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## Precalculus (6th Edition) Blitzer The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]$ Consider the given matrix $A=\left[ \begin{matrix} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$ Compute matrix in the form of: $\left[ \left. A \right\|I \right]=\left[ \left. I \right\|B \right]$ Augment matrix with identity matrix: $\left[ \left. A \right\|I \right]=\left[ \begin{matrix} 1 & 2 & -1 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operation we will reduce the matrix in row echelon form for the inverse as below: \begin{align} & {{R}_{1}}\leftrightarrow {{R}_{2}}, \\ & {{R}_{2}}\to {{R}_{2}}+\frac{1}{2}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}+\frac{1}{2}\times {{R}_{_{1}}}, \\ & {{R}_{3}}\to {{R}_{3}}+\frac{1}{2}\times {{R}_{2}} \\ \end{align} The resulting matrix is: \begin{align} & \left[ \left. A \right\|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 & 3 & 4 \\ \end{matrix} \right] \\ & =\left[ \left. I \right\|B \right] \end{align} Therefore, the inverse of the matrix is given by: ${{A}^{-1}}=\left[ \begin{matrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]$ Where $B={{A}^{-1}}$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 1+2+\left( -2 \right) & 1+2+\left( -3 \right) & 2+2+\left( -4 \right) \\ \left( -2 \right)+0+2 & \left( -2 \right)+0+3 & \left( -4 \right)+0+4 \\ 1+\left( -1 \right)+0 & 1+\left( -1 \right)+0 & 2+\left( -1 \right)+0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Now, evaluate the product ${{A}^{-1}}A={{I}_{3}}$ \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & -1 \\ -2 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1+\left( -2 \right)+2 & 2+0+-2 & \left( -1 \right)+1+0 \\ 1+\left( -2 \right)+1 & 2+0+\left( -1 \right) & \left( -1 \right)+1+0 \\ 2+\left( -6 \right)+4 & 4+0+\left( -4 \right) & \left( -2 \right)+3+0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.
# Therefore the prime p is not among p 1 p k which • Notes • 74 This preview shows pages 9–12. Sign up to view the full content. 1, which is impossible. Therefore, the prime p is not among p 1 , . . . , p k , which contradicts our assumption that these are the only primes. 4 This preview has intentionally blurred sections. Sign up to view the full version. Chapter 2 Congruences This chapter reviews the notion of congruences. 2.1 Definitions and Basic Properties For positive integer n and for a, b Z , we say that a is congruent to b modulo n if n \| ( a - b ), and we write a b (mod n ). If n - ( a - b ), then we write a 6≡ b (mod n ). The number n appearing in such congruences is called the modulus . A trivial observation is that a b (mod n ) if and only if there exists an integer c such that a = b + cn . Another trivial observation is that if a b (mod n ) and n 0 \| n , then a b (mod n 0 ). A key property of congruences is that they are “compatible” with integer addition and multi- plication, in the following sense: Theorem 2.1 For all positive integers n , and all a, a 0 , b, b 0 Z , if a a 0 (mod n ) and b b 0 (mod n ) , then a + b a 0 + b 0 (mod n ) and a · b a 0 · b 0 (mod n ) . Proof. Suppose that a a 0 (mod n ) and b b 0 (mod n ). This means that there exist integers c and d such that a 0 = a + cn and b 0 = b + dn . Therefore, a 0 + b 0 = a + b + ( c + d ) n, which proves the first equality of the theorem, and a 0 b 0 = ( a + cn )( b + dn ) = ab + ( ad + bc + cdn ) n, which proves the second equality. 2 2.2 Solving Linear Congruences For a positive integer n , and a Z , we say that a is a unit modulo n if there exists a 0 Z such that aa 0 1 (mod n ), in which case we say that a 0 is a multiplicative inverse of a modulo n . Theorem 2.2 An integer a is a unit modulo n if and only if a and n are relatively prime. 5 Proof. This follows immediately from the fact that a and n are relatively prime if and only if there exist s, t Z such that as + bt = 1. 2 We now prove a simple a “cancellation law” for congruences: Theorem 2.3 If a is relatively prime to n , then ax ax 0 (mod n ) if and only if x x 0 (mod n ) . More generally, if d = gcd( a, n ) , then ax ax 0 (mod n ) if and only if x x 0 (mod n/d ) . Proof. For the first statement, assume that gcd( a, n ) = 1, and let a 0 be a multiplicative inverse of a modulo n . Then, ax ax 0 (mod n ) implies a 0 ax a 0 ax 0 (mod n ), which implies x x 0 (mod n ), since a 0 a 1 (mod n ). Conversely, if x x 0 (mod n ), then trivially ax ax 0 (mod n ). That proves the first statement. For the second statement, let d = gcd( a, n ). Simply from the definition of congruences, one sees that in general, ax ax 0 (mod n ) holds if and only if ( a/d ) x ( a/d ) x 0 (mod n/d ). Moreover, since a/d and n/d are relatively prime, the first statement of the theorem implies that ( a/d ) x ( a/d ) x 0 (mod n ) holds if and only if x x 0 (mod n/d ). That proves the second statement. 2 We next look at solutions x to congruences of the form ax b (mod n ), for given integers n, a, b . Theorem 2.4 Let n be a positive integer and let a, b Z . If a is relatively prime to n , then the congruence ax b (mod n ) has a solution x ; moreover, any integer x 0 is a solution if and only if x x 0 (mod n ) . This preview has intentionally blurred sections. Sign up to view the full version. This is the end of the preview. Sign up to access the rest of the document. • Spring '13 • MRR {[ 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
What is Half of 3/4: You can calculate “half” of a fraction by doubling the denominator (bottom number * 2), so half of 3/4 is 3/8 (formula: half of a/b is the same as a/(b2), for example, half of 3/4 equals 3/(42) which equals 3/8). 1/2 of 3/4 = 3/4 divided by 2. (¾)/2= 3/(42) = 3/8. (31)/(42) = 3/8. Half of 3/4 is expressed as 3/4 divided by 2. Dividing 3/4 by 2 gives us a value of 3/8. To divide a fraction by a whole number, you simply invert the number and multiply it by the fraction. So, in this case, we inverted 2 to get 1/2 and multiply 3/4 by 1/2 to get 3/8. This means that half of 3/4 is equal to 3/8. It’s important to understand this concept of dividing fractions, as it is a fundamental operation in mathematics and is widely used in real-life problems and applications. The ability to simplify fractions, perform operations with them, and convert them to decimals or percents is essential in fields such as finance, engineering, and science. So half of 3/4 is 3/8 ## What is Half of 3/4 in Cups? If you want to add half of 3/4 cup ingredients (3/8 cup) to a recipe, you can use these different ways to calculate this amount. Measure half of 3/4 cup of ingredients by using a tablespoon. The number of tablespoons that add up to 3/4 cup is 12, so divide 12 in half and add 6 tablespoons of elements to your recipe for half of 3/4 cup. ### What is half of 3/4 on a tape measure? We know it’s more than half of 3/4 of an inch and less than one full inch. The marking is halfway between 3/4 (6/8) and 7/8. Therefore, the marking is half of 1/8, or 1/16. #### Related Question to What is Half of 3/4? 1. Question: What is half of 1/3 cup in cups? Reducing the Size of Recipes 1/3 cup 2 tablespoons + 2 teaspoons 1/2 cup 1/4 cup 2/3 cup 1/3 cup 3/4 cup 6 tablespoons 2. Question: What is half of 3/4 inches? Answer: what is half of 3/4? It is 3/8 or 0.375. 3. Question: How many cups are 3/4 cups? Volume Equivalents (liquid) 8 tablespoons 1/2 cup 4 fluid ounces 12 tablespoons 3/4 cup 6 fluid ounces 16 tablespoons 1 cup 8 fluid ounces 2 cups 1 pint 16 fluid ounces 4. Question: What is half of 1 4 cups? Answer: We are to get half of 114 1 1 4 cups. 5. Question: What’s half of 1/4 cup in tablespoons? Answer: 1/4 cup = 4 tablespoons. 1/3 cup = 5 tablespoons plus 1 teaspoon. 3/8 cup = 6 tablespoons. 1/2 cup = 8 tablespoons. 6. Question: What is half of 5/8 on a tape measure?
# What percent of 500 is 120? Nov 16, 2016 24% of 500 is 120. #### Explanation: To rephrase this problem: 120 is what percent of 500. Let's call the percent we are looking for $p$. Then we can say: $p$ of 500 = 120 or $p \times 500 = 120$ Solving for $p$ while keeping the equation balanced gives: $p \times \frac{500}{500} = \frac{120}{500}$ p = 0.24 or 24% Nov 16, 2016 24% #### Explanation: To find out what percentage one quantity is of another quantity: color(red)("Fraction " xx 100%) As soon as you write a fraction it also represents a decimal fraction. Multiplying by 100% is the same as multiplying by 1, so you are not changing the value of the fraction, only the form of the fraction. Fraction = decimal = percent What percent is 120 of 500? color(red)(120/500 xx 100%) =24% In this case you can find the percent by making an equivalent fraction with a denominator of 100. (120div5)/(500div5) = 24/100 = 24%#
# Lesson: Symmetric Matrices 1 point ## Transcript A matrix A is symmetric if A-transpose = A, or equivalently, if aij = aji for all i and j. Note that the definition of a symmetric matrix requires that it be a square matrix, since if A is an m-by-n matrix, then A-transpose is an n-by-m matrix, so A-transpose = A would mean that m = n. Here are some examples of symmetric matrices. A matrix A is said to be orthogonally diagonalizable if there exists an orthogonal matrix P and a diagonal matrix D such that (P-transpose)AP = D. Now, recall that if P is an orthogonal matrix, then P-transpose = P-inverse, so this definition is just the same as saying that there is an orthogonal matrix P that diagonalizes A. But before we prove that every symmetric matrix is orthogonally diagonalizable, we will do some examples and some practice problems of diagonalizing symmetric matrices, as this hands-on work will help us understand the theoretical proof. Let’s diagonalize the symmetric matrix A = [1, 2; 2, -2]. To do this, we first need to find the eigenvalues, which means we need to find the roots of the characteristic polynomial (determinant of (A – (lambda)I)). So the matrix A – (lambda)I is [1 – lambda, 2; 2, -2 – lambda]. If we take the determinant of this matrix, it will be (1 – lambda)(-2 – lambda) – 2(2). Multiplying this out, this equals -6 + lambda + lambda-squared, which factors as (2 – lambda)(-3 – lambda). So we see that the eigenvalues for A are lambda1 = 2 and lambda2 = -3. Our next step is to find a basis for the eigenspaces. For lambda1 = 2, we’ll need to find the nullspace of (A – lambda1)I, which is [1 – lambda1, 2; 2, -2 – lambda1], which equals [-1, 2; 2, -4]. Row reducing, we see that this matrix will be row equivalent to the matrix [1, -2; 0, 0], so our nullspace consists of all solutions to the equation x1 – 2x2 = 0. If we replace the variable x2 with the parameter s, then we have that x1 = 2s and x2 = s, so the general solution is (s times the vector [2; 1]), which means that the single vector set {[2; 1]} is a basis for the eigenspace for lambda1. For lambda2 = -3, we need to find the nullspace of (A – (lambda2)I), which is [1 – lambda2, 2; 2, -2 – lambda2], which equals [4, 2; 2, 1]. Row reducing, we can quickly see that this matrix is row equivalent to the matrix [2, 1; 0, 0], so our nullspace consists of all solutions to the equation 2x1 + x2 = 0. If we replace the variable x1 with the parameter s, then we have that x1 = s and x2 = -2s, so the general solution to our equation is s[1; -2]. This means that the set containing only the vector [1; -2] is a basis for the eigenspace for lambda2. Now, our study of diagonalization—specifically, the Diagonalization Theorem—tells