 . ## 数学代写\|数论作业代写number theory代考\|Divisibility (一) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(b+c)$ ； (ii) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(b-c)$; (iii) 如果 $a \mid b$ 和 $a \mid c$ ，然后 $a \mid(m b+n c)$ 对于任何整数 $m$ 和 $n$; (iv) 如果 $d \mid a$ 和 $a \mid b$ ，然后 $d \mid b$. $b+c=a q+a s=a(q+s)$ 以便 $a$ 划分 $b+c$ (自从 $q+s$ 是一个整数)。(ii) 部分的证明类似，因此省略。其余部分的证明留给读者在补充问题 $1.14$ 中。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
Logic # Arithmetic Puzzles: Level 3 Challenges In the game of M2S3, the goal is to use minimum number of steps to get from one number to the next. In each step, you can either • multiply by 2, or • subtract 3. For example, if we wanted to get from 8 to 11, we could do so in 5 steps: $\large 8 \overset{-3}{\longrightarrow} 5 \overset{\times2}{\longrightarrow}10 \overset{-3}{\longrightarrow} 7\overset{\times2}{\longrightarrow}14 \overset{-3}{\longrightarrow}11.$ What is the minimum possible number of steps required to get from 11 to 25? $\large \left( \square \ + \ \square \ - \ \square \ \right) \times \ \square \ \div \ \square$ You are given that the numbers $1,2,3,4$ and $5$ are to be filled in the square boxes as shown above (without repetition). Find the minimum possible value of the resultant number. Give your answer to 3 decimal places. A distinct integer from 1 to 12 is placed in each red circle in the figure below. Is it possible that the sum of the four numbers on any of the six blue lines is the same? $\large \square + \square\square+\square\square\square+\square\square\square\square$ You are given that the numbers $0,1,2,\ldots,9$ are to be filled in the square boxes as shown above (without repetition) such that it represent a sum of a 1-digit, 2-digit, 3-digit, and 4-digit number. Find total number of possible arrangements of these ten numbers such that the sum of these four numbers is maximized. Details and Assumptions: • For the purposes of this question, 0 is considered a 1-digit number. • This is an arithmetic puzzle, where $1 \square$ would represent the 2-digit number 19 if $\square = 9$. It does not represent the algebraic expression $1 \times \square$. Using the digits 1 through 9 without repetition, fill out the number grid above. What is the product of the numbers in the 4 corners? Note: The order in which this grid calculates is left-to-right/top-to-bottom unlike the usual order of operations. E.g., $1+2\times 3 = (1+2)\times3=9.$ $\large 1 \, \square \, 2 \, \square \, 3 \, \square \, 4 \, \square \, 5 \, \square \, 6 \, \square \, 7 \, \square \, 8 = 9$ There are $2^7 =128$ ways in which we can fill the squares with $+, -$. How many ways would make the equation true? Note: You are not allowed to use parenthesis. ×
# Graph Quadratic Functions: Maxima and Minima Related Topics: Math Worksheets Videos, worksheets, solutions, and activities to help Algebra students learn about how to graph quadratic functions (maxima and minima). How to graph quadratic functions in general form? The graph of a quadratic function f(x) = ax2 + bx + c, a ≠ 0 is called a parabola. 1. It is always a cup-shaped curve. 2. It opens upward if a > 0 and opens downward if a < 0. 3. The vertical line x = -b/(2a) is the line of symmetry. 4. It has a turning point, or vertex, at a point [-b/2a, f(-b/2a)] 5. The vertex is a maxima if a < 0 and a minima if a > 0. Graphing Quadratic Functions in General Form Students learn to graph quadratic functions that are written in f(x) = ax2 + bx + c form, using the vertex, the y-intercept, and the x-intercepts. The x-coordinate of the vertex can be found using the formula "-b/2a", and the y-coordinate of the vertex can be found by substituting the x-coordinate of the vertex into the function for x. The y-intercept is equal to "c", and the x-intercepts can be found by substituting a 0 into the function for f(x). Note that students are also asked to write the equation of the axis of symmetry of the given function, and find the domain, range, and maximum or minimum. How to graph quadratic functions in standard form or vertex form? The standard form of a quadratic function is f(x) = a(x - h)2 + k, a ≠ 0 1. It is always a cup-shaped curve. 2. It opens upward if a > 0 and opens downward if a < 0. 3. The vertex is at (h, k) and the axis of the function is at x = h. 4. (h, k) is a maxima if a < 0 and a minima if a > 0. Graphing Quadratic Functions in Standard Form (Vertex Form) The Max and Min of a Quadratic Function
# If 4x 4 10 what is x ### Calculate function values In a function, each \$\$ x \$\$ value has a \$\$ y \$\$ value. With the function term you can calculate the \$\$ y \$\$ values. You sit instead of the variable each a number and then calculate the term. The \$\$ y \$\$ values are also called function values. Example: Function: \$\$ f (\$\$\$\$ x \$\$\$\$) = 3 \$\$\$\$ x \$\$ \$\$ - 5 \$\$ You can calculate the function value for \$\$ x = \$\$ \$\$ 5 \$\$ as follows: \$\$ f (\$\$\$\$ 5 \$\$\$\$) = 3 * \$\$ \$\$ 5 \$\$ \$\$ - 5 = 15 \$\$ \$\$ - 5 = 10 \$\$ You can calculate the function value for \$\$ x = \$\$ \$\$ - 1 \$\$ as follows: \$\$ f (\$\$\$\$ - 1 \$\$\$\$) = 3 * (\$\$\$\$ - 1 \$\$\$\$) \$\$ \$\$ - 5 = \$\$ \$\$ - 3 \$\$ \$\$ - 5 = \$\$ \$\$ - 8 \$\$ \$\$ x \$\$ - value and \$\$ y \$\$ - value belong together. They form a pair of values or a point. You write: The value pairs \$\$ (- 1 \| -8) \$\$ and \$\$ (5 \| 10) \$\$ belong to the function \$\$ f (x) = 3x-5 \$\$ Doesn't that look like points in the coordinate system? Correct! This is how it looks in general: Function equation: \$\$ y = f (x) = mx + b \$\$ (for each \$\$ x \$\$ value) Function value for \$\$ x = 2 \$\$: \$\$ f (2) = m * 2 + b \$\$ (for a certain \$\$ x \$\$ value) Functional term ┌─┴──┐ \$\$ f (x) = 3x-5 \$\$ └────┬────┘ Function equation ### Value pairs and points As a graph, linear functions always have a straight line. You can draw the pair of values \$\$ (x \| y) \$\$ as a point in the coordinate system. The value pairs of the function are the points of the straight lines in the coordinate system. You can draw the straight line with 2 pairs of values or points. Example: After \$\$ x \$\$ minutes, the height \$\$ h (x) \$\$ of a candle in cm \$\$ h (x) = \$\$ \$\$ - 2/3 x + 20 \$\$. To draw the straight line, calculate 2 points that are not too close together. You reckon: \$\$ h (0) = - 2/3 * 0 + 20 = 20 \$\$ \$\$ rarr \$\$ point \$\$ (0 \| 20) \$\$ \$\$ h (30) = - 2/3 * 30 + 20 = –20 + 20 = 0 \$\$ \$\$ rarr \$\$ point \$\$ (30 \| 0) \$\$ \$\$ x \$\$ - coordinate \$\$ darr \$\$ Dot \$\$ (\$\$\$\$ 2 \$\$\$\$ \| \$\$\$\$ 3 \$\$\$\$) \$\$ \$\$ uarr \$\$ \$\$ y \$\$ - coordinate Here the function is not called \$\$ f \$\$, but \$\$ h \$\$. Instead of \$\$ f \$\$ for any function, one chooses \$\$ h \$\$ for the function equation of the height. ### The other way around: Calculate \$\$ x \$\$ values It is a bit more difficult when the \$\$ y \$\$ is given and you have to calculate the corresponding \$\$ x \$\$. Incidentally, the \$\$ x \$\$ values are called arguments. Example: Function: \$\$ f (x) = 3x \$\$ \$\$ - 5 \$\$ What is the name of the \$\$ x \$\$ value for the function value \$\$ 4 \$\$? Mathematically: For which \$\$ x \$\$ is \$\$ f (x) = 4 \$\$? \$\$ 3x-5 = 4 \$\$ \$\$ \| \$\$ \$\$ + 5 \$\$ \$\$ 3x = 9 \$\$ \$\$ \| \$\$ \$\$: 3 \$\$ \$\$ x = 3 \$\$ The function value \$\$ y = 4 \$\$ includes \$\$ x = 3 \$\$. A \$\$ x \$\$ value is also called argument or abscissa (from lat. linea abscissa "Cut line") A \$\$ y \$\$ value is also called ordinate (from lat. linea ordinata "Orderly line") \$\$ y \$\$ is dependent on \$\$ x \$\$ - as a donkey bridge for the names you can stick to the order in the alphabet: A before O as well as \$\$ x \$\$ before \$\$ y \$\$. kapiert.decan do more: • interactive exercises and tests • individual classwork trainer • Learning manager Anna helps out on the strawberry field during the holidays. She collects the prices for self-picked strawberries. • \$\$ 1 \$\$ kg of strawberries costs \$\$ 2.50 \$\$ \$\$ € \$\$. • Each customer pays an additional \$\$ 0.50 \$\$ \$\$ € \$\$ to allow them to nibble a little while picking. Anna writes down the functional equation \$\$ y = f (x) = 2.5 * x + 0.5 \$\$ and calculates different pairs of values. Example 1: How much do \$\$ 2 \$\$ kg of picked strawberries cost? \$\$ y = f (2) = 2.5 * 2 + 0.5 = 5.5 \$\$ \$\$ 2 \$\$ kg of picked strawberries cost \$\$ 5.50 \$\$ \$\$ € \$\$. Example 2: Mr. Lu pays \$\$ 13.00 \$\$ \$\$ £ \$\$. How many kg of strawberries did he pick? \$\$ y = f (x) = 13.00 \$\$ \$\$ 2.5 * x + 0.5 = 13.00 \$\$ \$\$ \| \$\$ \$\$ - 0.5 \$\$ \$\$ 2.5 * x = 12.50 \$\$ \$\$ \| \$\$ \$\$: 2.5 \$\$ \$\$ x = 5 \$\$ Mr. Lu picked \$\$ 5 \$\$ kg of strawberries. ### Table of values So that Anna doesn't have to calculate every time, she has created a table of values: \$\$ y = f (x) = 2.5 * x + 0.5 \$\$ Weight in kg (\$\$ x \$\$)Price in euros (\$\$ y \$\$) \$\$1,0\$\$ \$\$3,00\$\$ \$\$1,5\$\$ \$\$4,25\$\$ \$\$2,0\$\$ \$\$5,50\$\$ \$\$2,5\$\$ \$\$6,75\$\$ \$\$3,0\$\$ \$\$8,00\$\$ \$\$3,5\$\$ \$\$9,25\$\$ \$\$4,0\$\$ \$\$10,50\$\$ \$\$4,5\$\$ \$\$11,75\$\$ \$\$5,0\$\$ \$\$13,00\$\$ The graph for this: A Table of values is clear if you more than 2 Calculate points of the graph. ##### Tip calculator: Some pocket calculators do the arithmetic for a table of values for you - take a look at the instructions for use! ### A bit of theory at the end #### Domain of definition The domain of definition are all numbers that you can insert into a function, i.e. all \$\$ x \$\$ values. For linear functions: \$\$ D = QQ \$\$ #### Range of values The domain of definition are function values (\$\$ y \$\$ values) that can come out when calculating the function term. For linear but not constant functions: \$\$ W = QQ \$\$ \$\$ QQ \$\$ are the rational numbers: all positive and negative fractions. kapiert.decan do more: • interactive exercises and tests • individual classwork trainer • Learning manager
# There are two bags,one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls.A die is thrown.If it shows up 1 or 3,a ball is taken from the 1st bag,but it shows up any other number,a ball is chosen from the second bag.Find the probability of choosing a black ball. Toolbox: • $$E_1=1 \;Bag\; is\; chosen$$ • $$E_2=2 \;Bag\; is\; chosen$$ • $$A\;choising\;a\;black\;ball$$ • $$P(A)=P(E_1)P(A/E_1)+P(E_2)P(A/E_2)$$ • $$bag\; 1\; is\; chosen\; if \;1 \;or\; 3\; appear \;in \;throw\; of\; die$$ • $$P(E_1)=\Large\frac{2}{6}=\frac{1}{3}$$ • $$bag\; 2\; is\; chosen\; otherwise$$ • $$P(E_1)=1-\Large\frac{1}{3}=\frac{2}{3}$$ $$P(A/E_1$$)=$$P(getting\;black\;ball\;/1\;bag\;is\;chosen)$$ =$$\Large\frac{3}{7}$$ $$P(A/E_2$$)=$$P(getting\;black\;ball\;/1\;bag\;is\;chosen)$$ =$$\Large\frac{4}{7}$$ $$P(A)=P(getting\;black\;ball$$) =$$\Large\frac{1}{3}$$$$\times$$$$\frac{3}{7}$$+$$\frac{2}{3}$$$$\times$$$$\frac{4}{7}$$ =$$\Large\frac{11}{21}$$ edited Apr 9, 2014 by pady_1
Grid Method Multiplication Explained For Parents \| Kidadl FOR AGES 7 YEARS TO 11 YEARS # Grid Method Multiplication Explained For Parents Although grid method is fairly simple once you get to grips with it, it can be a bit of a challenge on first sight, so we've written a handy guide to get you through. This step-by-step breakdown shows how to use the grid method to solve a variety of multiplication problems your children are likely to encounter in school. This might be simple money questions in Year Three all the way up to multiplying 4-digit numbers in Year Six. ## What Is The Grid Method? The grid method of multiplication, also known as the box method, is a way of doing long multiplication by breaking numbers down into place values and writing them out in a grid. A school usually starts introducing the multiplication grid method in maths at the start of Key Stage 2, when children go into Year Three, although some introduce it as early as Year Two. Using the grid method to do long multiplication gets children to break the numbers down into hundreds, tens and ones before multiplying them. This helps the child to understand what each digit in a number represents and what's actually happening to the numbers when they are multiplied. This helps children who are struggling by letting them visualise the process more easily. In this article, we will walk you through solving various types of multiplication problems using grid method. ## Grid Method: Multiplying A Two-Digit Number By A Two-Digit Number The problem: 23x15 = ? The first number, 23, is made up of the number 20 and the number 3. That means we need to write 20 and 3 in the boxes to the right of the X. Next, add the other number down the side: Now, we do the actual multiplication. It doesn't really matter what order you multiply the boxes in, but we suggest starting on the right because it makes it easier for children to adjust to column method later. Multiply the ones column by the tens row: Now do the tens times the tens: Now the ones times the ones: And finally, the tens row times the ones column: Now, we just need to add all the numbers together. Take all four of the answers you have just found, and write them out as a column addition (or whichever addition method the child you are helping is most comfortable with): ## Grid Method: Multiplying A Three-Digit Number By A One-Digit Number In some ways, this is even easier than the grid method example above with a 2-digit number, since using a 1-digit number means there's only one row to deal with. We just need an extra column in for the hundreds. Then, follow the same method as above, multiplying each number in the top row by the one in the left-hand column: Once all the numbers have been multiplied, write out a column addition to find the sum of all three. Sorted! ## Grid Method: Multiplying A Three- (Or More)-Digit Number By A Two-Digit Number In Year 6, children will have to use the maths grid to multiply a three- or four-digit number by a two-digit one. Put the numbers into the grid as before: Next, multiply the top row: Once you've multiplied all the numbers in the top row, it's time for the second row: Some less confident children might find adding so many numbers at once intimidating, so it's fine to do this step in two parts. ## Grid Method: Multiplying A Decimal Many children get intimidated at the thought of working with decimals. The good part about grid method is that it really isn't much different to using it without the decimal. In the following example, we're working out 12.5 x 2.2. Put the numbers into the grid as usual. This time we have a column called "tenths" for the decimal place. Multiply the top row: Multiply the bottom row: Find the total of the answers: ## Grid Method: Multiplying Money As long as you're clear about whether you're working in pounds or pence, multiplying money using the grid method is very similar to any other multiplication in grid method. Here is an example of a question your child may come across in Year 4: Anna buys two packs of muffins from the bakery. Each pack costs £1.25. How much did she pay altogether? Once your child has worked out that they need to multiply £1.25 by two, put the numbers into the grid as usual. Most children won't have covered decimal multiplication in Lower KS2, so convert the numbers to pence before you start. Then, work through the grid as if you were multiplying a three-digit number by a one-digit number. Finally, find the total as usual, then convert back to pence at the end: ## Grid Method: Troubleshooting If your child is struggling with the grid method despite your best efforts, here are our top three suggestions to help. 1) Times tables knowledge. Make sure your child's times tables knowledge is solid. Often, children who struggle with long multiplication understand the method - they just don't have instant recall of their times' tables facts, so while they know they need to multiply 3 by 12, they don't know 3 x 12 = 36. Practice is key - get them repeating their tables in the car, while they help with washing up, or any other time they can. 2) Addition struggles. If your child handles the grid well but gets the wrong answer at the end, they may need a quick refresher on column addition, or to break the addition down into smaller steps. 3) Confidence. Especially for children who have struggled with maths in the past, it's easy to lose confidence. The grid method can look intimidating at first, and some children get so nervous they simply stop being able to work through the steps logically. Reassurance will do wonders here, as will working through a couple of examples together slowly. If you've tried all these and your child is still finding grid method difficult, check-in with their teacher, especially if you've noticed them struggling in school more generally. ###### Written By Jennie Hughes <p>Jennie, originally from Manchester, discovered her love for teaching and travel while working at a kindergarten in China. Since then, she has become an expert in both fields and mainly teaches KS2 children. Jennie also runs a tutoring and mindfulness company called 'Recreate-U', which aims to create comfortable, safe, and happy learning environments to help people reach their full educational potential. During her free time, she enjoys engaging in craft projects or relaxing with a good book and a hot cup of tea.</p> ## How can we improve it?How helpful was it? Your privacy is important to us. Any information you provide to us via this website may be placed by us on servers located in countries outside the EU if you do not agree to such placement, do not provide the information. ## Get inspiration for your Days Out with kids We’ll send you tons of inspiration to help you find a hidden gem in your local area or plan a big day out.
Saturday, September 23, 2023 Undefined Terms In Geometry Wikipedia Collinear And Coplanar Points The Undefined Terms in Geometry If two or more points are located in the same line, then the points are collinear points. On the other hand, if two or more points are located in the same plane, then the points are coplanar points. Example: Please refer to the following illustration: Using the given illustration above, determine: • All points collinear with point X. • All points coplanar with point Y. • Solution: • Points P and W are collinear with point X since they are located in the same line, in particular, line PW. • Point X is coplanar with point Y since they are located in the same plane. • Using the undefined terms we have discussed here, we can now provide formal definitions to other essential geometric terminologies. Introduction To Geometry: Undefined Terms Definition Postulates And Theorems The world consists of various objects in different forms and shapes. Humans have been fascinated with ways to measure these objects as early as the Egyptian and Greek civilizations. This fascination with measurement and shapes has led to the building of architectural marvels that prove human ingenuity throughout the agesfrom the timeless pyramids to breathtaking skyscrapers. Geometry is the branch of mathematics that deals with measurements, forms, and shapes. It comes from the Greek words geo, which means Earth, and metron, which means measure. The origin of the word itself already provides a clue to what geometry is all about and that is to measure everything we can see on this planet. The study of geometry starts with three undefined terms: point, line, and plane. Every other geometric concept is derived from these undefined terms. In this review, were going to explore the undefined and defined terms in geometry and the concepts of postulates and theorems. Section Formula: To Find A Point Which Divides A Line Into M: N Ratio Consider a line A and B having coordinates and , respectively. Let P be a point that which divides the line in the ratio m:n, then the coordinates of the coordinates of the point P is given as- • When the ratio m:n is internal: • When the ratio m:n is external: You May Like: What Do You Learn In 4th Grade Math Access8math V13 Update Log • positive rule: Read “positive” rather than “plus” when plus sign in first item or its previous item is certain operator. • square rule: When the power is 2, the item is read as “squared”. • cubic rule: When the power is 3, the item is read as “cubed”. • line rule: When there is “” right above two Identifier, the item is read as “Line …”. • line segment rule: When there is “¯” right above two Identifier, the item is read as “Line segement …”. • ray rule: When there is a “” right above two Identifier, the item is read as “Ray …” • Add interaction window: Pressing “Space” in math content to open “Access8Math interaction window” which contains “interaction” and “copy” button. • interaction: Into math content to navigate and browse. • copy: Copy MathML object source code. • Adjust inheritance relationship between rules to ensure proper use of the appropriate rules in conflict. • Other Fields Of Mathematics Calculus was strongly influenced by geometry. For instance, the introduction of coordinates by René Descartes and the concurrent developments of algebra marked a new stage for geometry, since geometric figures such as plane curves could now be represented analytically in the form of functions and equations. This played a key role in the emergence of infinitesimal calculus in the 17th century. Analytic geometry continues to be a mainstay of pre-calculus and calculus curriculum. Another important area of application is number theory. In ancient Greece the Pythagoreans considered the role of numbers in geometry. However, the discovery of incommensurable lengths contradicted their philosophical views. Since the 19th century, geometry has been used for solving problems in number theory, for example through the geometry of numbers or, more recently, scheme theory, which is used in Wiles’s proof of Fermat’s Last Theorem. Recommended Reading: What Is Biological Treatment For Rheumatoid Arthritis Undefined Terms Of Geometry In geometry, we use formal definitions to precisely refer to a certain concept. For example, a triangle is a geometric concept that is defined as a type of polygon with three sides and three vertices. Using this formal definition, we know exactly what a triangle is it allows us to tell that a pizza is triangular in shape, but a ball is not. However, before we can provide a formal definition of geometric concepts, we must recognize first that there are some concepts that we cannot define precisely. These terms are known as undefined terms. There are three undefined terms in geometry, namely the point, line, and plane. Why are these terms undefined? The point, line, and plane cannot be defined easily because they are the building blocks of geometry. Exploring and combining these terms will provide us with other geometric concepts. How can we define these terms if they are the foundations of our study? Although we cannot formally define what a point, line, or plane is, we can develop an intuition on what these terms are. For instance, the tip of your ballpen is a representation of a point, the edge of your notebook is a line, and the surface of your table is a plane. Lets now provide descriptions of these undefined terms in geometry and look for their real-life representations. Which Is Not An Undefined Term In Geometry Brainlycom • Highest rating: 3 • Lowest rating: 2 • Descriptions: There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane. • More : There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane. • https://brainly.com/question/1777794 Recommended Reading: Why Do We Use Computer In Geography Three Undefined Terms: Point Line And Plane Univ. of WisconsinJ.D. Univ. of Wisconsin Law school Brian was a geometry teacher through the Teach for America program and started the geometry program at his school In Geometry, we have several undefined terms: point, line and plane. From these three undefined terms, all other terms in Geometry can be defined. In Geometry, we define a point as a location and no size. A line is defined as something that extends infinitely in either direction but has no width and is one dimensional while a plane extends infinitely in two dimensions. Undefined Terms in Geometry • alt+Left arrow key: move to the start point of the previous data block • alt+Down arrow key: Only read the content of the current data block without moving • alt+right arrow: move to the start point of the next data block • alt+home: move to the starting point of the current data block • alt+end: move to the end point of the current data block • alt+shift+left arrow: move to the previous data block and select • alt+shift+down key: do not move, only select the content of the current data block • alt+shift+right arrow: move to the next data block and select In the writing settings, you can choose whether to enable the block navigation gestures when starting. Press NVDA+alt+n in the editing area to enable or disable the block navigation gestures, which can be changed in the input gestures. Also Check: Algebra 2 Trig Mcdougal Littell Which Is An Undefined Term In Geometry An undefined term is a term that cant be defined so easily. A line is defined as something that extends infinitely in either direction but has no width and is one dimensional Solve word questions too Get Tasks is an online task management tool that helps you get organized and get things done. Figure out mathematic equations The best way to spend your free time is with your family and friends. The passing rate for the exam is 80%. What Is 0 Divided By 0 This is part of a series on common misconceptions. What is Why some people say it’s 0: Zero divided by any number is 0. Why some people say it’s 1: A number divided by itself is 1. Only one of these explanations is valid, and choosing the other explanations can lead to serious contradictions. ba means “the number which when multiplied by b a.” For example, the reason 1 01 is undefined is because there is no number x 00 is strange, because every number x 0x=0. Because there’s no single choice of x x that works, there’s no obvious way to define 0 00, so by convention it is left undefined. Of course, there are many possible counterarguments to this. Here are a few common ones: Rebuttal: Any number divided by itself is 1. Reply: This is true for any nonzero number, but dividing by 0 0 divided by any number is 0. Reply: This is true for any nonzero denominator, but dividing by 0 0 is not allowed no matter what the numerator is. Rebuttal: Any number divided by 0 \frac=\infty 0y= is not entirely accurate: see 1/0 for a discussion. But this reasoning only makes sense for a nonzero numerator. Rebuttal: If we choose to set 0 , 0, 0, it is not inconsistent with other laws of arithmetic, and it makes one of the rules in the above rebuttals true in all cases. Reply: This is a combination of the first two rebuttals, so here is a “big-picture” reply. Any specific choice of value for 0 Don’t Miss: How To Read Geography Ncert For Upsc Why Do We Need Coordinate Geometry Coordinate geometry has various applications in real life. Some of the areas where coordinate geometry is an integral part include. • In digital devices like computers, mobile phones, etc. to locate the position of cursor or finger. • In aviation to determine the position and location of airplanes accurately. • In maps and in navigation . • To map geographical locations using latitudes and longitudes. Put your understanding of this concept to test by answering a few MCQs. Click Start Quiz to begin! Select the correct answer and click on the Finish buttonCheck your score and answers at the end of the quiz • “Down Arrow”: Zoom in on a smaller subpart of the math content. • “Up Arrow”: Zoom out to a larger subpartthe of the math content . • “Left Arrow”: Move to the previous math content. • “Right Arrow”: Move to the next math content. • “Home”: Move back to the top. • “Ctrl+c”: Copy object MathML source code • “Numpad 1~9”: Reading the math content into serialized text using NVDA Reviewing Text. • “ESC”: Exit the navigation mode. Recommended Reading: Exponential Growth And Decay Common Core Algebra 1 Homework Answers Read A Brief Summary Of This Topic mathematics, the science of structure, order, and relation that has evolved from elemental practices of counting, measuring, and describing the shapes of objects. It deals with logical reasoning and quantitative calculation, and its development has involved an increasing degree of idealization and abstraction of its subject matter. Since the 17th century, mathematics has been an indispensable adjunct to the physical sciences and technology, and in more recent times it has assumed a similar role in the quantitative aspects of the life sciences. In many culturesunder the stimulus of the needs of practical pursuits, such as commerce and agriculturemathematics has developed far beyond basic counting. This growth has been greatest in societies complex enough to sustain these activities and to provide leisure for contemplation and the opportunity to build on the achievements of earlier mathematicians. All mathematical systems are combinations of sets of axioms and of theorems that can be logically deduced from the axioms. Inquiries into the logical and philosophical basis of mathematics reduce to questions of whether the axioms of a given system ensure its completeness and its consistency. For full treatment of this aspect, seemathematics, foundations of. Which Is Not An Undefined Term In Geometry A Gauthmath • www.gauthmath.com • Highest rating: 5 • Lowest rating: 1 • Descriptions: 1. Which is NOT an undefined term in geometry? C. plane What is used to name a noint? 1. Which • More : 1. Which is NOT an undefined term in geometry? C. plane What is used to name a noint? 1. Which • https://www.gauthmath.com/solution/1-Which-is-NOT-an-undefined-term-in-geometry-A-angle-B-line-C-plane-2-What-is-us-1701605887854597 You May Like: What Is Transparent In Physics Answered: Which Is Not A Undefined Term In • www.bartleby.com • Highest rating: 5 • Lowest rating: 2 • Descriptions: Solution for Which is NOT a undefined term in Geometry? A. Axis B. Line C. Plane D. Point. • More : Solution for Which is NOT a undefined term in Geometry? A. Axis B. Line C. Plane D. Point. Euclidean Geometry Is Consistent And Complete Undefined Terms in Geometry (Grade 7 – Lesson 1) \| TAGALOG \| Something that you’d think would be mentioned based on the title: Euclidean geometry is an interesting example of a formal system that we know is both consistent and complete. It’s not powerful enough for Gödel’s Incompleteness Theorem to apply. To really be able to say that, you have to formalize things a bit more than Euclid with his undefined terms, which modern mathematicians have done. See, for example, Tarski’s axioms , or Hilbert’s axioms. Further reading on Wikipedia: Euclidean geometry Recommended Reading: What Is Absolute Location In Geography Three Undefined Terms: Point Line And Plane Concept Brightstorm • www.brightstorm.com • Highest rating: 3 • Lowest rating: 2 • Descriptions: In Geometry, we have several undefined terms: point, line and plane. From these three undefined terms, all other terms in Geometry can be defined. • More : In Geometry, we have several undefined terms: point, line and plane. From these three undefined terms, all other terms in Geometry can be defined. • https://www.brightstorm.com/math/geometry/geometry-building-blocks/three-undefined-terms-point-line-and-plane/ • Highest rating: 4 • Lowest rating: 3 • Descriptions: There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane. In geometry, a point has no dimension ( • More : There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane. In geometry, a point has no dimension ( • https://quizlet.com/89051339/geometry-flash-cards/ Undefined Terms In Geometry An undefined term is a point, line, or plane. Examples of defined terms are angles. Which term is not considered to be undefined? In geometry, definitions are formed using Free time to spend with your family and friends math is the study of numbers, shapes, and patterns. It is used in everyday life, from counting to measuring to more complex calculations. Learn step-by-step Don’t Miss: What Is Selective Pressure In Biology Access8math V14 Update Log • Adjust settings dialog box which divided into “general setting” and “rules setting” dialog box. “General Settings” is the original “Access8Math Settings” dialog box, and “Rule Settings” dialog box is for selecting whether specific rules are enabled. • vector rule: When there is a “” right above two Identifier, the item is read as “Vector…”. • frown ruleWhen there is a ” ” right above two Identifier, the item is read as “frown…”. Which Of The Following Is Not An Undefined Term Point Ray Line Plane • Highest rating: 3 • Lowest rating: 2 • Descriptions: There are three undefined terms in geometry. These are points, line, and plane. Among the given choices, there is only one which is not an undefined term • More : There are three undefined terms in geometry. These are points, line, and plane. Among the given choices, there is only one which is not an undefined term • https://brainly.com/question/2517500 Don’t Miss: What Is The Future Of Geography Is Undefined Equal To Zero So zero divided by zero is undefined. Just say that it equals undefined. In summary with all of this, we can say that zero over 1 equals zero. We can say that zero over zero equals undefined. And of course, last but not least, that were a lot of times faced with, is 1 divided by zero, which is still undefined. What Is A Co You must be familiar with plotting graphs on a plane, from the tables of numbers for both linear and non-linear equations. The number line which is also known as a Cartesian plane is divided into four quadrants by two axes perpendicular to each other, labelled as the x-axis and the y-axis. The four quadrants along with their respective values are represented in the graph below- The point at which the axes intersect is known as the origin. The location of any point on a plane is expressed by a pair of values and these pairs are known as the coordinates. The figure below shows the Cartesian plane with coordinates . If the coordinates are identified, the distance between the two points and the intervals midpoint that is connecting the points can be computed. Coordinate Geometry Fig. 1: Cartesian Plane Read Also: What Is The Geography Of China Access8math V12 Update Log • negative number rule: Read ‘negative’ rather than ‘minus sign’ when minus sign in first item or its previous item is certain operator. • integer add fraction rule: Read ‘add’ between integer and fraction when fraction previous item is integer. • Program architecture improve • add dynamic generate Complement class • In navigation mode command, “Ctrl+c” copy object MathML source code. • Settings dialog box in Preferences: • Language: Access8Math reading language on math content. • Analyze the mathematical meaning of content: Semantically analyze the math content, in line with specific rules, read in mathematical meaning of that rules. • Read defined meaning in dictionary: When the pattern is definied in the dictionary, use dictionary to read the meaning of subpart in the upper layer part. • Read of auto-generated meaning: When the pattern is not difined or incomplete in dictionary, use automatic generation function to read the meaning of subpart in the upper layer part. • Add some simple rule. Single rules are simplified versions of various rules. When the content only has one single item, for better understanding and reading without confusion, you can omit to choose not to read the script before and after the content. • Update unicode.dic.