us that A can be diagonalized by a matrix P whose columns are the basis vectors for the eigenspaces. That is, we know that the matrix P = [2, 1; 1, -2] is an invertible matrix such that (P-inverse)AP = D, where D will be the matrix [lambda1, 0; 0, lambda2], which, in our case, is [2, 0; 0, -3]. So that settles the issue of A being diagonalizable, but what about being orthogonally diagonalizable? A quick check shows that the columns of P are orthogonal, but not orthonormal. What if we normalize these vectors? Well, since the set containing the vector [2; 1] is a basis for the eigenspace for lambda1, we also have that the set containing s[2; 1] would be a basis for the eigenspace for lambda1 for any scalar s, which means, for example, that the set containing the vector [2/(root 5); 1/(root 5)] is a basis for the eigenspace for lambda1. And similarly, the set containing the vector [1/(root 5); -2/(root 5)] is a basis for the eigenspace for lambda2. And going back to our knowledge of diagonalization, this means that the matrix Q whose first column is [2/(root 5); 1/(root 5)], our basis vector for lambda1, and whose second column is [1/(root 5); -2/(root 5)], our basis vector for lambda2, is an invertible matrix such that (Q-inverse)AQ = D, where D is still [2, 0; 0, -3]. Moreover, we see that Q is orthogonal, so we have shown that A is orthogonally diagonalizable. Let’s look at a bigger example. We’ll try to diagonalize the symmetric matrix A = [1, -2, -2; -2, 1, 2; -2, 2, 1]. To do this, we first need to find the eigenvalues, which means we need to find the roots of the characteristic polynomial, of the determinant of (A – (lambda)I). So first, we’ll subtract lambda from our diagonal elements, and then proceed to compute the determinant. By expanding along the first row, we’ll see that we get (1 – lambda) times the determinant of the submatrix [1 – lambda, 2; 2, 1 – lambda], and then we get minus a -2 times the determinant of the submatrix [-2, 2; -2, 1 – lambda], and then plus -2 times the determinant of the submatrix [-2, 1 – lambda; -2, 2]. If we multiply all of this out, we’ll see that our characteristic polynomial is 5 + 9(lambda) + 3(lambda-squared) – (lambda-cubed). A quick check shows that lambda = -1 is a root of this equation. If we factor out a (-1 – lambda) term, we are left with the polynomial -5 – 4(lambda) + (lambda-squared). This factors as (-1 – lambda)(5 – lambda). So the characteristic polynomial for A is ((-1 – lambda)-quantity-squared)(5 – lambda), and this means that the eigenvalues for A are lambda1 = -1 and lambda2 = 5. Now, the next step is to find a basis for the eigenspaces. For lambda1 = -1, we need to find the nullspace of our matrix (A – (lambda1)I), which is the matrix [2, -2, -2; -2, 2, 2; -2, 2, 2]. If we row reduce this matrix, we see that it is row equivalent to the matrix [1, -1, -1; 0, 0, 0; 0, 0, 0]. The nullspace of this matrix will consist of all solutions to the equation x1 – x2 – x3 = 0. If we replace the variable x2 with the parameter s, and the variable x3 with the parameter t, then we have that x1 = s + t, x2 = s, and x3 = t, so our general solution can be written as (s times the vector [1; 1; 0]) + (t times the vector [1; 0; 1]). This means that the set containing the vectors [1; 1; 0] and [1; 0; 1] is a basis for the eigenspace for lambda1. For our other eigenvalue, lambda2 = 5, we again need to find the nullspace of the matrix (A – (lambda2)I), which, in this case, is the matrix [-4, -2, -2; -2, -4, 2; -2, 2, -4]. This row reduction is a little more complicated, so I’ll show some of the intermediate steps, but we will find that this matrix is row equivalent to the matrix [1, 0, 1; 0, 1, -1; 0, 0, 0]. The nullspace of this matrix will consist of all solutions to the system x1 + x3 = 0, x2 – x3 = 0. If we replace the variable x3 with the parameter s, then we have that x1 = -s, x2 = s, and x3 = s. So we can write the general solution of this system as (s times the vector [-1; 1; 1]), which means that the set containing the vector [-1; 1; 1] is a basis for the eigenspace for lambda2. Again, our study of diagonalization tells us that A can be diagonalized by a matrix P whose columns are the basis vectors for the eigenspaces. That is, we know that this matrix P = [1, 1, -1; 1, 0, 1; 0, 1, 1] is an invertible matrix such that (P-inverse)AP = D, where D is the matrix whose diagonal entries are lambda1, lambda1, and lambda2, or in our case, [-1, 0, 0; 0, -1, 0; 0, 0, 5]. Now, again, we’ve shown that A is diagonalizable, but what about being orthogonally diagonalizable? In this particular case, we do not even have that the columns of P are orthogonal, much less orthonormal. We could apply the Gram-Schmidt procedure to the columns of P, but our theory of diagonalization requires that the columns of P be basis vectors for the corresponding eigenspaces, not simply any basis vector for R3. So as we did in the previous example, we will go back to our eigenspaces and find orthonormal bases for them. Looking at the eigenspace for lambda2 first, since it is the span of a single vector, we simply need to normalize this vector to get an orthonormal basis. And so we see that the set containing only the vector [-1/(root 3); 1/(root 3); 1/(root 3)] is an orthonormal basis for the eigenspace for lambda2. But the eigenspace for lambda1 has two basis vectors, and these basis vectors were not orthogonal, so we will need to use the Gram-Schmidt procedure to find first an orthogonal, and then an orthonormal, basis for our eigenspace. We keep our first vector as [1; 1; 0], and then using the Gram-Schmidt procedure, our second vector will become [1; 0; 1] – ((([1; 0; 1] dotted with [1; 1; 0])/((the norm of [1; 1; 0])-squared)) times the vector [1; 1; 0]). That is to say, it’s the vector [1; 0; 1] – (1/2)[1; 1; 0], which equals [1/2; -1/2; 1]. So we know that the set {[1; 1; 0], [1/2; -1/2; 1]} is an orthogonal basis for our, the eigenspace, and then by normalizing the vectors, we see that the set {[1/(root 2); 1/(root 2); 0], [1/(root 6); -1/(root 6); 2/(root 6)]} is an orthonormal basis for the eigenspace for lambda1. Now, again, returning to our knowledge of diagonalization, we know that the matrix Q whose columns are these new basis vectors, so the first column can be the basis vector [1/(root 2); 1/(root 2); 0], the second column can be our second basis vector for lambda1, which is [1/(root 6); -1/(root 6); 2/(root 6)], and lastly, the third column can be our basis vector from lambda2, the [-1/(root 3); 1/(root 3); 1/(root 3)]. So again, our theory tells us that this will diagonalize A, but have we, in fact, ended up with a Q that is orthonormal? Well, a quick check verifies that Q is, in fact orthonormal. We’ve already made sure that every, all the vectors are normal vectors, and a quick look shows that they are, in fact, orthogonal as well. So we have shown that A is orthogonally diagonalizable. Again, I will be assigning the problems on showing that various similar matrices are orthogonally diagonalizable before we prove that this is always the case, as I feel that working hands-on with these matrices will help motivate the steps in the proof. At this point, all you need to know is that if you find an orthonormal basis for all of the eigenspaces, then you will end up with an orthogonal matrix P to diagonalize A.
# How do you find the conditional density function of X given Y? ## How do you find the conditional density function of X given Y? First, to find the conditional distribution of X given a value of Y, we can think of fixing a row in Table 1 and dividing the values of the joint pmf in that row by the marginal pmf of Y for the corresponding value. For example, to find pX\|Y(x\|1), we divide each entry in the Y=1 row by pY(1)=1/2. ## How do you find the conditional probability of a density function? The conditional density for X given R = r equals h(x \| R = r) = ψ(x, r) g(r) = 1 π √ r2 − x2 for \|x\| < r and r > 0. What is the conditional distribution of X given Y Y? For any random variables X and Y, the conditional distribution of Y given X = x specifies how Y varies when X = x. We have already seen instances of conditional distributions when X and Y are independent. In that case, Y varies just as it usually does, regardless of the values of X. What is the probability density function of X Y? The function fXY(x,y) is called the joint probability density function (PDF) of X and Y. In the above definition, the domain of fXY(x,y) is the entire R2. We may define the range of (X,Y) as RXY={(x,y)\|fX,Y(x,y)>0}. ### What is the area under the conditional C * * * * * * * * * density function? Explanation: Area under any conditional CDF is 1. ### How do you calculate conditional CDF? The conditional CDF of X given A, denoted by FX\|A(x) or FX\|a≤X≤b(x), is FX\|A(x)=P(X≤x\|A)=P(X≤x\|a≤X≤b)=P(X≤x,a≤X≤b)P(A). What is the area under conditional probability density function? How do you find conditional density? The conditional density function is f((x,y)\|E)={f(x,y)/P(E)=2/π,if(x,y)∈E,0,if(x,y)∉E. #### What is the conditional probability of Y given X X? Seen as a function of y y y for given x x x, P ( Y = y ∣ X = x ) P(Y = y \| X = x) P(Y=y∣X=x) is a probability, so the sum over all y y y (or integral if it is a conditional probability density) is 1. #### What is the conditional variance of Y given X X? Similar to the conditional expectation, we can define the conditional variance of X, Var(X\|Y=y), which is the variance of X in the conditional space where we know Y=y. If we let μX\|Y(y)=E[X\|Y=y], then Var(X\|Y=y)=E[(X−μX\|Y(y))2\|Y=y]=∑xi∈RX(xi−μX\|Y(y))2PX\|Y(xi)=E[X2\|Y=y]−μX\|Y(y)2. What is the probability distribution of X Y? This is referred to as the joint probability of X = x and Y = y. If X and Y are discrete random variables, the function given by f (x, y) = P(X = x, Y = y) for each pair of values (x, y) within the range of X is called the joint probability distribution of X and Y . What is the probability density function of thermal noise? 1. What is the probability density function of thermal noise? Explanation: Thermal noise is approximately white, it means that its power spectral density is nearly equal throughout the frequency spectrum. The amplitude of the signal has a Gaussian probability density function. ## What are real life examples of a probability density function? One very important probability density function is that of a Gaussian random variable, also called a normal random variable. The probability density function looks like a bell-shaped curve. One example is the density ρ(x) = 1 √2πe − x2 / 2 , which is graphed below. ## What does conditional probability distribution mean? A conditional probability distribution is a probability distribution for a sub-population. That is, a conditional probability distribution describes the probability that a randomly selected person from a sub-population has the one characteristic of interest. How does probability density function work? In probability theory, a probability density function (PDF), or density of a continuous random variable, is a function, whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would equal that What is the integral of probability density function? In mathematics, a probability density function (pdf) serves to represent a probability distribution in terms of integrals. A probability density function is non-negative everywhere and its integral from −∞ to +∞ is equal to 1.