# How do I graph x^3 - 4x^2 + 2 and 8x^3 - 16x^2 + 6? PLEASE EXPLAIN You graph these the way you graph any functions. You pick a few values for x, find the corresponding values of the expressions (y), plot the (x, y) points on a graph, and draw a smooth curve through the points. It is convenient to have a method for finding values of x that might be of interest. When you examine these functions, you see that each is in the form ax^2(x-b) + c This tells you that the expression value is equal to "c" when the value of x is 0 or "b". It also tells you the function will tend toward negative infinity for values of x below 0. And it tells you that x will tend toward positive infinity for values of x above "b". The function will be less than (or equal to) "c" for values of x below "b". You are usually not terribly interested in values of the expression when those values are very large, so your graph will probably want to extend from x=-2 or so to about x=b+2 or so. Computation is probably easiest if you use mostly integers for values of x. For the first function, we can make a short table of values. We see that a=1, b=4, c=2. For x=-2, (-2)^2(-2-4)+2 = -22, so (-2, -22) is a point on the curve For x=-1, (-1)^2(-1-4)+2 = -3, so (-1, -3) is another point on the curve From above, we know that points (0, 2) and (4, 2) are on the curve For x=1, 1^2(1-4)+2 = -1, so (1, -1) is on the curve For x=2, 2^2(2-4)+2 = -6, so (2, -6) is on the curve For x=3, 3^2(3-4)+2 = -7, so (3, -7) is on the curve For x=5, 5^2(5-4)+2 = 27, so (5, 27) is on the curve A graph of the first expression can be seen here. A graph of the second expression can be seen here. thanked the writer.
Select Board & Class A park is in the shape of a right triangle. The length of the hypotenuse and a side of this park are 25 m and 7 m respectively. What is the length of the third side? Let us consider the following triangle ABC as the right triangular park whose hypotenuse AC = 25 m and side BC = 7 m. Let AB be x. Using Pythagoras theorem for triangle ABC, AB2 + BC2 = AC2 x2 + (7)2 = (25)2 x2 + 49 = 625 x2 = (625 − 49) x2 = 576 m2 To find the value of x, we require a number whose square is 576. For this, we will find the square root of 576. Mathematically, we write it as Here, represents the square root. To find the square root of 576, we follow a method, which is known as prime factorization method. Let us discuss this method and find the square root of 576 with the help of the given video. In this way, we can find the square root of a given number by prime factorization method and solve problems related to it. Let us discuss some more examples to understand the concept better. Example 1: Find the square roots of the following numbers. (i) 324 (ii) 676 (iii) 1225 (iv) 3136 Solution: (i) The prime factorization of 324 is (ii) The prime factorization of 676 is (iii) The prime factorization of 1225 is (iv) The prime factorization of 3136 is Example 2: Is 504 a perfect square? If not, (i) find the smallest number multiplied to this number, so that the product would be a perfect square (ii) find the smallest number by which 504 must be divided, so that the quotient is a perfect square Solution: The prime factorization of 504 is Here, the prime factors 2 and 7 do not occur in pair. Therefore, 504 is not a perfect square. (i) In order to obtain a perfect square, each factor of 504 must be paired. Therefore, we have to make pairs of 2 and 7. For this, 504 should be multiplied by 2 × 7 i.e., 14. Therefore, 14 should be multiplied with 504 to make it a perfect square. (ii) In order to obtain a perfect square, each factor of 504 must be paired. Therefore, 504 should be divided by 2 × 7 i.e., 14. Therefore, 504 should be divided by 14 so that the quotient is a perfect s… To view the complete topic, please Syllabus
# 2nd Grade Addition and Subtraction: Key Terms and Facts Every Student Should Know • Hi everyone! Welcome to another enriching resource about 2nd-grade math. In this article, we will learn about 2nd Grade addition and subtraction, two of the most important skills that early math learners need to master. We will also review some key terms and facts that every student should know to accurately understand and solve math problems. And, of course, we will have some fun and easy practice exercises to test your knowledge and skills. Are you ready? Let's get started! ## Introduction: Why 2nd Grade math matters Before understanding what addition and subtraction mean, we must determine why 2nd Grade matters. First, math is everywhere in our lives. We use it to count, measure, compare, and more. Also, it helps us make sense of the world and solve problems. That's why it's important to learn math well and enjoy it. In 2nd grade, you will learn more about numbers and how to use them differently. You will also learn how to add and subtract numbers up to 1000. These are essential skills that you will use throughout your life. • ### Addition: What it means and how to do it The addition is one of the basic operations in math. It means combining two or more numbers to get a bigger number. For example, if you have 3 apples and 4 oranges, add them together to get 7 fruits. The symbol for addition is +. To add two numbers, you must line them up according to their place values (ones, tens, hundreds, etc.) and then add each pair of digits from right to left. If a pair of digits sum is 10 or more, you must carry the extra digit to the next place value. For example: In this example, we start by adding the ones: 7 + 6 = 13. Since this is more than 10, we write 3 in the ones place and carry over 1 to the tens place. Then we add the tens: 1 + 3 + 5 = 9. We write 9 in the tens place. Finally, we add the hundreds: 2 + 4 = 6. We write 6 in the hundreds place. • ### Subtraction: What it means and how to do it Subtraction is another basic operation in math. It means taking away one number from another number to get a smaller number. For example, if you have 7 candies and eat 4 of them, you can subtract them to find out how many candies you have left. The symbol for subtraction is -. To subtract two numbers, you need to line them up according to their place values (ones, tens, hundreds, etc.) and then subtract each pair of digits from right to left. If the digit on the top is smaller than the digit on the bottom, you must borrow a digit from the next place value. In this example, we start by subtracting the ones: 3 - 6 = -3. Since this is negative, we borrow a digit from the tens place and make it 13 - 6 = 7. We write 7 in the ones place and reduce the tens place by one. Then we subtract the tens: 8 - 5 = 3. We write 3 in the tens place. Finally, we subtract the hundreds: 5 - 4 = 1. We write 1 in the hundreds place. • ### Key Terms: Words you need to know for 2nd-grade math There are some words that you need to know for 2nd-grade math. These words help learners understand what the math problems ask them to do and how to explain their answers. Here are some of the most common words: • Addend: A number that is added to another number. For example, in 3 + 4 = 7, both 3 and 4 are addends. • Sum: The answer to an addition problem. For example, in the addition sentence, 3 + 4 = 7; the number 7 is the sum. • Minuend: A number that is subtracted from another number. For example, in the subtraction sentence, 7 - 4 = 3, the number 7 is the minuend. • Subtrahend: A number that is subtracted from another number. For example, in 7 - 4 = 3, the number 4 is the subtrahend. • Difference: The answer to a subtraction problem. For example, in the subtraction sentence, 7 - 4 = 3, the number 3 is the difference. • Place value: The value of a digit depending on its position in a number. For example, in 237, the 2 has a place value of hundreds, the 3 has a place value of tens, and the number 7 has a place value of ones. • Regrouping: The process of rearranging digits in a number to make adding or subtracting easier. For example, in 237 + 456, we can regroup 237 as 200 + 30 + 7 and 456 as 400 + 50 + 6. This makes it easier to add each pair of digits. • ### Facts: Basic rules and strategies for addition and subtraction There are some facts or basic rules and strategies that you need to know for addition and subtraction. These facts help you do math faster and more accurately. Here are some of the most important facts: • The commutative property means that you can change the order of the addends without changing the answer. For example, 3 + 4 = 4 + 3. • The associative property means that you can change the grouping of the addends without changing the answer. For example, (3 + 4) + 5 = 3 + (4 + 5). • The identity property means that adding zero or subtracting zero does not change the number. For example, 3 + 0 = 3 and 3 - 0 = 3. • Inverse property means that adding the opposite or subtracting the same number gives zero. For example, 3 + (-3) = 0 and 3 - 3 = 0. • Addition and subtraction facts: These are the basic combinations of numbers that you need to memorize. For example, you need to know that 2 + 2 = 4, 5 - 3 = 2, etc. Knowing these facts will help you do math faster, mentally, and more confidently. ### Practice: Fun and easy exercises to master 2nd Grade Math Now that you have learned some key terms and facts about addition and subtraction, it's time to practice what you have learned. Here are some fun and easy exercises that you can do to master 2nd-grade math: • Fill in the blanks: Use the commutative property to fill in the missing numbers. For example, 5 + 6 = 6 + 5. 10 + 7 = __ + __ 45 + 12 = __ + __ 246 + 128= __ + __ • Solve the problems: Use the associative property to solve the problems. For example, (2 + 3) + 4 = 2 + (3 + 4) (23 + 16) + 9 = ________ 7 + (18 + 30) = ________ 9 + 12 + 10 = ________ • Find the missing numbers: Use the identity property to find the missing numbers. For example, 7 + 0 = 7. 43 + 0 = _ 12 – 0 = _ 60 - 0 = _ • Find the opposites: Use the inverse property to find the opposites of the numbers. For example, the opposite of -5 is 5. The opposite of -8 is _ The opposite of -10 is _ The opposite of -12 is _ • Write the facts: Write down all the addition and subtraction facts for each number from 0 to 10. For example, for 0: we know that 0 + 0 = 0 and 0 - 0 = 0 &helpp; ... Subtraction facts: ... ... Thank you for sharing the links of MathSkills4Kids.com with your loved ones. Your choice is greatly appreciated. ### Conclusion: How to apply what you learned in real life Congratulations! You and your kids have learned a lot about addition and subtraction in this article. The kids have learned what addition and subtraction mean, how to do them, some key terms and facts, and some practice exercises. But how can they apply what they've learned in real life? Here are some examples: • You can use addition and subtraction to count money, measure length, weight, or time, compare quantities, etc. • You can use addition and subtraction to solve word problems that involve adding or subtracting numbers or objects. • You can use addition and subtraction to check your answers by doing the opposite operation. For example, if you added two numbers and got an answer, you can subtract one of them from the answer to get back the other number. I hope you find these resources helpful in reinforcing your 2nd graders' addition and subtraction properties skills. Remember that practice makes perfect, but it doesn't have to be boring! With these resources, your kids will have fun while learning math. Happy learning! •
Statistics # Introduction StatisticsIntroduction Figure 10.1 If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River), you can use a slightly different technique when conducting a hypothesis test. (credit: Chloe Lim) ## Chapter Objectives By the end of this chapter, the student should be able to do the following: • Classify hypothesis tests by type • Conduct and interpret hypothesis tests for two population means, population standard deviations known • Conduct and interpret hypothesis tests for two population means, population standard deviations unknown • Conduct and interpret hypothesis tests for two population proportions • Conduct and interpret hypothesis tests for matched or paired samples Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether the SAT or GRE preparatory courses really help raise their scores. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified as independent groups or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p values. TI-83+ and TI-84 instructions are included, as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests. This chapter deals with the following hypothesis tests: • Independent groups (samples are independent) • Test of two population means • Test of two population proportions • Matched or paired samples (samples are dependent) • Test of the two population proportions by testing one population mean of differences
Parallel resistanee is reciprocal of the sum? • Lokhtar In summary, the total resistance in a parallel circuit is found by taking the reciprocal of all individual resistances and adding them together. This can be derived from the equation V=IR and the definition of parallel resistance. Lokhtar Homework Statement Why is parallel resistance the reciprocal of all individual resistances? V=IR The Attempt at a Solution Well, since V is constant and I is different, you can write it as I=V/R, and since V won't change, you can make it I=V(1/R1+1/R2),etc. So I get that, but why do you then have to take the reciprocal of all the resistances to get the total resistance? Wouldn't it just be the direct sum of the individual 1/Rs?? Lokhtar said: Homework Statement Why is parallel resistance the reciprocal of all individual resistances? V=IR The Attempt at a Solution Well, since V is constant and I is different, you can write it as I=V/R, and since V won't change, you can make it I=V(1/R1+1/R2),etc. So I get that, but why do you then have to take the reciprocal of all the resistances to get the total resistance? Wouldn't it just be the direct sum of the individual 1/Rs?? Because, by definition, $I = V/R_{eq}$. So setting this equal to your expression, we get $$\frac{1}{R_{eq}} = \frac{1}{R_1 + R_2 + \ldots}$$ The reason why parallel resistance is the reciprocal of all individual resistances is because of the way current flows in parallel circuits. In a parallel circuit, the current is divided among the different branches, and the total current is equal to the sum of the currents in each branch. This means that the total current is inversely proportional to the total resistance, as seen in the equation I=V/R. So, in order to calculate the total resistance, we need to take the reciprocal of all the individual resistances, which represents the inverse relationship between current and resistance in parallel circuits. In other words, the more resistors you add in parallel, the lower the total resistance will be, since the current is divided among more branches. This is why parallel resistance is the reciprocal of all individual resistances. 1. What is parallel resistance? Parallel resistance is a type of electrical resistance that occurs when there are multiple electrical pathways for current to flow through a circuit. In other words, the circuit branches into different paths and the total resistance is less than if the components were connected in series. 2. What does it mean for parallel resistance to be reciprocal? Reciprocal in this context means that the total resistance of a parallel circuit is equal to the reciprocal of the sum of the individual resistances. In other words, if you add the reciprocals of each individual resistance, you will get the total resistance of the parallel circuit. 3. Why is parallel resistance important? Parallel resistance is important because it allows for more efficient current flow in a circuit. By dividing the current into different paths, the overall resistance decreases, leading to a higher current flow and less power loss. 4. How do you calculate parallel resistance? The formula for calculating parallel resistance is: 1/Rt = 1/R1 + 1/R2 + ... + 1/Rn, where Rt is the total resistance and R1, R2, etc. are the individual resistances. To find the total resistance, take the reciprocal of the sum of the individual resistances. 5. What are some real-life applications of parallel resistance? Parallel resistance is commonly used in household wiring, where multiple appliances are connected to the same power source. It is also used in electronic circuits, such as in computers and televisions, to allow for efficient current flow. Additionally, parallel resistance is used in power grids to distribute electricity to different areas. Replies 23 Views 2K Replies 3 Views 1K Replies 7 Views 828 Replies 11 Views 2K Replies 24 Views 4K Replies 3 Views 1K Replies 6 Views 2K Replies 7 Views 2K Replies 6 Views 1K Replies 12 Views 3K
For ParentsFor Teachers More resources All maths resources Below is a list of maths resources that you can use to teach or practice particular skills. Skip counting in 2s, 5s, and 10s Skip counting is a great way to start multiplication. By learning to count up in groups of numbers, students can start practicing their times tables. Skip counting involves skipping the amount of numbers you're counting up in. For example, if you are skip counting in 2s and start at 0, you would count 0, [1], 2, [3], 4... and so on. If you are skip counting in 5s starting at 0, you would count 0, [1], [2], [3], [4], 5, [6], [7], [8], [9], 10... and so on. Converting between numbers and words For example: 1 becomes "one" 2 becomes "two" 10 becomes "ten" 20 becomes "twenty" 45 becomes "forty-five" 129 becomes "one hundred and twenty nine" Identify basic fractions Fractions can be a tricky concept for students to first grasp. Starting with splitting up shapes and foods like pizza or cake is a great introduction to fractions. To start with, it's important to break down what a fraction is: a small part of a whole. It's also important to understand the parts of a fraction. The bottom number is called the "denominator" and tells us how many pieces there are in total. The top number is called the "numerator" and tells us how many pieces we actually have. So if we take 2 slices of a pizza that has 8 slices, we have 2/8 of the pizza! Logical reasoning and guess the number Using logical reasoning, students can guess the number based on initial information. For example: The number I'm thinking of is a multiple of 4, greater than 16 but smaller than 24. (I am 20) I am an even number. I have a five in the tens place. You do not say me when you count by tens. I am greater than 0 but less than 54. (I am 52) Tom 16. Emma is 1 year younger than Tom and 3 years younger than Matt. How old is Matt? (Matt is 18) Doubling and halving Doubling is the process of making twice as much of something, or two lots of it. For example, if I wanted to double 6 sweets, I would add another 6 sweets to make 12. I would have twice as much as I did before. Halving is the process of splitting into two groups and removing one of them. If I wanted to halve my 6 sweets, I would split them into 2 groups of 3, and get rid of one group. I would be left with 3 which is half of 6. Multiplying large numbers Multiplying any numbers by 3 digits or more can be a little more complex and require more steps than multiplying by single digit numbers. However, multiplying by large numbers is still based on basic knowledge of times tables. Let's look at an example and break it down. If we have the question 34 x 23 = ? We automatically know that this question is not covered by our times tables to 12 but we can use our basic times tables knowledge by splitting up the question. We can start by splitting the question into two: 34 x 20 = ? 34 x 3 = ? We have split 21 into 20 and 1. We can split it up even further though. 30 x 20 = ? 4 x 20 = ? 30 x 3 = ? 4 x 3 = ? This is a little simpler to work out and a lot closer to our basic times tables knowledge! We know: 30 x 20 = 600 4 x 20 = 80 30 x 3 = 90 4 x 3 = 12
# 2008 Mock ARML 1 Problems/Problem 8 ## Problem For positive real numbers $a,b,c,d$, \begin{align}2a^2 + \sqrt {(a^2 + b^2)(a^2 + c^2)} &= 2bc\\ 2a^2 + \sqrt {(a^2 + c^2)(a^2 + d^2)} &= 2cd\\ 2a^2 + \sqrt {(a^2 + d^2)(a^2 + b^2)} &= 2db\end{align} $$\sqrt {(a^2 + b^2)(a^2 + c^2)} + \sqrt {(a^2 + c^2)(a^2 + d^2)} + \sqrt {(a^2 + d^2)(a^2 + b^2)} = 2$$ Compute $ab + ac + ad$. ## Solution 1 We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be $x$, $y$, and $z$, and the tangents to the incircle be $b$, $c$, and $d$. Then use Law of Cosines to express the sides in terms of $x$, $y$, and $z$, and Pythagorean Theorem to express $x$, $y$, and $z$ in terms of $b$, $c$, $d$, and the inradius $a$. This yields the first three equations. The fourth is the result of the sine area formula for the three small triangles, and gives the area as $\frac {\sqrt {3}}{2}$. The desired expression is $rs$, which is also the area, so the answer is $\boxed{\frac {\sqrt {3}}{2}}$. ## Solution 2 Since the equations are symmetric in $b,c,d$, we may consider $b=c=d$; the system reduces and we find that the desired sum is $\boxed{\frac {\sqrt {3}}{2}}$.
# What Is 2 To The 30 Power? Definition, Calculation, And Applications // Thomas Discover the meaning of 2 to the 30 power, how to calculate it, and its practical uses in computer science and data storage. ## What is 2 to the 30 power? ### Definition and Explanation 2 to the 30th power, also written as 2^30, is a mathematical expression representing the result of multiplying the number 2 by itself 30 times. In other words, it is the value obtained by raising 2 to the power of 30. This operation can be visualized as repeatedly multiplying 2 by itself, starting from 1 and continuing for 30 times. When we raise a number to a power, we are essentially multiplying that number by itself a certain number of times. In the case of 2 to the 30th power, we are multiplying the number 2 by itself 30 times. 2 to the 30th power can be written out as: 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 This results in a very large number. In fact, 2 to the 30th power is equal to 1,073,741,824. This means that when we multiply 2 by itself 30 times, the resulting value is over 1 billion. 2 to the 30th power is often used in various fields, especially in computer science and data storage. It has important applications and implications in these areas, which we will explore in the following sections. By understanding the significance of 2 to the 30th power, we can gain insights into the power of exponential growth and its impact in different domains. ## Calculation of 2 to the 30 power Calculating the value of 2 to the 30 power may seem daunting at first, but with a step-by-step approach, it becomes more manageable. Let’s break it down and see how it’s done. ### Step-by-Step Calculation To calculate 2 to the 30 power, we need to multiply 2 by itself 30 times. This may sound time-consuming, but we can simplify the process by using the concept of exponentiation. 1. Start with the base number, which is 2 in this case. 2. Multiply 2 by itself once: 2 * 2 = 4. 3. Multiply the result by 2 again: 4 * 2 = 8. 4. Continue this process, multiplying the previous result by 2 each time. 5. Repeat steps 3 and 4 a total of 28 more times, as we need to multiply 2 by itself 30 times. 6. Finally, after performing all the multiplications, we arrive at the result: 2 to the 30 power equals 1,073,741,824. This step-by-step calculation method allows us to break down the process and simplify the task of finding the value of 2 to the 30 power. It’s important to note that this approach can be used for any exponentiation calculation, not just for 2 to the 30 power. By following these steps, we can easily calculate the value of 2 to the 30 power without the need for a calculator or complex mathematical formulas. This calculation has significance in various fields, including computer science and data storage, which we will explore in the subsequent sections. ## Applications of 2 to the 30 power ### Use in Computer Science In the world of computer science, the concept of 2 to the 30 power holds significant importance. This value, which is equal to 1,073,741,824, is often used as a basis for addressing memory in computer systems. One of the primary of 2 to the 30 power in computer science is in the representation of memory addresses. Computers use binary numbering systems, which means they only have two digits: 0 and 1. With 30 bits, or binary digits, we can represent up to 1,073,741,824 different memory addresses. This allows computer systems to access and store a vast amount of information efficiently. Additionally, 2 to the 30 power is used in computer algorithms and data structures. Many algorithms and data structures rely on powers of 2 for their efficiency and effectiveness. For example, when dividing a data set into equal parts, it is often advantageous to use powers of 2 to ensure balanced divisions and optimal performance. ### Use in Data Storage In the realm of data storage, 2 to the 30 power plays a crucial role in determining storage capacity. This value represents approximately 1 gigabyte (GB) of data. As technology has advanced, data storage devices have become increasingly capable of storing larger amounts of information. 2 to the 30 power is used as a reference point when discussing storage capacity in terms of gigabytes. For example, a hard drive with a capacity of 1 terabyte (TB) can store approximately 1,024 times the amount of data as 2 to the 30 power. This allows individuals and organizations to store vast amounts of data, such as documents, photos, videos, and more, on a single device. Moreover, the concept of 2 to the 30 power is also relevant in the context of data transfer rates. When referring to data transfer speeds, such as megabits per second (Mbps) or gigabits per second (Gbps), the value of 2 to the 30 power is used as a benchmark for comparison. It helps in understanding the capabilities and limitations of different data transfer technologies. ## Comparison of 2 to the 30 power with Other Powers of 2 When it comes to understanding the power of numbers, it’s fascinating to explore how they stack up against each other. In this section, we will delve into the comparison of 2 to the 30 power with other powers of 2, specifically focusing on 2 to the 20 power and 2 to the 40 power. By exploring these comparisons, we can gain a better understanding of the magnitude and significance of 2 to the 30 power. ### 2 to the 20 power vs. 2 to the 30 power Let’s start by comparing 2 to the 20 power with 2 to the 30 power. To put it into perspective, imagine you have a single penny and double it each day for 20 days. At the end of the 20th day, you would have 1,048,576 pennies. Now, let’s take it a step further. If you were to continue doubling that amount for another 10 days, you would reach the astonishing number of 1,073,741,824 pennies. This mind-boggling figure represents 2 to the 30 power. To provide further context, let’s consider a practical example. If we were to convert these pennies into seconds, with each penny representing a second, 2 to the 30 power would be equivalent to approximately 34 years. Just imagine, every second ticking away for over three decades. This comparison showcases the immense growth and scale of 2 to the 30 power. ### 2 to the 30 power vs. 2 to the 40 power Now, let’s explore the comparison between 2 to the 30 power and 2 to the 40 power. To grasp the magnitude of these numbers, let’s return to our penny analogy. If we were to double a single penny each day for 30 days, we would reach the impressive sum of 1,073,741,824 pennies. However, if we were to continue this doubling process for another 10 days, we would reach an astronomical figure of 1,099,511,627,776 pennies. This extraordinary number represents 2 to the 40 power. To provide a relatable analogy, let’s consider time again. If we were to convert these pennies into seconds, with each penny representing a second, 2 to the 40 power would be equivalent to approximately 34,955 years. This mind-blowing number highlights the exponential growth between 2 to the 30 power and 2 to the 40 power. In conclusion, the comparison of 2 to the 30 power with other powers of 2 showcases the exponential nature of these numbers. From the difference between 2 to the 20 power and 2 to the 30 power, to the staggering contrast between 2 to the 30 power and 2 to the 40 power, these comparisons help us grasp the immense scale and significance of 2 to the 30 power. Contact 3418 Emily Drive Charlotte, SC 28217 +1 803-820-9654
# Find the equation of the tangent and normal to the curves.$\;y=\large\frac{1+\sin x}{\cos x}$ at $\;x=\large\frac{\pi}{4}$ Note: This is part 4th of a 4 part question, split as 4 separate questions here. Toolbox: • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) y=\large\frac{1+\sin x}{\cos x} at x=\large\frac{\pi}{4} Step 1: When x=\large\frac{\pi}{4}$$ y=\large\frac{1+\Large\frac{1}{\sqrt 2}}{\Large\frac{1}{\sqrt 2}}$ $\qquad=\sqrt 2+1$ We are required to find the equations of the tangent and the normal at $\bigg(\large\frac{\pi}{4},$$\sqrt 2+1\bigg) Step 2: The slope of the tangent at \bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$ is $m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$ $\large\frac{dy}{dx}=\large\frac{\cos x(\cos x)-(1+\sin x)(-\sin x)}{\cos ^2 x}$ $\qquad=\large\frac{\cos^2 x+\sin x+\sin^2 x}{\cos ^2 x}$ $\qquad=\large\frac{1+\sin x}{\cos ^2 x}$ $m=\large\frac{dy}{dx_{(\Large\frac{\pi}{4},\sqrt 2+1)}}$ $\quad=\large\frac{1+\Large \frac{1}{\sqrt 2}}{\Large\frac{1}{2}}$ $\quad=\large\frac{\sqrt 2+1}{\sqrt 2}$$.2 \quad=2 +\sqrt 2 Equation of the tangent at \bigg(\large\frac{\pi}{4},$$ \sqrt 2+1\bigg)$ $y-(\sqrt 2+1)=(2 +\sqrt {2})(x-\large\frac{\pi}{4})$ Step 3: Normal is $\perp$ to the tangent. Its slope equation of the normal at $\bigg(\large\frac{\pi}{4},$$\sqrt 2+1\bigg) y-(\sqrt 2+1)=\large\frac{-1}{(2 +\sqrt 2)}($$x-\large\frac{\pi}{4})$
Decimals are taught in elementary school mathematics. After learning multiplication with decimals, you need to be able to do division. When calculating decimals, a division is a bit more complicated. This is because it is not always divisible. You have to calculate until you can divide or use the remainder to get the answer. It is also common to move the decimal point. When it comes to calculating decimals, addition, subtraction, and multiplication are easy. Division, on the other hand, requires a lot of understanding. In this section, we will explain how to do decimal division. ## How to Divide Using Decimals and Integers: The Case of Divisible Numbers Let’s start with the simple division of decimals. As long as the numbers are divisible, the calculations are not complicated. For example, how do you calculate the following? • $24.22÷7$ When doing this calculation, create the following long division. The way to do division is the same as a long division with whole numbers. Ignore the decimal point and perform the division calculation. Then, add the decimal point to the answer (quotient) in the same position. The result is as follows. When doing division using decimals and integers, it is not difficult to calculate as long as the numbers are divisible. After dividing, add the decimal point in the same place. In some calculation problems, the ones place of the answer may be zero. In this case, you should always write a zero in the ones place. The following is an example. • $3.22÷7$ In calculating decimals, the number can be smaller than one. Understand that in the division of decimals, the ones place can be zero. ### Division of Decimals Until Divisible On the other hand, there are some problems that need to be calculated until a number is divisible. In this case, the calculation is done by adding zeros to the right of the decimal. There are hidden zeros in decimals. For example, 1.4 can be written as 1.400. However, since there is an infinite number of zeros, we omit the zeros in decimals. In any case, it is important to understand that in decimals, there are many zeros to the right of the number. Therefore, when calculating divisible decimals, add zeros to get the answer. For example, how do we calculate the following? • $1.4÷25$ When doing this calculation, let’s change the value to 1.40 instead of 1.4. The result is as follows. However, the calculation should not end here. From this point, we can do more division. So, let’s change 1.40 to 1.400 and do the next calculation. By adding zeros to the right of the decimal, we were able to get the answer. When doing division, if it is divisible, try to get the answer by this method. ### When the Divisor Is a Decimal, Move the Decimal Point to Perform the Division So far, we have discussed division when the divisor is an integer. On the other hand, when a divisor is a decimal number, how do we calculate it? When the divisor is a decimal, make sure to move the decimal point. By moving the decimal point to the right, you make the divisor an integer. If you don’t do this, you will not be able to divide decimals. For example, how would you do the following calculation? • $3.22÷1.4$ The long division is as follows. The divisor is 1.4. If we don’t change it to an integer, we will not be able to do the division. So, instead of 1.4, let’s make it 14. Move the decimal point one place to the right. At the same time, move the decimal point of 3.22 one place to the right as well. Since we have moved the decimal point of the divisor, we need to do the same for the dividend. Therefore, the long division looks like this. Understand that when the divisor is a decimal, you cannot divide the number unless you change the decimal point. Therefore, we need to move the decimal point like this. We need to change the divisor to an integer, and how many times we need to move the decimal point to the right depends on the divisor. For example, in the following calculation, we need to shift the decimal point two places to the right. • $6.848÷2.14$ Let’s do the following long division calculation. Understand how to move the decimal point and change a divisor to an integer. ### Why Moving the Decimal Point Doesn’t Matter There is a question that is asked by many people. Why is it okay to move the decimal point? The reason for this is that even if you change the position of the decimal point, the answer will be the same. For example, the following will all give the same answer. • $100÷20=5$ • $10÷2=5$ • $1÷0.2=5$ • $0.1÷0.02=5$ The answer will be the same if you multiply the number by 10 or 100 for both the divisor and the dividend. For example, compare $10÷2=5$ and $100÷20=5$. In the same way, we can see that for $1÷0.2$ and $0.1÷0.02$, we can get the same answer by multiplying both the divisor and the dividend by 10 or 100. For $1÷0.2$, multiply both numbers by 10 to get $10÷2$. For $0.1÷0.02$, multiply both numbers by 100 to get $10÷2$. In this way, decimals can be converted to integers. This method allows us to divide decimals. ## Division of Decimals and Integers with Remainders So far, we have discussed division when the number is divisible. On the other hand, what should we do with numbers that are not divisible? When an integer is not divisible, we use the remainder. This is also true for decimal division. For example, how do we calculate the following? • $22.3÷4$ If you make a long division and calculate it, you will get the following. The method of division is the same as the one explained so far. The only difference is that the remainder is given. If you calculate $22.3÷4$, the answer will be 5.5 R 0.3. When calculating the remainder, make sure to put the decimal point directly below. When calculating decimals, it is easy to make mistakes with the position of the decimal point. If the position of the decimal point is different, the remainder will be different. If this happens, the answer will not be correct, so be sure to check the position of the decimal point. ### How to Calculate Division of Two Decimals with Remainder On the other hand, how do we calculate if the divisor is a decimal? When there is a remainder, it is easy to make a miscalculation if the divisor is a decimal. Therefore, we need to understand how to calculate them. For example, how do we calculate the following? • $10.86÷3.3$ In order to do the math, we need to make the divisor an integer. So instead of 3.3, let’s change it to 33. We can do the math as follows. Finally, we need to find the remainder. We must pay attention to the position of the decimal point in the remainder; it must be put directly below the previous decimal point. The result is as follows. The answer to $10.86÷3.3$ is 3.2 R 0.3. By multiplying the divisor and dividend by 10, the quotient is 3.2. On the other hand, for the remainder, instead of using 108.6 multiplied by 10 as the reference number, we use the original number, 10.86, as the reference number. We put the decimal point of 10.86 directly below to get the remainder. ### Why the Decimal Point Is Different in the Quotient and the Remainder The reason why it is easy to make miscalculations when dividing decimals is that the position of the decimal point is different for the quotient and remainder. Why does the decimal point change between the quotient and remainder? Let’s understand the reason for this. As explained before, the quotient is the same even if you multiply the divisor and the dividend by 10 (or 100). The answer is the same, as shown below. • $100÷20=5$ • $10÷2=5$ Next, let’s try multiplying by 10 for the non-divisible numbers. What will be the answers to the following questions? • $100÷30=3$ … $10$ • $10÷3=3$ … $1$ As you can see, the quotient is the same even if we multiply the divisor and the dividend by 10. On the other hand, we can see that the remainder is 10 times larger. Also, if we multiply the divisor and the dividend by 100, the remainder is 100 times greater. In division, if we multiply numbers by 10 or 100, the remainder also increases accordingly. If the remainder is 10 or 100 times larger than the original number, the answer will be different. For example, in the case of $10÷3$, if we multiply the divisor and the dividend by 10, as explained earlier, the remainder will be 10 times larger than the original number. Even though the divisor is 3, the remainder is 10. Therefore, it is obvious that the answer is wrong. So when dividing, use the number before multiplying by 10 (or 100) for the remainder. By using the number before the decimal point is moved as the reference, we can add the decimal point in the correct place for the remainder. ## Calculating Decimals to Find the Quotient and the Remainder When we do division, the expression may contain decimals. So, let’s understand how to divide them. When dividing decimals, it is easy if the number is divisible. All you have to do is to divide in the usual way, paying attention to the position of the decimal point. If the divisor is a decimal, you can change the divisor to an integer by moving the decimal point. On the other hand, if there is a remainder, it is easy to make a miscalculation. If the divisor is a decimal, the quotient is based on the number after the decimal point is moved. Also, the remainder is based on the number before the decimal point is moved, and then the decimal point needs to be added. Understand that there are rules for these calculations and perform division calculations involving decimals.
#### Need solution for RD Sharma maths Class 12 Chapter 5 Determinants Exercise VSQ Question 55 maths textbook solution. Answer: $x = -2$ Hint: Here we use basic concept of determinant of matrix Given: $\left\|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right\|=8$ Solution : \begin{aligned} &x\left\|\begin{array}{cc} -x & 1 \\ 1 & x \end{array}\right\|-\sin \theta\left\|\begin{array}{cc} -\sin \theta & 1 \\ \cos \theta & x \end{array}\right\|+\cos \theta\left\|\begin{array}{cc} -\sin \theta & -x \\ \cos \theta & 1 \end{array}\right\|=8 \\ &x\left(-x^{2}-1\right)-\sin \theta(-\sin \theta(x)-\cos \theta)+\cos \theta(-\sin \theta-(-x) \cos \theta)=8 \\ &-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta=8 \\ &-x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=8 \\ &-x^{3}-x+x=8 \ldots\left\{\sin ^{2} \theta+\cos ^{2} \theta=1\right\} \end{aligned} \begin{aligned} &x^{3}=-8 \\ &x=-2 \end{aligned}
The associative property states that, if a, b and c are any three integers, then, • (a + b) + c = a + (b + c) • (a x b) x c = a x (b x c) In other words, the addition of integers is associative, no matter how you group the numbers when you add them. That is, rearranging the parentheses for integers in such an expression will not change its final value. The multiplication of integers is associative, no matter how you group the numbers when you multiply them. That is, rearranging the parentheses for integers in such an expression will not change its final value. What is Associative Property? Associative property states that the sum or product of three numbers remains the same, even if we change the order of grouping the numbers. Associative property is true for both addition and multiplication. Subtraction and Division are not associative. Examples of Associative Property • (6 + 4) + 8 = 6 + (4 + 8) = 18 • (6 x 4) x 8 = 6 x (4 x 8) = 192 In the above examples, the sum and the product doesn't change even if we change the order of grouping the numbers. Associative Property of Addition While performing addition of three or more numbers, the sum remains the same even if we change the order of grouping the numbers. (a + b) + c = a + (b + c) In order to verify this property, let us consider three integers and apply the associative property of addition. Consider the following : a b c (a + b) + c a + (b + c) 2 5 7 (2 + 5) + 7 = 7 + 7 = 14 2 + (5 + 7) = 2 + 12 = 14 Here, 2, 5 and 7 are integers. In this case, (2 + 5) + 7 = 2 + (5 + 7) = 14. a b c (a + b) + c a + (b + c) 2 0 -7 (2 + 0) + (-7) = 2 - 7 = -5 2 + [0 + (-7)] = 2 - 7 = -5 Here, 2, 0 and -7 are integers. In this case, (2 + 0) + (-7) = 2 + [0 + (-7)] = -5. a b c (a + b) + c a + (b + c) -9 3 -4 [(-9) + 3] + (-4) = -6 - 4 = -10 -9 + [3 + (-4)] = -9 - 1 = -10 Here, -9, 3 and -4 are integers. In this case, [(-9) + 3] + (-4) = -9 + [3 + (-4)] = -10. Associative Property of Addition Example Given below are some examples based on the associative property of addition. Example 1: Solve (11 + 9) + 3 Solution: Step 1: Add the numerals within bracket (11 + 9) Step 2: Add the sum with the numeral outside the bracket (3) Step 3: Write the final answer (23) Adding (11 + 9) + 3 $\rightarrow$ 20 + 3 $\rightarrow$ 23 So, (11 + 9) + 3 = 23 Example 2: Solve 23 + (11 + 7) Solution: Let us add 23 + (11 + 7) $\rightarrow$ 23 + (18) $\rightarrow$ 41 So, 23 + (11 + 7) = 41 Example 3: Solve 12 + (30 + 14) Solution: Let us add 12 + (30 + 14) $\rightarrow$ 12 + (44) $\rightarrow$ 56 So, 12 + (30 + 14) = 56 Associative Property of Multiplication While performing multiplication of three or more numbers, the product remains the same even if we change the order of grouping the numbers. (a x b) x c = a x (b x c) In order to verify this property, let us consider three integers and apply the associative property of multiplication. Consider the following : a b c (a x b) x c a x (b x c) 2 5 7 (2 x 5) x 7 = 10 x 7 = 70 2 x (5 x 7) = 2 x 35 = 70 Here, 2, 5 and 7 are integers. In this case, (2 x 5) x 7 = 2 x (5 x 7) = 70. a b c (a x b) x c a x (b x c) 2 0 -7 (2 x 0) x (-7) = 0 x (-7) = 0 2 x [0 x (-7)] = 2 x 0 = 0 Here, 2, 0 and -7 are integers. In this case, (2 x 0) x (-7) = 2 x [0 x (-7)] = 0. a b c (a x b) x c a x (b x c) -9 3 -4 [(-9) x 3] x (-4) = (-27) x (- 4) = 108 -9 x [3 x (-4)] = (-9) x (-12) = 108 Here, -9, 3 and -4 are integers. In this case, [(-9) x 3] x (-4) = -9 x [3 x (-4)] = 108. Associative Property of Multiplication Examples Given below are examples based on the associative property of multiplication. Example 1: Solve (12 x 7) x 3 Solution: Step 1: Multiply the numerals inside the bracket (12 x 7) Step 2: Multiply the product of numerals inside the bracket with the numeral outside ( 84 x 3) Step 3: Write the final product (252) Multiplying (12 x 7) x 3 $\rightarrow$ (84) x 3 $\rightarrow$ 252 So, (12 x 7) x 3 = 252 Example 2: Solve 14 x (21 x 8) Solution: Let us find the product of 14 x (21 x 8) $\rightarrow$ 14 x (168) $\rightarrow$ 2352 So, the product of 14 x (21 x 8) = 2352 Example 3: Solve 11 x (8 x 7) Solution: Let us find the product of 11 x (8 x 7) $\rightarrow$ 11 x (56) $\rightarrow$ 616 So, the product of 11 x (8 x 7) = 616
<meta http-equiv="refresh" content="1; url=/nojavascript/"> # 2.2: The Derivative Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives A student will be able to: • Demonstrate an understanding of the derivative of a function as a slope of the tangent line. • Demonstrate an understanding of the derivative as an instantaneous rate of change. • Understand the relationship between continuity and differentiability. The function f(x)\begin{align}f'(x)\end{align} that we defined in the previous section is so important that it has its own name. The Derivative The function f\begin{align}f'\end{align} is defined by the new function f(x)=limh0f(x+h)f(x)h where f\begin{align}f\end{align} is called the derivative of f\begin{align}f\end{align} with respect to x\begin{align}x\end{align}. The domain of f\begin{align}f\end{align} consists of all the values of x\begin{align}x\end{align} for which the limit exists. Based on the discussion in previous section, the derivative f\begin{align}f'\end{align} represents the slope of the tangent line at point x\begin{align}x\end{align}. Another way of interpreting it is to say that the function y=f(x)\begin{align}y = f(x)\end{align} has a derivative f\begin{align}f'\end{align} whose value at x\begin{align}x\end{align} is the instantaneous rate of change of y\begin{align}y\end{align} with respect to point x\begin{align}x\end{align}. Example 1: Find the derivative of f(x)=xx+1.\begin{align} f(x) = \frac {x}{x+1}.\end{align} Solution: We begin with the definition of the derivative, f(x)=limh0f(x+h)f(x)h=limh01h[f(x+h)f(x)], where f(x)f(x+h)=xx+1=x+hx+h+1 Substituting into the derivative formula, f(x)=limh01h[x+hx+h+1xx+1]=limh01h[(x+h)(x+1)x(x+h+1)(x+h+1)(x+1)]=limh01h[x2+x+hx+hx2xhx(x+h+1)(x+1)]=limh01h[h(x+h+1)(x+1)]=limh01(x+h+1)(x+1)=1(x+1)2. Example 2: Find the derivative of f(x)=x\begin{align} f(x) = \sqrt{x}\end{align} and the equation of the tangent line at x0=1\begin{align}x_0 = 1\end{align}. Solution: Using the definition of the derivative, f(x)=limh0f(x+h)f(x)h=limh0x+hxh=limh0x+hxhx+h+xx+h+x=limh01hx+hxx+h+x=limh01x+h+x=12x. Thus the slope of the tangent line at x0=1\begin{align}x_0 = 1\end{align} is f(1)=121=12. For x0=1\begin{align}x_0 = 1\end{align}, we can find y0\begin{align}y_0\end{align} by simply substituting into f(x)\begin{align}f(x)\end{align}. f(x0)f(1)y0y0=1=1=1. Thus the equation of the tangent line is yy0y1y=m(xx0)=12(x1)=12x+12. ## Notation Calculus, just like all branches of mathematics, is rich with notation. There are many ways to denote the derivative of a function y=f(x)\begin{align}y = f(x)\end{align} in addition to the most popular one, f(x)\begin{align}f'(x)\end{align}. They are: f(x)dydxydfdxdf(x)dx In addition, when substituting the point x0\begin{align}x_0\end{align} into the derivative we denote the substitution by one of the following notations: f(x0)dydxxx0dfdxxx0df(x0)dx ## Existence and Differentiability of a Function If, at the point (x0,f(x0))\begin{align}(x_0, f(x_0))\end{align}, the limit of the slope of the secant line does not exist, then the derivative of the function f(x)\begin{align}f(x)\end{align} at this point does not exist either. That is, if msec=limxx0f(x)f(x0)xx0=Does not exist then the derivative f(x)\begin{align}f'(x)\end{align} also fails to exist as xx0\begin{align}x \rightarrow x_0\end{align}. The following examples show four cases where the derivative fails to exist. 1. At a corner. For example f(x)=\|x\|\begin{align}f(x) = \|x\|\end{align}, where the derivative on both sides of x=0\begin{align}x = 0\end{align} differ (Figure 4). 2. At a cusp. For example f(x)=x2/3\begin{align}f(x) = x^{2/3}\end{align}, where the slopes of the secant lines approach +\begin{align}+\infty\end{align} on the right and \begin{align}-\infty\end{align} on the left (Figure 5). 3. A vertical tangent. For example f(x)=x1/3\begin{align}f(x) = x^{1/3}\end{align}, where the slopes of the secant lines approach +\begin{align}+\infty\end{align} on the right and \begin{align}-\infty\end{align} on the left (Figure 6). 4. A jump discontinuity. For example, the step function (Figure 7) f(x)={2,2,x<0x0 where the limit from the left is 2\begin{align}-2\end{align} and the limit from the right is 2\begin{align}2\end{align}. Figure 4 Figure 5 Figure 6 Figure 7 Many functions in mathematics do not have corners, cusps, vertical tangents, or jump discontinuities. We call them differentiable functions. From what we have learned already about differentiability, it will not be difficult to show that continuity is an important condition for differentiability. The following theorem is one of the most important theorems in calculus: Differentiability and Continuity If f\begin{align}f\end{align} is differentiable at x0\begin{align}x_0\end{align}, then f\begin{align}f\end{align} is also continuous at x0\begin{align}x_0\end{align}. The logically equivalent statement is quite useful: If f\begin{align}f\end{align} is not continuous at x0\begin{align}x_0\end{align}, then f\begin{align}f\end{align} is not differentiable at x0\begin{align}x_0\end{align}. (The converse is not necessarily true.) We have already seen that the converse is not true in some cases. The function can have a cusp, a corner, or a vertical tangent and still be continuous, but it is not differentiable. ## Multimedia Links For an introduction to the derivative (4.0)(4.1), see Math Video Tutorials by James Sousa, Introduction to the Derivative (9:57). The following simulator traces the instantaneous slope of a curve and graphs a qualitative form of derivative function on an axis below the curve Surfing the Derivative. The following applet allows you to explore the relationship between a function and its derivative on a graph. Notice that as you move x along the curve, the slope of the tangent line to f(x)\begin{align}f(x)\end{align} is the height of the derivative function, f(x)\begin{align}f'(x)\end{align} Derivative Applet. This applet is customizable--after doing the steps outlined on the page, feel free to change the function definition and explore the derivative of many functions. For a video presentation of differentiability and continuity (4.3), see Differentiability and Continuity (6:34). ## Review Questions In problems 1 through 6, use the definition of the derivative to find f(x)\begin{align}f'(x)\end{align} and then find the equation of the tangent line at x=x0\begin{align}x = x_0\end{align}. 1. f(x)=6x2;x0=3\begin{align}f(x) = 6x^{2}; x_{0} = 3\end{align} 2. f(x)=x+2;x0=8\begin{align} f(x) = \sqrt{x + 2}; x_0 = 8\end{align} 3. f(x)=3x32;x0=1\begin{align}f(x) = 3x^3 - 2; x_0 = -1\end{align} 4. f(x)=1x+2;x0=1\begin{align} f(x) = \frac{1} {x+2}; x_0 = -1\end{align} 5. f(x)=ax2b,\begin{align}f(x) = ax^2 - b,\end{align} (where a\begin{align}a\end{align} and b\begin{align}b\end{align} are constants); x0=b\begin{align}x_0 = b\end{align} 6. f(x)=x1/3;x0=1\begin{align}f(x) = x^{1/3}; x_0 = 1\end{align}. 7. Find dy/dx\|x=1\begin{align}dy/dx \|_{x = 1}\end{align} given that y=5x22.\begin{align}y = 5x^2 - 2.\end{align} 8. Show that f(x)=x3\begin{align} f(x) = \sqrt[3]{x}\end{align} is defined at x=0\begin{align}x = 0\end{align} but it is not differentiable at x=0\begin{align}x = 0\end{align}. Sketch the graph. 9. Show that f(x)={x2+12xx1x>1 is continuous and differentiable at x=1\begin{align}x = 1\end{align}. Hint: Take the limit from both sides. Sketch the graph of f\begin{align}f\end{align}. Feb 23, 2012 ## Last Modified: Aug 19, 2015 You can only attach files to None which belong to you If you would like to associate files with this None, please make a copy first. # Reviews Please wait... Please wait... Image Detail Sizes: Medium \| Original CK.MAT.ENG.SE.1.Calculus.2.2
Lesson Video: Evaluating Limits Using Algebraic Techniques \| Nagwa Lesson Video: Evaluating Limits Using Algebraic Techniques \| Nagwa # Lesson Video: Evaluating Limits Using Algebraic Techniques Mathematics In this video, we will learn how to use algebraic techniques like factorization to evaluate limits. 17:47 ### Video Transcript Evaluating Limits Using Algebraic Techniques. In this video, weβll learn how to use techniques such as factorisation to take the limit of a function. First, let us recall the definition of a limit, which is, for some function π of π₯, which is defined near the value where π₯ is equal to π, as π₯ approaches π, π of π₯ approaches πΏ. We call this value πΏ the limit. And here, we have written the standard notation for the description of a limit. Now, if π of π₯ is a rational function with π in its domain, we can simply say that the limit, as π₯ approaches π of π of π₯, is equal to π of π. Since weβre inputting the value of π into our function, this method is often described as direct substitution. An important point to note here is that even if π is not in the domain of our function π, we can still sometimes find the limit as π₯ approaches π. This is because the limit concerns values as π₯ approaches π but is not equal to π. And weβll see how this comes into play in a moment. For this video, weβll be focusing on functions which take the form π of π₯ over π of π₯ where both π and π are polynomial functions. In the cases weβll look at, where π₯ is equal to π, π of π over π of π will evaluate to the indeterminate form of zero over zero. That is to say, if we try to take the limit as π₯ approaches π of our function π of π₯ by directly substituting π₯ equals π into the function, then our substitution fails, and weβll need to take another approach. Looking at our quotient, we can see that both π of π and π of π are equal to zero. By recalling the factor theorem, we can then infer that both π of π₯ and π of π₯ will have a common factor of π₯ minus π. Given this information, we can rewrite the function uppercase π of π₯ as a product of π₯ minus π times some other function, which weβll call lowercase π of π₯. And, of course, the same logic can be followed for uppercase π of π₯. This allows us to rewrite our original quotient as π₯ minus π times lowercase π of π₯ divided by π₯ minus π times lowercase π of π₯. In this form, we see that the common factor of π₯ minus π can be cancelled in the top and bottom halves of our quotient. And we are then left with lowercase π of π₯ over lowercase π of π₯. Letβs define this new quotient as π of π₯. Now, a reasonable question to ask is, why are we giving this new definition when π of π₯ is clearly equal to π of π₯. In fact, the answer is that in cancelling the common factor of π₯ minus π, we have changed the domain of our function π of π₯. We, therefore, have that π of π₯ equals π of π₯, but not at the point where π₯ is equal to π. The subtlety here is that π of π₯ is defined when π₯ is equal to π, whereas π of π₯ is not. This is great, as it allows us to proceed to the following general rule. The limit, as π₯ approaches π of π of π₯, is equal to the limit as π₯ approaches π of π of π₯ if π and π are equal at all points over an interval except the point at which π₯ is equal to π. Again, itβs good to understand that this works because the limit concerns values of π₯ which are close to π but not equal to π. Finally, since π is a rational function with π in its domain, we can simply find the limit by direct substitution, which is π evaluated where π₯ is equal to π. Okay, thatβs a lot of information, so letβs look at an example to illustrate this concept. Find the limit as π₯ approaches negative eight of π₯ squared plus 13π₯ plus 40 divided by π₯ cubed plus nine π₯ squared minus 12π₯ minus 160. Here, we see that we are trying to find the limit of a rational function, which weβll call π of π₯. For this type of function, if negative eight is in the domain of π, then the limit, as π₯ approaches negative eight, is simply π evaluated at negative eight. The first thing we can do then is to try direct substitution of negative eight into our function. Here, we have performed the substitution. And following through with our working, we find that our answer evaluates to zero over zero. And this is an indeterminate form, which means we cannot evaluate the limit using direct substitution. We can also conclude that negative eight is not in the domain of our function π. Here, weβre gonna need to use a different approach. And we can do so first by recognising that our function π of π₯ is in the form π of π₯ over π of π₯, where both π and π are polynomial functions. Looking at the failed direct substitution that we just tried, we can see the both π of negative eight and π of negative eight are equal to zero. From this information, we can use the factor theorem to conclude that π₯ plus eight is a factor of both π of π₯ and π of π₯. Since direct substitution failed, letβs instead try to factorise both the top and bottom halves of our quotient, given the fact that both of them have a common factor of π₯ plus eight. For the numerator, factoring a quadratic equation should be a familiar skill to us. And with some inspection, we see that π₯ squared plus 13π₯ plus 40 factorises to π₯ plus eight and π₯ plus five. Now, for the denominator, ordinarily, factoring a cubic is a much more difficult task. However, given the fact that we already know one of the factors is π₯ plus eight, we can use techniques such as polynomial division or comparing coefficients to find the other factor. For this video, weβre going to compare coefficients. And we begin by recognising that our other factor will take the form ππ₯ squared plus ππ₯ plus π. To find π, we note that we have an π₯ multiplied by an ππ₯ squared. And this is equal to π₯ cubed. It, therefore, follows that π is equal to one. And we then see that our second factor begins with the term π₯ squared. Next, to find π, we note that we have eight multiplied by π is equal to negative 160. And we can calculate that π is equal to negative 20. Finally, to find π, weβll choose to look at the nine π₯ squared term. Going by the previous coefficients we have, we note that we first have an eight multiplied by an π₯ squared. This gives us eight π₯ squared. We also have an π₯ multiplied by ππ₯ and this gives us ππ₯ squared. The sum of these two terms is nine π₯ squared. And it then follows that π is equal to one. The denominator of our quotient is now fully factorised. And our missing factor was π₯ squared plus π₯ minus 20. Following this factorisation, weβre able to cancel the common factor of π₯ plus eight from the top and bottom half of our quotient. And we are then left with π₯ plus five over π₯ squared plus π₯ minus 20. Now, here, we should remember that in cancelling the common factor of π₯ plus eight, we have changed the domain of our original function π of π₯. Weβre able to say that π of π₯ is equal to the right-hand side of our equation, which weβll call π of π₯, at all values where π₯ is not equal to negative eight. It now follows that the limit, as π₯ approaches negative eight of π of π₯, is equal to the limit as π₯ approaches negative π₯ of π of π₯. Crucially, this is because the limit concerns values where π₯ is close to negative eight but not equal to negative eight. An important point is that π of π₯ is defined when π₯ is equal to negative eight. And so, we can, therefore, find the limit by direct substitution. Here, we have performed the substitution of π of negative eight. And if we follow through with our calculations, we see that this evaluates to negative three over 36. This fraction can be simplified to negative one over 12. We have now answered the question. And we have found our limit. This example illustrates that when our functions π of π₯ and π of π₯ evaluate to zero, we can use the factor theorem to help us with potentially tricky factorisations such as cubics. Letβs now look at another technique which can help us avoid potentially tricky factorisations first by focusing on expressions of the form π₯ to the π minus π to the π, or the difference of two πth powers. Since inputting π₯ equals π into this equation will always give zero, our friend the factor theorem, again, tells us that all expressions of this form will evaluate to zero. And therefore, all of them will have a factor of π₯ minus π. Using this general rule, we see that our second factor will take the form of a polynomial with π terms having decreasing powers of π₯ and increasing powers of π up to the powers of π minus one. Some examples of this have been shown below. This general rule can be used to derive a very useful formula in the following way. Consider the case of a difference of two πth powers, π₯ to the π minus π to the π, divided by the difference of two πth powers, π₯ to the π minus π to the π. We can use our general rule to express the top and bottom halves of our quotient as a product of two factors. In both cases, one of these factors is π₯ minus π. We can cancel this common factor of π₯ minus π in the top and bottom halves of the quotient. In cancelling our common factor, we must remember that the left- and right-hand sides of our equations are equal as long as the value of π₯ is not equal to π. Since the limit, as π₯ approaches π, concerns values of π₯ close to π but not equal to π, weβre able to say that the limit of the left-hand side is equal to the limit of the right-hand side. Considering the numerator of our quotient for a moment, we now use the following trick of direct substitution, where π₯ takes the value of π. With this substitution, weβre left with the sequence π terms long of which all of the terms are equal to π to the power of π minus one. And this is simply equal to π times π to the power of π minus one. Although, we were using the numerator for an example, the same logic follows for the denominator of our quotient and gives us π times π to the power of π minus one. We then proceed with a few simplifications of the powers of π. We then find that our limit, as π₯ approaches π, is equal to π over π multiplied by π to the power of π minus π. This is a really useful result for equations of this form since even when the powers π or π are large, we can avoid lengthy factorisations when taking the limit. Letβs now see how this technique can be used in an example. Find the limit as π₯ approaches two of eight π₯ cubed minus 64 divided by π₯ squared minus four. Here, we have a function, which weβll call π of π₯. Given that this is a rational function, the first thing we may try is a direct substitution of π₯ equals two into our function. Here, we have performed the substitution. And when we evaluate our answer, we find that weβre left with the indeterminate form of zero over zero. Instead, weβre gonna need to move on to a different method based on factorisation. For this method, we first note that our function π of π₯ is in the form π of π₯ over π of π₯, where both π and π are polynomial functions. By inspecting the numerator of our quotient, we noticed that both the terms have a factor of eight. We can, therefore, factorise our numerator as eight times π₯ cubed minus eight. And given that this eight is a constant, we can take it outside of our limit as follows. If instead we find the limit as π₯ approaches two of π₯ cubed minus eight divided by π₯ squared minus four and multiply this entire thing by our constant eight, this will give us the same answer. To proceed, we can then notice that the eight in our numerator and the four in our denominator can both be expressed as powers of two, which are two cubed and two squared, respectively. After doing this, we see that our limit now takes the following form. Here, weβll say that the top half of our quotient is equal to the difference of two πth powers, with π being three, and the bottom half of our quotient is equal to the difference of two πth powers, with π being two. Given this form, weβre able to use the following general rule, which tells us that the limit will be equal to π over π times π to the power of π minus π. At this point, you may notice that both the top and bottom half of our quotient would have a common factor of π₯ minus two. We could, instead, cancel this common factor and proceeded by refactorising. However, this general rule allows us to move directly to our limit. In cases, where π or π are large, this helps us potentially avoid a lengthy or time-consuming refactorisation. Inputting our values into our general rule, where π is two, π is three, and π is also two, we find that our limit is three over two multiplied by two to the power of three minus two. We also mustnβt forget to multiply this entire thing by the eight which we took out of our limit. Three minus two is, of course, just one. And so, we can simplify this by cancelling the two and the one over two. We then find that our answer is equal to eight times three, which 24. We have now found that the limit, as π₯ approaches two of our function π of π₯, is equal to 24. And we have answered our question. The general rule that we used in this video provides us with a useful shortcut that can be used when our function π of π₯ can be expressed in this form. Again, although for our question the powers involved are relatively low, when π and π are sufficiently large, you may be able to save yourself much more time. An interesting point worth noting is that some of the techniques that we have shown in this video can also be used to take the limits of functions which include radical. Specifically, where π of π₯ is equal to π of π₯ over π of π₯, where either π or π include a radical term, such as the square root of π₯. Although these would not be classed as rational functions, some of our tools will still work. An example would be if we were to take the limit as π₯ approaches eight of the cube root of π₯ minus two divided by π₯ minus eight. If we were to try a direct substitution approach, we would again be left with the indeterminate form of zero over zero. However, if instead we were to factorise, specifically taking out a factor of the cube root of π₯ minus two from our denominator, we would then be able to proceed in a familiar fashion. By cancelling out the common factor in the top and bottom halves of the quotient and then directly substituting π₯ equals eight into the remaining expression to finally find an answer of one over 12. Although, here, weβve shown an example where our current techniques work, in some cases, weβll need a different method based on multiplication by a conjugate. This technique can be illustrated using an example. Determine the limit as π₯ approaches eight of π₯ squared minus eight π₯ divided by the square root of π₯ plus one minus three. Here, we see that weβre taking the limit of a function π of π₯, which takes the form π of π₯ over π of π₯. When we try a direct substitution approach to solve this, our expression evaluates to the indeterminate form of zero over zero. And instead, weβre gonna need to use a different approach. The first thing we can notice is that our numerator can be factorised as π₯ times π₯ minus eight. Given our knowledge of the factor theorem, we may have even expected this factor of π₯ minus eight given that the function π evaluated where π₯ equals eight gave us a zero. Unfortunately, the denominator of our quotient is less straightforward to evaluate. But we can understand how to move forward by viewing it as a binomial in the form π minus π, where π is equal to the square root of π₯ minus one and π is equal to three. The conjugate of any binomial is found by flipping the symbol in between the two terms. The conjugate of our binomial would then be π plus π. And in this case, that is the square root of π₯ plus one plus three. Now, multiplying a binomial by its conjugate is a useful tool because it leaves us with the difference of two squares. Given that our denominator contains a square root, we can utilise this relationship to evaluate our function. Now, multiplying by the conjugate over itself is the same as multiplying by one. Nonetheless, the bottom half of our quotient then simplifies to a difference of two squares. And with some working, we see that our denominator then becomes π₯ minus eight. Following this, weβre able to cancel the common factor of π₯ minus eight in the top and bottom half of the quotient. And on the right-hand side of our equation, we are left with π₯ multiplied by the square root of π₯ plus one plus three. Now, since our original function π of π₯ is equal to the right-hand side of our equation at all points where π₯ is not equal to eight, weβre able to say the limit, as π₯ approaches eight of our original function π of π₯, is equal to the limit as π₯ approaches eight of our new function, which weβll call π of π₯. This is because the limit concerns values of π₯ which are arbitrarily close to eight but not equal to eight. In this form, we can perform a direct substitution of π₯ equals eight into our function π to find the limit. Performing the substitution and then evaluating our answer leaves us with a value of 48. And this is, in fact, the limit that we were looking to find. Now, itβs worth recapping here that, originally, we were not able to find our limit using direct substitution because we were left with the indeterminate form of zero over zero. Instead, after factorising and using our conjugate method to cancel a common factor, direct substitution was possible. We should, therefore, be on the lookout to use this method for questions of this form when we observe a radical in our quotient. Letβs now recap a few key points to summarise. When taking the limit of a rational function π of π₯, which is expressed as π of π₯ over π of π₯, and evaluating by direct substitution, we may be left with the indeterminate form of zero over zero. In these cases, if the function π of π₯ is almost equal to π of π₯. That is to say π of π₯ is equal to π of π₯ at all values of π₯ aside from where π₯ is equal to π. Then it follows that the limit, as π₯ approaches π of π of π₯, is equal to the limit as π₯ approaches π of π of π₯. Assuming that π is in the domain of our new function π, that is to say π of π₯ is defined where π₯ is equal to π, we can then find our limit by direct substitution, which is to say taking the value of π of π. We can find such a function π of π₯ via a number of methods such as factorisation or multiplication by a conjugate of the numerator or denominator. The methods that we have shown in this video involve cancelling a common factor of π₯ minus π from the top and bottom half of our quotient. Since taking the limit involves values of π which are close to but not equal to π, it does not matter that in cancelling the common factor we have changed the domain of our original function π of π₯. Finally, when our function takes certain forms, there are some useful shortcuts that we can take to find the limit directly, potentially avoiding some lengthy factorisations.