## Percentage For SBI PO : Set – 15 1) The total number of students in a school is 5600 out of which 60% are boys what is the total number of girls in this school? a) 2240 b) 3360 c) 2860 d) 3240 e) None of these a) Total number of students in a class=5600 Total number of boys = 60% So total number of girls = 40% 40% of (5600) = 2240 2) If a man after spending 85% of the income he saves Rs. 4560 per month, his monthly income is a) Rs. 32800 b) Rs. 31600 c) Rs. 30400 d) Rs. 29600 e) None of these C) Man spending income = 85% So savings = 15% 15% of x = 4560 X= 30400 3) It is known that 6% of the mangoes are rotten. If the number of rotten mangoes is 54, then the total number of mangoes is a) 900 b) 950 c) 1010 d) 1040 e) 1080 a) 6% of rotten mangoes are 54 So total number of mangoes = 6% of (x) =54 X = 900 4) A fruit seller had some oranges. He sells 35% oranges and still has 390 oranges. Originally he had a) 620 b) 550 c) 600 d) 520 e) None of these C) Sold oranges = 35% Total number of oranges = 100 – 35 = 65% Remaining oranges =390 65% of X = 390 X=600 5) In an examination 90% of the students passed and 240 failed. How many students appeared for the examination? a) 2360 b) 2400 c) 2450 d) 2520 e) 2580 b) Percentage of passed students = 90% Number of failed students = 240 10 % of x = 240 X = 2400 6) In a class of 60 students, each student got sweets that are 15% the total number of students. How many sweets were there? a) 540 b) 520 c) 550 d) 580 e) 510 a) Total number of students = 60 Each student got 15% of total number of students So we get 15% of (60) = 9 7) In a class of 60 students and 6 teachers, each student got sweets that are 20% of the total number of students and each teacher got sweets that are 25% of the total number of students. How many sweets were there? a) 875 b) 840 c) 810 d) 860 e) 895 c) Number of sweets students got = 60(20 % of (60)) = 720 Number of sweets Teachers got = 6 (25% of (60)) = 90 Total number of sweets = 720 + 90 = 810 8) 720 sweets were distributed equally among children in such a way that number of sweets received by each child is 20% of the total number of children. How many sweets did each child receive? a) 8 b) 10 c) 9 d) 12 e) None of these d) 9) A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? a) 45% b) 54 6/11% c) 55% d) 48% e) 45 5/11% e) Explanation 10) The difference between a number and its two-fifth is 510. What is 10% of that number? a) 12.75 b) 85 c) 204 d) 78 e) None of these
So in that example the degree is 1. derivative dy dx, Here we look at a special method for solving "Homogeneous Differential Equations". If and are two real, distinct roots of characteristic equation : y er 1 x 1 and y er 2 x 2 b. \dfrac{k\text{cabbage}}{kt} = \dfrac{\text{cabbage}}{t}, Homogenous Diffrential Equation. x2is x to power 2 and xy = x1y1giving total power of 1+1 = 2). Let's rearrange it by factoring out z: f (zx,zy) = z (x + 3y) And x + 3y is f (x,y): f (zx,zy) = zf (x,y) Which is what we wanted, with n=1: f (zx,zy) = z 1 f (x,y) Yes it is homogeneous! It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. \), Solve the differential equation $$\dfrac{dy}{dx} = \dfrac{x(x - y)}{x^2}$$, Homogeneous Differential Equation A differential equation of the form f (x,y)dy = g (x,y)dx is said to be homogeneous differential equation if the degree of f (x,y) and g (x, y) is same. Example 6: The differential equation is homogeneous because both M (x,y) = x 2 – y 2 and N (x,y) = xy are homogeneous functions of the same degree (namely, 2). If = then and y xer 1 x 2. c. If and are complex, conjugate solutions: DrEi then y e Dx cosEx 1 and y e x sinEx 2 Homogeneous Second Order Differential Equations M(x,y) = 3x2+ xy is a homogeneous function since the sum of the powers of x and y in each term is the same (i.e. Second Order Linear Differential Equations – Homogeneous & Non Homogenous v • p, q, g are given, continuous functions on the open interval I ... is a solution of the corresponding homogeneous equation s is the number of time The equation is a second order linear differential equation with constant coefficients. \end{align} The general solution of this nonhomogeneous differential equation is In this solution, c1y1 (x) + c2y2 (x) is the general solution of the corresponding homogeneous differential equation: And yp (x) is a specific solution to the nonhomogeneous equation. v + x \; \dfrac{dv}{dx} &= 1 + v\\ Therefore, if we can nd two so it certainly is! v + x\;\dfrac{dv}{dx} &= \dfrac{xy + y^2}{xy}\\, to tell if two or more functions are linearly independent using a mathematical tool called the Wronskian. \begin{align} laplace\:y^'+2y=12\sin\left (2t\right),y\left (0\right)=5. y′ = f ( x y), or alternatively, in the differential form: P (x,y)dx+Q(x,y)dy = 0, where P (x,y) and Q(x,y) are homogeneous functions of the same degree. A first order Differential Equation is Homogeneous when it can be in this form: dy dx = F ( y x ) We can solve it using Separation of Variables but first we create a new variable v = y x. v = y x which is also y = vx. (1 - 2v)^{-\dfrac{1}{2}} &= kx\\ An equation of the form dy/dx = f(x, y)/g(x, y), where both f(x, y) and g(x, y) are homogeneous functions of the degree n in simple word both functions are of the same degree, is called a homogeneous differential equation. -\dfrac{2y}{x} &= k^2 x^2 - 1\\ In our system, the forces acting perpendicular to the direction of motion of the object (the weight of the object and … Linear inhomogeneous differential equations of the 1st order; y' + 7y = sin(x) Linear homogeneous differential equations of 2nd order; 3y'' - 2y' + 11y = 0; Equations in full differentials; dx(x^2 - y^2) - … substitution $y = vx. \begin{align} \end{align} Then a homogeneous differential equation is an equation where and are homogeneous functions of the same degree. Then \begin{align} The order of a diﬀerential equation is the highest order derivative occurring. $, , $$\dfrac{dy}{dx} = v\; \dfrac{dx}{dx} + x \; \dfrac{dv}{dx} = v + x \; \dfrac{dv}{dx}$$, Solve the differential equation $$\dfrac{dy}{dx} = \dfrac{y(x + y)}{xy}$$, \end{align}, \begin{align} homogeneous if M and N are both homogeneous functions of the same degree. Added on: 23rd Nov 2017. \int \dfrac{1}{1 - 2v}\;dv &= \int \dfrac{1}{x} \; dx\\ \( \dfrac{d \text{cabbage}}{dt} = \dfrac{\text{cabbage}}{t}, \ln (1 - 2v)^{-\dfrac{1}{2}} &= \ln (kx)\\ Then. The two main types are differential calculus and integral calculus. It is considered a good practice to take notes and revise what you learnt and practice it. A diﬀerential equation (de) is an equation involving a function and its deriva-tives. -2y &= x(k^2x^2 - 1)\\ \end{align} Differential Equations are equations involving a function and one or more of its derivatives. A first order Differential Equation is Homogeneous when it can be in this form: We can solve it using Separation of Variables but first we create a new variable v = y x. Homogeneous Differential Equations in Differential Equations with concepts, examples and solutions. \end{align} Therefore, we can use the substitution $y = ux, $$y’ = u’x + u.$$ As a result, the equation is converted into the separable differential … Applications of differential equations in engineering also have their own importance. f(kx,ky) = \dfrac{(kx)^2}{(ky)^2} = \dfrac{k^2 x^2}{k^2 y^2} = \dfrac{x^2}{y^2} = f(x,y). In the special case of vector spaces over the real numbers, the notion of positive homogeneity often plays a more important role than homogeneity in the above sense. The value of n is called the degree. He's modelled the situation using the differential equation: First, we need to check that Gus' equation is homogeneous. $, $$\dfrac{1}{1 - 2v}\;dv = \dfrac{1}{x} \; dx$$, This Video Tells You How To Convert Nonhomogeneous Differential Equations Into Homogeneous Differential Equations. … \begin{align} Set up the differential equation for simple harmonic motion. For example, we consider the differential equation: (x 2 + y 2) dy - xy dx = 0 Online calculator is capable to solve the ordinary differential equation with separated variables, homogeneous, exact, linear and Bernoulli equation, including intermediate steps in the solution. Diﬀerential equations are called partial diﬀerential equations (pde) or or-dinary diﬀerential equations (ode) according to whether or not they contain partial derivatives. Abstract. Multiply each variable by z: f (zx,zy) = zx + 3zy. &= \dfrac{x(vx) + (vx)^2}{x(vx)}\\ It is easy to see that the given equation is homogeneous. y &= \dfrac{x(1 - k^2x^2)}{2} \end{align} x\; \dfrac{dv}{dx} &= 1, Martha L. Abell, James P. Braselton, in Differential Equations with Mathematica (Fourth Edition), 2016. A second order differential equation involves the unknown function y, its derivatives y' and y'', and the variable x. Second-order linear differential equations are employed to model a number of processes in physics. But the application here, at least I don't see the connection. Homogeneous vs. Non-homogeneous. Let's consider an important real-world problem that probably won't make it into your calculus text book: A plague of feral caterpillars has started to attack the cabbages in Gus the snail's garden. Using y = vx and dy dx = v + x dv dx we can solve the Differential Equation. \begin{align} FREE Cuemath material for JEE,CBSE, ICSE for excellent results!, \begin{align} \end{align} f (tx,ty) = t0f (x,y) = f (x,y). \text{cabbage} &= Ct. \int \;dv &= \int \dfrac{1}{x} \; dx\\, $$A simple way of checking this property is by shifting all of the terms that include the dependent variable to the left-side of an … Differential equation with unknown function () + equation. A first order differential equation is homogeneous if it can be written in the form: \( \dfrac{dy}{dx} = f(x,y),$$ where the function $$f(x,y)$$ satisfies the condition that $$f(kx,ky) = f(x,y)$$ for all real constants $$k$$ and all $$x,y \in \mathbb{R}$$. And N are both homogeneous functions of the same degree n't see the connection y=x^3y^2, y\left 0\right!, ty ) = 5 are differential calculus and integral calculus step 2: Integrate both sides of the degree! 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# 2004 AMC 10B Problems/Problem 24 In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$? $\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$ ## Solution 1 Set $\overline{BD}$'s length as $x$. $\overline{CD}$'s length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length(because $\angle BAD=\angle DAC$). Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\frac{5}{3}\implies\boxed{\text{(B)}}$ ## Solution 2 $[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("A", A, N); label("B", B, SW); label("C", C, SE); label("D", D, S); label("E", E, NE); markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); [/asy]$ Let $E = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC \cong \angle ADC$ because they subtend the same arc, $\overarc{AC}$. Furthermore, $\angle BAE \cong \angle BCD$ because they both subtend $\overarc{BD}$, so $\triangle ABE \sim \triangle ADC$ by $\text{AA}$ similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$. By the Angle Bisector Theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$, so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$. This in turn gives $BE = \frac{21}{5}$. Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}$.