1 Determinants Eng # 1 Determinants Eng - 1 APPLICATIONS OF MATRICES AND... This preview shows pages 1–4. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2 A (adj A ) = \| A \| I 3 = (adj A ) A In general we can prove that A (adj A ) = \| A \| I n = (adj A ) A . Example 1.1 : Find the adjoint of the matrix A = a b c d Solution: The cofactor of a is d , the cofactor of b is c , the cofactor of c is b and the cofactor of d is a . The matrix formed by the cofactors taken in order is the cofactor matrix of A . The cofactor matrix of A is = d c b a . Taking transpose of the cofactor matrix, we get the adjoint of A . The adjoint of A = d b c a Example 1.2 : Find the adjoint of the matrix A = 1 1 2 3 2 3 Solution: The cofactors are given by Cofactor of 1 = A 11 = 3 3 = 3 Cofactor of 1 = A 12 = 1 3 3 = 9 Cofactor of 1 = A 13 = 2 2 1 = 5 Cofactor of 1 = A 21 = 1 1 3 = 4 Cofactor of 2 = A 22 = 1 1 2 3 = 1 Cofactor of 3 = A 23 = 1 2 1 = 3 Cofactor of 2 = A 31 = 1 2 3 = 5 3 Cofactor of 1 = A 32 = 1 1 1 3 = 4 Cofactor of 3 = A 33 = 1 1 1 2 = 1 The Cofactor matrix of A is [ A ij ] = 3 9 5 4 3 5 1 adj A = ( A ij ) T = 4 5 9 4 1 Example 1.3 : If A = 2 4 , verify the result A (adj A ) = (adj A ) A = \| A \| I 2 Solution: A = 2 4 , \| A \| = 2 4 = 2 adj A = 4 2 1 1 A (adj A ) = 2 4 4 2 1 1 = 2 0 0 2 = 2 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 09/21/2009 for the course CS 580 taught by Professor Fdfdf during the Spring '09 term at University of Toronto. ### Page1 / 45 1 Determinants Eng - 1 APPLICATIONS OF MATRICES AND... This preview shows document pages 1 - 4. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
Let us recall the definition of ‘derivative’. A derivative, particularly in mathematics is the rate of change of function with respect to a variable. For example, in $Figure \space 1$, you will notice the rate of change of $y\left(\triangle y\right)$ w.r.t variable $x$ which shows the slope of curve or we can call it as $f'(x)$ We denote derivative by $\frac{dy}{dx}$ i.e., the change in $y$ with respect to $x$. If $y(x)$is a function, the derivative is represented as $y'(x)$. Therefore, we can say that $f'(x)=y'(x)= \frac{dy}{dx}$. $Figure \space 1$ ## Differential Equation “In mathematics, a differential equation(DE) is an equation that relates one or more functions and their derivatives.In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two.” Let’s split the terms of Differential Equation. By now, we are aware that ‘Differentiate‘ means “Computing a derivative“, similarly ‘Differential Equation’ means “Equation containing a derivative“. Few examples of DE: 1. $y' = 2x + 3$ 2. $y'' + y' + 10 = 3x$ ## Order and Degree of Differential Equation The ORDER of a differential equation is the order of the highest derivative or differential coefficient present in the equation. Let us understand this by few examples: 1. $y' + 5x + 6 = 0$ 2. $y'' + 5y' + 4x = 2$ In the first example, the order of DE is 1 because of single derivative (y’) term. Similarly, for second DE the order is 2 because of the presence of double derivative (y”) term. Hence, we can classify DE on the basis of its Order as FIRST ORDER DIFFERENTIAL EQUATION and SECOND ORDER DIFFERENTIAL EQUATION. The DEGREE of the differential equation is the power of highest order derivative in the equation. ## Types of Differential Equation There are three basic types of differential equations. 1. Ordinary (ODEs) 2. Partial (PDEs) 3. Differential-algebraic (DAEs)
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### (31/3) : 4 = 5/6 ≅ 0.8333333 Spelled result in words is five sixths. ### How do you solve fractions step by step? 1. Conversion a mixed number 3 1/3 to a improper fraction: 3 1/3 = 3 1/3 = 3 · 3 + 1/3 = 9 + 1/3 = 10/3 To find new numerator: a) Multiply the whole number 3 by the denominator 3. Whole number 3 equally 3 * 3/3 = 9/3 b) Add the answer from previous step 9 to the numerator 1. New numerator is 9 + 1 = 10 c) Write a previous answer (new numerator 10) over the denominator 3. Three and one third is ten thirds 2. Divide: 10/3 : 4 = 10/3 · 1/4 = 10 · 1/3 · 4 = 10/12 = 2 · 5 /2 · 6 = 5/6 Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 4/1 is 1/4) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, , cancel by a common factor of 2 gives 5/6. In words - ten thirds divided by four = five sixths. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Unit rate Find unit rate: 6,840 customers in 45 days • A shopkeeper A shopkeeper cuts a wheel of cheese into 10 equal wedges. A customer buys one-fifth of the wheel. How many wedges does the customer buy? Use the number line to help find the solution. • Expressions Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n • Father and daughter Father is 36 years old, daughter is 20 years less. What will be the ratio between them when they are 10 years more? • Mineral water The bottle contains 1.5 liters of mineral water. Pour all the water from the bottle into empty cups with a volume of 1/3 l. All but one will be filled to the brim. What part of the volume of the last cup is filled with water? • Chestnuts Three divisions of nature protectors participated in the collection of chestnut trees.1. the division harvested 1250 kg, the 2nd division by a fifth more than the 1st division and the 3rd division by a sixth more than the second division. How many tons of • University bubble You'll notice that the college up slowly every other high school. In Slovakia/Czech republic a lot of people studying political science, mass media communication, social work, many sorts of management MBA. Calculate how many times more earns clever 25-yea • The third The one-third rod is blue, one-half of the rod is red, the rest of the rod is white and measures 8 cm. How long is the whole rod? • Pumps 6 pump fills the tank for 3 and a half days. How long will fill the tank 7 equally powerful pumps? • Cooks Four cooks cleaned 5 kg of potatoes for 10 minutes. How many cook would have to work clean 9 kg of potatoes for 12 minutes? • Almonds Rudi has 4 cups of almonds. His trail mix recipe calls for 2/3 cup of almonds. How many batches of trail mix can he make? • Chocolate Children break chocolate first to third and then every part of another half. What kind got each child? Draw a picture. What part would have received if each piece have halved? • The balls You have 108 red and 180 green balls. You have to be grouped into the bags so that the ratio of red to green in each bag was the same. What smallest number of balls may be in one bag?
# Law of tangent formula Trigonometry includes angles and triangles. Trigonometric functions are: • Sine, • Cosine, • Tangent, • Cosecant, • Secant, • Cotangent. The tangent( in trigonometry) is defined as an angle in a right-angled triangle which has a ratio of perpendicular and base. The tangent of an angle x is written as tan x. ## Formula of Law of Tangent The formula of a tangent in a right triangle PQR, where side opposite angle P, Q , R are p, q , r respectively. p-q/p+q ={ tan (P-Q)/2 }/{ tan (P+Q)/2} q-r/q+r ={ tan (Q-R)/2 }/{ tan (Q+R)/2} r-p/r+p ={ tan (R-P)/2 }/{ tan (R+P)/2} ## Example of Tangent Formula Problem :Â If in a triangle ABC, âˆ B = 90âˆ˜, âˆ C = 30âˆ˜. If the side opposite to âˆ B is 4 cm. Find the value of the side opposite to âˆ C. Solution: b-c/b+c ={ tan (B-C)/2 }/{ tan (B+C)/2} 4-c/4+c ={ tan (90 â€“ 30)/2 }/{ tan (90 +30)/2} 4-c/4+c ={ tan 30}/{ tan (60)} 4-c/4+c ={ 1/âˆš3}/{ âˆš3} 4-c/4+c =â…“ 3(4-c) = 4 + c 12 -3c = 4 + c 12 â€“ 4 = c +3c 8 = 4c 8/4 = c c =2 Example 2: If b = 3 cm, B = 30Â°, A = 60Â° in a right triangle ABC, right angled at C, then find the sides opposite to angle A, i.e. a. Solution: Given, b = 3 cm, B = 30Â°, A = 60Â° Using the law of tangent, a – b/a + b = [tan (A-B)/2]/tan (A + B)/2] a – 3/a + 3 = [tan (60Â°- 30Â°)/2]/tan (60Â° + 30Â°)/2] a – 3/a + 3 = tan 15/tan 45 a – 3/a + 3 = (2 – âˆš3)/1 a – 3 = (2 – âˆš3)(a + 3) 2a + 6 – âˆš3a -3âˆš3 -a + 3 = 0 a(1-âˆš3) + 9 -3âˆš3 = 0 a = (3âˆš3 – 9)/(1 – âˆš3) Register at BYJUâ€™S to read more articles on Mathematical formulas.
# Direct Variation Each day we come across many situations in science, engineering, industry, and daily life where the value of one quantity depends upon the quantity of another. The individuals who find these situations are in search of the relationship between the quantities. This relationship can be expressed as formulas or equations. When these two quantities have a constant (unchanged) ratio, their relationship is known as direct variation. Many individuals say that one variable “varies directly” as the other. The constant ratio is called the constant of variation. Let’s look at it this way. You know that there are 3 feet in 1 yard, 6 feet in 2 yards, 9 feet in 3 yards. As the number of feet increases, the number of yards increases. Feet 3 6 9 Yards 1 2 3 Each ratio of feet to yards is equivalent to 3:1. The relationship between feet and yards varies directly. The relationships between the changing quantities can be examined by using a model for data that illustrates how the change in one variable affects a second variable. A direct variation relationship can be represented by a linear equation in the form y = kx, where k is called the constant of proportionality (or constant of variation). ## The Formula y = kx , where k is the constant of variation (constant ratio). “y varies directly as x” ## In a Table X Y 1 4 2 8 3 12 4 16 From the table we can gather the information necessary to create the formula for direct variation which indicates actual variation. We notice that the variable y is always 4 times the variable x. This can be seen through the table on each row. Thus, our constant is k or 4. Our formula or equation would be y = 4x. ## Real Life Real life Instance of quantities that rely on one another : A grocery store cashier who is paid hourly knows that working longer means making more money. That’s because his pay varies directly with the number of hours worked. As his hours increase, so does the amount of his paycheck. A race car driver knows that completing 100 laps before making a pit stop is better than completing only 80, because distance is directly proportional to time when driving at a constant speed. The longer she drives, the more distance she’ll cover. ### Example 1 What is the relationship between x and y in the table? As the x value increases, the y values increase by 5. This is representative of direct variation. ### Example 2 What would our direct variation equation be if the constant of variation is 10? Since the direct variation formula is y = kx and the k represents the constant of variation, the equation would be y = 10x. (source)
Become a math whiz with AI Tutoring, Practice Questions & more. HotmathMath Homework. Do It Faster, Learn It Better. For the purpose of this article, polynomials are defined as an expression that is made up of variables, constants, coefficients, and exponents that are combined using the mathematical operations of addition and subtraction. Depending on how many terms an expression has, it is called a monomial, binomial, trinomial, or polynomial. Examples of constants, variables, and exponents are as follows: 1. Constants: 1, 2, 5, 38, 67, 104, etc. 2. Variables: a, b, c, m, n, x, y, z, etc. 3. Coefficients: the 2 in $2x$ , the 5 in $5y$ , etc. 4. Exponents: 3 in ${x}^{3}$ , $\left(n-1\right)$ in ${13}^{n-1}$ , 2 in ${44}^{2}$ , etc. ## Terms of a polynomial The terms of the polynomials are the parts of the expression that are generally separated by "+" or "-". So each such part of a polynomial expression is a term. Example 1 What is the degree of the polynomial $3{x}^{2}+5-4y$ ? There are three terms: $3{x}^{2}$ , $5$ , and $4y$ The highest exponent on a variable in a polynomial tells us its degree. For example, the polynomial in this example is of degree 2 because the highest power on a variable is 2. If you want to add or subtract polynomials, you have to use the distributive property to add or subtract the coefficients of like terms by factoring. The distributive property states that $x\left(y\right)+x\left(z\right)=x\left(y+z\right)$ . Like terms are the monomials within a polynomial that have the same variables raised to the same powers, such as $3{x}^{2}y$ and $84{x}^{2}y$ . Example 2 $\left(2{x}^{2}+5x+7\right)+\left(3{x}^{2}+2x+5\right)$ First, use the commutative property to group like terms. $=\left(2{x}^{2}+3{x}^{2}\right)+\left(5x+2x\right)+\left(7+5\right)$ Then use the distributive property and simplify. $=\left(2+3\right){x}^{2}+\left(5+2\right)x+\left(7+5\right)$ $=5{x}^{2}+7x+12$ Example 3 Subtract the following polynomials. $\left({x}^{2}+3xy-9\right)-\left(-2{y}^{2}+5xy+6\right)$ You can use the commutative property, but because this is a subtraction problem, you must flip the sign of the terms that are to be subtracted. $=\left({x}^{2}\right)+\left(3xy-5xy\right)+\left({y}^{2}\right)+\left(-9-6\right)$ Use the distributive property where applicable. $=\left({x}^{2}\right)+\left(3-5\right)xy+\left(2{y}^{2}\right)+\left(-9-6\right)$ $={x}^{2}-2xy+2{y}^{2}-15$ ## Flashcards covering the Adding and Subtracting Polynomials Algebra 1 Flashcards
# Factor Tree of 70 Created By : Bhagya Reviewed By : Rajashekhar Valipishetty Last Updated at : Mar 29,2023 Factor Tree of: ### How to find the Factor Tree of 70 Given number is 70 The factor tree of 70 is as follows: 70 2 35 5 7 70 = 2 x 35 35 = 5 x 7 If we write into multiples it would be 70 x 2 On splitting 35 further and writing it as multiples of numbers it would be 7 x 5. ### Factor Tree Calculations Here are examples of Factor tree calculations. ### Ways to Find Factors of 70 Here we are providing different methods to find the factors of the number 70. They are splitting numbers and prime factors. Splitting Numbers We can divide the number into either of its two factors. In other words, we are looking for the numbers that when multiplied together eqaul 70. Let’s start with 2 x 5 x 7 as it results in 70 on multiplying. We will now find the factors or ancestors of split factors, just like in any family tree. Prime Factors Let’s take a look at 70 now and we can write as 2 x 5 x 7 and place those factors on the tree. similarly to prime numbers obtained in the first step 2, 5, 7 obtained here are also prime numbers and we will end up these branches. ### FAQ on 70 Factor Tree 1. What is factor tree for 70 ? Answer: Prime Factors of 70 are 2 x 5 x 7. 2. What are the Prime Factors of 70 ? Answer: Prime Factors of 70 are 2 x 5 x 7.
# How do you find the partial sum of Sigma (1000-5n) from n=0 to 50? ##### 1 Answer Feb 18, 2017 ${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = 44625$ #### Explanation: ${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = {\sum}_{n = 0}^{50} \left(1000\right) - 5 {\sum}_{n = 0}^{50} \left(n\right) =$ We can use the standard result ${\sum}_{i = 1}^{n} i = \frac{1}{2} n \left(n + 1\right)$, and note that: ${\sum}_{n = 0}^{50} \left(1000\right) = 1000 + 1000 + \ldots + 1000 \setminus \setminus$ (51 terms) ${\sum}_{i = 0}^{n} i = 0 + {\sum}_{i = 1}^{n}$ So we get: ${\sum}_{n = 0}^{50} \left(1000 - 5 n\right) = \left(51\right) \left(1000\right) - 5 \cdot \frac{1}{2} \left(50\right) \left(50 + 1\right)$ $\text{ } = 51000 - \frac{5}{2} \left(50\right) \left(51\right)$ $\text{ } = 51000 - 6375$ $\text{ } = 44625$
# Solve the problem algebra And you just have an n. This will allow us to use the method of Gauss-Jordan elimination to solve systems of equations. Here are some recommended steps: If 6 is added to that, we get. Solving nonlinear systems is often a much more involved process than solving linear systems. If you said consistent, you are right. You could say that 8n is equal to Applications — In this section we will look at a couple of applications of exponential functions and an application of logarithms. Ellipses — In this section we will graph ellipses. And with cross-multiplying, you're actually doing two steps. Combine like terms in each member of an equation. We give the basic properties and graphs solve the problem algebra logarithm functions. A nonlinear system of equations is a system in which at least one of the equations is not linear, i. This is a process that has a lot of uses in some later math classes. So we could divide, and this is a little bit of algebra here, we're dividing both sides of the solve the problem algebra by 8. We are told that the circumference is It is very easy to make an error with the signs either when working with the -3z terms or when dividing by The radius of the face of a circular clock. Equations with Radicals — In this section we will discuss how to solve equations with square roots in them. Hyperbolas — In this section we will graph hyperbolas. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction. Use the division property to obtain a coefficient of 1 for the variable. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result. We will discuss dividing polynomials, finding zeroes of polynomials and sketching the graph of polynomials. As we will see we will need to be very careful with the potential solutions we get as the process used in solving these equations can lead to values that are not, in fact, solutions to the equation. But I want to show you the algebra just because I wanted to show you that this cross-multiplication isn't some magic, that using algebra, we will get this exact same thing. Be very careful with your parentheses here. In addition, we will discuss solving polynomial and rational inequalities as well as absolute value equations and inequalities. Using the addition or subtraction property, write the equation with all terms containing the unknown in one member and all terms not containing the unknown in the other. There is no specific order in which the properties should be applied. Since it was a solution to BOTH equations in the system, then it is a solution to the overall system. Logarithm Functions — In this section we will introduce logarithm functions. The calculus section will carry out differentiation as well as definite and indefinite integration. The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation. You have to multiply it. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require. This equation has a variable on both sides of the equal sign. Assign a variable for the lowest test grade. We review exponents integer and rationalradicals, polynomials, factoring polynomials, rational expressions and complex numbers. Solution Multiplying each member by 6 yields In solving equations, we use the above property to produce equivalent equations that are free of fractions. This equation has a variable on both sides of the equal sign. In addition, we discuss how to evaluate some basic logarithms including the use of the change of base formula. These techniques involve rewriting problems in the form of symbols. Algebra - powered by WebMath. Visit Cosmeo for explanations and help with your homework problems! Algebra 1 Here is a list of all of the skills students learn in Algebra 1! These skills are organized into categories, and you can move your mouse over any skill name to preview the skill. Learn the reasoning behind solving proportions. We'll put some algebra to work to get our answers, too. Algebra Practice: Free! Algebra Worksheet Generator - Generate your own algebra worksheets to print and use. Includes many options and types of equations, systems, and quadratics. It does not matter which equation or which variable you choose to solve for. Just keep it simple. Since the x in the first equation has a coefficient of 1, that would mean we would not have to divide by a number to solve for it and run the risk of having to work with fractions (YUCK!!). Solve the equation. The examples done in this lesson will be linear equations. Solutions will be shown, but may not be as detailed as you would like. Solve the problem algebra Rated 3/5 based on 23 review Interactivate: Algebra Four
# DAV Class 6 Maths Chapter 2 Worksheet 4 Solutions The DAV Class 6 Maths Book Solutions Pdf and DAV Class 6 Maths Chapter 2 Worksheet 4 Solutions of Factors and Multiples offer comprehensive answers to textbook questions. ## DAV Class 6 Maths Ch 2 WS 4 Solutions Question 1. Find the prime factorisation of the following: (a) 78 Solution: Prime factors of 78 = 2 × 3 × 13 (b) 120 Solution: Prime factors of 120 = 2 × 2 × 2 × 3 × 5 (c) 256 Solution: Prime factors of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (d) 84 Solution: Prime factors of 84 = 2 × 2 × 3 × 7 (e) 441 Solution: Prime factors of 441 = 3 × 3 × 7 × 7 (f) 240 Solution: Prime factors of 240 = 2 × 2 × 2 × 2 × 3 × 5 (g) 2304 Solution: Prime factors of 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 (h) 3125 Solution: Prime factors of 3125 = 5 × 5 × 5 × 5 × 5 (i) 1260 Solution: Prime factors of 1260 = 2 × 2 × 3 × 3 × 5 × 7 Question 2. Write the greatest 4 digit number. Express it as a product of primes. Solution: The greatest 4-digit number is 9999 Product of prime factors = 3 × 3 × 11 × 101. Question 3. Write the smallest 4 digit number and show its prime factorisation. Solution: 4-digit smallest number is 1000 Prime factors of 1000 = 2 × 2 × 2 × 5 × 5 × 5. Question 4. Express each of the following numbers as sum of two odd primes: (a) 18 Solution: 18 = 5 + 13 (b) 32 Solution: 32 = 13 + 19 (c) 66 Solution: 66 = 29 + 37 (d) 90 Solution: 90 = 53 + 37 Question 5. Express the following as sum of three odd primes: (a) 41 Solution: 41 = 3 + 7 + 31 (b) 23 Solution: 23 = 3 + 7 + 13 (c) 75 Solution: 75 = 3 + 11 + 61 (d) 59 Solution: 59 = 3 + 13 + 43
Courses Courses for Kids Free study material Offline Centres More Store # What percent is 1 min 48 seconds of 1 hour? Last updated date: 24th Jul 2024 Total views: 350.1k Views today: 8.50k Verified 350.1k+ views Hint: We need to find what is 1 min 48 seconds as the percent of 1 hour. Firstly, we start to solve the problem by expressing 1 min 48 seconds and 1 hour in the form of seconds. Then, we divide the value of 1 min 48 seconds in seconds by the value of 1 hour in seconds. Lastly, we multiply the result of division by 100 to get the desired result. Complete step-by-step solution: A percentage is a value expressed as a fraction of a hundred or $\dfrac{1}{100}$ . It is denoted by the symbol ’$\%\;$’. The value expressed as a percentage does not have any dimension or unit of measurement. For Example: $\Rightarrow 2\%,3\%$ According to the question, We are required to express 1 min 48 seconds in the form of seconds. The value of 1 min 48 seconds can be written as follows, $\Rightarrow 1\min \text{48seconds}=1\min +48\text{seconds}$ We know that one minute has 60 seconds. Following the same, we get, $\Rightarrow 1\min \text{48seconds}=60\text{seconds}+48\text{seconds}$ Simplifying the above equation, we get, $\therefore 1\min \text{48seconds}=108\text{seconds}$ We know that one hour has 60 minutes. Following the same, we get, $\Rightarrow 1\text{hour}=60\text{minutes}$ We know that one minute has 60 seconds. Following the same, we get, $\Rightarrow 1\text{hour}=60\times 60 \text{seconds}$ Simplifying the above equation, we get, $\therefore 1\text{hour}=3600 \text{seconds}$ Now, we need to divide the value of 1 min 48 seconds in seconds by the value of 1 hour in seconds. Dividing the value of 1 min 48 seconds in seconds by the value of 1 hour in seconds, we get, $\Rightarrow \dfrac{108}{3600}$ We need to multiply the above expression by 100 to get the value in terms of percentage. Multiplying the above expression by 100, we get, $\Rightarrow \dfrac{108}{3600}\times 100\%$ Canceling the common factors, we get, $\Rightarrow 3\%$ $\therefore$ $3\%\;$ of 1 hour is $1\min \text{48seconds}$ Note: The given question can be solved alternatively as follows, A particular quantity can be expressed as the percentage of the other by simply multiplying the fraction of the quantity by $100\%\;$ In our case, Fraction $=\dfrac{1\min \text{ 48seconds}}{1\text{hour}}$ Converting the numerator and the denominator of the fraction into seconds, we get, $Fraction=\dfrac{\text{108}}{3600}$ $\Rightarrow fraction\times 100\%$ Substituting the value of the fraction in the above expression, we get, $\Rightarrow \dfrac{108}{3600}\times 100\%$ $\Rightarrow 3\%$
# The Area of a Quadrilateral: Understanding the Basics Quadrilaterals are fascinating geometric shapes that have four sides and four angles. They can come in various forms, such as rectangles, squares, parallelograms, trapezoids, and rhombuses. Calculating the area of a quadrilateral is an essential skill in geometry and has practical applications in fields like architecture, engineering, and design. In this article, we will explore the concept of finding the area of a quadrilateral, discuss different methods for calculating it, and provide real-world examples to illustrate its significance. ## Understanding the Basics: What is the Area of a Quadrilateral? The area of a quadrilateral refers to the amount of space enclosed within its four sides. It is measured in square units, such as square centimeters (cm²) or square meters (m²). The formula for calculating the area of a quadrilateral depends on its shape and properties. Let’s delve into some common types of quadrilaterals and their respective area formulas. ### 1. Rectangles and Squares A rectangle is a quadrilateral with four right angles, where opposite sides are equal in length. A square is a special type of rectangle with all sides of equal length. To find the area of a rectangle or square, we multiply the length of one side (base) by the length of an adjacent side (height). The formula for the area of a rectangle or square is: Area = Base × Height For example, consider a rectangle with a base of 5 cm and a height of 8 cm. The area of this rectangle would be: Area = 5 cm × 8 cm = 40 cm² ### 2. Parallelograms A parallelogram is a quadrilateral with opposite sides that are parallel and equal in length. To calculate the area of a parallelogram, we multiply the length of the base by the perpendicular distance between the base and the opposite side. The formula for the area of a parallelogram is: Area = Base × Height For instance, let’s consider a parallelogram with a base of 6 cm and a height of 10 cm. The area of this parallelogram would be: Area = 6 cm × 10 cm = 60 cm² ### 3. Trapezoids A trapezoid is a quadrilateral with one pair of parallel sides and another pair of non-parallel sides. To find the area of a trapezoid, we multiply the sum of the lengths of the parallel sides (base1 and base2) by the height and divide the result by 2. The formula for the area of a trapezoid is: Area = (Base1 + Base2) × Height ÷ 2 For example, let’s say we have a trapezoid with base1 measuring 4 cm, base2 measuring 8 cm, and a height of 6 cm. The area of this trapezoid would be: Area = (4 cm + 8 cm) × 6 cm ÷ 2 = 36 cm² ### 4. Rhombuses A rhombus is a quadrilateral with all sides of equal length. To calculate the area of a rhombus, we multiply the lengths of the diagonals and divide the result by 2. The formula for the area of a rhombus is: Area = (Diagonal1 × Diagonal2) ÷ 2 For instance, consider a rhombus with diagonal1 measuring 6 cm and diagonal2 measuring 10 cm. The area of this rhombus would be: Area = (6 cm × 10 cm) ÷ 2 = 30 cm² ## Methods for Calculating the Area of a Quadrilateral While the formulas mentioned above are specific to certain types of quadrilaterals, there are general methods that can be used to find the area of any quadrilateral. Let’s explore two such methods: the shoelace formula and the Brahmagupta’s formula. ### 1. The Shoelace Formula The shoelace formula, also known as Gauss’s area formula or the surveyor’s formula, is a method for finding the area of any polygon, including quadrilaterals. It involves using the coordinates of the vertices of the quadrilateral to calculate its area. The steps for using the shoelace formula are as follows: 1. Write down the coordinates of the vertices of the quadrilateral in a clockwise or counterclockwise order. 2. Multiply each x-coordinate by the y-coordinate of the next vertex in the sequence. 3. Multiply each y-coordinate by the x-coordinate of the previous vertex in the sequence. 4. Sum up all the products obtained in steps 2 and 3. 5. Take the absolute value of the sum obtained in step 4 and divide it by 2. Let’s illustrate the shoelace formula with an example. Consider a quadrilateral with the following coordinates: A(2, 3), B(5, 7), C(8, 4), and D(4, 1). Using the shoelace formula, we can calculate its area as follows: 1. (2 × 7) + (5 × 4) + (8 × 1) + (4 × 3) = 14 + 20 + 8 + 12 = 54 2. Absolute value of 54 ÷ 2 = 27 Therefore, the area of the given quadrilateral is 27 square units. ### 2. Brahmagupta’s Formula Brahmagupta’s formula is a method for finding the area of any cyclic quadrilateral, which is a quadrilateral that can be inscribed in a circle. The formula involves using the lengths of the sides of the quadrilateral to calculate its area. The steps for using Brahmagupta’s formula are as follows: 1. Measure the lengths of the four sides of the quadrilateral (a, b, c, and d). 2. Calculate the semiperimeter of the quadrilateral by summing up the lengths of all four sides and dividing the result by 2 (s = (a + b + c + d) ÷ 2). 3. Calculate the area of the quadrilateral More from this stream ### Transform Your Space with Truly Red White and Tru Design Discover how to infuse sophistication and boldness into your living spaces with Truly Red White and Tru. Learn how 85% of designers are utilizing Tru to elevate interiors through accent walls, striking furniture, and subtle decor touches. Dive into the world of design with this impactful color trio. ### Unlock Hidden Gems: Trick or Trade 2023 Card List Revealed Discover the untapped power of obscure cards in the "Trick or Trade 2023" list! Unveil unique gameplay mechanics and seize the opportunity to boost your wins by 10%. 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Greatest Common Factor of Monomial In this chapter, we will learn to find greatest common factor for given set of monomial using solved examples. To understand the chapter fully, you should have basic knowledge about the concept of HCF and monomials. Greatest common monomial factor In this topic, we will try to find the common factors present in the given monomials and then combine them. To find the greatest common factor, follow the below steps; (a) First find HCF of coefficients. Separate all the coefficient from given monomial and then find the HCF. The HCF tells the highest factor present in the given coefficient. (b) Find common variable terms. Select the particular variable and compare its exponents present in all monomials. Select the variable will lowest power as it ensures that the factor is common among all the monomials. Now combine the calculated coefficients and variables for final result. I hope you got the basic idea of the process. Let us see some examples for further clarity. GCF of Monomial – Solved examples Example 01 Find greatest common factor of following monomials. \mathtt{( a) \ \ 6x^{2} y}\\\ \\ \mathtt{( b) \ \ 2x^{3} y^{3}}\\\ \\ \mathtt{( c) \ \ 3x^{2} y^{2}} Solution Given above are three monomials. To find the GCF, follow the below steps; (a) Find HCF of coefficient. HCF (6, 2, 3) = 1 Hence, 1 is the highest number which is present in all the three monomial. (b) Now find GCF of variables. To find the GCF, select the variable with lowest coefficient. Variable x Among given monomials, the lowest exponent is 2. This means that \mathtt{x^{2}} is present in all the three monomials. Variable y Among given monomial, the lowest exponent is 1. Combining all the components we get; \mathtt{\Longrightarrow \ 1.x^{2} .y}\\\ \\ \mathtt{\Longrightarrow \ x^{2} y} Hence, \mathtt{x^{2} y} is the highest common factor among given variables. Example 02 Find the greatest common factors of given monomials. \mathtt{a) \ \ 8m^{5} n^{4}}\\\ \\ \mathtt{( b) \ \ 16\ m^{8} n^{5}}\\\ \\ \mathtt{( c) \ \ 4\ m^{9} n^{6}} Solution To find the greatest common factor, follow the below steps; (a) Find HCF of coefficient. HCF (8, 16, 4) = 4 Hence, number 4 is the common factor among given monomials. (b) Find highest common variable factors Here you have to choose variable with the lowest power. Variable m Lowest power is 5 Variable n Lowest power is 4 Combining all the components, we get; \mathtt{\Longrightarrow \ 4.m^{5} .n^{4}}\\\ \\ \mathtt{\Longrightarrow \ 4.m^{5} .n^{4}} Hence, \mathtt{\Longrightarrow \ 4.m^{5} .n^{4}} is the highest common factor among given monomials. Example 03 Find GCF of below monomials \mathtt{( a) \ \ 15\ x^{2} y^{8} z^{3}}\\\ \\ \mathtt{( b) \ -27\ xy^{2} z} Solution (a) Find HCF of coefficient HCF (-27, 15) = 3 (b) Find highest common variable factors. Choose the variables with lowest power. Variable x Lowest power is 1 Variable y Lowest power is 2 Variable z Lowest power is 1 Combining all the components we get; \mathtt{\Longrightarrow \ 3\ .x\ .\ y^{2} .\ z}\\\ \\ \mathtt{\Longrightarrow 3\ x\ y^{2} \ z\ } Hence, \mathtt{3\ x\ y^{2} \ z} is the highest common factor. Example 04 Find the highest common factor of below monomials \mathtt{( a) \ \ 60\ x\ y}\\\ \\ \mathtt{( b) \ 12\ x^{2} y^{3}}\\\ \\ \mathtt{( c) \ \ -16\ x^{3}} Solution (a) Find HCF of coefficient HCF (60, 12, -16 ) = 4 Hence, 4 is the greatest number present in all the monomial. (b) Highest common factor among monomial. Here you have to select the lowest possible exponent. Variable x Lowest given exponent is 1. Variable y Lowest exponent is 0 Combining all the exponents we get 4x as the highest common factor. Example 05 Find the highest common factor among given monomial. \mathtt{( a) \ \ -20\ x^{7} \ y^{6}}\\\ \\ \mathtt{( b) \ -12\ x^{8} y^{5}} Solution (a) Find HCF of coefficient. HCF (-20, -12) = -4 Hence, -4 is the highest common factor present in both monomials. (b) GCF of variables. Here you have to select the lowest exponent value. Variable x Lowest exponent is 7. Variable y Lowest exponent is 5. Combining all the components we get; \mathtt{\Longrightarrow \ 4.x^{7} .y^{5}}\\\ \\ \mathtt{\Longrightarrow \ 4x^{7} y^{5}}
# 2021 JMPSC Sprint Problems/Problem 2 ## Problem Brady has an unlimited supply of quarters (\$0.25), dimes (\$0.10), nickels (\$0.05), and pennies (\$0.01). What is the least number (quantity, not type) of coins Brady can use to pay off \$$2.78$? ## Solution It is generally best to use the smallest number of coins with the most value, specifically the quarters, for taking away a big chunk of the problem. We are able to fit $11$ quarters, or $2.75$ into $2.78$. That only leaves $3$ cents. We cannot put any nickels nor dimes, therefore we require three pennies to get a total of $2.78$. The least number of coins Brady can use to pay off $2.78$ will be $14$ coins. -OofPirate ## Solution 2 You want as many quarters in order to cut down on the number of coins. The most amount of quarters you can have is $11$. Since you can't use three cents on anything other than pennies, the remaining coins are $3$ pennies. Therefore $11+3=14$ - kante314 -
# Trigonometrical Ratios of any Angle We will learn how to find the trigonometrical ratios of any angle using the following step-by-step procedure. Step I: To find the trigonometrical ratios of angles (n ∙ 90° ± θ); where n is an integer and θ is a positive acute angle, we will follow the below procedure. First we need to determine the sign of the given trigonometrical ratio. Now to determine the sign of the given trigonometrical ratio we need to find the quadrant in which the angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies. Now, using the rule “All, sin, tan, cos” we will find the sign of the given trigonometrical ratio.Therefore, (i) All trigonometrical ratios are positive if the given angle (n ∙ 90° + θ) or (n .90° + θ) lies in the I quadrant (first quadrant); (ii) Only sin and csc ratios is positive if the given angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies in the II quadrant (second quadrant); (iii) Only tan and cot ratios is positive if the given angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies in the III quadrant (third quadrant); (iv) Only cos and sec ratios is positive if the given angle (n ∙ 90° + θ) or (n ∙ 90° - θ) lies in the IV quadrant (fourth quadrant). Step II: Now determine whether n is an even or odd integer. (i) If n is an even integer the form of the given trigonometrical ratio will remain the same i.e., sin (n ∙ 90° + θ) = sin θ sin (n ∙ 90° - θ) = - sin θ; cos (n ∙ 90° + θ) = cos θ; cos (n ∙ 90° - θ) = - cos θ; tan (n ∙ 90° + θ) = tan θ; tan (n ∙ 90° - θ) = - tan θ. csc (n ∙ 90° + θ) = csc θ csc (n ∙ 90° - θ) = - csc θ; sec (n ∙ 90° + θ) = sec θ; sec (n ∙ 90° - θ) = - sec θ; cot (n ∙ 90° + θ) = cot θ; cot (n ∙ 90° - θ) = - cot θ. (ii) If n is an odd integer then the form of the given trigonometrical ratio is altered i.e., sin changes to cos; i.e., sin (n ∙ 90° + θ) = cos θ or, sin (n ∙ 90° - θ) = - cos θ csc changes to sec; i.e., csc (n ∙ 90° + θ) = sec θ or, csc (n ∙ 90° - θ) = - sec θ cos changes to sin; i.e., cos (n ∙ 90° + θ) = sin θ or, cos (n ∙ 90° - θ) = - sin θ sec changes to csc; i.e., sec (n ∙ 90° + θ) = csc θ or, sec (n ∙ 90° - θ) = - csc θ tan changes to cot; i.e., tan (n ∙ 90° + θ) = cot θ or, tan (n ∙ 90° - θ) = - cot θ cot changes to tan; i.e., cot (n ∙ 90° + θ) = tan θ or, cot (n ∙ 90° - θ) = - tan θ ` Trigonometric Functions ### New! Comments Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
# Teaching Factors and Divisibility Would You Rather Listen to the Lesson? When seventh graders dive into units of factors, fractions, and exponents, one of the earliest lessons focuses on divisibility and factors. This includes the ability to define divisibility rules, factors, as well as GCF and LCM. To make these early lessons as accessible as possible, math teachers can turn to awesome, diverse teaching strategies. We’ll share a few strategies with you today. Use them and never worry about teaching factors and divisibility again! ## Strategies to Teach Factors and Divisibility ### Divisibility You can start your lesson on factors and divisibility by explaining what divisibility means. Divisibility rules are a collection of general rules that we use to check if a given number is entirely divisible by another number. Point out that “divisible by” means that when we divide one number by another, the result is a whole number. That is, a number ‘a’ is divisible by another number ‘b’ if the division a ÷ b is exact (no remainder). For example, 12 ÷ 4 = 3; therefore, 12 is divisible by 4. Also, 12 is divisible by 3, because we can write the other division 12 ÷ 3 = 4. So, 12 is divisible by both 4 and 3. We can also say 4 and 3 are divisors or factors of 12. Provide a few more examples. • 15 is divisible by 5, because 15 ÷ 5 = 3 (there is no remainder) • 15 is not divisible by 7, because 15 ÷ 7 = 2.17 (there is no remainder) • 0 is divisible by 7, because 0 ÷ 7 = 0 exactly (there is no remainder, and remind students that 0 is a whole number) #### Divisibility Rules As mentioned, we use divisibility rules to determine if a number ‘a’ is divisible by another number ‘b’. You can write each rule on the whiteboard or prepare a chart with the rules and hang it on a wall in the classroom. Provide an example as you explain each rule: 1. If the last digit of a number is even, then this number is divisible by 2. For example, in 104 the last digit is even (4), so the number is divisible by 2. 104 ÷ 2 = 52. 2. If the sum of all digits of a given number is divisible by 3, then this number is divisible by 3. If we take 813 as an example, we can see that the sum of all its digits is 12 (8 + 1 + 3 = 12). 12 is a number divisible by 3, so we can infer that 813 is also divisible by 3. 3. If the number that the last two digits in a given number form is divisible by 4, then this number is also divisible by 4. For instance, in 2,152, the number formed by the last two digits is 52. 52 is divisible by 4, i.e. 52 ÷ 4 = 13; therefore, 2,152 is also divisible by 4. 2,152 ÷ 4 = 538. 4. If the last digit in a number is 0 or 5, then this number is divisible by 5. For example, in 495 the last digit is 5, so we can conclude that 495 is divisible by 5. By quickly checking on a calculator, 495 ÷ 5 = 99. 5. If the number is divisible by both 2 and 3, then this number is also divisible by 6. Let’s take 3,672. 3,672 ÷ 2 = 1,863 and 3,672 ÷ 3 = 1,224. Since 3,672 is divisible by 2 and 3, we can say that it’s also divisible by 6. 6. If we take a number, double the last digit, subtract it from the rest of the number (not including the last digit) and the number we end up with by doing so is divisible by 7, the original number is also divisible by 7. For example, if we double the last digit in 623, we’ll get 32 = 6. Then, by subtracting 6 from the rest of the number we’ll get 62 – 6 = 56. 56 is divisible by 7, thus 623 is also divisible by 7. 7. If the number formed by the last three digits in a given number is divisible by 8, then the original number is also divisible by 8. For instance, in 16,512, the number formed by the last three digits is 512. This is a number divisible by 8, i.e. 512 ÷ 8 = 64. From here, we can arrive at the conclusion that 16,512 is also divisible by 8. 8. If in a given number the sum of all digits is divisible by 9, then this number is also divisible by 9. You can take 1,377 as an example. The sum of all digits is 18 (1 + 3 + 7 + 7 = 18). Since we know that 18 is divisible by 9, we can infer that 1,377 is also divisible by 9. 9. If the last digit of a given number is 0, then we know that this number is divisible by 10. For example, in 260, the last digit is 0, therefore 260 is divisible by 10. If you want to quickly check on a calculator 260 ÷ 10 = 26. ### Factors Once you have provided an overview of divisibility and divisibility rules, you can continue with explaining what factors are. Simply put, a factor is a number that divides exactly into another number, i.e. with no remainder. Provide a few examples of this, such as: • 3 and 6 are factors of 12, because 12 ÷ 6 = 2 (no remainder) and 12 ÷ 3 = 4 (no remainder) • 2 and 5 are factors of 10, because 10 ÷ 5 = 2 (no remainder) and 10 ÷ 2 = 5 (no remainder) • 2 and 4 are factors of 8, because 8 ÷ 2 = 4 (no remainder) and 8 ÷ 4 = 2 (no remainder) Add that a prime factor is a number that has exactly two different factors, the number itself and 1. For example, the numbers 2, 3, 5, 7, 11, 13, 17, 19, and 23 are considered prime numbers. These are the first few, but there are many others of course. Finally, explain that a common factor is a number that is a factor of two or more numbers. For instance, if we look at the factors of 12 and 16, we can observe that the factors of 12 are 1, 2, 3, 4, 6, and 12, whereas for 16 they are 1, 2, 4, 8, and 16. Their common factors are 1, 2, and 4. ### Greatest Common Factor & Least Common Multiple Point out that the greatest common factor (GCF) is the greatest number that divides exactly into two or more numbers or we can say greatest of the common factors of two or more numbers. Add that the greatest common factor (GCF) is also known as the greatest common divisor (G.C.D) and highest common factor (HCF). The Least Common Multiple (LCM) is the smallest number that is a multiple of two or more numbers. The least common multiple is also known as the lowest common denominator (LCD) and lowest common multiple (LCM). Explain the steps for finding GCF and LCM: 1. First, we’ll make a factor tree for each number. This means that we’ll write all the factors of each number. 2. Then, we’ll determine the prime factors that these numbers have in common. 3. To find the GCF, we’ll look at the factors common for both numbers and simply identify which of these has the greatest value. The answer represents the greatest common factor. 4. To find LCM, we’ll multiply the common factors along with any numbers that are not in common and the answer is LCM. However, we will only multiply each factor the greatest number of times it occurs in either number. In simpler terms, if the number 2 occurs two times in one number, and in the other number it occurs three times, we’ll multiply 2 three times. #### Example 1 – Finding GCF: Provide an example of how we apply the above steps in order to find the greatest common factor (GCF). Write two numbers on the whiteboard, such as 10 and 15. Add the factors of 10 (1, 2, 5, 10), as well as of 15 (1, 3, 5, 15). Point out that now we’ll determine the factors that 10 and 15 have in common. These are 1 and 5. The only thing left to do is identify the greatest one. This is 5. So 5 is the greatest common factor for 10 and 15. We can also say that GCF (10, 15) = 5 #### Example 2 – Finding LCM After having provided an example of how to identify the greatest common factor, proceed with an example of how to find the least common multiple (LCM). Write two numbers on the whiteboard, such as 18 and 12. Explain that by using the prime factorization of these two numbers, we’ll construct the smallest number whose prime factorization has all of the ingredients of both these numbers. This number will be the least common multiple. In simpler terms, after determining the prime factors, we will multiply each factor the greatest number of times it occurs in either number. You can write that the prime factors of 18 = 233, In terms of 12, you can write that 12 = 223. So LCM will need to have enough prime factors to cover both of these numbers. We see that 18 has one 2 and two 3s, whereas 12 has two 2s and one 3. We’ll multiply the factors the greatest number of times they occur in both 18 and 12, so two 2s and two 3s. In other words: LCM (18, 12) = 2233 LCM (18, 12) = 36 If you have the technical means in your classroom, you can also enrich your lesson by including videos. For example, use this video to illustrate to students how to check whether random numbers are divisible by 2, 3, 4, 5, 6, 9, and 10 by using the divisibility rules. In addition, this video contains step-by-step instructions and examples on finding the least common multiple (LCM). Finally, use this video to demonstrate to students how to find the greatest common factor (GCF). ## Activities to Practice Factors and Divisibility ### LCM Game This is an online game that will help students practice finding the least common multiple of numbers, and thus developing their multiplication and division skills. To implement this game in your classroom, the only thing you’ll need is suitable devices (one per student). This is an individual game, which makes it a great resource for homeschooling parents as well. Provide instructions for the game to students. The game consists of two rows, there are different pairs of numbers in the upper row and the corresponding LCMs are in the bottom row. Explain that each pair of numbers should be paired with their correct LCM. Once the child manages to match all numbers with their LCM, the game will continue with another round. Keep playing as long as time allows. ### GCF Jeopardy Game This online game is in jeopardy game format. In it, students will get to practice their skills at finding the greatest common factor. Make sure each child has a suitable device and explain the game. Pair students up. Instruct students to choose the 2-player mode from the menu (homeschooling parents can choose one player). Each player chooses their avatar. Then, they take turns answering questions with their avatar. Students click on their avatar before selecting a question. Depending on the difficulty of the question, they can score up to 400 points. Each question is related to finding the greatest common factor of a set of given numbers. There are multiple-choice answers from which the student must select one. In the end, students compare their scores. In each pair, the student with the highest score wins the game. You can also introduce symbolic prizes for the winners. ### ‘Am I Divisible By Him?’ Game This is a game where students will sharpen their ability to quickly check whether a number is divisible by another number with the help of the divisibility rules. To use the game in your classroom, you’ll need to prepare plenty of task cards (around 20 cards per pair). Divide students into pairs and place the cards in the middle, facing down. Each card contains two numbers, for example, 21; 872; 4. In each of the numbers on the cards, without doing actual division, students have to determine whether the first number is divisible by the second number. Player one draws the first card and has 1 minute to answer the question on the card they selected. If they answer correctly, they score one point; if they answer incorrectly, they lose two points. Player two repeats the procedure. Make sure there is a ‘checker’ with the answer sheet in each group. In the end, the student with the greatest score wins the game. A new round begins. The winner becomes the ‘checker’, whereas the student that was a ‘checker’ thus far joins the new round as a player. Create space for discussion and reflection at the end of the game. How did students know whether a certain number was divisible by another one? How did they know that a certain number wasn’t divisible by another one? Which divisibility rules did they apply to check this? ## Before You Leave… If you enjoyed these teaching strategies and activities on teaching factors and divisibility, and you’re looking for more math materials for children of all ages, sign up for our emails and get loads of free lessons and content! Make sure to also head over to our blog – you’ll find plenty of awesome resources for your class!
{[ promptMessage ]} Bookmark it {[ promptMessage ]} microbook_3e-page11 # microbook_3e-page11 - 4 Graph these equations(placing Q on... This preview shows page 1. Sign up to view the full content. Homework #1B More Math Review Problems 1. Graph these equations (placing Y on the vertical axis and X on the horizontal axis): x Y = 2X + 2 x Y = 4X + 2 Comparing the two equations, which is different: the slope or the Y-intercept? How is it different? Are the lines parallel or do they intersect? 2. Graph these equations (placing Y on the vertical axis and X on the horizontal axis): x Y = 2 + 2X x Y = 2 – 2X Comparing the two equations, which is different: the slope or the Y-intercept? How is it different? Are the lines parallel or do they intersect? 3. Graph these equations (placing Q on the vertical axis and P on the horizontal axis): x Q = 4 + 2P x Q = 2 + 2P Comparing the two equations, which is different: the slope or the Q-intercept? How is it different? Are the lines parallel or do they intersect? This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 4. Graph these equations (placing Q on the vertical axis and P on the horizontal axis): x Q = 4 – 2P x Q = 2 + 2P These two equations have different slopes and different Q-intercepts. Do the lines intersect? If so, can you find the value of P and Q at which they intersect? 5. Solve these two equations for P. Then graph the new equations by placing P on the vertical axis and Q on the horizontal axis: x Q = 4 – 2P x Q = 2 + 2P Do the lines intersect? If so, can you find the value of P and Q at which they intersect? X Y (1st) Y (2nd) X Y (1st) Y (2nd) P Q (1st) Q (2nd) P Q (1st) Q (2nd) Q P (1st) P (2nd) If demand curves slope down and supply curves slope up, then which of these two equations resembles a demand curve? Which resembles a supply curve? Page 11... View Full Document {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern
# Division Questions That Will Test Your Elementary School Students Division questions will appear throughout elementary school as division is one of the four mathematical operations- addition, subtraction, multiplication and division. Children will encounter all four operations from as early as third grade. However, concepts of division such as sharing are exposed to children as early as kindergarten. From experience, children seem to find division the trickiest to grasp. Here are some questions to use for division practice with children, from basic division to tricky word problems to challenge your student’s division skills, plus answer keys and expert guidance on how to solve them. ### What are division questions? Division questions can be standard arithmetic (e.g. 25 ÷ 5 or 3,426 ÷ 2) or can be presented as division word problems which present a math problem in a ‘real-life’ context. In this case of division, this will involve sharing or equal grouping and may require children to interpret the remainders appropriately, either by rounding up or down. ### How to solve division questions In a division problem: • The amount being divided is called the dividend • Whatever the dividend is being divided by is called the divisor; • The answer is called the quotient. Division questions may be solvable mentally, perhaps by recalling multiplication facts or place value knowledge to divide into equal groups. For example: a) Mrs. Patel has 16 candies to share between 4 children. How many should each child get? 16 candies ÷ 4 = 4 candies each. Children should be able to use their multiplication and division fact families to solve this. b) A $283 bill at a restaurant has to be split between 10 people. How much should each person pay?$283 ÷ 10 = 28.30 each. Children should be able to use their place value knowledge for this. Other division questions may require the use of a formal written method, such as the partial product method. c) A length of ribbon measures 28.8cm. It is cut into 6 equal pieces. How long is each piece? 28.8cm ÷ 6 = 4.8cm Teaching Long Division Worksheets Division can be tough, but it doesn't have to be! Download this free worksheet to help your students better understand long division specifically. ### Division questions elementary Division questions will appear throughout elementary school. However, as the formal method of division (commonly known as the partial product method) isn’t introduced until upper elementary school, the division questions students will encounter will reflect this. In 3rd grade, students will fluently divide within 100 relating division to their multiplication facts. By 4th grade, students will divide up to four-digit whole number dividends by one-digit divisors using strategies based on place value, such as partial quotients, area models, and arrays. Students in 4th grade should also work with division of whole numbers involving remainders. By 5th grade, students will divide up to four-digit whole number dividends by two-digit divisors using strategies based on place value. Students at this level will also begin to divide decimals to the hundredths place with up to four-digit dividends and two-digit divisors using similar strategies. The standard algorithm for division (also known as long division) is introduced at different times based on what standards your state follows. Common Core states that students should fluently divide multi-digit numbers using the standard algorithm in 6th grade. TEKS states that students start to use the standard algorithm in 4th grade. See below the table, which has been adapted from the Common Core and aligns with TEKS as well, that details the development of skills in division throughout elementary school. If you’re looking for more information on how to teach division or how to fill student’s division knowledge gaps using interventions, read our articles written by math specialists. ### Division questions for 3rd grade 1. 75 children take part in a game. There are 5 children on each team. How many teams are there altogether? Answer: 45 children ÷ 5 = 9 teams 2. David wants to decorate cakes. Each cake will have 3 cherries. David has 18 cherries. How many cakes can he decorate? Answer: 18 cherries ÷ 3 = 6 cakes 3. John needs 48 balloons. The store sells balloons in packs of 6. How many packs does he need to buy? Answer: 48 balloons ÷ 6 = 8 packs ### Division questions for 4th grade 1. Riley works at an apple orchard and is putting 496 apples into bags. Each bag holds 4 apples. How many bags of apples can she make? Answer: 496 apples ÷ 4 = 124 bags of apples 2. Bryan made4,932 selling albums at a concert. Each album was sold for $9. How many albums did he sell? Answer:$4,932 ÷ 9 = 548 albums 3. Sam wants to walk a long distance, for charity, over 6 weekends. The total distance Sam wants to walk is 293km. Approximately how far should he walk each weekend? Answer: 293km ÷ 6 = 48r5 km. We can round this up to 49km – if he walks 49km each weekend he will definitely meet his target! ### Division questions for 5th grade 1. I have 1 ½ liters of juice. I need to share it all equally between 6 glasses. How many milliliters of juice should I pour into each glass? Answer: 1 ½ liters = 1,500ml. 1,500ml ÷ 6 = 250ml of juice in each glass 2. A school fair raises $5,168. The school keeps ¼ of the money for new playground equipment. It gives the rest to charity. How much money does it give to charity? Answer:$5,168 ÷ 4 = $1,292 kept by the school. They give the rest to charity, which is$5,168 – $1,292 =$3,876 to charity 3. A charity organization raised $4,743 at their monthly fundraiser. They share this equally amongst their 12 branches. How much does each branch get? Answer:$4743 ÷ 12 = $395.2 ### Division Questions for 6th grade 1. 8827 ÷ 97 Answer: 91. The multiples here could be listed more quickly (than by using repeated addition or partitioning) by adding 100 and subtracting 3 each time. 2. 3 pineapples cost the same as 2 mangoes. One mango costs$1.35. How much does one pineapple cost? Answer: 2 mangoes = $1.35 x 2 =$2.70 (the same as 3 pineapples). $2.70 ÷ 3 = 90¢, which is the cost of 1 pineapple. 3. Amina posts three large letters. The postage costs the same for each letter. She pays with a$20 note. Her change is $14.96. What is the cost of posting one letter? Answer:$20 – $14.96 =$5.04 (the cost of three letters). $5.04 ÷ 3 =$1.68, which is the cost of one letter. ### Simple division questions 1. 8 meters of rope is cut into 4 equal lengths to make jump ropes. How long is each jump rope? Answer: 8m ÷ 4 = 2m 2. Circle two numbers which divide by 5 with no remainder: 9, 50, 12, 35, 31, 57 Answer: 50 and 35 – both are multiples of 5 as they end in either a 0 or 5 3. 28 children go swimming in groups of 4. How many groups will there be? Answer: 28 children ÷ 4 = 7 groups ### Difficult division questions 1. 999 will divide exactly by 37. There is no remainder. a) Write down the remainder when 1000 is divided by 37. b) Write down the remainder when 998 is divided by 37. c) Write down a multiple of 37 that is greater than 1,000. a) As 1,000 is only 1 more than 999, the remainder will be 1. b) As 998 is only 1 less than 999, the remainder will be 1 less than the divisor (37), so the remainder is 36. c) As 999 is a multiple of 37, if we add 37 we will get another multiple of 37. We can work out that 999 + 37 = 1,036, which is a multiple of 37 greater than 1,000 2. Write the missing number: 12.5 ÷ ___ = 7.5 ÷ 1.5 Answer: 7.5 ÷ 1.5 = 5 – we can figure this out by counting in multiples of 1.5 until we reach 7.5 (5 of 1.5 is 7.5). We could also make each number 10 times bigger to make it ‘easier’ to solve as the answer will be the same: 75 ÷ 15 = 5. We are now left with 12.5 ÷ ___ = 5. When the divisor is missing, we can divide the dividend by the quotient instead to reach the answer. 12.5 ÷ 5 = 2.5, so the missing number is 2.5. 3. A bus company has 62 buses. On average, each bus travels 19 miles on a gallon of fuel and goes 284 miles each day. The bus company says it needs about 900 gallons of fuel every day. Make an estimate to show whether what the company says is reasonable. Answer: To make an estimate, we could round 19 miles to 20 and 284 miles to 280. 280 miles per day ÷ 20 miles per gallon = 14 gallons per bus per day. For approximately 60 buses (we’ve rounded 62 to 60), this would be 14 x 60, which would be 840 gallons. This is close to 900 gallons so the company is fairly accurate. Looking for more resources to support your teaching of division? See our blog on printable division worksheets. Do you have students who need extra support in math? Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor. Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way. Personalized one-on-one math tutoring programs are available for:
Circle A circle is easy to make: Draw a curve that is "radius" away from a central point. And so: All points are the same distance from the center. You Can Draw It Yourself Put a pin in a board, put a loop of string around it, and insert a pencil into the loop. Keep the string stretched and draw the circle! Play With It Try dragging the point to see how the radius and circumference change. (See if you can keep a constant radius!) The Radius is the distance from the center outwards. The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle. And here is the really cool thing: When we divide the circumference by the diameter we get 3.141592654... which is the number π (Pi) So when the diameter is 1, the circumference is 3.141592654... We can say: Circumference = π × Diameter Example: You walk around a circle which has a diameter of 100m, how far have you walked? Distance walked = Circumference = π × 100m = 314m (to the nearest m) Also note that the Diameter is twice the Radius: And so this is also true: Circumference = 2 × π × Radius In Summary: × 2 × π Remembering • Radius is the shortest word and shortest measure • Diameter is longer • Circumference is the longest Definition The circle is a plane shape (two dimensional), so: Circle: the set of all points on a plane that are a fixed distance from a center. Area The area of a circle is π times the radius squared, which is written: A = π r2 Where • A is the Area To help you remember think "Pie Are Squared" (even though pies are usually round): Example: What is the area of a circle with radius of 1.2 m ? Area= πr2 = π × 1.22 = 3.14159... × (1.2 × 1.2) = 4.52 (to 2 decimals) Or, using the Diameter: A = (π/4) × D2 Area Compared to a Square A circle has about 80% of the area of a similar-width square. The actual value is (π/4) = 0.785398... = 78.5398...% And something interesting for you: Names Because people have studied circles for thousands of years special names have come about. Nobody wants to say "that line that starts at one side of the circle, goes through the center and ends on the other side" when they can just say "Diameter". So here are the most common special names: Lines A line that "just touches" the circle as it passes by is called a Tangent. A line that cuts the circle at two points is called a Secant. A line segment that goes from one point to another on the circle's circumference is called a Chord. If it passes through the center it is called a Diameter. And a part of the circumference is called an Arc. Slices There are two main "slices" of a circle. The "pizza" slice is called a Sector. And the slice made by a chord is called a Segment. Common Sectors The Quadrant and Semicircle are two special types of Sector: Quarter of a circle is called a Quadrant. Half a circle is called a Semicircle. Inside and Outside A circle has an inside and an outside (of course!). But it also has an "on", because we could be right on the circle. Example: "A" is outside the circle, "B" is inside the circle and "C" is on the circle. Ellipse A circle is a "special case" of an ellipse.
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 2/6 + 2/6 = 2/3 ≅ 0.6666667 Spelled result in words is two thirds. ### How do you solve fractions step by step? 1. Add: 2/6 + 2/6 = 2 + 2/6 = 4/6 = 2 · 2/2 · 3 = 2/3 For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 6) = 6. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 6 = 36. In the next intermediate step, , cancel by a common factor of 2 gives 2/3. In words - two sixths plus two sixths = two thirds. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • In the cafeteria There are 18 students in Jacob’s homeroom. Six students bring their lunch to school. The rest eat lunch in the cafeteria. In simplest form, what fraction of students eat lunch in the cafeteria? • A newspaper's A newspaper's cover page is 3/8 text, and photographs fill the rest. If 2/5 of the text is an article about endangered species, what fraction of the whole cover page is the article about endangered species? Add this two mixed numbers: 1 5/6 + 2 2/11= • An electrician An electrician needs 1 1/3 rolls of electrical wire to wire each room in a house. How many rooms can he wire with 6 2/3 rolls of wire? • King's birthday To celebrate the king's birthday workers fire 1/5 all purchased rockets. To celebrate the Queen's birthday fire 1/6 of the remaining rockets and to celebrate the birthday of king's son remaining 15,000 rockets. How many rockets they purchased? • Bitoo and Reena Bitoo ate 3/5 part of an apple and the remaining part was eaten by his sister Reena. How much part of an apple did Renna eat? Who had the larger share? By how much? • The tap For one day flows 148 l of water out of the tap. How much water will flow out for 3/4 day? • A laundry Mr. Green washed 1/4 of his laundry. His son washed 3/7 of it. Who washed most of the laundry? How much of the laundry still needs to be washed? • Earning 2 Rahul earns 5000 Rs, he saves 500 Rs . What's percentage of his earnings does he save? * • Two numbers The sum of the two numbers is 1. Find both numbers if you know that half of the first is equal to one seventh of the second number. • Expressions Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n • Regrouping Subtract mixed number with regrouping: 11 17/20- 6 19/20 • Mountain Mountain has an elevation of 7450 meters and in the morning is the middle portion thereof in the clouds. How many meters of height is in the sky if below the clouds are 2,000 meters, and above clouds are two-fifths of the mountain's elevation?
# Like Terms – Definition, Simplification, Examples \| How to find Like Terms? Like terms are those that include the same variables raised to the same power. The general and only difference is numerical coefficients. Only we can combine like terms in a phrase. To make algebraic formulas easier to deal with, we combine like terms to shorten and simplify them. Like terms in mathematics refer to quantities that have the same variables and exponents. Letâ€™s learn the definitions of similar phrases and apply your maths abilities to appropriate sample problems. Students of 6th Grade Math can get ultimate grip on the concept by referring to this article. ## What are Like Terms? Like terms are the terms that contain the same variable and are even raised to the same power. 3x + 12x, for example, is an algebraic expression having similar terms. We can add like terms to simplify this algebraic statement. Similarly, we may do all mathematical operations on equivalent words. Only like terms can be joined in an expression. To make algebraic formulas easier to deal with, we combine like terms to shorten and simplify them. To combine like terms, add the coefficients while keeping the variables constant. Example:Â Individual components of an equation or expression separated by ‘+’ or ‘-‘ signs are referred to as algebraic terms. Consider the following formula: 3x + 8y. We can’t make it any simpler because ‘x’ and ‘y’ are unknown. Consider another example. 3xÂ² + 11x + 7y + 5x + 6xÂ² This expression may be rewritten as follows if it is rearranged: 6xÂ² + 3xÂ²+ 11x + 5x + 7y By includingLike Terms, =9xÂ² + 16x + 7y As seen, algebraic terms with the same variables are added to one another. The addition of specific words was only feasible since the variables in both situations are the same, even though the numerical coefficients varied, which may be added as regular numbers, and the variable factor stays unchanged. See More: ### Like Terms Examples 1. In a mathematical expression 6xÂ²y+ 3xyÂ² â€“ xy â€“ 8yxÂ², Find Like and unlike terms? Because they both have the same numerical coefficients xÂ²y, the like terms are 6xÂ²y, âˆ’ 8yxÂ². And the unlike terms are 3xyÂ², â€“ xy since they each have distinct numerical coefficients. 2. Combine the Like Terms 5xÂ³yÂ³zÂ³ + 11xÂ³yÂ³zÂ³ – 7xÂ³yÂ³zÂ³ The three polynomial expressions have the same variables (xyz) raised to the same power. The one and only difference are in the numerical coefficients. As a result, the polynomials are summed together thus resulting in 9xÂ³yÂ³zÂ³. 3. Combine the Like Terms 8xÂ² – 3xÂ² + 4xÂ²? We see that the three terms of the trinomial ( 8xÂ², 3xÂ², and 4xÂ²) have the same variables (xÂ²) raised to the same power (2). The main difference is the numerical coefficients. As a result, the above expression can be simplified as 9xÂ². ### FAQs on Like Terms 1. In an algebraic equation, can we simplify like terms? Yes, we can simplify Like terms in an algebraic equation. Like terms have the unique virtue of being able to be simplified while performing an Algebraic Operation. 2. Why do we group like terms together? We combine like terms in algebraic equations because doing so simplifies expressions to their simplest form, requiring no further work. Expressions can be answered quickly by merging like terms. 3. What is meant by Like terms? Like terms are the terms that have similar variables and exponent powers. The coefficients of these factors might vary. Algebraic-like words are terms that are related to one another. The algebraic expression’s similar terms can be combined to simplify the equation and obtain the solution in a straightforward manner. Scroll to Top Scroll to Top
Equally Inclined Lines By the meaning of equally inclined lines, we mean that the lines which make equal angles with both the co-ordinate axes. The above diagram shows that PQ and RS are the two equally inclined lines. From the above diagram it is clear that; For PQ: Inclination θ = 45°, Therefore, slope = tan 45° = 1. For RS: Inclination θ = -45°, Therefore, slope = tan (-45°) = -1. Solved example on equally inclined lines: Find the equation of the lines which is passes through the point (-2. 3) and equally inclined to the co-ordinate axes. Solution: From the above diagram it is clear that; there are two lines PQ and RS, equally inclined to the co-ordinate axes. For line PQ: m = tan 45° = 1 and (x$$_{1}$$, y$$_{1}$$) = (-2, 3) Therefore, its equation: y – y$$_{1}$$ = m(x – x$$_{1}$$) ⟹ y – 3 = 1(x + 2) ⟹ y - 3 = x + 2 ⟹ y = x + 5 For line RS: m = tan (-45°) = -1 and (x$$_{1}$$, y$$_{1}$$) = (-2, 3) Therefore, its equation: y – y$$_{1}$$ = m(x – x$$_{1}$$) ⟹ y – 3 = -1(x + 2) ⟹ y - 3 = -x - 2 ⟹ y = -x + 1 Therefore, the required equations are y = x + 5 and y = -x + 1 Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. Recent Articles 1. Constructing a Line Segment \|Construction of Line Segment\|Constructing Aug 14, 24 09:52 AM We will discuss here about constructing a line segment. We know how to draw a line segment of a certain length. Suppose we want to draw a line segment of 4.5 cm length. 2. Construction of Perpendicular Lines by Using a Protractor, Set-square Aug 14, 24 02:39 AM Construction of perpendicular lines by using a protractor is discussed here. To construct a perpendicular to a given line l at a given point A on it, we need to follow the given procedure 3. Construction of a Circle \| Working Rules \| Step-by-step Explanation \| Aug 13, 24 01:27 AM Construction of a Circle when the length of its Radius is given. Working Rules \| Step I: Open the compass such that its pointer be put on initial point (i.e. O) of ruler / scale and the pencil-end be… 4. Practical Geometry \| Ruler \| Set-Squares \| Protractor \|Compass\|Divider Aug 12, 24 03:20 PM In practical geometry, we study geometrical constructions. The word 'construction' in geometry is used for drawing a correct and accurate figure from the given measurements. In this chapter, we shall…
# Question: What Does It Mean To Multiply When Would You Need To Multiply 2 Numbers? ## Why do you add a zero when multiplying two digits? No. There is one trick. When you multiply the second part, add a “0” to that answer, because you are multiplying the value from the tens column (the 2). If it’s from the tens, add a zero.. ## Do you add first or multiply first? Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right. Next, add and subtract from left to right. ## What is the rule for subtracting exponents? If a number is raised to a power. You simply compute the result and then perform the normal subtraction. If both the exponents and the bases are the same, you can subtract them like any other like terms in algebra. For example, 3y – 2xy = x y. ## What do you get when you multiply two numbers? When we multiply two numbers, the answer we get is called ‘product’. The number of objects in each group is called ‘multiplicand,’ and the number of such equal groups is called ‘multiplier’. ## What does it mean to multiply a number by itself? In mathematics, we call multiplying a number by itself “squaring” the number. We call the result of squaring a whole number a square or a perfect square. A perfect square is any number that can be written as a whole number raised to the power of 2. ## How many two digit numbers are there? 90The total number of two digit numbers is 90. From 1 to 99 there are 99 numbers, out of which there are 9 one-digit numbers, i.e., 1, 2, 3, 4, 5, 6, 7, 8 and 9. ## What does the multiply sign look like? the symbol (⋅), (×), or (∗) between two mathematical expressions, denoting multiplication of the second expression by the first. In certain algebraic notations the sign is suppressed and multiplication is indicated by immediate juxtaposition or contiguity, as in ab. ## How is Bodmas calculated? According to BODMAS rule, in a given mathematical expression that contains combination of signs: brackets ((), {}, [], −), multiplication, of, addition, subtraction, division, we must first solve or simplify the brackets followed by of (powers and roots etc.), of, then division, multiplication, addition and subtraction … ## How do you multiply numbers with powers? First, multiply the bases together. Then, add the exponent. Instead of adding the two exponents together, keep it the same. ## Why do you add a zero when multiplying by 10? Why ‘adding zero’ is a problem From the outset, pupils need to grasp the idea that when you multiply by 10, 100, 1000 and so on, each digit shifts to the left on a place value table because they’re adding another place to the number. ## Why do we multiply before we add? It’s a centuries-old convention that allows us to write expressions without so many parentheses. Under that convention, multiplication has a higher order of precedence than addition or subtraction, so rather than fully parenthesizing an expression like , we can write it as . ## What are the four rules of maths? The four basic mathematical operations–addition, subtraction, multiplication, and division–have application even in the most advanced mathematical theories. Thus, mastering them is one of the keys to progressing in an understanding of math and, specifically, of algebra. ## Do you add the powers or multiply? The exponent “product rule” tells us that, when multiplying two powers that have the same base, you can add the exponents. In this example, you can see how it works. Adding the exponents is just a short cut! The “power rule” tells us that to raise a power to a power, just multiply the exponents. ## Does multiply mean? In math, to multiply means to add equal groups. When we multiply, the number of things in the group increases. The two factors and the product are parts of a multiplication problem. … Here is another example of a multiplication fact that shows multiplication is also repeated addition. ## Does times mean multiply? The word “times” doesn’t mean anything to students. Often a student will say the multiplication symbol means “times.” But when pushed further, they can only define it as a synonym for multiplication. (An informal canvas of friends at dinner revealed the same level of awareness.)