You are Here: Home >< Maths # STEP Maths I, II, III 1989 solutions Watch 1. STEP I (Mathematics) 1: 2: Solution by kabbers 3: Solution by Dystopia 4: Solution by Dystopia 5: Solution by Dystopia 6: Solution by Swayam 7: Solution by squeezebox 8: Solution by squeezebox 9: Solution by nota bene 10: 11: Solution by Glutamic Acid 12: 13: 14: 15: 16: STEP II (F.Maths A) 1: Solution by squeezebox 2: Solution by kabbers 3: Solution by squeezebox 4: 5: Solution by Dystopia 6: Solution by bobo 7: 8: 9: 10: 11:Solution by bobo 12: 13: 14:Solution by squeezebox 15: 16: STEP III (F.Maths B) 1: Solution by squeezebox 2: Solution by bobo 3: 4: 5: 6: 7: 8: Solution by Dystopia 9: Solution by squeezebox 10: Solution by squeezebox 11: 12: 13: 14: 15: 16: Solutions written by TSR members: 1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007 2. I am no expert at STEP, so don't hesitate to correct me, it is more than likely made that I have made a mistake. STEP I - Question 8 Using good old de Moivre's theorem; (using and similary; Now, consider: Where c is . Now let , so: Which gives; (). These values of give distinct values of or () Hence the roots of the equation are: and . 3. STEP III - Question 10 Lets assume that the result is true for n = k; () This is the same as () except k+1 replaces k. Hence if the result is true for n=k, its true for n= k+1. When n=1, LHS of () = 1x2x3x4x5 = 120 RHS of () = (1/6)x1x2x3x4x5x6 = 120. So () is true for n=1. Hence, by induction; . Since; . Using (); In this case, From the previous parts, we know that: and clearly, also: as , and So, Using a similar arguement, we can show that: We have shown that the limits exist and are equal to . Hence; 4. STEP I, Q2 The For , find Using integration by parts (): By approximating the area corresponding to by n rectangles of equal width and with their top right hand vertices on the curve , show that, as , Firstly, note that (by the above indefinite integral) The sum of the areas of our rectangles is going to be Claim: So for k + 1, we must prove that is the equation we arrive at. Basis case (let k=1): Inductive step: So, As , our sum tends to the integral . So, is this ok? have i missed anything out or not explained each step in enough detail? 5. Question 4, STEP I With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points. Diagram showing five points is attached. Attached Images 6. Question 5, STEP I a) Let x = 1. Let x = -1 As n is even. So As their sum is b) Suppose that Then Note that and that So, upon adding these two expressions, we get Also, So true by induction. 7. STEP II Q1 Substituting x= y-a into the equation: Notice that the reduced form shown in the question has no term, so we need to find the value of a which will get rid of the term. Expanding out and collectiing terms we get: So the value of a to get rid of the is -1. So we now have: () Substituting y=z/b into (); dividing both sides by 2 and multiplying by ; . And so to make this equal to; we make b = . Lets take b = and let z = so; and (these give distinct values of , and hence, three distinct solutions.) we know that; x = y - a = (z/b) - a. Hence the solutions of the equation are: and 8. Question 11, STEP I. Vertical component = Using s = ut + 1/2at^2. Quadratic in t. Ignoring the negative root. R = horizontal component x time. Horizontal component = Multiply both numerator and denominator by 2. Using We can simplify it. Splitting up Mr. Big Fraction. Ignore the first fraction for the moment, we'll work on the second one. Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. \dfrac{2v^2 \cos \theta \sin \theta}{2g} \times \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta} We can use our old friend Mr. Double Angle Formula to simplify it: The whole fraction looks like: Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. R = (\dfrac{v^2\sin 2\theta}{2g}) + (\dfrac{v^2 \sin 2\theta}{2g}) \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta} And lo and behold, there's a common factor to both fractions. Let's take it out: And that's what we're looking for. I'll finish off the last small bit in a few minutes 9. STEP I Question 6 The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be (where x' is the general x coordinate of the normal). At the point Q, x' = 0 as it cuts the y axis. The distance PQ can be worked out using Pythagoras' theorem. It's given that So (use a substitution of u = x^2 if you can't see why) 10. STEP III, Q8 Let Then Using an integrating factor, , we get Applying boundary conditions: --- I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want. 11. Hang on a sec, mine was mislabelled, should be Step I not step II.. sorry about that! edit: thanks.. now typing up II/2 12. STEP II/2 Therefore, by (), we can also use () to express 2cot2x as Therefore, we can compare the coefficients of the respective summations and say for the second part, note that we have to find an identity for 2csc2x (using partial fractions) Also, , so: Comparing coefficients of the polynomials (in x) of the expansion, Rearranging and substituting in the earlier definition of , any corrections welcome ! 13. STEP II, (11) attached attachments 2&3 are in the wrong order Attached Files 14. step1989q11.doc (540.0 KB, 248 views) 15. step1989q11p3.doc (205.5 KB, 235 views) 16. step1989q11p2.doc (158.5 KB, 251 views) 17. Step III (2) Attached Files 18. step1989q2p1.doc (113.5 KB, 280 views) 19. step1989q2p2.doc (79.0 KB, 262 views) 20. step1989q2p3.doc (71.5 KB, 256 views) 21. STEP II (6) attached Attached Files 22. step1989q6p1.doc (79.5 KB, 249 views) 23. step1989q62.doc (95.0 KB, 294 views) 24. STEP II Question 3 Equating real and imaginary parts: () () if x=a, then: (1) (2) Now, subbing (1) and (2) into: We get: As required. Substituting y=b into () and (*): (3) (4). This time, using , we end up with: . Since and , . Hence, () corresponds to a single hyeprbola branch, which is above the line v=0. One point of intersection is: and . Differentiating implicitly we get: . Which is equal to: at and . Differentiating implicitly we get: . Which is, when and . X Hence the curves intersect and right angles at this point. Regarding the sketch, I think the elipse becomes a vertical straight line from v = -1 to v = 1, and the hyperbola becomes the line v = 1. 25. STEP 1, Q3 (By the ratio theorem.) These vectors are perpendicular, so the scalar product is zero. (Note that the scalar product is commutative and distributive.) Let Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2. So Note that Where The maximum value is occurs when Suppose that Then So By the cosine rule, Let F be the midpoint of AB. By Pythagoras, 26. bump* (just to remind people that this thread is still here ) - I have solutions that I can type up if no one else wants to, but I really don't want to take over the thread.. 27. STEP I - Question 7 Graph Attached If we work out the area of a quarter of the rectangle, which is in the top right hand bit of the axes, and maximise this, it make calculation soooo much eaiser. (Since we can forget about modulus signs and y being negative) : Let the area of the rectangle be . Base of smaller rectangle = 1-x height = so which we want to maximise. Which is zero when Finding the second derivative and subsituting x = in shows is a maximum at this point. So . ________ We can do the second bit in the same way, since its a closed curve that is symmetrical in both axes. Let the area of the second rectangle be . Working in the top right quadrant again; Base = 1-x height = . So . Which is zero when x = 0 or x = . Clearly the area is a minimum when x= 0, and so is a maximum when x =. . Attached Files 28. STEP I Q7 graph.doc (196.0 KB, 212 views) 29. I've got a solution to I question 9 , but don't have a scanner available so will add graphs when I get the opportunity. Question 9 STEP I and Setting y'=0 f(2)=-2, f(-2)=2, also for graphing purposes it may be worth to note that f(0)=0 and f(4)=2 and f(-4)=-2. y=0 has solutions x=0 or f''(-2)=- i.e. local max f''(2)=+ i.e. local min Graph, see attachment (a) i.e. therefore the equation of y in this X-Y plane is setting it equal to 0 gives f(1)=-2 ans f(-1)=2 For graphing it can also be good to see that f(2)=2, f(-2)=-2 and Graph, see attachment (b) i.e. therefore the equation of y in this X-Y plane is Setting this equal to 0 gives . f(2)=-1 and f(-2)=1 For graphing, also see that , f(4)=1 and f(-4)=-1 Graph, see attachment. (c) i.e. therefore the equation in the X-Y plane is Setting this equal to 0 gives f(0)=2, f(2)=-2. For graphing, also see that and f(-1)=-2, f(3)=2 (d) i.e. therefore the equation in the X-Y plane is setting this equal to 0 gives f(2)=0 f(-2)=2 For graphing I'll check f(4) and f(-4) too, which give 2 and 0 respectively. For the last part, we're looking for a graph in the X-Y plane with local min (0,0) and local max (1,1). To obtain this on the form X=ax+b and Y=cy+d, first observe that for a graph of a cubic, changing the sign means a reflection in the x-axis i.e. local max and min swap places, this is what we need to do. That means either a or c is negative (but not both!). a squeezes the graph along the X-axis and c squeezes it along the Y-axis, c moves the graph to the left/right and d moves the graph up/down. Knowing these we can see that and will produce the desired graph. Okay, so I hope I've interpreted everything correct in the question. Seems pretty nice as a question. 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# JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers. ## JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2 Question 1. The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Solution: The class interval having the maximum frequencies is 35 – 45. f1 = 23, l = 35, h = 10, f0 = 21, f2 = 14 Maximum number of patients admitted in the hospital are of the age 36.8 years. The average age of the patient admitted to the hospital is 35.37 years. Question 2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical Determine the modal lifetimes of the components. Class interval having the maximum frequency is 60 – 80. f1 = 61, f0 = 52, f2 = 38, l = 60, h = 20. Question 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: Expenditure (in Rs.) No. of families 1000 – 1500 24 1500 – 2000 40 2000 – 2500 33 2500 – 3000 28 3000 – 3500 30 3500 – 4000 22 4000 – 4500 16 4500 – 5000 7 Solution: Step deviation method: $$\bar{x}$$ = a + $$\left[\frac{\sum f_i u_i}{\sum f_i}\right]$$h = 3250 + $$\left[\frac{-235}{200}\right]$$ × 500 = 3250 – $$\left[-\frac{1175}{2}\right]$$ = 3250 – 587.5 = 2662.5. Mean expenditure Rs. 2662.50. l = 1500, f1 = 40, f2 = 33, f0 = 24, h = 500 Modal monthly expenditure = 1847.83. Question 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher No. of states/U.T. 15 – 20 3 20 – 25 8 25 – 30 9 30 – 35 10 35 – 40 3 40 – 45 0 45 – 50 0 50 – 55 2 Solution: l = lower limit of the CI = 30, f1 = 10, f0 = 9, f2 = 3, h = 5 Mode = $$l+\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right] \times \mathrm{h}$$ = $$=30+\left[\frac{10-9}{20-9-3}\right] \times 5$$ = $$30+\left[\frac{1}{8} \times 5\right]=30+\frac{5}{8}$$ = 30 + 0.625 = 30.625. Most states/UT’s have a student-teacher ratio of 30.6. On an average this ratio is 29.2. Question 5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored Number of batsmen 3000 – 4000 4 4000 – 5000 18 5000 – 6000 9 6000 – 7000 7 7000 – 8000 6 8000 – 9000 3 9000 – 10000 1 10000 – 11000 1 Find the mode of the data. Solution: l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000 Question 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data. Solution: Class interval having the maximum frequency is 40 – 50. f1 = 20, f0 = 12, f2 = 11, l = 40, h = 10.