+0 # PLS HELP ASAP 0 19 1 +355 Find the number of ordered pairs $$a, b$$ of integers such that$$\frac{a + 2}{a + 5} = \frac{b}{4}.$$ Apr 18, 2024 #1 +193 0 Multiply both sides of the equation by (a+5): a+2=b⋅4a+5 Multiply both sides by 4: 4(a+2)=b(a+5) Expand the left side: 4a+8=b(a+5) This equation represents the following: 4a+8 is a multiple of b (since it's equal to b times (a+5)) a+5 is a divisor of 4a+8 (since the right side is a multiple of the left side) Since we're looking for integer solutions for (a,b), let's analyze the divisibility conditions for both sides: For 4a+8 to be divisible by b, b must be a divisor of 8. The divisors of 8 are 1, 2, 4, and 8. For a+5 to be a divisor of 4a+8, we need to check divisibility for each possible value of b (divisors of 8): If b=1, then a+5 must divide 4a + 8. In this case, the only integer solution for a is -5 (since 4a + 8 = -a + 8, which is divisible by -1 when a = -5). If b=2, then a+5 must divide 4a + 8. This doesn't have any integer solutions for a, because the left side (always even) cannot be equal to the right side (always odd). If b=4, then a+5 must divide 4a + 8. Here, a = -3 is the only integer solution (since 4a + 8 = a + 8, which is divisible by 4 when a = -3). If b=8, then a+5 must divide 4a + 8. The only integer solution for a is -1 (since 4a + 8 = 3a + 8, which is divisible by 8 when a = -1). So we found possible integer solutions for (a,b) as: (-5, 1) (-3, 4) (-1, 8) There are a total of 3 such ordered pairs. Apr 18, 2024
# Big Ideas Math Answers Grade 7 Chapter 10 Surface Area and Volume Area and Volume are the important topics in maths. This will be helpful in the real-time environment. Know how and where to apply the formulas from this page. Get a detailed explanation for all the questions here. Download Big Ideas Math Answers Grade 7 Chapter 10 Surface Area and Volume pdf for free of cost. As per your convenience, we have provided the solutions in pdf format so that you can prepare offline. ## Big Ideas Math Book 7th Grade Answer Key Chapter 10 Surface Area and Volume Get the guided notes for Chapter 10 Surface Area and Volume from here. This will be the best resource to enhance your math skills. The topics covered in this chapter are Surface Areas of Prisms, Cylinders, Pyramids, Volume of Prism, Pyramids, Cross Sections of Three-Dimensional Figures. Test yourself by solving questions given at the end of the chapter. Lesson: 1 Surface Areas of Prisms Lesson: 2 Surface Areas of Cylinders Lesson: 3 Surface Areas of Pyramids Lesson: 4 Volumes of Prisms Lesson: 5 Volumes of Pyramids Lesson: 6 Cross Sections of Three-Dimensional Figures Chapter 10 – Surface Area and Volume ### Surface Area and Volume STEAM Video/Performance Task STEAM Video Paper Measurements The thickness of a single piece of paper cannot be precisely measured using a ruler. What another method can you use to measure the thickness of a piece of paper? Watch the STEAM Video “Paper Measurements.” Then answer the following questions. 1. A stack of 500 pieces of paper is 2 inches tall. How tall is a stack of 250 pieces? 100 pieces? 10 pieces? How thick is a single piece of paper? 2. You have a circular notepad. How can you find the volume of one piece of paper in the notepad? 1. for 250 pieces = 46750 cubic inches 2. volume = 187 cubic inches Explanation: 1. for 250 pieces = l w x h 250 pieces = 250 x 17 x 11 pieces = 46750 100 pieces = 100 x 17 x 11 pieces = 18,700 10 pieces = 10 x 17 x 11 pieces = 1870 piece = 1 x 17 x 11 piece = 187 2. volume = l x w x h volume = 8.5 x 11 x 2 volume = 187 Volumes and Surface Areas of Small Objects After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the dimensions of a shipping box and the number of bouncy balls that fit in the box. You will be asked to use the box to estimate the volume of each bouncy ball. Why might it be helpful to use the volume of a container of objects to estimate the volume of one of the objects? The volume of the object =11,809.8 cubic centimeters Explanation: The volume of rectangular prism = l x w x h where l = length, w = width, h = height rectangular prism = 27 x 27 x 16.2 volume = 11,809.8 cubic centimeters ### Surface Area and Volume Getting Ready for Chapter 10 Chapter Exploration Question 1. Work with a partner. Perform each step for each of the given dimensions. • Use 24 one-inch cubes to form a rectangular prism that has the given dimensions. • Make a sketch of the prism. • Find the surface area of the prism. a . 52 inches b . 98 inches c . 76 inches d . 70 inches e . 68 inches f . 64 inches g . 52 inches prism : The surface area of the square prism = 2lw + 2lh + 2wh The surface area of the rectangle prism = 2(lw + lh + wh) Explanation: a . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 4 w = 3 h = 2 rectangular prism =2(4 x 3) +(2 x 3) +(4 x 2) prism = 2(12) + (6) +(8) surface area =2( 26) surface area = 52 b . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 1 w = 1 h = 24 rectangular prism =2(1 x 1) +(1 x 24) +(1 x 24) prism = 2(1) + (24) +(24) surface area =2( 49) surface area = 98 c . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 1 w = 2 h = 12 rectangular prism =2(1 x 2) +(2 x 12)+(1 x 12) prism = 2(2) + (24) +(12) surface area =2( 38) surface area = 76 d. The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 1 w = 3 h = 8 rectangular prism =2(1 x 3) +(3 x 8)+(1 x 8) prism = 2(3) + (24) +(8) surface area =2( 35) surface area = 70 e . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 1 w = 4 h = 6 rectangular prism =2(1 x 4) +(4 x 6)+(1 x 6) prism = 2(4) + (24) +(6) surface area =2( 34) surface area = 68 f . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 2 w = 2 h = 6 rectangular prism =2(2 x 2) +(2 x 6)+(1 x 6) prism = 2(4) + (12) +(6) surface area =2( 32) surface area = 64 g . The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 2 w = 4 h = 3 rectangular prism =2(2 x 4) +(4 x 3)+(2 x 3) prism = 2(8) + (12) +(6) surface area =2( 26) surface area = 52 Question 2. REASONING Work with a partner. If two blocks of ice have the same volume, the block with the greater surface area will melt faster. The blocks below have equal volumes. Which block will melt faster? Explain your reasoning. The first block will melt faster. Explanation: The surface area of the square prism = 2lw + 2lh + 2wh where l = 1ft w = 1ft h = 1 ft surface area = 2(1 x 1)+2(1 x 1) + 2(1 x 1) surface area = 2(1)+2(1) + 2(1) surface area = 2+2+2 surface area = 6 The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 2 w = 1 h = 0.5 rectangular prism =2(2 x 1) +(2 x 0.5) +(1 x 0.5) prism = 2(2) + (1) +(0.5) surface area = 4 + 1 + 0.5 surface area = 5.5 Vocabulary The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts. lateral surface area the slant height of a pyramid regular pyramid cross section Lateral Surface Area = The lateral surface of an object is all of the sides of the object, excluding its base and top. Slant Height of a Pyramid = The slant height of an object is the distance measured along a lateral face from the base to the apex along the center of the face. Regular Pyramid = In geometry, a pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. Each base edge and apex form a triangle. Cross Section = a surface or shape exposed by making a straight cut through something, especially at the right angles to an axis. ( The cross-section of an octahedron is a square.) ### Lesson 10.1 Surface Areas of Prisms EXPLORATION 1 Writing a Formula for Surface Area Work with a partner. a. Use the diagrams to write a formula for the surface area of a rectangular prism. Explain your reasoning. b. Choose dimensions for a rectangular prism. Then draw the prism and use your formula in part(a) to find the surface area. a : b : The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height. EXPLORATION 2 Surface Areas of Prisms Work with a partner. a. Identify the solid represented by the net. Then find the surface area of the solid. b. Describe a method for finding the surface area of any prism. a : The surface area of the solid = 94 b : The surface area of the solid = 2(lw + lh + wh) Explanation: a : b : The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height. Try It Find the surface area of the prism. Question 1. The surface area of the prism = 52 ft Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(2 x 3) + 2(3 x 4) +2(2×4) surface area = 2(6) + 2(12) + 2(8) surface area = 12 + 24 + 16 surface area = 52 ft Question 2. The surface area of the prism = 288 sq. m Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(5 x 8) + 2(8 x 8) +2(5×8) where l = 8m,w = 8m, h= 8m surface area = 2(40) + 2(64) + 2(40) surface area = 80 + 128 + 80 surface area = 288 sq. m Question 3. Find the surface area of the prism at the left. The surface area of the prism at the left = 288 sq. m Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(5 x 12) + 2(12 x 13) +2(5×13) where l = 13m,w = 5m, h= 12m surface area = 2(60) + 2(156) + 2(65) surface area = 120 + 312 + 130 surface area = 562 sq. m Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 4. WRITING Explain the meaning of each term in the formula for the surface area of a rectangular prism. The surface area of the rectangular prism = 2(lw + lh + wh) Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height Question 5. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers. The surface area of the triangular prism = 26 sq. cm The surface area of the bases of the prism = 26 sq. cm Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 3) + 4 (5) surface area = 6 + 20 surface area = 26 sq. cm The surface area of the bases of the prism = ph + 2B surface area = 4(5) + 2(3) surface area = 26 sq. cm The area of the net of the prism = The sum of the areas of the bases and the lateral faces of the prism. Question 6. You want to stain the lateral faces of the wooden chest shown. Find the area that you want to stain in square inches. The area that you want to stain in square inches = 3456 square inches. Explanation: The surface area of the lateral faces of the prism = ph + 2B surface area = 4(4) + 2(4) surface area = 24 ft 1 feet = 144 square inches 144 x 24 = 3456 square inches. Question 7. One can of frosting covers about 280 square inches. Is one can of frosting enough to frost the cake? Explain. no one can of frosting is not enough to frost the cake. Explanation: The cake piece is in the shape of a rectangle The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(39 + 117 + 27) 183 sq. inches. Question 8. DIG DEEPER! The surface area of the bench = 16 sq. ft Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(1 x 1.5) +(1.5 x 2) +(2 x 5) surace area = 2(1.5) + 3 + 10 surface area = 16 sq. ft ### Surface Areas of Prisms Homework & Practice 10.1 Review & Refresh Classify the pair of angles. Then find the value of x. Question 1. The value of x = 146 Explanation: The above-shown angle = obtuse angle given that 34 degrees 180 – 34 = 146 Question 2. The value of x = 106 Explanation: The above-shown angle = right angle given that 74 degrees 180 – 74 = 106Question 3. The value of x = 121 Explanation: The above-shown angle = acute angles. given that 59 degrees x + 10 = 59 x = 49 180 – 49 = 121 Question 3. Find the area of a circle with the indicated dimensions. Use 3.14 or $$\frac{22}{7}$$ for π. Question 4. The area of the circle = 1,384.74 sq. in Explanation: The area of the circle = πr2 area = 3.14 x 21 x 21 where π = 3.14 radius = 21 in area = 1,384.74 sq. in Question 5. diameter: 36mm The area of the circle = 530.66 sq. mm Explanation: The area of the circle = πr2 area = 3.14 x 13 x 13 where π = 3.14 radius = 13 mm area = 530.66 sq. mm Question 6. The area of the circle = 226.865 sq. m Explanation: The area of the circle = πr2 area = 3.14 x 8.5 x 8.5 where π = 3.14 radius = 8.5 m area = 226.865 sq. m Concepts, Skills, & Problem Solving SURFACE AREA OF A PRISM Identify the solid represented by the net. Then find the surface area of the solid. (See Explorations 1 & 2, p. 409.) Question 7. The surface area of the prism = 182 Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(6 x 8) +(9 x 6) +(10 x 8) surace area = 2(48) + 54 + 80 surface area = 182 Question 8. The surface area of the prism = 44 Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(5 x 2) +(2 x 3) +(3 x 2) surace area = 2(10) + 6+ 6 surface area = 44 sq. units FINDING THE SURFACE AREA OF A PRISM Find the surface area of the prism. Question 9. The surface area of the prism = 324 sq. m Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(6 x 16) +(16 x 3) +(3 x 6) surace area = 2(96) + 48+ 18 surface area = 324 sq. m Question 10. The surface area of the prism = 294 sq. yd Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(7 x 7)+ 2(7 x 7) +2(7×7) surface area = 2(49) + 2(49) + 2(49) surface area = 98 + 98 + 98 surface area = 294 sq. yd Question 11. The surface area of the triangular prism = 32 sq. m Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 6) + 4 (5) surface area = 12 + 20 surface area = 32 sq. m Question 12. The surface area of the triangular prism = 166 sq. ft Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 15) + 17 (8) surface area = 30 + 136 surface area = 166 sq. ft Question 13. The surface area of the prism = 46.4 yds Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(1.2 x 5) +(5 x 3) +(1.2 x 6) surace area = 2(1) + 15+ 7.2 surface area = 23.2 x 2 surface area = 46.4 sq. yds Question 14. The surface area of the prism = 693 sq. in Explanation: The surface area of the paralleloram prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(9 x 10) +(9 x 13.5) +(13.5 x 10) surace area = 2(90) +121.5 + 135 surface area = 346.5 x 2 surface area = 693 sq. inches Question 15. YOU BE THE TEACHER yes my friend is correct. Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(5 x 3)+ 2(3 x 4) +2(5×3) surface area = 2(15) + 2(12) + 2(15) surface area = 30+ 24 + 30 surface area = 84 sq. cm Question 16. MODELING REAL LIFE A cube-shaped satellite has side lengths of 10 centimeters. What is the least amount of aluminum needed to cover the satellite? The least amount of aluminum needed to cover the satellite = 90 centimeters. Explanation: Given that cube has the side lengths of 10 and 10 centimeters 10 x 10 = 100 100 -10 = 90 so the least amount of aluminum needed to cover = 90 centimeters. FINDING SURFACE AREA Find the surface area of the prism. Question 17. The surface area of the triangular prism = 80 in Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 16) + 4 (12) surface area = 32 + 48 surface area = 80 sq. in Question 18. The surface area of the prism = 58 sq. m Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(4 x 4) +(4 x 2) +(2.5 x 2) surace area = 2(16) + 8+ 5 surface area = 29 x 2 surface area = 58 sq. m Question 19. OPEN-ENDED Draw and label a rectangular prism that has a surface area of 158 square yards. Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(4 x 5) +(5x 6) +(4 x 7) surace area = 2(20) + 30+ 29 surface area = 79 x 2 surface area = 158 square yards Question 20. DIG DEEPER! A label that wraps around a box of golf balls covers 75% of its lateral surface area. What is the value of x? The value of x = 25% Explanation: Given that the label that wraps around a box of golf covers 75% That means for 100% 100 – 75 = 25% Question 21. STRUCTURE You are painting the prize pedestals shown(including the bottoms). You need 0.5 pint of paint to paint the red pedestal. a. The edge lengths of the green pedestal are one-half the edge lengths of the red pedestal. How much paint do you t do you need to paint the green pedestal? b. The edge lengths of the blue pedestal are triple the edge lengths of the green pedestal. How much paint do you need to paint the blue pedestal? c. Compare the ratio of paint volumes to the ratio of edge lengths for the green and red pedestals. Repeat for the green and blue pedestals. What do you notice? a. The paint you need to paint the green pedestal =1024 in b. The paint you need to paint the blue pedestal = 5798464 in c. 1 : 2 for green and red , 1 : 3 for green and blue. b. Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) where l = length , w= width , h = height surface area = 2(16 x 16) +(16x 24) +(16 x 24) surace area = 2(256) + 384+ 384 surface area = 1024 x 2 surface area = 2048 sq. in Question 22. NUMBER SENSE A key chain-sized puzzle cube is made up of small cubes. Each small cube has a surface area of 1.5 square inches. a. What is the edge length of each small cube? b. What is the surface area of the entire puzzle cube? The edge length of each small cube = 1 The surface area of the entire puzzle cube = 81 Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(0 x 0)+ 2(0 x 0) +1.5(1x 1) surface area = 2(0) +2(0) +1.5(1) surface area = 0 + 0 + 1.5 surface area = 1.5 The surface area of the each small cube = 81 ### Lesson 10.2 Surface Areas of Cylinders A is a solid that has two parallel, identical circular bases. EXPLORATION 1 Finding the Surface Area of a Cylinder Work with a partner. a. Make a net for the can. Name each shape in the net. b. How are the dimensions of the paper related to the dimensions of the can? c. Write a formula that represents the surface area of a cylinder with h a height of and bases with a radius of r. d. Estimate the dimensions of each can. Then use your formula in part(c) to estimate the surface area of each can. a. b. The dimensions of the paper = the dimensions of the can c . d. The surface area of the cylinder = 2πr (h + r) Try It Find the surface area of the cylinder. Round your answer to the nearest tenth if necessary. Question 1. The surface area of the cylinder = 34138.08 sq. yd Explanation: The surface area of the cylinder = 2πr2 + 2πrh where r = radius, h = height 2 x 3.14 x 36 + 2 x 3.14 x 6 x 9 226.08 + 33912 34138.08 Question 2. The surface area of the cylinder = 395.64 sq. cm Explanation: The surface area of the cylinder = 2πr2 + 2πrh where r = radius, h = height 2 x 3.14 x 9+ 2 x 3.14 x 3 x 18 56.52 + 339.12 395.64 sq. cms Find the lateral surface area of the cylinder. Round your answer to the nearest tenth. Question 3. The lateral surface of the cylinder = 75.36 sq. cms Explanation: The lateral surface area of the cylinder = 2πrh area = 2 x 3.14 x 3 x 4 area = 75.36 sq. cms Question 4. The lateral surface of the cylinder = 150.72 sq. yds Explanation: The lateral surface area of the cylinder = 2πrh area = 2 x 3.14 x 3 x 8 area = 150.72 sq. yds Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 5. WRITING Which part of the formula S = 2πr2 + 2πrh represents the lateral surface area of a cylinder? the areas of the bases? The area of the bases of the cylinder. Explanation: The area of the bases of the cylinder = 2πr2 + 2πrh where r = radius , h = height Question 6. CRITICAL THINKING You are given the height of a cylinder and the circumference of its base. Describe how to find the surface area of the cylinder. The surface area of the cylinder = 2πr2 + 2πrh Explanation: The area of the bases of the cylinder =2πr2 + 2πrh where r = radius, h = height Question 7. FINDING A SURFACE AREA Find the surface area of the cylinder at the left. Round your answer to the nearest tenth. The lateral surface of the cylinder = 351.68 sq. in Explanation: The lateral surface area of the cylinder = 2πrh area = 2 x 3.14 x 16 + 2 x 3.14x 4 x 10 area = 100.48 + 251.2 area = 351.68 sq. in Question 8. FINDING A LATERAL SURFACE AREA Find the lateral surface area of the cylinder at the right. Round your answer to the nearest tenth. The lateral surface of the cylinder = 351 sq. in Explanation: The lateral surface area of the cylinder = 2πrh area = 2 x 3.14 x 16 + 2 x 3.14x 4 x 10 area = 100.48 + 251.2 area = 351.68 sq. in IN the question given that round to the nearest tenth area = 351 sq. in Question 9. You remove the lid of the can. What is the percent of change in the surface area of the can? The percent of the change in the surface area of the can = 22,019.25 mm Explanation: The surface area of the can = 2π rx r+ 2πrh 2 x 3.14 x 1806.25 + 2 x 3.14 x 42.5 x 40 22,019.25 mm The surface area of the triangular prism = 166 sq. ft Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 15) + 17 (8) surface area = 30 + 136 surface area = 166 sq. ft Question 10. After burning half of a cylindrical candle, the surface area is 176 square inches. The radius of the candle is 2 inches. What was the original height of the candle? The original height of the Question 11. DIG DEEPER! The area of the sheet of wrapping paper is equal to the lateral surface area of a cylindrical tube. The tube is 14 inches tall. What is the surface area of the tube, including the bases? Explain your reasoning. The surface area of the tube including the bases = 42.96 inches Explanation: The lateral surface area of the cylinder = 2 x 3.14 x r x h surface area = 2 x 3.14 x 13 x 14 surface area = 1142.96 inchs ### Surface Areas of Cylinders Homework & Practice 10.2 Review & Refresh Find the surface area of the prism. Question 1. The surface area of the prism = 142 sq. cms Explanation: The surface area of the prism = 2lw + 2lh + 2wh surface area = 2(7 x 3)+ 2(3 x 5) +2(5×7) surface area = 2(21) + 2(15) + 2(35) surface area = 42+ 30 + 70 surface area = 142 sq. cm Question 2. The surface area of the triangular prism = 649 sq. ft Explanation: The surface area of the triangular prism = 2B + ph where B = area of the base, p = perimeter of the base, h = height surface area = 2( 20) + 29 (21) surface area = 40+ 609 surface area = 649 sq. ft Question 3. Which of the following is equivalent to 0.625? A. $$\frac{5}{8}$$ B. $$\frac{625}{100}$$ C. 0.625% D. 6.25% Explanation: (5/8) = 0.625 Concepts, Skills, & Problem Solving FINDING SURFACE AREA Find the surface area of the cylinder. (See Exploration 1, p. 415.) Question 4. a can with a radius of 60 millimeters and a height of 160 millimeters The surface area of the cylinder = 82,896 millimeters Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 60 millimeters , height = 160 given surface area = 2 x 3.14 x 60 x 60 + 2 x 3.14 x 60 x 160 surface area = 22608 + 60,288 surface area = 82,896 sq. millimeters Question 5. a hay bale with a diameter of 30 inches and a height of 72 inches The surface area of the cylinder = 8195.4 inches Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 15 inches , height = 72 inches given surface area = 2 x 3.14 x 15 x 15 + 2 x 3.14 x 15 x 72 surface area = 1413 + 6,782.4 surface area = 8195.4 sq. inches FINDING SURFACE AREA Find the surface area of the cylinder. Round your answer to the nearest tenth if necessary. Question 6. The surface area of the cylinder = 94.2 ft Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 3 ft , height = 2 ft given surface area = 2 x 3.14 x 3 x 3 + 2 x 3.14 x 3 x 2 surface area = 56.52 + 37.68 surface area = 94.2 sq. ft Question 7. The surface area of the cylinder = 31.4 sq. m Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 1 m , height = 4 m given surface area = 2 x 3.14 x 1 x 1 + 2 x 3.14 x 1 x 4 surface area = 6.28+ 25.12 surface area = 31.4 sq. m Question 8. The surface area of the cylinder = 527.52 sq. ft Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 7 ft , height = 5 ft given surface area = 2 x 3.14 x 7 x 7 + 2 x 3.14 x 7 x 5 surface area = 307.72+ 219.8 surface area = 527.52 sq. ft Question 9. The surface area of the cylinder = 87.92 sq. mm Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 2mm , height = 5 mm given surface area = 2 x 3.14 x 2 x 2 + 2 x 3.14 x 2 x 5 surface area = 25.12+ 62.8 surface area = 87.92 sq. mm Question 10. The surface area of the cylinder = 489.84 sq. ft Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 6 ft , height = 7 ft given surface area = 2 x 3.14 x 6 x 6 + 2 x 3.14 x 6 x 7 surface area = 226.08+ 263.76 surface area = 489.84 sq. ft Question 11. The surface area of the cylinder = 678.24 sq. cm Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 6 cm , height = 12 cm given surface area = 2 x 3.14 x 6 x 6 + 2 x 3.14 x 6 x 12 surface area = 226.08+ 452.16 surface area = 678.24 sq. cm FINDING LATERAL SURFACE AREA Find the lateral surface area of the cylinder. Round your answer to the nearest tenth if necessary. Question 12. The lateral surface area of the cylinder = 376.8 sq. ft Explanation: The surface area of the cylinder = 2πrh where radius = 10 ft , height = 6 ft given surface area = 2 x 3.14 x 10 x 6 surface area = 376.8 sq. ft Question 13. The lateral surface area of the cylinder = 226.08 sq. in Explanation: The surface area of the cylinder = 2πrh where radius = 4 in , height = 9 in given surface area = 2 x 3.14 x 4 x 9 surface area = 226.08 sq. in Question 14. The lateral surface area of the cylinder = 87.92 sq. m Explanation: The surface area of the cylinder = 2πrh where radius = 7 m , height = 2 m given surface area = 2 x 3.14 x 7 x 2 surface area = 87.92 sq. m Question 15. YOU BE THE TEACHER No my friend is not correct. Explanation: The surface area of the cylinder = 2πr2 + 2πrh surface area = 2 x 3.14 x 5 x 5 + 2 x 3.14 x 5 x 10.6 surface area = 157 + 332.84 surface area = 489.84 sq. yds Question 16. MODELING REAL LIFE The tank of a tanker truck is a stainless steel cylinder. Find the surface area of the tank. The surface area of the tank = 1356.48 sq. ft Explanation: The surface area of the cylinder = 2πr2 + 2πrh surface area = 2 x 3.14 x 4 x 4 + 2 x 3.14 x 4 x 50 surface area = 100.48+ 1256 surface area = 1,356.48 sq. ft Question 17. MODELING REAL LIFE The Petri dish shown has no lid. What is the surface area of the outside of the Petri dish? The surface area of the outside of the petri dish= 20,410 sq. mm Explanation: The surface area of the outside of the petri dish = 2πr2 + 2πrh surface area = 2 x 3.14 x 50 x 50 + 2 x 3.14 x 50x 15 surface area = 15700+ 4710 surface area = 20410 sq. mm Question 18. REASONING You have two 8.5-by-11-inch pieces of paper. You form the lateral surfaces of two different cylinders by taping together a pair of opposite sides on each piece of paper so that one cylinder has a height of 8.5 inches and the other has a height of 11 inches. Without calculating, compare the surface areas of the cylinders (including the bases). Explain. The surface area of the cylinder 1= 44.826012 sq. inches The surface area of the cylinder 2= 56.915012 sq. inches Explanation: The surface area of the cylinder 1 = 2πr2 + 2πrh surface area = 2 x 3.14 x 0.77 x 0.77 + 2 x 3.14 x 0.77x 8.5 surface area = 3.723412+ 41.1026 surface area = 44.826012 inches The surface area of the cylinder 1 = 2πr2 + 2πrh surface area = 2 x 3.14 x 0.77 x 0.77 + 2 x 3.14 x 0.77x 11 surface area = 3.723412+ 53.1916 surface area = 56.915012 inches Question 19. DIG DEEPER! A ganza is a percussion instrument used in samba music. a. Find the surface area of each of the two labeled ganzas. b. The smaller ganza weighs 1.1 pounds. Assume that the surface area is proportional to the weight. What is the weight of the larger ganza? a. The surface area of the smaller ganza= 296.73 sq. cm The surface area of the larger ganza= 1036.2 sq. cm Explanation: The surface area of the smaller ganza = 2πr2 + 2πrh surface area = 2 x 3.14 x 3.5 x 3.5 + 2 x 3.14 x 3.5x 10 surface area = 76.93+ 219.8 surface area = 296.73 sq. cm The surface area of the larger ganza = 2πr2 + 2πrh surface area = 2 x 3.14 x 5.5 x 5.5 + 2 x 3.14 x 5.5x 24.5 surface area = 189.97+846.23 surface area = 1036.2 sq. cm Question 20. PROBLEM SOLVING The wedge is one-eighth of the wheel of cheese. a. Find the surface area of the cheese before it is cut. b. Find the surface area of the remaining cheese after the wedge is removed. Did the surface area increase, decrease, or remain the same? a. The surface area of the cheese before it is cut= 75.36 sq. in b. The surface area of the cheese after the wedge is removed = 58.875 sq. in The surface area decreases. Explanation: The surface area of the cheese after the wedge is removed = 2πr2 + 2πrh surface area = 2 x 3.14 x 3 x 3 + 2 x 3.14 x 3 x 1 surface area = 56.52+ 18.84 in surface area = 75.36 sq. in The surface area of the cheese after the wedge is removed = 2πr2 + 2πrh surface area = 2 x 3.14 x 3 x 3 + 2 x 3.14 x 3 x (1/8) (1/8) = 0.125 surface area = 56.52+ 2.355 surface area = 58.875 in The surface area decreases Question 21. REPEATED REASONING A cylinder has radius r and height h. a. How many times greater is the surface area of a cylinder when both dimensions are multiplied by2? 3? 5? 10? b. Describe the pattern in part(a). Write an expression for the surface area of the cylinder when both dimensions are multiplied by a number. a. 2 times greater, 3 times greater, 5 times greater,10 times greater. b. The expression for the surface area of the cylinder when both dimensions are multiplied 2 = 2r + 2h,3r + 3h,5r + 5h, 10r + 10 h ### Lesson 10.3 Surface Areas of Pyramids Many well-known pyramids have square bases, however, the base of a pyramid can be any polygon. EXPLORATION 1 Making a Scale Model Work with a partner. Each pyramid below has a square base. a. Draw a net for a scale model of one of the pyramids. Describe the scale factor. b. Find the lateral surface area of the real-life pyramid that you chose in part(a). Explain how you found your answer. c. Draw a net for a pyramid with a non-rectangular base and find its lateral surface area. Explain how you found your answer. a. b. The lateral surface of the real life pyramid = A + (1/2 ) ps where A = area of base ,p= perimeter of base , s = slant height. c. The lateral surface area of the rectangular pyramid = A + (1/2)ps where A = area of base ,p= perimeter of base , s = slant height. A regular pyramid is a pyramid whose base is a regular polygon. The slant height lateral faces are triangles. The height of each triangle is the of the pyramid. Try It Question 1. What is the surface area of a square pyramid with a base side length of 9 centimeters and a slant height of 7 centimeters? The surface area of the square pyramid = 94.5 sq. centimeters Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 63 +(1/2) 9 x 7 surface area = 63 + 0.5 x 63 surface area = 31.5 + 63 94.5 sq. centimeters. Question 2. Find the surface area of the regular pyramid at the left. The surface area of the triangle pyramid = 86 sq. ft Explanation: The surface area of the triangle pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 60 +(1/2) 5.2 x 10 surface area = 60 + 2.6 x 10 surface area = 60+ 26 surface area = 86 sq. ft Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 3. VOCABULARY Can a pyramid have rectangles as lateral faces? Explain. Yes, pyramids have rectangles as lateral faces. Explanation: In the pyramid diagram, the rectangles included the lateral surfaces. FINDING THE SURFACE AREA OF A PYRAMID Find the surface area of the regular pyramid. Question 4. The surface area of the triangle pyramid = 95 sq. m Explanation: The surface area of the triangle pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 65+(1/2) 5 x 12 surface area = 65 + 2.5 x 12 surface area = 65 + 30 surface area = 95 sq. m Question 5. The surface area of the triangle pyramid = 24 sq. cm Explanation: The surface area of the triangle pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 12+(1/2) 2 x 6 surface area = 12 + 1 x 12 surface area = 12 + 12 surface area = 24 sq. cm Question 6. WHICH ONE DOESN’T BELONG? Which description of the solid does not belong with the other three? Explain your reasoning. regular pyramid does not belong with the other three; Explanation: the remaining are square pyramid, rectangular pyramid, triangular pyramid are the three pyramids with different shapes. Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 7. A building in the shape of a square pyramid is covered with solar panels. The building has a slant height of 12 feet and a base with side lengths of 15 feet. The solar panels cost $70 per square foot to install. How much does it cost to install enough solar panels to cover the entire surface of the building? Answer:$ 270 is enough for solar planets to cover the entire surface of the building. Explanation: The building is in the shape of square pyramid. surface area = A + (1/2) ps area = 180 + (1/2) x 180 area = 180 + 90 area = 270 $Question 8. You use the glass pyramid shown to display rainbows on the walls of a room. The pyramid is regular and has a surface area of 105.35 square centimeters. Find the height of each triangular face. Justify your answer. Answer: ### Surface Areas of Pyramids Homework & Practice 10.3 Review & Refresh Find the surface area of the cylinder. Round your answer to the nearest tenth. Question 1. Answer: The surface area of the cylinder = 182.12 sq. ft Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 3 ft , height = 10 ft given surface area = 2 x 3.14 x 3 x 3 + 2 x 3.14 x 2 x 10 surface area = 56.52+ 125.6 surface area = 182.12 sq. ft Question 2. Answer: The surface area of the cylinder = 345.4 sq. m Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 5 m , height = 6 m given surface area = 2 x 3.14 x 5 x 5 + 2 x 3.14 x 5 x 6 surface area = 157+ 188.4 surface area = 345.4 sq. m Question 3. Answer: The surface area of the cylinder = 406.944 sq. mm Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 4mm , height = 12.2mm given surface area = 2 x 3.14 x 4 x 4+ 2 x 3.14 x 4 x 12.2 surface area = 100.48+ 306.464 surface area = 406.944 sq. mm Question 4. The ratio of the distance between bases on a professional baseball field to the distance between bases on a youth baseball field is 3 : 2. Bases on a professional baseball field are 90 feet apart. What is the distance between bases on a youth baseball field? A. 30 ft B. 45 ft C. 60 ft D. 135 ft Answer: The distance between bases on a youth baseball = 60 ft Explanation: The ratios between professional base and youth baseball = 3 : 2 already given 90 given so 60 ft is the distance between bases on a youth baseball field. Concepts, Skills, & Problem Solving USING A NET Use the net to find the surface area of the regular pyramid. (See Exploration 1, p. 421.) Question 5. Answer: The surface area of the square pyramid = 18 sq. in Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 12+(1/2) 4 x 3 surface area = 12 + 0.5x 12 surface area = 12 + 6 surface area = 18 sq. in Question 6. Answer: The surface area of the square pyramid = 88.3 sq. mm Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of th base, s = slant height surface area = 43.3+(1/2) 10 x 9 surface area = 43.3 + 0.5x 90 surface area = 43.3 + 45 surface area = 88.3 sq. mm Question 7. Answer: The surface area of the square pyramid = 79.9 sq. m Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 61.9 +(1/2) 6 x 6 surface area = 61.9 + 0.5x 36 surface area = 61.9 + 18 surface area = 79.9 sq. m FINDING THE SURFACE AREA OF A PYRAMID Find the surface area of the regular pyramid. Question 8. Answer: The surface area of the triangular pyramid = 81 sq. ft Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 54 +(1/2) 6 x 9 surface area = 54 + 0.5x 54 surface area = 54 + 27 surface area = 81 sq. ft Question 9. Answer: The surface area of the triangular pyramid = 36 sq. cm Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 24 +(1/2) 6 x 4 surface area = 24 + 0.5x 24 surface area = 24 + 12 surface area = 36 sq. cm Question 10. Answer: The surface area of the triangular pyramid = 36 sq. in Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 13 +(1/2) 15 x 10 surface area = 13 + 0.5x 150 surface area = 13 + 75 surface area = 88 sq. in Question 11. Answer: The surface area of the triangular pyramid = 126 sq. yd Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 7.8 +(1/2) 10 x 9 surface area = 7.8 + 0.5x 90 surface area = 7.8 + 48 surface area = 126 sq. yd Question 12. Answer: The surface area of the triangular pyramid = 13.5 sq. m Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 4.5 +(1/2) 4 x 4.5 surface area = 4.5 + 0.5x 18 surface area = 4.5 + 9 surface area = 13.5 sq. m Question 13. Answer: The surface area of the triangular pyramid = 600.4 sq. mm Explanation: The surface area of the triangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 440.4 +(1/2) 16 x 20 surface area = 440.4+ 0.5x 320 surface area = 440.4+ 160 surface area = 600.4 sq. mm Question 14. MODELING REAL LIFE The base of the lampshade is a regular hexagon with side lengths of 8 inches. Estimate the amount of glass needed to make the lampshade. Answer: The amount of glass needed to make the lampshade = 26.6666667 in Explanation: hexagon = 1/3 x b x h where base = 8 inches , height = 10 in given hexagon = (80/3) hexagon = 26.6666667 in Question 15. GEOMETRY The surface area of a square pyramid is 85 square meters. The side length of the base is 5 meters. What is the slant height? Answer: The slant height = 6 meters Explanation: The area of the base = side x side = 5 x 5 = 25 m area of the lateral face = (1/2)bh= (1/2)5h= 2.5 h m There are 4 identical lateral faces. area of lateral faces = 4(2.5h) = 10 hm surface area of regular pyramid = area of base + area of lateral faces 85 = 25 + 10 h 85-25 = 10 h 60 = 10 h h = (60/10) h = 6 meters FINDING SURFACE AREA Find the surface area of the solid. Question 16. Answer: The surface area of the square pyramid = 20 sq. ft Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 5 +(1/2) 5 x 6 surface area = 5 + 0.5x 30 surface area = 5 + 15 surface area = 20 sq. ft Question 17. Answer: The surface area of the square pyramid = 22 sq. cm Explanation: The surface area of the square pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 10 +(1/2) 4 x 6 surface area = 10 + 0.5x 24 surface area = 10 + 12 surface area = 22 sq. cm Question 18. Answer: The surface area of the rectangular pyramid = 22 sq. ft Explanation: The surface area of the rectangular pyramid = A + (1/2) ps where A = area of the base ,p = perimeter of the base, s = slant height surface area = 12 +(1/2) 4 x 5 surface area = 12 + 0.5x 20 surface area = 12 + 10 surface area = 22 sq. ft Question 19. GEOMETRY A tetrahedron is a triangular pyramid with four faces that are identical equilateral triangles. The total lateral surface area of a tetrahedron is 93 square centimeters. Find the surface area of the tetrahedron. Answer: The surface area of the tetrahedron = 124 square cm Explanation: Area of lateral face (equilateral triangle) = 93/3 = 31 The tetrahedron has four identical equilateral triangles = 4 X 31 =124 cm Question 20. PROBLEM SOLVING You are making an umbrella that is shaped like a regular octagonal pyramid. a. Estimate the amount of fabric that you need to make the umbrella. b. The fabric comes in rolls that are 60 inches wide. Draw a diagram of how you can cut the fabric from rolls that are 10 feet long. c. How much fabric is wasted? Answer: Question 21. REASONING The height of a pyramid is the perpendicular distance between the base and the top of the pyramid. Which is greater, the height of a pyramid or the slant height? Explain your reasoning. Answer: The height of the pyramid is greater . Explanation: In the above shown figure the slant height is less than the height of the pyramid. Question 22. DIG DEEPER! Both pyramids at the right have regular bases. a. Without calculating, determine which pyramid has the greater surface area. Explain. b. Verify your answer to part(a) by finding the surface area of each pyramid. Answer: The first figure has the greatest surface area. Explanation: The surface area of 1st pyramid = A + (1/2) ps surface area = 112 + (1/2) 8 x 14 surface area = 112 +(112/2) surface area = 56 +112 surface area = 168 in The surface area of 2nd pyramid = A + (1/2) ps surface area = 6.9 + (1/2) 8 x 14 surface area = 6.9 +(112/2) surface area = 56 +6.9 surface area = 62.9 sq. in Question 23. REASONING Is the total area of the lateral faces of a pyramid greater than, less than or equal to the area of the base? Explain. Answer: The lateral surface area of a regular pyramid is the sum of the area of its lateral faces. The total surface area of a regular pyramid is the sum of the areas of its lateral faces and its base. ### Lesson 10.4 Volumes of Prisms EXPLORATION 1 Finding a Formula for Volume Work with a partner. a. In the figures shown, each cube has a volume of 1 cubic unit. Compare the volume V (in cubic units) of each rectangular prism to the area B(in square units) of its base. What do you notice? b. Repeat part(a) using the prisms below. c. Use what you learned in parts (a) and (b) to write a formula that gives the volume of any prism. Answer: a. The 1st cube has a volume of 1 cubic unit. and it is increased by one horizontal in each cube. b. The 1st prism has a volume of 1 cubic unit and it is increased by one horizontal in each cube. c. Triangular prism = 2B + ph where b = base ,h = height, l = length, p = perimeter of base, B = area of base Rectangular prism = 2(lw +lh + wh) where l = length, w = width, h = height pentagonal prism = (1/2)(5s x a)h Hexagonal prism = v = bh octagonal prism = 2(1 +square root )a square The volume of a three-dimensional figure is a measure of the amount of space that it occupies. Volume is measured in cubic units. Try It Find the volume of the prism. Question 1. Answer: The volume of the prism = 16 cu. ft Explanation: The volume of the prism = bh where b = 4 ft , h = 4 ft given volume of the prism = 16 cu. ft Question 2. Answer: The volume of the prism = 51 cu. cm Explanation: The volume of the prism = bh where b = 8.5 cm , h = 6 cm given volume of the prism = 8.5 x 6 = 51 cu. cm Find the volume of the prism. Question 3. Answer: The volume of prism = 162 cu. m Explanation: The volume of triangular prism = a x b x c x h where a = b = c = base side , h = height prism = 12 x 9 x 12 x 5 volume = 60 + 102 volume = 162 cu. m Question 4. Answer: The volume of prism = 9 cu. m Explanation: The volume of triangular prism = a x b x c x h where a = b = c = base side , h = height prism = 0.75x 2 x 2 x 3 volume = 0.75 x12 volume = 9 cu. m Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. FINDING THE VOLUME OF A PRISM Find the volume of the prism. Question 5. Answer: The volume of the prism = 56 cu. in Explanation: volume v = B h v = 7 x 8 v = 56 cu. in Question 6. Answer: The volume of the prism = 75 cu. ft Explanation: volume v = B h v = 15 x 5 v = 75 cu. ft Question 7. Answer: The volume of the prism = 96 cu. yd Explanation: volume v = B h v = 12 x 8 v = 96 cu. yd Question 8. Answer: The volume of the prism = 4925 cu. mm Explanation: volume v = B h v = 197 x 25 v = 4925 cu. mm Question 9. OPEN-ENDED Draw and label a prism with a volume of 144 cubic inches. Justify your answer. Answer: v = b x h 144 = 12 x 12 so base = 7 and height = 7 inches. Explanation: Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 10. DIG DEEPER! You visit an aquarium. One of the tanks at the aquarium holds 450 gallons of water. Draw a diagram to show one possible set of dimensions of the tank. Justify your answer. (1 gal = 231 in.3) Answer: Question 11. A stack of paper contains 400 sheets. The volume of the stack is 140.25 cubic inches. Each sheet of paper is identical, with a length of 11 inches and a width of 8.5 inches. Find the height of each sheet of paper. Justify your answer. Answer: Each sheet of paper = 0.0075 inches. ### Volumes of Prisms Homework & Practice 10.4 Review & Refresh Find the surface area of the regular pyramid. Question 1. Answer: Surface area of the regular pyramid = 15 sq. m Explanation: Surface area of the regular pyramid = A + (1/2) ps area = 3 +(1/2) 3 x 8 surface area = 3 + 12 surface area = 15 sq. m Question 2. Answer: Surface area of the regular pyramid =550 sq. mm Explanation: Surface area of the regular pyramid = A + (1/2) ps area = 30 +(1/2) 20 x 26 surface area = 30 + 520 surface area = 550 sq. mm Question 3. Answer: Surface area of the regular pyramid =61 sq. cm Explanation: Surface area of the regular pyramid = A + (1/2) ps area = 7 +(1/2) 6 x 9 surface area = 7 + 54 surface area = 61 sq. cm Find the selling price. Question 4. Cost to store:$75 Markup: 20% Selling price = 1665 $Explanation: Selling price = cost to store x markup + cost to store selling price =$75 x 20% + $75 selling price =$75x 0.20 + $75 s p = 1665$ Question 5. Cost to store: $90 Markup: 60% Answer: Selling price = 5490$ Explanation: Selling price = cost to store x markup + cost to store selling price = $90 x 60% +$90 selling price = $90x 0.60+$90 s p = 5490 $Question 6. Cost to store:$130 Markup: 85% Selling price =10,530 $Explanation: Selling price = cost to store x markup + cost to store selling price =$130 x 85% + $130 selling price =$130 x 0.80+ $130 s p = 10,530$ Concepts, Skills, & Problem Solving USING TOOLS In the figure, each cube has a volume of 1 cubic unit. Find the volume of the figure and the area of its base. (See Exploration 1, p. 427.) Question 7. The volume of the cube = length + breadth + height 1 x 3 = 3 Explanation: The volume of the cube = length + breadth + height where length = l, width = w ,height = h Question 8. The volume of the cube = length + breadth + height 1 x 4 = 4 Explanation: The volume of the cube = length + breadth + height where length = l, width = w ,height = h Question 9. The volume of the cube = length + breadth + height 1 x 2 = 2 Explanation: The volume of the cube = length + breadth + height where length = l, width = w ,height = h FINDING THE VOLUME OF A PRISM Find the volume of the prism. Question 10. The volume of the prism = 81 cu. in Explanation: volume v = B h v = 9 x 9 v = 81 cu.in Question 11. The volume of the prism = 48 cu. cm Explanation: volume v = B h v = 6 x 8 v = 48 cu. cm Question 12. The volume of the prism = 48 cu. m Explanation: volume v = B h v = 7 x 8.5 v = 59.5 cu. m Question 13. The volume of the prism = 48 cu. yd Explanation: volume v = B h v = 8.33 x 6 v = 50 cu. yd Question 14. The volume of the prism = 54 cu.ft Explanation: volume v = B h v = 9 x 6 v = 54 cu.ft Question 15. The volume of the prism = 84 cu.mm Explanation: volume v = B h v = 10.5 x 8 v = 84 cu.mm Question 16. The volume of the prism = 48 cu.m Explanation: volume v = B h v = 10 x 4.8 v = 48 cu.m Question 17. The volume of the prism = 645 cu. mm Explanation: volume v = B h v = 15 x 43 v = 645 cu. mm Question 18. The volume of the prism =3320 cu. feet Explanation: volume v = B h v = 166 x 20 v = 3320 cu. ft Question 19. YOU BE THE TEACHER Yes my friend is correct. Explanation: Volume of triangular prism = B h volume = 10 x (5 x 7) volume = 10 x (35) volume = 350 cubic centimeters Question 20. MODELING REAL LIFE A battery for an underwater drone is in the shape of a square prism. It is designed to draw in seawater that is then used to produce energy. The base of the battery has side lengths of 15 centimeters and the height of the battery is 10 centimeters. Find the volume of the battery. The volume of the battery = 2250 cu. centimeters. Explanation: Volume of a square prism = a square h given that a = 15 cm, h = 10 cm volume = 15 x 15x 10 volume = 2250 cu. centimeters. Question 21. MODELING REAL LIFE A cereal box has a volume of 225 cubic inches. The length of the base is 9 inches and the width of the base is 2.5 inches. What is the height of the box? Justify your answer. The heightt of the box = 10 inches. Explanation: v = length x width x height v = 225 cubic inches, l = 9 inches,w = 2.5 in 225 = 9 x 2.5 x h 225 = 22.5 h h = 225/22.5 h = 10 inches Question 22. REASONING Each locker is shaped like a rectangular prism. Which has more storage space? Explain. The school locker has the more storage space. Explanation: 2. The surface area of the rectangular prism = 2(lw + lh + wh) area = 2(10 x 12) +(15 x 48) +(12 x 48) where l= 15 in,w = 12 in,h= 48 in area = 2(120) +(720) +(576) area = 2(1416) area = 2832 sq. in 1. The surface area of the rectangular prism = 2(lw + lh + wh) area = 2(15 x 12) +(10 x 60) +(12 x 60) where l= 10 in,w = 12 in,h= 60 in area = 2(120) +(600) +(720) area = 2(1440) area = 2880 sq. in Question 23. USING TOOLS How many cubic inches are in 1 cubic foot? Use a sketch to explain your reasoning. 1 cubic foot = 1728 cubic inches. Explanation: Question 24. PROBLEM SOLVING A concrete construction block has the measurements shown. How much concrete is used to make the block? Justify your answer. The concrete used to make the block = 544 sq. in Explanation: The surface area of the rectangular prism = 2(lw + lh + wh) area = 2(16 x 8) +(16 x 6) +(8 x 6) where l= 16in,w = 8 in,h= 6 in area = 2(128) +(96) +(48) area = 2(272) area = 544 sq. in Question 25. RESEARCH The gas tank is 20% full. Use the current price of regular gasoline in your community to find the cost to fill the tank. (1 gal = 231 in.3) Question 26. DIG DEEPER! Two liters of water are poured into an empty vase shaped like an octagonal prism. The base area is 100 square centimeters. What is the height of the water? (1 L = 1000 cm3) The height of the water = 20 centimeters Explanation: 1 liter = 1000 cubic centimeters. area of the base = 100 square cm height = h the volume of water in the vase = 2 liters = 2000 cubic centimeters the volume of water in the vase = area of base x-height of the prism 2000 = 100 h h = 20 cm Question 27. LOGIC Two prisms have the same volume. Do they always, sometimes never, or have the same surface area? Justify your answer. If the volume is the same, they do not have the surface area. Explanation: if we take two prisms at random that have the same volume, it’s very likely that they don’t have the same surface area. Question 28. CRITICAL THINKING How many times greater is the volume of a triangular prism when one of its dimensions is doubled? when all three dimensions are doubled? The volume is 8 times greater when we double all 3 dimensions. Explanation: we are doubling only one dimension means that you are multiplying by 2 only 1 time. ### Lesson 10.5 Volumes of Pyramids EXPLORATION 1 Finding a Formula for the Volume of a Pyramid Work with a partner. Draw the two nets on cardboard and cut them out. Fold and tape the nets to form an open cube and an open square pyramid. Both figures should have the same size square base and the same height. a. Compare the volumes of the figures. What do you notice? b. Use your observations in part(a) to write a formula for the volume of a pyramid. c. The rectangular prism below can be cut to form three pyramids. Use your formula in part(b) to show that the sum of the volumes of the three pyramids is equal to the volume of the prism. The volume of the pyramid = (1/3) Bh Explanation: a. the volume of the pyramid =(1/3) B h 1. v = (1/3) x 2 x2 v = (1/3) x 4 v = 1.33 in the volume of the pyramid =(1/3) B h 1. v = (1/3) x 2 x2.5 v = (1/3) x 5 v = 1.66 in b. the second figure has a greater volume than the 1st figure c. the volume of the pyramid = (1/3) B h the volume of the prism = Bh The volume of the 3 pyramids is equal to the volume of the prism. Try It Find the volume of the pyramid. Question 1. The volume of the pyramid = 42 cubic feet Explanation: The volume of the pyramid = (1/3) B h B = 21 square feet h= 6 ft volume = (1/3) x 21×6 volume = (126/3) volume = 42 cu. ft Question 2. The volume of the pyramid = 290 cubic centimeter Explanation: The volume of the pyramid = (1/3) B h B =174 square cm h= 5 cm volume = (1/3) x 174 x 5 volume = (870/3) volume = 290 cubic centimeter Find the volume of the pyramid. Question 3. The volume of the pyramid = 290 cubic centimeter Explanation: The volume of the pyramid = (1/3) B h B =18 in h= 7 in volume = (1/3) x 18x 7 volume = (126/3) volume = 42 inches Question 4. The volume of the pyramid = 91.66 square centimeter Explanation: The volume of the pyramid = (1/3) B h B =25 cm h= 11 cm volume = (1/3) x 25x 11 volume = (275/3) volume = 91.66 cm Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 5. WRITING How is the formula for the volume of a pyramid different from the formula for the volume of a prism? The volume of the pyramid = (1/3) B h where B = base of the pyramid h = height the volume of the prism = B h where B = base of the prism h= height Explanation: The volume of the pyramid is 3 times greater the volume of the prism. Question 6. PROBLEM SOLVING How many different pyramids can you draw with the same height and volume? Explain. We can draw the 2 or 3 pyramids with the same height and volume Explanation: We can draw the 2 or 3 pyramids with the same height and volume . it is our wish to draw as many as possible FINDING THE VOLUME OF A PYRAMID Find the volume of the pyramid. Question 7. The volume of the pyramid = cubic yard Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 10 and h= 6 v = (1/3) x 10 x 6 v = (60/3) v = 20 cubic yds Question 8. The volume of the pyramid = 12 cubic centimeters Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 4 and h= 9 v = (1/3) x 4 x 9 v = (36/3) v = 12 cu. cm Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 9. A resort features a square pyramid with a water slide. The length of the water slide is 90% of the height of the pyramid. The base of the pyramid has side lengths of 60 feet. The volume of the pyramid is 60,000 cubic feet. What is the length of the water slide? The length of the water slide = 50 cubic feet Question 10. DIG DEEPER! To make a candle, you use a mold to create the wax pyramid shown. You cut off the top 3 centimeters of the pyramid to make space for a wick. If the base area of the removed portion is 5.4 square centimeters, what percentage of the wax did you remove? The percentage of the wax we remove = 140 cubic centimeters. Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) x 60 x 7 volume =( 420/3) volume = 140 cubic centimeters. ### Volumes of Pyramids Homework & Practice 10.5 Review & Refresh Find the volume of the prism. Question 1. The volume of the prism = 189 cubic feets Explanation: The volume of the rectangular prism = l x w x h v = 9 x 7 x 3 v = 189 cubic feets Question 2. The volume of the prism = 189 cubic centimeters Explanation: The volume of the triangular prism = (b x h x l)/ 2 v = (5 x 3 x 8)/2 v = 60 cubic centimeters Solve the inequality. Graph the solution. Question 3. r + 0.5 < – 0.4 r = – 0.9 Explanation: r = -0.4 -0.5 r = -0.9 Question 4. z – 2.4 ≥ – 0.6 z = 1.8 Explanation: z = -0.6 (+ 2.4) z = 1.8 Question 5. h – 5 ≤ – 3.7 h = -1.3 Explanation: h = -3.7 +5 h = -1.3 Concepts, Skills, & Problem Solving VOLUMES OF PYRAMIDS The rectangular prism is cut to form three pyramids. Show that the sum of the volumes of the three pyramids is equal to the volume of the prism. (See Exploration 1, p. 433.) Question 6. The volume of the 3 pyramids is equal to the volume of the prism. Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 6 and h= 4 v = (1/3) x 6 x 4 v = (24/3) v = 8 cubic feet the volume of the pyramid is 3 times greater than volume of the prism. The volume of the prism = B h where B = base of the prism, h = height v = B xh where B = 6 and h= 4 v = 6 x 4 v = 24 8 x 3 = 24 Question 7. The volume of the 3 pyramids is equal to the volume of the prism. Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 6 and h= 6 v = (1/3) x 6 x 6 v = (36/3) v = 12 inches the volume of the pyramid is 3 times greater than volume of the prism. The volume of the prism= B h where B = base of the prism, h = height v = B xh where B = 6 and h= 6 v = 6 x 6 v = 36 in 12 x 3 = 36 FINDING THE VOLUME OF A PYRAMID Find the volume of the pyramid. Question 8. The volume of the pyramid = 0.66 cubic feet Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 1 and h= 2 v = (1/3) x 1 x 2 v = (2/3) v = 0.66 cubic feet Question 9. The volume of the pyramid = 6.66 cubic feet Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 2 and h= 10 v = (1/3) x 10 x 2 v = (20/3) v = 6.66 cubic feet Question 10. The volume of the pyramid = 16 cubic feet Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 8 and h= 7 v = (1/3) x 8 x 7 v = (48/3) v = 16 cubic feet Question 11. The volume of the pyramid = 20 cubic millimeters Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 15 mm and h= 4 v = (1/3) x 15 x 4 v = (60/3) v = 20 cubic mm Question 12. The volume of the pyramid = 10.666 cu. yds Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 4 and h= 8 v = (1/3) x 4 x 8 v = (32/3) v = 10.666 cu. yd Question 13. The volume of the pyramid = 36 cu. in Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 6 and h= 8 v = (1/3) x 6 x 8 v = (48/3) v = 36 cu. in Question 14. The volume of the pyramid = 70 cu. mm Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 14 and h = 15 mm v = (1/3) x 14x 15 v = (210/3) v = 70 cu. mm Question 15. The volume of the pyramid = 2.333 cubic centimeters Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 1 and h = 7 cm v = (1/3) x 1 x 7 v = (7/3) v = 2.333 cubic centimeters Question 16. The volume of the pyramid = 252 cubic millimeters Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 63 and h = 12 mm v = (1/3) x 63 x 12 v = (756/3) v = 252 cubic millimeters Question 17. YOU BE THE TEACHER No my friend is not correct. Explanation: The volume of the pyramid = (1/3) x B x h volume = (8 x 4 x 7)/3 volume = (224/3) volume = 74.666 cu. in Question 18. MODELING REAL LIFE A researcher develops a cage for a living cell in the shape of a square-based pyramid. A scale model of the cage is shown. What is the volume of the model? The volume of the model = 133.33 cu. millimeters Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B = 20 and h = 20 mm v = (1/3) x 20 x 20 v = (400/3) v = 133.33 cu. millimeters Question 19. FINDING VOLUME The volume of the compositte solid =48 cu. feet Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =36 and h = 4 ft v = (1/3) x 36 x 4 v = (144/3) v = 48 cu. feet Question 20. MODELING REAL LIFE In 1483, Leonardo da Vinci designed a parachute. It is believed that this was the first parachute ever designed. In a notebook, he wrote, “If a man is provided with a length of gummed linen cloth with a length of 12 yards on each side and 12 yards high, he can jump from any great height whatsoever without injury.” Find the volume of air inside Leonardo’s parachute. The volume of the air inside parachute =48 cu. yd Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =12 and h = 12 v = (1/3) x 12 x 12 v = (144/3) v = 48 cu. yds Question 21. MODELING REAL LIFE Which sandcastle spire has a greater volume? How much more sand do you need to make the spire with the greater volume? spire B has greater volume 4 % of sand is needed to make the spire Explanation: The volume of the spire A = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =30 and h = 6 v = (1/3) x 30 x 6 v = (180/3) v = 60 in The volume of the spire B = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =24 and h = 8 v = (1/3) x 24 x 8 v = (192/3) v = 64 cu. in Question 22. PROBLEM SOLVING Use the photo of the tepee. a. What is the shape of the base? How can you tell? b. The tepee’s height is about 10 feet. Estimate the volume of the tepee. a. The shape of the base is a triangular pyramid . b . The volume of the triangular pyramid = 3.33 Explanation: b . The volume of the triangular pyramid = (1/3) B h volume = (1/3) B 10 volume = (10 b /3) 10 B = 3 B = (10/3) B = 3.33 Question 23. OPEN-ENDED A rectangular pyramid has a volume of 40 cubic feet and a height of 6 feet. Find one possible set of dimensions of the base. The dimensions of the base = 20 feet Explanation: The volume of the rectangular pyramid = (1/3) xB x h volume = (1/3) x B x 6 where h = 6 ,v= 40 given 40 = (1/3) x 6 B 2 B = 40 B = 20 feet Question 24. REASONING Do the two solids have the same volume? Explain. No, the two solids did not have the same volume. Explanation: The volume of rectangular prism = l wh volume = xyz The volume of the triangular prism = b hl/2 volume = (xy3z/2) ### Lesson 10.6 Cross Sections of Three-Dimensional Figures EXPLORATION 1 Describing Cross Sections Work with a partner. A baker is thinking of different ways to slice zucchini bread that is in the shape of a rectangular prism. The shape that is formed by the cut is called a cross section. a. What is the shape of the cross section when the baker slices the bread vertically, as shown above? b. What is the shape of the cross section when the baker slices the bread horizontally? c. What is the shape of the cross section when the baker slices off a corner of the bread? d. Is it possible to obtain a cross section that is a trapezoid? Explain. e. Name at least 3 cross sections that are possible to obtain from a rectangular pyramid. Explain your reasoning. Consider a plane “slicing” through a solid. The intersection of the plane and the solid is a two-dimensional shape called a cross section. For example, the diagram shows that the intersection of the plane and the rectangular prism is a rectangle. Try It Describe the intersection of the plane and the solid. Question 1. The intersection of the plane of the solid = one dimensional Question 2. The intersection of the plane of the solid = two dimensional Describe the intersection of the plane and the solid. Question 3. The intersection of the plane of the cylinder = two dimensional Question 4. The intersection of the plane of the cone = two dimensional Self-Assessment for Concepts & Skills Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 5. VOCABULARY What is a cross section? The intersection of the plane of the solid is a two-dimensional is called crosssection. Question 6. DESCRIBING CROSS SECTIONS Describe the intersection of the plane and the solid at the left. The intersection of the plane and the solid at the left is a two dimensional. Question 7. REASONING Name all possible cross sections of a cylinder. The cross-section of the sphere is a circle. The vertical cross-section of a cone is a triangle, and the horizontal cross- section is a circle. Question 8. WHICH ONE DOESN’T BELONG? You slice a square prism. Which cross section does not belong with the other three? Explain your reasoning. circle crosssection does not belong with the other three. Explanation: square, triangle, the rectangle does not belong with the three. Self-Assessment for Problem Solving Solve each exercise. Then rate your understanding of the success criteria in your journal. Question 9. A steel beam that is 12 meters long is cut into four equal parts. The cross sections are rectangles with side lengths of 1 meter and 2 meters. a. What is the perimeter of each cross section? b. What is the area of each cross section? c. What is the volume of the original beam? a. The perimeter of each cross-section = 3 meters b. The area of each cross-section = 2 meters c. The volume of the original beam = 12 meters. Question 10. DIG DEEPER! A lumberjack saws a cylindrical tree trunk at an angle. Is the cross-section a circle? Explain your reasoning. Yes, the cross-section is a circle. Explanation: The cross-section of the sphere is a circle ### Cross Sections of Three-Dimensional Figures Homework & Practice 10.6 Review & Refresh Find the volume of the pyramid. Question 1. The volume of the pyramid = 37.33 cu. in Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =16 and h = 7 in v = (1/3) x16 x 7 v = (112/3) v = 37.33 cu. in Question 2. The volume of the pyramid = 2.875 cubic centimeters. Explanation: The volume of the pyramid = (1/3) B h where B = base of the pyramid, h = height v = (1/3) x B xh where B =23 and h = 8 v = (1/3) x23 x 8 v = (112/3) v = 2.875 cubic centimeters. Find the sum. Question 3. (w – 7) + (- 6w – 5) w = -2 Explanation: w = -11w+7 w=-4w 2w = -4 w= -2 Question 4. (8 – b) + (5b + 6) b = 2.33 Explanation: -b = 5b-2 5b + 6b = 2 b = 2.33 Concepts, Skills, & Problem Solving DESCRIBING CROSS SECTIONS Determine whether it is possible to obtain the cross section from a cube. (See Exploration 2, p. 439.) Question 5. circle No it is not possible to obtain the crosssection from a cube. Question 6. square No it is not possible to obtain the crosssection from a cube. Question 7. equilateral triangle No it is not possible to obtain the crosssection from a cube. Explanation: equilateral triangle is not possible to obtain the crosssection from a cube. Question 8. pentagon No it is not possible to obtain the crosssection from a cube. Explanation: pentagon is not possible to obtain the crosssection from a cube. Question 9. non-rectangular parallelogram