• +91 9971497814 • info@interviewmaterial.com # Pa.Linear Eq Ex-3.4 Interview Questions Answers ### Related Subjects Question 1 : Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) x/2+ 2y/3 = -1 and x-y/3 = 3 Solutions: (i) x + y = 5 and 2x – 3y = 4 By the method of elimination. x + y = 5 ……………………………….. (i) 2x – 3y = 4 ……………………………..(ii) When the equation (i) is multiplied by 2, we get 2x + 2y = 10 ……………………………(iii) When the equation (ii) is subtracted from (iii) we get, 5y = 6 y = 6/5 ………………………………………(iv) Substituting the value of y in eq. (i) we get, x=5−6/5 = 19/5 ∴x = 19/5 , y = 6/5 By the method of substitution. From the equation (i), we get: x = 5 – y………………………………….. (v) When the value is put in equation (ii) we get, 2(5 – y) – 3y = 4 -5y = -6 y = 6/5 When the values are substituted in equation (v), we get: x =5− 6/5 = 19/5 ∴x = 19/5 ,y = 6/5 (ii) 3x + 4y = 10 and 2x – 2y = 2 By the method of elimination. 3x + 4y = 10……………………….(i) 2x – 2y = 2 ………………………. (ii) When the equation (i) and (ii) is multiplied by 2, we get: 4x – 4y = 4 ………………………..(iii) When the Equation (i) and (iii) are added, we get: 7x = 14 x = 2 ……………………………….(iv) Substituting equation (iv) in (i) we get, 6 + 4y = 10 4y = 4 y = 1 Hence, x = 2 and y = 1 By the method of Substitution From equation (ii) we get, x = 1 + y……………………………… (v) Substituting equation (v) in equation (i) we get, 3(1 + y) + 4y = 10 7y = 7 y = 1 When y = 1 is substituted in equation (v) we get, A = 1 + 1 = 2 Therefore, A = 2 and B = 1 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 By the method of elimination: 3x – 5y – 4 = 0 ………………………………… (i) 9x = 2y + 7 9x – 2y – 7 = 0 …………………………………(ii) When the equation (i) and (iii) is multiplied we get, 9x – 15y – 12 = 0 ………………………………(iii) When the equation (iii) is subtracted from equation (ii) we get, 13y = -5 y = -5/13 ………………………………………….(iv) When equation (iv) is substituted in equation (i) we get, 3x +25/13 −4=0 3x = 27/13 x =9/13 ∴x = 9/13 and y = -5/13 By the method of Substitution: From the equation (i) we get, x = (5y+4)/3 …………………………………………… (v) Putting the value (v) in equation (ii) we get, 9(5y+4)/3 −2y −7=0 13y = -5 y = -5/13 Substituting this value in equation (v) we get, x = (5(-5/13)+4)/3 x = 9/13 ∴x = 9/13, y = -5/13 (iv) x/2 + 2y/3 = -1 and x-y/3 = 3 By the method of Elimination. 3x + 4y = -6 …………………………. (i) x-y/3 = 3 3x – y = 9 ……………………………. (ii) When the equation (ii) is subtracted from equation (i) we get, -5y = -15 y = 3 ………………………………….(iii) When the equation (iii) is substituted in (i) we get, 3x – 12 = -6 3x = 6 x = 2 Hence, x = 2 , y = -3 By the method of Substitution: From the equation (ii) we get, x = (y+9)/3…………………………………(v) Putting the value obtained from equation (v) in equation (i) we get, 3(y+9)/3 +4y =−6 5y = -15 y = -3 When y = -3 is substituted in equation (v) we get, x = (-3+9)/3 = 2 Therefore, x = 2 and y = -3 Question 2 : Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction? Solution: Let the fraction be a/b According to the given information, (a+1)/(b-1) = 1 => a – b = -2 ………………………………..(i) a/(b+1) = 1/2 => 2a-b = 1…………………………………(ii) When equation (i) is subtracted from equation (ii) we get, a = 3 …………………………………………………..(iii) When a = 3 is substituted in equation (i) we get, 3 – b = -2 -b = -5 b = 5 Hence, the fraction is 3/5. (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Solution: Let us assume, present age of Nuri is x And present age of Sonu is y. According to the given condition, we can write as; x – 5 = 3(y – 5) x – 3y = -10…………………………………..(1) Now, x + 10 = 2(y +10) x – 2y = 10…………………………………….(2) Subtract eq. 1 from 2, to get, y = 20 ………………………………………….(3) Substituting the value of y in eq.1, we get, x – 3.20 = -10 x – 60 = -10 x = 50 Therefore, Age of Nuri is 50 years Age of Sonu is 20 years. (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Solution: Let the unit digit and tens digit of a number be x and y respectively. Then, Number (n) = 10B + A N after reversing order of the digits = 10A + B According to the given information, A + B = 9…………………….(i) 9(10B + A) = 2(10A + B) 88 B – 11 A = 0 -A + 8B = 0 ………………………………………………………….. (ii) Adding the equations (i) and (ii) we get, 9B = 9 B = 1……………………………………………………………………….(3) Substituting this value of B, in the equation (i) we get A= 8 Hence the number (N) is 10B + A = 10 x 1 +8 = 18 (iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received. Solution: Let the number of Rs.50 notes be A and the number of Rs.100 notes be B According to the given information, A + B = 25 ……………………………………………………………………….. (i) 50A + 100B = 2000 ………………………………………………………………(ii) When equation (i) is multiplied with (ii) we get, 50A + 50B = 1250 …………………………………………………………………..(iii) Subtracting the equation (iii) from the equation (ii) we get, 50B = 750 B = 15 Substituting in the equation (i) we get, A = 10 Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Solution: Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B. According to the information given, A + 4B = 27 …………………………………….…………………………. (i) A + 2B = 21 ……………………………………………………………….. (ii) When equation (ii) is subtracted from equation (i) we get, 2B = 6 B = 3 …………………………………………………………………………(iii) Substituting B = 3 in equation (i) we get, A + 12 = 27 A = 15 Hence, the fixed charge is Rs.15 And the Charge per day is Rs.3 krishan
# NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.2 ## Chapter 13 Ex.13.2 Question 1 If $${\text{P}}\left( {\text{A}} \right) = \frac{3}{5}{\text{ and P}}\left( {\text{B}} \right) = \frac{1}{5}$$, find $${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)$$if A and B are independent events. ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = \frac{3}{5}$$ and $${\text{P}}\left( {\text{B}} \right) = \frac{1}{5}$$ A and B are independent events. Thus, $${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{{25}}$$ ## Chapter 13 Ex.13.2 Question 2 Two cards are drawn at random and without replacement from a pack of $$52$$ playing cards. Find the probability that both the cards are black. ### Solution Since there are $$26$$ black cards in a deck of $$52$$ cards. Let $${\text{P}}\left( {\text{A}} \right)$$be the probability of getting a black card in the first draw. $$\therefore {\text{ P}}\left( {\text{A}} \right) = \frac{{26}}{{52}}\; = \frac{1}{2}$$ Let $${\text{P}}\left( {\text{B}} \right)$$be the probability of getting a black card on second draw. Since the card is not replaced, $$\therefore {\text{ P}}\left( {\text{B}} \right) = \frac{{25}}{{51}}$$ Therefore, probability of getting both the cards black$$= \frac{1}{2} \times \frac{{25}}{{51}} = \frac{{25}}{{102}}$$ ## Chapter 13 Ex.13.2 Question 3 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale. ### Solution Let A, B, and C be the respective events that the first, second, and the third drawn orange is good. Thus, probability that first drawn orange is good, $${\text{P}}\left( {\text{A}} \right) = \frac{{12}}{{15}}$$ The oranges are not replaced. Thus, probability of getting second orange good, $${\text{P}}\left( {\text{B}} \right) = \frac{{11}}{{14}}$$ Similarly, probability of getting third orange good, $${\text{P}}\left( {\text{C}} \right) = \frac{{10}}{{13}}$$ Since the box is approved for sale, if all the three oranges are good. Thus, probability of getting all the oranges good $$= \frac{{12}}{{15}} \times \frac{{11}}{{14}} \times \frac{{10}}{{13}} = \frac{{44}}{{91}}$$ Thus, the probability that the box is approved for sale is $$= \frac{{44}}{{91}}$$ ## Chapter 13 Ex.13.2 Question 4 A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘$$3$$ on the die’. Check whether A and B are independent events or not. ### Solution The sample space is given by, {\text{S = }}\left\{ \begin{align}{l}\left( {{\text{H,1}}} \right),\left( {{\text{H,2}}} \right),\left( {{\text{H,3}}} \right),\left( {{\text{H,4}}} \right),\left( {{\text{H,5}}} \right),\left( {{\text{H,6}}} \right)\\\left( {{\text{T,1}}} \right),\left( {{\text{T,2}}} \right),\left( {{\text{T,3}}} \right),\left( {{\text{T,4}}} \right),\left( {{\text{T,5}}} \right),\left( {{\text{T,6}}} \right)\end{align} \right\} Let A: Head appears on the coin \begin{align}&{\text{A = }}\left\{ {\left( {{\text{H,1}}} \right),\left( {{\text{H,2}}} \right),\left( {{\text{H,3}}} \right),\left( {{\text{H,4}}} \right),\left( {{\text{H,5}}} \right),\left( {{\text{H,6}}} \right)} \right\}\\ &\Rightarrow {\text{P}}\left( {\text{A}} \right) = \frac{6}{{12}} = \frac{1}{2}\end{align} B: $$3$$ on die$$= \left\{ {\left( {{\text{H,3}}} \right),\left( {{\text{T,3}}} \right)} \right\}$$ \begin{align}& \Rightarrow\; {\text{P}}\left( {\text{B}} \right) = \frac{2}{{12}} = \frac{1}{6}\\&{\text{A}} \cap {\text{B}} = \left\{ {\left( {{\text{H,3}}} \right)} \right\}\\&{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{{12}}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{2}{6} = {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\end{align} Therefore, A and B are independent events. ## Chapter 13 Ex.13.2 Question 5 A die marked $$1, 2, 3$$ in red and $$4, 5, 6$$ in green is tossed. Let A be the events, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent? ### Solution The sample space (S) is $${\text{S}} = \left\{ {1,2,3,4,5,6} \right\}$$ \begin{align}&{\text{Let A: the number is even}} = \left\{ {2,4,6} \right\}\\ &\Rightarrow\; \; {\text{P}}\left( {\text{A}} \right) = \frac{3}{6} = \frac{1}{2}\\&{\text{B: the number is red}} = \left\{ {1,2,3} \right\}\\& \Rightarrow\; \; {\text{P}}\left( {\text{B}} \right) = \frac{3}{6} = \frac{1}{2}\\&\therefore {\text{A}} \cap {\text{B}} = \left\{ {\text{2}} \right\}\end{align} \begin{align}&{\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{6}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \ne \frac{1}{6}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align} Therefore, A and B are not independent events. ## Chapter 13 Ex.13.2 Question 6 Let E and F be the events with $${\text{P}}\left( {\text{E}} \right) = \frac{3}{5}{\text{, P}}\left( {\text{F}} \right) = \frac{3}{{10}}{\text{ and P}}\left( {{\text{E}} \cap {\text{F}}} \right) = \frac{1}{5}$$. Are E and F are independent? ### Solution Given $${\text{P}}\left( {\text{E}} \right) = \frac{3}{5}{\text{, P}}\left( {\text{F}} \right) = \frac{3}{{10}}$$ and $${\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right) = \frac{1}{5}$$ \begin{align}&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{3}{5} \times \frac{3}{{10}} = \frac{9}{{50}} \ne \frac{1}{5}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) \ne {\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right)\end{align} Therefore, E and F are not independent. ## Chapter 13 Ex.13.2 Question 7 Given that the events A and B are such that $${\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{5}$$ and $${\text{P}}\left( {\text{B}} \right) = {\text{p}}$$. Find p if they are (i) mutually exclusive (ii) independent. ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{5}$$ and $${\text{P}}\left( {\text{B}} \right) = {\text{p}}$$ (i) When A and B are mutually exclusive, $${\text{A}} \cap {\text{B}} = \phi$$ \begin{align}&\therefore {\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0\\&{\text{Since, P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + {\text{p}} - 0\\ &\Rightarrow\; \; {\text{p}} = \frac{3}{5} - \frac{1}{2} = \frac{1}{{10}}\end{align} (ii) When A and B are independent,$${\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2}{\text{p}}$$ \begin{align}{\text{Since, P}}\left( {{\text{A}} \cup {\text{B}}} \right)& = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + {\text{p}} - \frac{1}{2}{\text{p}}\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + \frac{{\text{p}}}{2}\\ &\Rightarrow\; \; \frac{{\text{p}}}{2} = \frac{3}{5} - \frac{1}{2} = \frac{1}{{10}}\\ &\Rightarrow\; \; \frac{{\text{p}}}{2} = \frac{2}{{10}} = \frac{1}{5}\end{align} ## Chapter 13 Ex.13.2 Question 8 Let $$A$$ and $$B$$ be independent events with $${\text{P}}\left( {\text{A}} \right) = 0.3$$ and $${\text{P}}\left( {\text{B}} \right) = 0.4$$. Find \begin{align}&\left( {\text{i}} \right){\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\&\left( {{\text{ii}}} \right){\text{ P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\&\left( {{\text{iii}}} \right){\text{ P}}\left( {\left. {\text{A}} \right\|{\text{B}}} \right)\\&\left( {{\text{iv}}} \right){\text{ P}}\left( {\left. {\text{B}} \right\|{\text{A}}} \right)\end{align} ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = 0.3$$ and $${\text{P}}\left( {\text{B}} \right) = 0.4$$. If $$A$$ and $$B$$ are independent events, then \begin{align}{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) &= {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &= 0.3 \times 0.4\\& = 0.12\end{align} Since, $${\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)$$ $$\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = 0.3 + 0.4 - 0.12 = 0.58$$ Since, $${\text{P}}\left( {\left. {\text{A}} \right\|{\text{B}}} \right) = \frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}}$$ $$\Rightarrow\; \; {\text{P}}\left( {\left. {\text{A}} \right\|{\text{B}}} \right) = \frac{{0.12}}{{0.4}} = 0.3$$ Since, $${\text{P}}\left( {\left. {\text{B}} \right\|{\text{A}}} \right) = \frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}$$ $$\Rightarrow\; \; {\text{P}}\left( {\left. {\text{B}} \right\|{\text{A}}} \right) = \frac{{0.12}}{{0.3}} = 0.4$$ ## Chapter 13 Ex.13.2 Question 9 If A and B are two events such that $${\text{P}}\left( {\text{A}} \right) = \frac{1}{4}{\text{, P}}\left( {\text{B}} \right) = \frac{1}{2}$$and $${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{8}$$, find P (not A and not B). ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = \frac{1}{4}{\text{, P}}\left( {\text{B}} \right) = \frac{1}{2}$$ and $${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{8}$$. \begin{align}&{\text{P (not on A and not on B)}} = {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right)\\&{\text{P (not on A and not on B)}} = {\text{P}}\left( {{{\left( {{\text{A}} \cup {\text{B}}} \right)}^\prime }} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{A'}} \cap {\text{B' = }}\left( {{\text{A}} \cup {\text{B}}} \right){\text{'}}} \right]\\ &= 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\ &= 1 - \left[ {{\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)} \right]\\& = 1 - \left[ {\frac{1}{4} + \frac{1}{2} - \frac{1}{8}} \right]\\ &= 1 - \frac{5}{8}\\ &= \frac{3}{8}\end{align} ## Chapter 13 Ex.13.2 Question 10 Events A and B are such that $${\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {\text{B}} \right) = \frac{7}{{12}}$$ and P (not A or not B) $$= \frac{1}{4}$$. State whether A and B are independent? ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {\text{B}} \right) = \frac{7}{{12}}$$and P (not A or not B)$$= \frac{1}{4}$$ \begin{align} &\Rightarrow\; \; {\text{P}}\left( {{\text{A'}} \cup {\text{B'}}} \right) = \frac{1}{4}\\& \Rightarrow\; \; {\text{P}}\left( {{{\left( {{\text{A}} \cap {\text{B}}} \right)}^\prime }} \right) = \frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{A'}} \cup {\text{B' = }}\left( {{\text{A}} \cap {\text{B}}} \right){\text{'}}} \right]\end{align} \begin{align}& \Rightarrow\; \; 1 - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{4}\\ &\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....\;\left( 1 \right)\\&{\text{But}},\;{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{7}{{12}} = \frac{7}{{24}}\;\;\;\;\;.....\;\left( 2 \right)\\&{\text{Here,}}\;\frac{3}{4} \ne \frac{7}{{24}}\\&\therefore \;{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) \ne {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\end{align} Therefore, A and B are not independent events. ## Chapter 13 Ex.13.2 Question 11 Given two independent events $$A$$ and $$B$$ such $${\text{P}}\left( {\text{A}} \right) = 0.3,\;{\text{P}}\left( {\text{B}} \right) = 0.6$$. Find (i) $${\text{P}}\left( {{\text{A and B}}} \right)$$ (ii) $${\text{P}}\left( {{\text{A and not B}}} \right)$$ (iii) $${\text{P}}\left( {{\text{A or B}}} \right)$$ (iv) $${\text{P}}\left( {{\text{neither}}\;{\text{A nor B}}} \right)$$ ### Solution Given, $${\text{P}}\left( {\text{A}} \right) = 0.3\;{\text{and}}\;{\text{P}}\left( {\text{B}} \right) = 0.6$$ $$A$$ and $$B$$ are independent events. (i)$${\text{P}}\left( {{\text{A and B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)$$ $$\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0.3 \times 0.6 = 0.18$$ (ii)$${\text{P}}\left( {{\text{A and not B}}} \right) = {\text{P}}\left( {{\text{A}} \cap {\text{B'}}} \right)$$ \begin{align} &= {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &= 0.3 - 0.18\\ &= 0.12\end{align} (iii)$${\text{P}}\left( {{\text{A or B}}} \right) = {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)$$ \begin{align}& = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &= 0.3 + 0.6 - 0.18\\ &= 0.72\end{align} (iv)$${\text{P}}\left( {{\text{neither}}\;{\text{A nor B}}} \right) = {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right)$$ \begin{align} &= {\text{P}}\left( {{{\left( {{\text{A}} \cup {\text{B}}} \right)}^\prime }} \right) \qquad \left[ {{\text{A'}} \cap {\text{B'}} = \left( {{\text{A}} \cup {\text{B}}} \right){\text{'}}} \right]\\ &= 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\& = 1 - 0.72\\ &= 0.28\end{align} ## Chapter 13 Ex.13.2 Question 12 A die tossed thrice. Find the probability of getting an odd number at least once. ### Solution Probability of getting an odd number in a single throw of a die $$= \frac{3}{6} = \frac{1}{2}$$ Similarly, probability of getting an even number$$= \frac{3}{6} = \frac{1}{2}$$ Probability of getting an even number three times $$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$ $$=1 −$$ Probability of getting an odd number in none of the throws $$=1 −$$ Probability of getting an even number thrice \begin{align} = 1 - \frac{1}{8}\\ = \frac{7}{8}\end{align} ## Chapter 13 Ex.13.2 Question 13 Two balls are drawn at random with replacement from a box containing $$10$$ black and $$8$$ red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red. ### Solution Given, Total number of balls$$= 18$$ Number of red balls$$= 8$$ Number of black balls$$= 10$$ (i) Probability of getting a red ball in the first draw $$= \frac{8}{{18}} = \frac{4}{9}$$ The ball is replaced after the first draw. $$∴$$ Probability of getting a red ball in the second draw $$= \frac{8}{{18}} = \frac{4}{9}$$ Therefore, probability of getting both the balls red $$= \frac{4}{9} \times \frac{4}{9} = \frac{{16}}{{81}}$$ (ii) Probability of getting first ball black$$= \frac{{10}}{{18}} = \frac{5}{9}$$ The ball is replaced after the first draw. Probability of getting second ball as red $$= \frac{8}{{18}} = \frac{4}{9}$$ Therefore, probability of getting first ball as black and second ball as red =$$= \frac{5}{9} \times \frac{4}{9} = \frac{{20}}{{81}}$$ (iii) Probability of getting first ball as red$$= \frac{8}{{18}} = \frac{4}{9}$$ The ball is replaced after the first draw. Probability of getting second ball as black $$= \frac{{10}}{{18}} = \frac{5}{9}$$ Therefore, probability of getting first ball as black and second ball as red $$= \frac{4}{9} \times \frac{5}{9} = \frac{{20}}{{81}}$$ Therefore, probability that one of them is black and other is red $$=$$ Probability of getting first ball black and second as red $$+$$ Probability of getting first ball red and second ball black $$= \frac{{20}}{{81}} + \frac{{20}}{{81}} = \frac{{40}}{{81}}$$ ## Chapter 13 Ex.13.2 Question 14 Probability of solving specific problem independently by A and B are $$\frac{1}{2}$$ and respectively.If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem. ### Solution Probability of solving the problem by A, $${\text{P}}\left( {\text{A}} \right) = \frac{1}{2}$$ Probability of solving the problem by B, $${\text{P}}\left( {\text{B}} \right) = \frac{1}{3}$$ Since the problem is solved independently by A and B, \begin{align}&{\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\\&{\text{P}}\left( {{\text{A'}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) = 1 - \frac{1}{2} = \frac{1}{2}\\&{\text{P}}\left( {{\text{B'}}} \right) = 1 - {\text{P}}\left( {\text{B}} \right) = 1 - \frac{1}{3} = \frac{2}{3}\end{align} (i)Probability that the problem is solved$$= {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)$$ \begin{align} &= {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{AB}}} \right)\\ &= \frac{1}{2} + \frac{1}{3} - \frac{1}{6}\\& = \frac{4}{6} = \frac{2}{3}\end{align} (ii) Probability that exactly one of them solves the problem $${\text{ = P}}\left( {\text{A}} \right) \cdot {\text{P}}\left( {{\text{B'}}} \right) + {\text{P}}\left( {\text{B}} \right) \cdot {\text{P}}\left( {{\text{A'}}} \right)$$ \begin{align} &= \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3}\\ &= \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\end{align} ## Chapter 13 Ex.13.2 Question 15 One card is drawn at random from a well shuffled deck of $$52$$ cards. In which of the following cases are the events E and F independent? (i) E: ‘the card drawn is a spade’ F: ‘the card drawn is an ace’ (ii) E: ‘the card drawn is black’ F: ‘the card drawn is a king’ (iii) E: ‘the card drawn is a king and queen’ F: ‘the card drawn is a queen or jack’ ### Solution (i) In a deck of $$52$$ cards, $$13$$ cards are spades and $$4$$ cards are aces. \begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a spade}}} \right) = \frac{{13}}{{52}} = \frac{1}{4}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in an ace}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\end{align} In the deck of cards, only $$1$$ card is an ace of spades. \begin{align} &{\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{the card drawn is spade and an ace}}} \right) = \frac{1}{{52}} \hfill \\& {\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{1}{4} \times \frac{1}{{13}} = \frac{1}{{52}} = {\text{P}}\left( {{\text{EF}}} \right) \hfill \\& \Rightarrow\; \; {\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{EF}}} \right) \hfill \\ \end{align} Therefore, the events E and F are independent. (ii) In a deck of $$52$$ cards, $$26$$ cards are black and $$4$$ cards are kings. \begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a black}}} \right) = \frac{{26}}{{52}} = \frac{1}{2}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in an ace}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\end{align} In the pack of $$52$$ cards, $$2$$ cards are black as well as kings. \begin{align}{\text{P}}\left( {{\text{EF}}} \right) &= {\text{P}}\left( {{\text{the card drawn is black king}}} \right) = \frac{2}{{52}} = \frac{1}{{26}}\\{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right)& = \frac{1}{2} \times \frac{1}{{13}} = \frac{1}{{26}} = {\text{P}}\left( {{\text{EF}}} \right)\end{align} Therefore, the given events E and F are independent. (iii) In a deck of $$52$$ cards, $$4$$ cards are kings, $$4$$ cards are queens, and $$4$$ cards are jacks. \begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a king or a queen}}} \right) = \frac{8}{{52}} = \frac{2}{{13}}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in a queen or a jack}}} \right) = \frac{8}{{52}} = \frac{2}{{13}}\end{align} There are $$4$$ cards which are king or queen and queen or jack. \begin{align}&{\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{the card drawn is king or a queen, or queen or a jack}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{2}{{13}} \times \frac{2}{{13}} = \frac{4}{{169}} \ne \frac{1}{{13}}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) \ne {\text{P}}\left( {{\text{EF}}} \right)\end{align} Therefore, the given events E and F are not independent. ## Chapter 13 Ex.13.2 Question 16 In a hostel, 60% of the students read Hindi newspaper, $$40\%$$ read English newspaper and $$20\%$$ read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper. ### Solution Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper. Given, \begin{align}{\text{P}}\left( {\text{H}} \right) &= 60\% = \frac{{60}}{{100}} = \frac{3}{5}\\{\text{P}}\left( {\text{E}} \right)& = 40\% = \frac{{40}}{{100}} = \frac{2}{5}\\{\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)& = 20\% = \frac{{20}}{{100}} = \frac{1}{5}\end{align} (i) Probability that a student reads Hindi or English newspaper \begin{align}{\text{P}}\left( {{\text{H}} \cup {\text{E'}}} \right) &= 1 - {\text{P}}\left( {{\text{H}} \cup {\text{E}}} \right)\\&= 1 - \left\{ {{\text{P}}\left( {\text{H}} \right) + {\text{P}}\left( {\text{E}} \right) - {\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)} \right\}\\& = 1 - \left( {\frac{3}{5} + \frac{2}{5} - \frac{1}{5}} \right)\\&= 1 - \frac{4}{5} = \frac{1}{5}\end{align} (ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by $${\text{P}}\left( {{\text{E}}\|{\text{H}}} \right)$$. $${\text{P}}\left( {\left. {\text{E}} \right\|{\text{H}}} \right) = \frac{{{\text{P}}\left( {{\text{E}} \cap {\text{H}}} \right)}}{{{\text{P}}\left( {\text{H}} \right)}} = \frac{{\frac{1}{5}}}{{\frac{3}{5}}} = \frac{1}{3}$$ (iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H\|E). $${\text{P}}\left( {\left. {\text{H}} \right\|{\text{E}}} \right) = \frac{{{\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)}}{{{\text{P}}\left( {\text{E}} \right)}} = \frac{{\frac{1}{5}}}{{\frac{2}{5}}} = \frac{1}{2}$$ ## Chapter 13 Ex.13.2 Question 17 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is \begin{align}&\left( {\text{A}} \right)\;0\\\\&\left( {\text{B}} \right)\frac{1}{3}\\\\&\left( {\text{C}} \right)\frac{1}{{12}}\\\\&\left( {\text{D}} \right)\frac{1}{{36}}\end{align} ### Solution When two dice are rolled, the number of outcomes is $$36$$. The only even prime number is $$2$$. Let $${\text{E}}$$ be the event of getting an even prime number on each die. \begin{align}&\therefore \;\;{\text{E}} = \left\{ {\left( {2,2} \right)} \right\}\\& \Rightarrow\; \; {\text{P}}\left( {\text{E}} \right) = \frac{1}{{36}}\end{align} Therefore, the correct answer is D. ## Chapter 13 Ex.13.2 Question 18 Two events A and B will be independent, if (A) A and B are mutually exclusive (B)$${\text{P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]$$ (C)$${\text{P}}\left( {\text{A}} \right) = {\text{P}}\left( {\text{B}} \right)$$ (D)$${\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) = 1$$ ### Solution Two events A and B are said to be independent, if $${\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)$$ Let’s take option B. \begin{align}&{\text{P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]\\& \Rightarrow\; \; {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {\text{B}} \right) + {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow\; \; 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {\text{B}} \right) + {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right) - {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\& \Rightarrow\; \; {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right){\text{ - P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\& \Rightarrow\; \; {\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\end{align} This implies that A and B are independent, if $${\text{ P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]$$ A.