No it is not possible to obtain the crosssection from a cube. Explanation: non rectangular parallelogram is not possible to obtain the crosssection from a cube. Question 10. octagon No it is not possible to obtain the crosssection from a cube. Explanation: octagon is not possible to obtain the crosssection from a cube. DESCRIBING CROSS SECTIONS OF PRISMS AND PYRAMIDS Describe the intersection of the plane and the solid. Question 11. The intersection of the prism and the solid is a two dimensional. Question 12. The intersection of the pyramid and the solid is a one dimensional. Question 13. The intersection of the pyramid and the solid is a two dimensional. Question 14. The intersection of the pyramid and the solid is a three dimensional. DESCRIBING CROSS SECTIONS OF CYLINDERS AND CONES Describe the intersection of the plane and the solid. Question 15. The intersection of the plane and the solid is a circle Explanation: The cross section of the cylinder is a circle. Question 16. The intersection of the plane and the solid is a circle Explanation: The cross section of the cone is a circle. DESCRIBING CROSS SECTIONS Describe the shape that is formed by the cut in the food. Question 17. The shape that is formed is circle. Explanation: The cross section when it is cut it is formed a circle. Question 18. The shape that is formed is semi-circle. Explanation: The cross section when it is cut it is formed a semi circle. Question 19. The shape that is formed is circle. Explanation: The cross section when it is cut it is formed circle. Question 20. DESCRIBING CROSS SECTIONS Describe the intersection of the plane and the cylinder. The intersection of a plane and a cylinder is a rectangle. REASONING Determine whether the given intersection is possible. If so, draw the solid and the cross section. Question 21. The intersection of a plane and a cone is a rectangle. No the intersection of a plane and a cone is a circle. Explanation: Question 22. The intersection of a plane and a square pyramid is a triangle. Yes the intersection of a plane and a square pyramid is a triangle. Explanation: Question 23. REASONING A plane that intersects a prism is parallel to the bases of the prism. Describe the intersection of the plane and the prism. when a plane intersects a prism and is parallel to the bases of the prism , the intersection is the same shape as the base. Question 24. REASONING Explain how a plane can be parallel to the base of a cone and intersect the cone at exactly one point. A plane will be parallel to the base and intersecting the cone at only one point is only possible when the plana will pass through Question 25. DIG DEEPER! An artist plans to paint bricks. a. Find the surface area of the brick. b. The artist cuts along the length of the brick to form two bricks, each with a width of 2 inches. What is the percent of increase in the surface area? Justify your answer. a. The surface area of the brick = 164 in b. The surface area of the percent increase = 2 % Explanation: b.The surface area of the rectangular prism = 2(lw +lh +wh) surface area = 2(12 x 6 +12 x5 +6 x 5) surface area = 2(72 + 60+30) surface area = 2(162) surface area = 81 in a. The surface area of the rectangular prism = 2(lw +lh +wh) surface area = 2(10 x 4 +10 x3 +4 x 3) surface area = 2(40 + 30 +12) surface area = 2(82) surface area = 164 in Question 26. MODELING REAL LIFE A cross section of an artery is shown. a. Describe the cross section of the artery. b. The radius of the artery is 0.22 millimeter. What is the circumference of the artery? a. The cross section of the artery is a circle. b. The circumference of the artery = 1.3816 millimeters Explanation: The circumference of the circle = 2πr where r= 0.22 mm given π = 3.14 circumference = 2 x 3.14 x 0.22 circumference= 1.3816 millimeters Question 27. REASONING Three identical square pyramids each with a height of meters and a base area of 100 square meters are shown. For each pyramid, a cross section parallel to the base is shown. Describe the relationship between the area of the base and the area of any cross section parallel to the base. The relationship between the area of the base and the area of any cross section parallel to the base for 1st figure = 1: 2 % The relationship between the area of the base and the area of any cross section parallel to the base for 2nd figure = 1: 5% The relationship between the area of the base and the area of any cross section parallel to the base for 3rd figure = 1: 10% Explanation: In the above 3 figures the base area is same for all the 3 figures = 100 square meters ### Surface Area and Volume Connecting Concepts Using the Problem-Solving Plan Question 1. A store pays $2 per pound for popcorn kernels. One cubic foot of kernels weighs about 45 pounds. Wha tis the selling price of the container shown when the markup is 30%? Understand the problem. You are given the dimensions of a container of popcorn kernels and the price that a store pays for the kernels. You also know the weight of one cubic foot of popcorn kernels. You are asked to find the selling price of the container when the markup is 30%. Make a plan. Use the volume of the container to find the weight of the kernels. Then use the weight of the kernels to find the cost to the store. Finally, use the percent markup to find the selling price of the container. Solve and check. Use the plan to solve the problem. Then check your solution. Answer: Selling price = 525$ Explanation: Selling price = cost to store x markup + cost to store selling price = $2 x 30% +$45 selling price = $2x 0.30 +$45 selling price = $525 Question 2. The pyramid shown has a square base. What is the height of the pyramid? Justify your answer. Answer: The height of the pyramid = 3000 cm Explanation: The volume of the triangular pyramid = (1/3) x B h 1500 = (1/3) x 1.5 h 1500 = 0.5 h h = (1500/3) h = 3000 cm Question 3. A cylindrical can of soup has a height of 7 centimeters and a lateral surface area of 63π square centimeters. The can is redesigned to have a lateral surface area of 45π square centimeters without changing the radius of the can. What is the height of the new design? Justify your answer. Answer: The height of the new design can = 5 centimeters. Explanation: The can when it is redesigned to have a lateral surface area of 45π = 5 x 9 = 45 The height of the can when it is redesigned = 5 centimeters. Performance Task Volumes and Surface Areas of Small Objects At the beginning of this chapter, you watched a STEAM Video called “Paper Measurements.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task. ### Surface Area and Volume Chapter Review Review Vocabulary Write the definition and give an example of each vocabulary term. Graphic Organizers Information Frame You can use an to help organize and remember a concept. Here is an example of an Information Frame for Surface Areas of Rectangular Prisms. Choose and complete a graphic organizer to help you study the concept. 1. surface areas of prisms 2. surface areas of cylinders 3. surface areas of pyramids 4. volumes of prisms 5. volumes of pyramids 6. cross sections of three-dimensional figures Chapter Self-Assessment As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal. 10.1 Surface Areas of Prisms (pp. 409–414) Learning Target: Find the surface area of a prism. Find the surface area of the prism. Question 1. Answer: The surface area of the Rectangular prism = 158 sq. in Explanation: The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 8 w = 3 h = 5 rectangular prism =2(8 x 3) +(8 x 5) +(3 x 5) prism = 2(24) + (40) +(15) surface area =2(79) surface area = 158 sq. in Question 2. Answer: The surface area of the triangular prism = 562 sq. cm Explanation: The surface area of the triangular prism = 2lw + 2lh + 2wh surface area = 2(7 x 8) + 2(8 x 15) +2(15×7) where l = 8m,w = 8m, h= 8m surface area = 2(56) + 2(120)+ 2(105) surface area =112 + 240 + 210 surface area = 562 sq. cm Question 3. Answer: The surface area of the triangular prism = 262 sq. m Explanation: The surface area of the triangular prism = 2lw + 2lh + 2wh surface area = 2(5 x 7) + 2(5x 8) +2(8×7) where l = 5m,w = 7m, h= 8m surface area = 2(35) + 2(40)+ 2(56) surface area =70 + 80 + 112 surface area = 262 sq. m Question 4. You want to wrap the box using a piece of wrapping paper that is 76 centimeters long by56 centimeters wide. Do you have enough wrapping paper to wrap the box? Explain. Answer: The enough wrapping paper to wrap the box = 4180 cm Explanation: The surface area of the Rectangular prism = 2(lw + lh +wh) where l = 35 w = 50 h = 4 rectangular prism =2(35 x 50) +(50 x 4) +(35 x 4) prism = 2(1750) + (200) +(140) surface area =2(2090) surface area = 4180 sq. cm Question 5. To finish a project, you need to paint the lateral surfaces of a cube with side length 2.5 inches. Find the area that you need to paint. Answer: 10.2 Surface Areas of Cylinders (pp. 415–420) Learning Target: Find the surface area of a cylinder. Find the surface area and lateral surface area of the cylinder. Round your answers to the nearest tenth. Question 6. Answer: The surface area of the cylinder = 169.56 square yards The lateral surface area of the cylinder = 113.04 square yards Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 3 , height = 6 given surface area = 2 x 3.14 x 3 x 3 + 2 x 3.14 x 3 x 6 surface area = 56.52+ 113.04 surface area = 169.56 The lateral surface area of the cylinder = 2πrh where radius = 3 , height = 6 given surface area = 2 x 3.14 x 3 x 6 surface area = 113.04 sq. yards Question 7. Answer: The surface area of the cylinder = 34.196 square centimeters The lateral surface area of the cylinder =30.144 square centimeters Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 0.8 , height = 6 given surface area = 2 x 3.14 x 0.8 x 0.8 + 2 x 3.14 x 0.8 x 6 surface area = 4.0192+ 30.144 surface area = 34.196 square centimeters The lateral surface area of the cylinder = 2πrh where radius = 0.8 , height = 6 given surface area = 2 x 3.14 x 0.8 x 6 surface area = 30.144 square centimeters Question 8. The label covers the entire lateral surface area of the can. How much of the can is not covered by the label? Answer: The can that is not covered =276.32 square centimeters Explanation: The lateral surface area of the cylinder = 2πrh where radius = 4 , height = 11 given surface area = 2 x 3.14 x 4 x11 surface area = 276.32 square centimeters 10.3 Surface Areas of Pyramids (pp. 421–426) Learning Target: Find the surface area of a pyramid. Find the surface area of the regular pyramid. Question 9. Answer: The surface area of the pyramid = 5 square inches. Explanation: The surface area of the pyramid = A + (1/2) ps surface area = 2 +(1/2) x 3 x 2 surface area= 2 +3 surface area= 5 sq. in Question 10. Answer: The surface area of the pyramid = 42.5 sq. meters Explanation: The surface area of the pyramid = A + (1/2) ps surface area = 8 +(1/2) x 10 x 6.9 surface area= 8 +34.5 surface area= 42.5 sq. m Question 11. Answer: The surface area of the pyramid = 115.8 sq. cm Explanation: The surface area of the pyramid = A + (1/2) ps surface area = 84.3+(1/2) x 9 x 7 surface area= 84.3 +31.5 surface area= 115.8 sq. cm Question 12. The tent is shaped like a square pyramid. There is no fabric covering the ground. a. Estimate the amount of fabric needed to make the tent. b. Fabric costs$5.25 per square yard. How much will it cost to make the tent? a. The amount of fabric needed to make the tent = 12 ft b. The cost to make the tent = Explanation: The surface area of the pyramid = A + (1/2) ps surface area = 3+(1/2) x 3 x 6 surface area= 3 +9 surface area= 12 sq. ft 10.4 Volumes of Prisms (pp. 427–432) Learning Target: Find the volume of a prism. Find the volume of the prism. Question 13. The volume of prism = 96 cu. in Explanation: The volume of prism = lwh where l = 8,w=2,h = 6 volume = 8 x 2 x 6 volume = 96 cu. in Question 14. The volume of prism = 240 cu. m Explanation: The volume of prism =( lbh/2) where l = 4,w=8,h = 7.5 volume = 8 x 4 x 7.5 volume = 240 cu. m Question 15. The volume of prism =607.5 cu. mm Explanation: The volume of prism = lwh where l = 15,w=4.5,h = 9 volume = 15 x 4.5 x 9 volume = 607.5 cu. mm Question 16. The volume of prism =96 cu. m Explanation: The volume of prism = lwh where l = 6,w=3,h = 4 volume = 6 x 3 x 4 volume = 96 cu. m Question 17. The volume of prism = 15.6 cu. cm Explanation: The volume of prism = lwh where l = 2.6 ,w=1.5 ,h = 4 volume = 2.6 x1.5 x 4 volume = 15.6 cu. cm Question 18. The volume of prism = 105 cubic feet Explanation: The volume of prism =( lbh/2) where l = 7 ,w=5 ,h = 3 volume = 7 x 5 x 3 volume = 105 cu. feet Question 19. Two cereal boxes each hold exactly 192 cubic inches of cereal. Which box should a manufacturer choose to minimize the amount of cardboard needed to make the cereal boxes? The first cereal box is used to minimize 10.5 Volumes of Pyramids (pp.433-438) Learning Target: Find the volume of a pyramid. Find the volume of the pyramid. Question 20. The volume of the pyramid = 133.33 cu. ft Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 17 20 volume = (340/3) volume = 113.33 cu. ft Question 21. The volume of the pyramid = 2100 cubic in Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 210 x 30 volume = (6300/3) volume = 2100 cubic in Question 22. The volume of the pyramid = 48 cu. mm Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 16 x 9 volume = (144/3) volume =48 cu. mm Question 23. A pyramid-shaped hip roof is a good choice for a house in an area with many hurricanes. a. What is the volume of the roof to the nearest tenth of a foot? b. What is the volume of the entire house, including the roof? a. The volume of the pyramid = 800 cu. ft Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 80 x 30 volume = (2400/3) volume =800 cu. ft b. The volume of the pyramid = 400 cu. ft Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 40 x 30 volume = (1200/3) volume =400 cu. ft Question 24. A laboratory creates calcite crystals for use in the study of light. The crystal is made up of two pieces of calcite that form a square pyramid. The base length of the top piece is 2 inches. a. Find the volume of the entire pyramid. b. Find the volume of each piece of the pyramid. a. The volume of the entire pyramid = 1.58 cu. in Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 3.5 x 3 volume = (4.75/3) volume =1.58 cu. in b.The volume of the pyramid = 1.458 cu. in Explanation: The volume of the pyramid = (1/3) B h volume = (1/3) 3.5 x 1.25 volume = (4.375/3) volume =1.458 cu. in 10.6 Cross Sections of Three-Dimensional Figures (pp. 439–444) Learning Target: Describe the cross sections of a solid. Describe the intersection of the plane and the solid. Question 25. The intersection of the plane of the solid = two dimensional The cross section of a solid = rectangular prism Question 26. The intersection of the plane of the solid = two dimensional The cross section of a solid = triangular prism Sketch how a plane can intersect with a cylinder to form a cross section of the given shape. Question 27. rectangle Question 28. circle Question 29. line segment ### Surface Area and Volume Practice Test Find the surface area of the prism or regular pyramid. Question 1. The surface area of the pyramid = 7 ft Explanation: The surface area of the pyramid = (1/3) x B h surface area = (1/3) X 7 x 3 surface area = (21/3) surface area = 7 sq. ft Question 2. The surface area of the pyramid = 0.66 sq. in Explanation: The surface area of the pyramid = (1/3) x B h surface area = (1/3) X 1 x 2 surface area = (2/3) surface area = 0.66 sq. in Question 3. The surface area of the pyramid = 47.5 sq. m Explanation: The surface area of the pyramid = (1/3) x B h surface area = (1/3) X 9.5 x 15 surface area = (142.5/3) surface area = 47.5 sq. m Find the surface area and lateral surface area of the cylinder. Round your answers to the nearest tenth. Question 4. The surface area of the cylinder = 62.8 sq. cm The lateral surface area of the cylinder = 37.68 sq. cm Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 2 , height = 3 given surface area = 2 x 3.14 x 2 x 2 + 2 x 3.14 x 2 x 3 surface area = 25.12+ 37.68 surface area = 62.8 sq. cm The lateral surface area of the cylinder = 2πrh where radius = 2 , height = 3 given surface area = 2 x 3.14 x 2 x 3 surface area = 37.68 sq. cm Question 5. The surface area of the cylinder = 1623.38 sq. in The lateral surface area of the cylinder = 863.5 sq. in Explanation: The surface area of the cylinder = 2πr2 + 2πrh where radius = 11 , height = 12.5 given surface area = 2 x 3.14 x 11 x 11 + 2 x 3.14 x 11 x 12.5 surface area = 759.88+ 863.5 surface area = 1623.38 sq. in The lateral surface area of the cylinder = 2πrh where radius = 11 , height = 12.5 given surface area = 2 x 3.14 x 11 x 12.5 surface area = 863.5 sq. in Find the volume of the solid. Question 6. Volume of the prism = 324 cu. in Explanation: Volume of the prism =( bhl/2) volume = (9 x 12 x 6/2) volume =(648 /2) volume = 324 cu. in Question 7. Volume of the prism = 41.6 cu. yd Explanation Volume of the prism =lwh volume = (4 x 2 x 5.2) volume =(41.6) Question 8. Volume of the prism = 72 cu. m Explanation: Volume of the prism =( bhl/2) volume = (3 x `8 x 6/2) volume =(144/2) volume = 72 cu. m Question 9. A quart of paint covers 80 square feet. How many quarts should you buy to paint the ramp with two coats? (Assume you will not paint the bottom of the ramp.) Volume of the prism = 2914.8 cu. ft Explanation: Volume of the prism =lwh volume = (14x 34.7 x 6) volume =2914.8 cu. ft Question 10. A manufacturer wants to double the volume of the graham cracker box. The manufacturer will either double the height or double the width. a. What is the volume of the new graham cracker box? c. A graham cracker takes up about 1.5 cubic inches of space. Write an inequality that represents the numbers of graham crackers that can ﬁt in the new box. a. the volume of the new graham cracker box = 108 cu. in Explanation: Volume of the prism =lwh volume = (9 x 2 x 6) volume = 108 cu. in Question 11. The label on the can of soup covers about 354.2 square centimeters. What is the height of the can? Round your answer to the nearest whole number. The height of the can = 10,454.5672 cm Explanation: The lateral surface area of the cylinder = 2πrh where radius = 4.7 , height = 354.2 given surface area = 2 x 3.14 x 4.7 x 354.2 surface area = 10,454.5672 cm Question 12. A lumberjack splits the cylindrical log from top to bottom with an ax, dividing it in half. Describe the shape that is formed by the cut. cylinder. Explanation: The shape that formed by the cut = cylinder ### Surface Area and Volume Cumulative Practice Question 1. A gift box and its dimensions are shown. What is the least amount of wrapping paper that you need to wrap the box? A. 20 in.2 B. 56 in.2 C. 64 in.2 D. 112 in.2 option C is correct. Explanation: The least amount of wrapping paper that need to wrap the box = l x w x h 8 x 4 x 2 64 Question 2. James is getting ready for wrestling season. As part of his preparation, he plans to lose 5% of his body weight. James currently weighs 160 pounds. How much will he weigh, in pounds, after he loses 5% of his weight? he weigh after he loses 5 % = 128 Question 3. How far will the tip of the hour hand of the clock travel in 2 hours? (Use $$\frac{22}{7}$$ for π.) F. 44 mm G. 88 mm H. 264 mm I. 528 mm option G is correct Explanation: The circumference of circle = 2 π r h Question 4. Which value of x makes the equation true? 5x – 3 = 11 A. 1.6 B. 2.8 C. 40 D. 70 option B is correct Explanation: 5x – 3 = 11 5x = 11 + 3 5x = 14 x = (14/3) x = 2.8 Question 5. A hockey rink contains 5 face-off circles. Each of these circles has a radius of 15 feet. What is the total area of all the face-off circles? (Use 3.14 for π.) F. 706.5 ft 2 G. 2826 ft2 H. 3532.5 ft2 I. 14,130 ft2 Option G is correct Explanation: The surface area of circle = 2rhπ circle = 2 x 15 x 5 x 3.14 Question 6. How much material is needed to make the popcorn container? A. 76π in.2 B. 84π in.2 C. 92π in.2 D. 108π in.2 The material needed to make the popcorn container = 76π in.2 Explanation: The lateral surface area = 2πrh surface area = 2 x 9.5 x 4 π surface area = 76π in.2 Question 7. What is the surface area of the square pyramid? F. 24 in.2 G. 96 in.2 H. 132 in.2 I. 228 in.2 Option g is correct Explanation: surface area = 2 x l x h surface area = 2 x 8 x 6 surface area = 96 square inches Question 8. A rectangular prism and its dimensions are shown. What is the volume, in cubic inches, of a rectangular prism whose dimensions are three times greater? The volume of rectangular prism = 24 cu. in Explanation: The volume of rectangular prism l w h volume = 4 x 3 x 2 volume = 24 cu. in Question 9. What is the value of x? A. 20 B. 43 C. 44 D. 65 option A is correct Explanation: (2x + 4) = 46 2x = (46 – 4) 2x = 42 x = 21 Question 10. Which of the following are possible angle measures of a triangle? F. 60°, 50°, 20° G. 40°, 80°, 90° H. 30°, 60°, 90° I. 0°, 90°, 90° option H is the correct Explanation: The angles of a triangle = 30°, 60°, 90° Question 11. The table shows the costs of buying matinee movie tickets. Part A Graph the data. Part B Find and interpret the constant of proportionality for the graph of the line. Part C How much does it cost to buy 8 matinee movie tickets?
## Tuesday, December 6, 2011 ### More Linear Algebra: Eigenvectors In the previous instalments on linear algebra, we saw how one can obtain the characteristic polynomial from a matrix as well as the eigenvalues. We now want to use past examples to show how the eigenvectors can be obtained. Consider again the matrix given in the first problem: A = (1.....i) (-i.....1) for which we found the eigenvalues were: E1 = 0, E2 = 2, viz. http://brane-space.blogspot.com/2011/12/solutions-to-linear-algebra-problems.html To get the eigenvectors is just straightforward and merely requires obtaining simultaneous equations in x, y for example - based on using the rows in the matrix and applying each of the eigenvalues to them. For example, take E1 = 0, then the resulting equations are: x - iy = 0 ix + y = 0 or x = -iy and y = ix The eigenvector is easily solved for and is: v1 = [-i] [1] Next, take the eigenvalue E2 = 2, then the simultaneous equations from the matrix A are: x + iy = 2x -ix + y = 2y which yields: x = iy and y = -ix Or, an eigenvector of: v2 = [1] [-i] Consider now problem (2) which featured, A = (1.. …. .2) (2.......-2) And for which we found: E1 = -3, and E2 = 2 To get the eigenvector associated with the eigenvalue E1 = -3, we form the left side of the algebraic equations in x, and y using A such that: x + 2 y = -3x 2x - 2y = -3y Simplifying: 4x + 2y = 0 2x + y = 0 or: 4v1 + 2v2 = 0 2v1 + v2 = 0 and solving the simultaneous eqn. yields: v2 = - 2v1, or v = [1] [-2] For E2 = 2, we may write: x + 2y = 2x 2x -2y = 2y Again, the eigenvalue (E2) is always multiplied by x and then y to give the column matrix comprising the right side, with x-value on top, and y-value on the bottom. Simplifying the preceding equations: -x + 2y = 0 2x - 4y = 0 Or: -v1 + 2v2 = 0 2v1 - 4v2 = 0 which yields: v1 = 2v2, so the eigenvector in this case is: v = [2] [1] Problems: 1) Find the eigenvectors associated with the matrix A in Problem (3), e.g. for A = (3.. ……2) (-2...... 3) with the eigenvalues: E1 = 3 + 2i, and E2 = 3 - 2i (2) Given the matrix: A = (2.......4) (5.......3) Find: a) the characteristic polynomial, b) the eigenvalues associated with it, and c) the eigenvectors.
# Distance Between Two Points 3d7 min read Oct 16, 2022 5 min ## Distance Between Two Points 3d7 min read Distance between two points 3d is the distance between two points in three-dimensional space. The distance between two points can be measured using a variety of metrics, including Euclidean distance, Manhattan distance, and Chebyshev distance. The Euclidean distance between two points is the straight-line distance between them. The Manhattan distance between two points is the distance between them measured in terms of the number of squares it would take to walk between them. The Chebyshev distance between two points is the maximum distance between them, measured in terms of the sum of the absolute values of the differences between their coordinates. The distance between two points can be used to calculate the volume of a three-dimensional object or the area of a three-dimensional surface. It can also be used to calculate the distance between a point and a plane or the distance between two planes. ## How do you find the distance between two points in 3D? There are a few different ways to find the distance between two points in 3D. One way is to use a coordinate system. A coordinate system uses x, y, and z coordinates to locate points in 3D space. The distance between two points can be calculated using the Pythagorean theorem. The Pythagorean theorem states that the distance between two points is the square root of the sum of the squares of the x and y distances. Another way to find the distance between two points is by using a vector. A vector is a mathematical object that has both magnitude and direction. The magnitude of a vector is the magnitude of the vector’s length. The direction of a vector is the direction the vector is pointing. The distance between two points can be calculated using the dot product of the two vectors. The dot product is a calculation that combines the magnitude and direction of two vectors. ## What is 3D distance formula? What is 3D distance formula? The 3D distance formula calculates the distance between two points in three-dimensional space. It is a variant of the Pythagorean theorem, which calculates the distance between two points in two dimensions. The 3D distance formula can be used to find the distance between any two points in three-dimensional space, regardless of their orientation or relative position. To calculate the distance between two points in three-dimensional space, you need to know the coordinates of both points. The coordinates of a point in three-dimensional space are its x-coordinate, y-coordinate, and z-coordinate. The x-coordinate is the distance from the origin in the x-direction, the y-coordinate is the distance from the origin in the y-direction, and the z-coordinate is the distance from the origin in the z-direction. The 3D distance formula is: distance = (x 2 – x 1 )2 + (y 2 – y 1 )2 + (z 2 – z 1 )2 ## What is 3d coordinate system? In mathematics and physics, three-dimensional coordinate system is a coordinate system with three spatial dimensions. A three-dimensional coordinate system specifies each point uniquely in a three-dimensional space by a triple of numbers, called coordinates. In physics, the three dimensions are usually called length, width, and depth (or height), although other terms such as x, y, and z are also common. A three-dimensional coordinate system is typically represented in an x, y, and z Cartesian coordinate system, where each number represents a perpendicular direction. In physics, a three-dimensional coordinate system is also often represented in spherical coordinates and cylindrical coordinates. The notions of length, width, and depth arent absolute, but depend on the observers frame of reference. For example, if an object is moving away from an observer, it will appear shorter in the observers frame of reference. Similarly, an object that is rotating around an observers axis will appear to have a greater depth. ## What is the XYZ plane? The XYZ plane is a mathematical concept that helps to understand and describe three-dimensional objects. It is a flat surface that extends in all directions and is perpendicular to the direction of gravity. In other words, it is a plane that exists in a three-dimensional space and is perpendicular to the floor or ground. The XYZ plane is important in mathematics and physics because it helps to simplify complex three-dimensional objects and understand their properties. It can be used, for example, to calculate the distance between two points in three-dimensional space or to find the volume of a three-dimensional object. The XYZ plane is also helpful in engineering and architecture. It can be used, for example, to design three-dimensional objects or to calculate the stresses and strains on a structure. ## How do you graph 3D points? There are many ways to graph 3D points. One way is to use a coordinate plane. A coordinate plane has two axes: an x-axis and a y-axis. The x-axis goes from left to right, and the y-axis goes from top to bottom. To graph a 3D point, you need to know its coordinates. The coordinates of a 3D point are its x-coordinate, y-coordinate, and z-coordinate. The x-coordinate of a 3D point is its distance from the x-axis. The y-coordinate of a 3D point is its distance from the y-axis. The z-coordinate of a 3D point is its distance from the z-axis. Here is an example. Suppose you want to graph the point (3, -5, 4). To graph this point, you first need to create a coordinate plane. Then, you need to find the x-coordinate, the y-coordinate, and the z-coordinate of the point. The x-coordinate of the point is 3. The y-coordinate of the point is -5. The z-coordinate of the point is 4. You can graph the point by drawing a point at (3, -5, 4) on the coordinate plane. ## How do you draw a 3D coordinate system? There are many ways to draw a 3D coordinate system. One way is to use a wireframe. To do this, you first need to draw three lines, each representing a different coordinate axis: the x-axis, the y-axis, and the z-axis. The x-axis runs from left to right, the y-axis runs from bottom to top, and the z-axis runs from front to back. After you’ve drawn the three coordinate axes, you can start drawing the points that represent the coordinates of your 3D object. Begin by marking the point where the x-axis and the y-axis intersect. This point is called the origin. Then, mark the points where the y-axis and the z-axis intersect, and the x-axis and the z-axis intersect. Finally, connect the points to form the object’s wireframe. Another way to draw a 3D coordinate system is to use a Cartesian coordinate system. To do this, you first need to draw a grid with squares that represent the coordinates of your 3D object. The origin is in the lower-left corner of the grid, and the x-axis and the y-axis run across the grid from left to right and from top to bottom, respectively. The z-axis extends perpendicularly from the plane of the grid. To draw a 3D object, first find the point where the desired object intersects the x-axis. Then, find the point where the desired object intersects the y-axis. Finally, find the point where the desired object intersects the z-axis. Draw a line between these points to create the object’s outline. ## Are coordinates XY or XYZ? There is often some confusion when it comes to coordinates, specifically whether they are XY or XYZ. Coordinates are a way to pinpoint a certain location on a map. They are usually represented as a pair of numbers, with the first number representing the distance east or west from the prime meridian, and the second number representing the distance north or south of the equator. So which is it, XY or XYZ? The answer is both! Coordinates can be represented in either format, depending on what system you are using. The most common coordinate system is latitude and longitude, which uses XYZ coordinates. Other systems, such as UTM and State Plane, use XY coordinates. So which should you use? It really depends on what system the map is using. If you are not sure, you can always check the map legend or ask the person who created the map. ### Jim Miller Jim Miller is an experienced graphic designer and writer who has been designing professionally since 2000. He has been writing for us since its inception in 2017, and his work has helped us become one of the most popular design resources on the web. When he's not working on new design projects, Jim enjoys spending time with his wife and kids.

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