$${\text{Let}}\;{\text{P}}\left( {\text{A}} \right) = {\text{m}},\;{\text{P}}\left( {\text{B}} \right) = {\text{n, 0}} < {\text{m}},{\text{n}} < {\text{1}}$$ A and B are mutually exclusive. \begin{align}&\therefore \;\;{\text{A}} \cap {\text{B}} = \phi \\& \Rightarrow\; \; {\text{P}}\left( {{\text{AB}}} \right) = 0\\&{\text{However, P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = {\text{mn}} \ne 0\\&\therefore \;{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align} C.Let A: Event of getting an odd number on throw of a die$$= \left\{ {1,3,5} \right\}$$ $$\Rightarrow\; \; {\text{P}}\left( {\text{A}} \right) = \frac{3}{6} = \frac{1}{2}$$ Let B: Event of getting an even number on throw of a die$$= \left\{ {2,4,6} \right\}$$ \begin{align} &\Rightarrow\; \; {\text{P}}\left( {\text{B}} \right) = \frac{3}{6} = \frac{1}{2}\\&{\text{Here,}}\;{\text{A}} \cap {\text{B}} = \phi \\&\therefore \;\;{\text{P}}\left( {{\text{AB}}} \right) = 0\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{4} \ne 0\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align} D. From the above example, it can be seen that, $${\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} + \frac{1}{2} = 1$$ However, it cannot be inferred that A and B are independent. Thus, the correct answer is B. Instant doubt clearing with Cuemath Advanced Math Program
# Eureka Math Precalculus Module 5 Lesson 13 Answer Key ## Engage NY Eureka Math Precalculus Module 5 Lesson 13 Answer Key ### Eureka Math Precalculus Module 5 Lesson 13 Exercise Answer Key Exercises 1–4 Exercise 1. Let X represent the monetary value of the prize that you win playing this game 1 time in dollars. Complete the table below, and calculate E(X). Exercise 2. Regarding the E(X) value you computed above, can you win exactly that amount on any 1 play of the game? No, you cannot win exactly $1.25 on any 1 play. On average, you are expected to win$1.25 per play if you were to play many times. Exercise 3. What is the least you could win in 4 attempts? What is the most you could win in 4 attempts? The least amount that could be won in 4 attempts is $2.00 (4∙$0.50), and the most is $20.00 (4∙$5.00). Exercise 4. How would you explain the E(X) value in context to someone who had never heard of this measurement? What would you expect for the total monetary value of your prizes from 4 attempts at this game? (To answer this question, check the introduction to this lesson; do not do anything complicated like developing a tree diagram for all of the outcomes from 4 attempts at the game.) On average, a person is expected to win $5.00 (4∙E(X)) for every 4 attempts. The winnings could be as little as$2.00 (4 small) or as large as $20.00 (4 larges). If 4 attempts were made many times, the winnings would average out to be about$5.00. Exercises 5–8 Exercise 5. Examining the outcomes for just 1 mouse, define the random variable, and complete the following table: Let X represent the number of left turns made by 1 mouse. Exercise 6. Using this value and the rule mentioned above, determine the expected number of left turns for 3 mice. (Remember, the value you compute may not be an exact, attainable value for 1 set of 3 mice. Rather, it is the average number of left turns by 3 mice based on many sets of 3 mice.) The expected value is 2.1 left turns. As stated in the student exercise, the value may not be an exact, attainable value for 1 set of 3 mice. Rather, it is the average number of left turns by 3 mice based on numerous attempts. The tree diagram below demonstrates the 8 possible outcomes for 3 mice where the first stage of the tree represents the decision made by the 1st mouse and the second stage represents the decision made by the 2nd mouse, and so on. Exercise 7. Use the tree diagram to answer the following questions: a. Complete the following table, and compute E(Y), the expected number of left turns for 3 mice. Let Y represent the number of left turns made by 1 mouse. b. Verify that the expected number of left turns for 3 mice, E(Y), is the same as 3 times the expected number of left turns for 1 mouse, 3∙E(X). Yes. E(Y) = 2.1 = 3∙E(X) = 3∙0.7 Exercise 8. Imagine that 200 mice are sent through the maze one at a time. The researchers believed that the probability of a mouse turning left due to the food is 0.7. How many left turns would they expect from 200 mice? 200∙0.7 = 140 They would expect 140 left turns. Exercises 9–13 Exercise 9. Compute a player’s net earnings for each of the 3 outcomes: small, medium, and large. Small: $0.50 –$2.00 = – $1.50 Medium:$1.50 – $2.00 = –$0.50 Large: $5.00 –$2.00 = + $3.00 Exercise 10. For two of the outcomes, the net earnings result is negative. What does a negative value of net earnings mean in context as far as the player is concerned? Answer: A negative expected value from the player’s perspective means a loss. Exercise 11. Let Y represent the net amount that you win (or lose) playing the duck game 1 time. Complete the table below, and calculate E(Y). Answer: Exercise 12. How would you explain the E(Y) value in context to someone who had never heard of this measurement? Write a sentence explaining this value from the perspective of a player; then, write a sentence explaining this value from the perspective of the charity running the game. Answer: To a player: The game is designed such that if you play it over and over again, the people running the game expect that you will lose$0.75 to them on average per attempt. To the people running the game: The game is designed such that if people play this game over and over again, you can expect to make about $0.75 on average per attempt. Exercise 13. How much money should the charity expect to earn from the game being played 100 times? Answer: 100∙$0.75 = $75.00 In 100 attempts, the charity should expect to earn$75.00. ### Eureka Math Precalculus Module 5 Lesson 13 Problem Set Answer Key Question 1. The Maryland Lottery Pick 3 game described in the Exit Ticket has a variety of ways in which a player can bet. Instead of the Front Pair bet of $0.50 described above with a payout of$25.00, a player could make a Front Pair bet of $1.00 on a single ticket for a payout of$50.00. Let Y represent a player’s net gain or loss from playing 1 game in this manner. a. Compute E(Y). E(Y) = 0.99∙ – $1.00 + 0.01∙$49.00 = – $0.99 +$0.49 = – $0.50 b. On average, how much does the Maryland Lottery make on each such bet? Answer: The Maryland Lottery makes on average$0.50 for each such bet. c. Assume that for a given time period, 100,000 bets like the one described above were placed. How much money should the Maryland Lottery expect to earn on average from 100,000 bets? 100 000∙$0.50 =$50000.00 Question 2. Another type of carnival or arcade game is a spinning wheel game. Imagine that someone playing a spinning wheel game earns points (payoff) as follows for each spin: • You gain 2 points 50% of the time. • You lose 3 points 25% of the time. • You neither gain nor lose any points 25% of the time. The results of each spin are added to one another, and the object is for a player to accumulate 5 or more points. Negative total point values are possible. a. Develop a model of a spinning wheel that would reflect the probabilities and point values. b. Compute E(X) where X represents the number of points earned in a given spin. E(X) = 2∙0.50 + ( – 3)∙0.25 + 0∙0.25 = 0.25 0.25 points will typically be earned in a given spin. c. Based on your computation, how many spins on average do you think it might take to reach 5 points? $$\frac{5}{0.25}$$ = 20. It would take on average 20 spins. d. Use the spinning wheel you developed in part (a) (or some other randomization device) to take a few spins. See how many spins it takes to reach 5 or more points. Comment on whether this was fewer spins, more spins, or the same number of spins you expected in part (c). Answers will vary. In most cases, a player can reach 5 or more points with very few spins, often far fewer than the average of 20. e. Let Y represent the number of spins needed to reach 5 or more points (like the number of spins it took you to reach 5 points in part (d)), and repeat the simulation process from part (d) many times. Record on a dot plot the various values of Y you obtain. After several simulations, comment on the distribution of Y. Answers will vary with students’ work. For parts (d) and (e), 1 simulation distribution (computer generated) of Y represents the number of spins needed to reach 5 or more points was as follows: The distribution is very skewed right, and sometimes it takes many spins to get to 5 or more points. The average number of spins needed based on these 33 simulations was 23.12 spins, close to the E(Y) = 20. In most cases, a player can reach 5 or more points with very few spins, often far fewer than the average of 20. (Note: This may be a good chance to remind students that it is important to see the center, shape, and spread of a distribution before drawing conclusions about a variable.) ### Eureka Math Precalculus Module 5 Lesson 13 Exit Ticket Answer Key Question 1. As posted on the Maryland Lottery’s website for its Pick 3 game, the chance of winning with a Front Pair bet is 0.01. (http://mdlottery.com/games/pick – 3/payouts/ accessed on November 17, 2013) A Front Pair bet is successful if the front pair of numbers you select match the Pick 3 number’s first 2 digits. For example, a bet of 12X would be a winner if the Pick 3 number is 120,121,122, and so on. In other words, 10 of the 1,000 possible Pick 3 numbers (1%) would be winners, and thus, the probability of winning is 0.01 or 1%. A successful bet of $0.50 pays out$25.00 for a net gain to the player of $24.50. a. Define the random variable X, and compute E(X). Answer: Let X represent a player’s NET gain or loss from playing 1 game in this manner. E(X) = 0.99∙ –$0.50 + 0.01∙$24.50 = -$0.495 + $0.245 = -$0.25 b. On average, how much does the Maryland Lottery make on each such bet? The Maryland Lottery makes on average $0.25 for each such bet. c. Assume that for a given time period, 100,000 bets like the one described above were placed. How much money should the Maryland Lottery expect to earn on average from 100,000 bets? Answer: 100 000∙$0.25 = $25000.00 Note: According to the Maryland Lottery Gaming and Control Agency’s Annual Report for Fiscal Year 2012, the Pick 3 game accounted for$254.60 million in net sales. (http://mlgca.com/annual – report/ accessed November 17, 2013)
# Percent Composition I Lesson In mathematics we not only sometimes need to find percentages of things, but also need to express amounts of things in percentages. For example we know there is a rainwater tank has $24$24L of water in it but can hold $50$50L, how do we know what percentage is full? ## Back to Fractions The key here is look at a situation carefully, gather all the relevant information and extract the correct fraction that we can use to convert into a percentage. Let's go back to our rainwater tank example. What do you think the fraction would be if we wanted to find how full the tank was? Remember that in a fraction the numerator represents how much there is and the denominator represents the total capacity. So here our fraction must be $\frac{24}{50}$2450! Another way this question might be asked without the tank is: what percentage is $24$24 of $50$50? To work out this question we would ALSO need the fraction $\frac{24}{50}$2450 Then to convert it into a percentage it is as easy as multiplying it by $100%$100%: $\frac{24}{50}\times100%$2450×100% $=$= $\frac{12}{25}\times100%$1225×100% simplify the fraction $=$= $\frac{12\times100}{25}$12×10025 $%$% write as one fraction $=$= $\frac{12\times4\times25}{25}$12×4×2525 $%$% split up the $100$100, so I can cancel out common factors $=$= $12\times4%$12×4% cancel the $25$25's $=$= $48%$48% evaluate! #### Worked Examples ##### Question 1 What percentage is $99$99 of $110$110? ##### Question 2 What percentage is $134$134 L of $536$536 L? ##### Question 3 When Bart looked at the bill from the mechanic, the total cost of repairs was $£800$£800. $£640$£640 of this was for labour and the rest was for replacement of parts. 1. What percentage of the cost of repairs was for labour? 2. What percentage of the cost of repairs was for replacement of parts?
# Is a zero matrix a diagonal matrix ## The identity matrix Further definitions in connection with matrices are briefly introduced below. For practical use, some special cases of a matrix must be listed here. In the matrix calculation some special cases of matrices are defined. For example, vectors, the zero matrix and the diagonal matrix are special cases of a matrix. A diagonal matrix is a square matrix and is characterized by the fact that all elements that are not on the main diagonal are zero. If all elements on the main diagonal are also 1, this matrix is called Identity matrix designated. Definition: [identity matrix] The identity matrix \$ {\ rm \ bf I} \$ is a square matrix whose elements on the main diagonal are 1 and the remaining elements are all 0. Example The (3,3) -unit matrix is given as an example of an identity matrix: \ begin {equation } \ mathbf {I} = \ begin {pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ end {pmatrix }. \ end {equation } The identity matrix is the neutral element in matrix multiplication. For arbitrary matrices \$ {\ rm \ bf A _ {(m, m)}} \$ the following applies: \ begin {equation } {\ rm \ bf A \ cdot I = I \ cdot A = A}. \ end {equation } Thus the matrix \$ {\ rm \ bf I} \$ corresponds to the number 1 when multiplying the matrix in \$ \ mathbb {R} \$. In Chapter 1, an inverse element was defined for the multiplication of real numbers. Compare definition \ ref {???}. In \$ \ mathbb {R} \$ there is an inverse element \$ \ alpha ^ {- 1} \$ for every \$ \ alpha \ neq 0 \$. This procedure can be used for the matrix calculation, but only for a given regular square matrix \$ \ rm \ bf {A} \$% (cf. definition% \ ref {regular matrix}) an inverse matrix \$ \ rm \ bf {A ^ {- 1}} \$ can be formed. Definition [Inverse Matrix] Let \$ {\ rm \ bf A} \$ be a square matrix. Then there is a (also square) matrix \$ {\ rm \ bf B} \$ for which the following applies: \ begin {equation } {\ rm \ bf A \ cdot B = B \ cdot A = I}, \ end {equation } so one calls \$ {\ rm \ bf B} \$ the inverse matrix of \$ {\ rm \ bf A} \$ and writes \$ {\ rm \ bf A ^ {- 1}} \$ for it. The following applies to the inverse matrix: \ begin {equation } {\ rm \ bf A ^ {- 1} \ cdot A = I}. \ end {equation } Give examples of square matrices without inverse. Note: Take (2,2) -matrices \$ \ ne \ rm \ bf O \$ with as many zeros as possible and try to construct an inverse. solution For example, we are looking for an inverse \$ \ mathbf {X ^ {- 1}} \$ to the matrix \$ X = \ begin {pmatrix} 1 & 0 \ 0 & 0 \ end {pmatrix}. \$ To do this, we have to solve the following relationship: \$\$ \ begin {pmatrix} 1 & 0 \ 0 & 0 \ end {pmatrix} \ cdot \ begin {pmatrix} x_ {11} & x_ {12} \ x_ {21} & x_ {22} \ end { pmatrix} = \ begin {pmatrix} 1 & 0 \ 0 & 1 \ end {pmatrix} \$\$ \$\$ \ begin {pmatrix} 1 \ cdot x_ {11} +0 \ cdot x_ {21} & 1 \ cdot x_ {12} +0 \ cdot x_ {22} \ 0 \ cdot x_ {11} +0 \ cdot x_ {21} & 0 \ cdot x_ {12} +0 \ cdot x_ {22} \ end {pmatrix} = \ begin {pmatrix} 1 & 0 \ 0 & 1 \ end {pmatrix} \$\$ \$\$ \ begin {pmatrix} x_ {11} & x_ {12} \ 0 & 0 \ end {pmatrix} = \ begin { pmatrix} 1 & 0 \ 0 & 1 \ end {pmatrix}. \$\$ This relationship cannot be resolved, however, since the lower right element of the right-hand matrix differs from that of the left-hand matrix. Remark: An inverse matrix exists if and only if: The rank of a \$ n \ times n \$ matrix Rg \$ \ rm \ bf {A} \$ equals \$ n \$, i.e. the number of linearly independent column vectors and at the same time the number of independent ones Line vectors. If the rows and columns of an existing \$ m \ times n \$ matrix \$ \ rm \ bf {A} \$ are swapped in such a way that a \$ n \ times m \$ matrix results, then the newly created \$ n \ times m is \$ -Matrix \$ \ rm \ bf {A '} \$ a matrix transposed to the output matrix \$ \ rm \ bf {A} \$. Definition [transposed / overturned matrix] A transposed matrix (also called overturned matrix) is created by exchanging rows and columns of a matrix. For a matrix \$ {\ rm \ bf A} \$ the transposed or overturned matrix is noted as \$ {\ rm \ bf A ^ T} \$. The following applies to the elements of the matrix \$ \ bf A \$: \ begin {equation } (a_ {ij}) ^ T = (a_ {ji}). \ end {equation } Note: By swapping the indices, every row of the output matrix \$ \ bf A \$ becomes columns of the matrix \$ \ bf A ^ T \$ and the columns of \$ \ bf A \$ become rows of the matrix \$ \ bf A ^ T \$. #### example Transpose a square matrix \ begin {equation } \ mathbf {A} = \ begin {pmatrix} 1 & 2 & 7 \ 8 & 3 & -5 \ 9 & 4 & 6 \ end {pmatrix} \ Longrightarrow \ mathbf {A ^ T} = \ begin {pmatrix} 1 & 8 & 9 \ 2 & 3 & 4 \ 7 & -5 & 6 \ end {pmatrix}. \ end {equation } In the case of square matrices, transposing corresponds to mirroring the elements on the main diagonal. Example transposing a non-square matrix \ begin {equation } \ mathbf {A} = \ begin {pmatrix} 1 & -2 \ 0 & 3 \ 5 & 7 \ end {pmatrix} \ Longrightarrow \ mathbf {A ^ T} = \ begin {pmatrix} 1 & 0 & 5 \ -2 & 3 & 7 \ end {pmatrix}. \ end {equation } Transpose the matrix! Transpose the matrix! Square matrices, which have the property that they are symmetrical with respect to their main diagonals, become symmetric matrices called. Symmetrical matrices are characterized by the fact that they correspond to their transposed matrix. Definition [symmetric matrix] A square matrix is called \$ \ rm \ bf A \$ symmetrical, if the following applies to the transposed matrix: \ begin {equation } {\ rm \ bf A ^ T = A}. \ end {equation } #### example The transposed matrix \$ \ bf A ^ T \$ corresponds to the original matrix \$ \ bf A \$: \ begin {equation } \ mathbf {A} = \ mathbf {A ^ T} = \ begin {pmatrix} 1 & 5 & 7 \ 5 & 2 & 3 \ 7 & 3 & 4 \ cr \ end {pmatrix}. \ end {equation }
# What are the two definitions of a derivative? ## What are the two definitions of a derivative? The definition of the derivative can be approached in two different ways. One is geometrical (as a slope of a curve) and the other one is physical (as a rate of change). ## What are the definitions of a derivative? The definition of the derivative is The slope of a line that lies tangent to the curve at the specific point. The limit of the instantaneous rate of change of the function as the time between measurements decreases to zero is an alternate derivative definition. ## Which is the best definition of a derivative? The derivative is The instantaneous rate of change of a function with respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a point. ## What is the first definition of derivative? A derivative is simply A measure of the rate of change. It can be the rate of change of distance with respect to time or the temperature with respect to distance. We want to measure the rate of change of a function y = f ( x ) y = f(x) y=f(x) with respect to its variable x x x. ## What are the 4 main types of derivatives? The four major types of derivative contracts are Options, forwards, futures and swaps. ## What is the limit definition of derivative? Limit Definition of the Derivative. We define the derivative of a function f(x) at x = x0 as. F (x0) = lim. H→0. F(x0 + h) − f(x0) ## What is first derivative and second derivative? Y = f ′ ( x ) . In other words, just as The first derivative measures the rate at which the original function changes, the second derivative measures the rate at which the first derivative changes. The second derivative will help us understand how the rate of change of the original function is itself changing. 🔗 ## What is the second derivative called? In calculus, the second derivative, or the Second order derivative, of a function f is the derivative of the derivative of f. ## What is the physical meaning of derivatives? The derivative is defined as An instantaneous rate of change at a given point. We usually differentiate two kinds of functions, implicit and explicit functions. Explicit functions are the functions in which the known value of the independent variable “x” directly leads to the value of the dependent variable “y”. ## What is the importance of slope in the definition of derivative? A derivative of a function is a representation of the rate of change of one variable in relation to another at a given point on a function. The slope Describes the steepness of a line as a relationship between the change in y-values for a change in the x-values. ## What are types of financial derivatives? In finance, there are four basic types of derivatives: Forward contracts, futures, swaps, and options. ## What are the features of derivative? Features of Derivatives: • Derivatives have a maturity or expiry date post which they terminate automatically. • Derivatives are of three types i.e. futures forwards and swaps and these assets can equity, commodities, foreign exchange or financial bearing assets. ## What are the major types of derivative securities? The main types of derivatives are Futures, forwards, options, and swaps. An example of a derivative security is a convertible bond. Such a bond, at the discretion of the bondholder, may be converted into a fixed number of shares of the stock of the issuing corporation. ## Is limit and derivative the same thing? The derivative is the slope of a function at some point on the function. The limit is your best guess at where the function will eventually end up when it approaches a particular number. The slope of a function could be 0 and it could be approaching 2 at x=0 if the function is y=2, for example. ## What is the formula for derivative? Derivative of the function y = f(x) can be denoted as F′(x) or y′(x). ## Whats the difference between dy dx and d dx? Derivative of the function y = f(x) can be denoted as F′(x) or y′(x).
The Educator's PLN The personal learning network for educators # Basic Concepts Of Trigonometry Trigonometry is a branch of mathematics. It is a central of mathematics. Trigonometry is all about the relation between the sides and angles of the triangles. The father of the trigonometry is “Hipparchus”, he introduced first trigonometry table. Trigonometry is the subject which is easy to understand if approach in a right manner. Go through the article to understand the basic concepts. Right Angle Triangle: Right angle triangle is a triangle in which one of its interior angle is 90 degree. The opposite side of the right angle triangle is hypothesis and the other two sides are adjacent and opposite. Basic sin, cos and tan functions are as follows, a) Sin θ = Opposite/Hypothesis b) Cos θ = Adjacent/Hypothesis c) Tan θ = Adjacent/Opposite Pythagoras Theorem: Pythagoras theorem is used to find the sides of the right angle. Pythagoras theorem is, (Hypotenuse)^2= (Adjacent)^2 + (Opposite)^2 Triangle Identities: Triangle identities are the function. There are some identities which are true for right angle triangles and some are true for all triangles. Understand more about trigonometry identities. Sine and cosine rule: Sine and cosine rule are very important in trigonometry. This is useful to solve triangle which is not a right angle. Sine rule: a/sin A = b/sin B = c/sin C Cosine rule: a^2 = b^2+c^2-2ab cos A b^2 = a^2+c^2-2ab cos B c^2 = a^2+b^2-2ab cos C Here a, b and c are the side of the triangle. And A, B and C are the angle of the triangle. Important points: • Understand the Trigonometry table. • To understand trigonometry you can have an online math tutor chat. • To know more about trigonometry basic take online trig help. Views: 18379 Comment ### You need to be a member of The Educator's PLN to add comments! Join The Educator's PLN Thomas Whitby created this Ning Network. ## Latest Activity Kevin Johnson posted a blog post ### How To Celebrate Your Graduation Differently? Graduation is the biggest event a student lives for, the pleasure of making it through the road is above anything in the world. You may be jumping on your feet, doing random dances and your joyous screams could be heard on Mars. Yes, it is totally understandable to be that crazy and happy for such a milestone you achieved. Although, take a moment thanking…See More Tuesday Yuna Buhrman, Deborah J. Oldman-Brown, Stephanie Cressman and 1 more joined The Educator's PLN Tuesday Kyle Kresge joined Jill Galloway's group ### Project-Based Learning (PBL) This group is for those who implement the project-based learning model or want to learn more about it. Saturday Kyle Kresge joined Shelly Terrell's group ### Edchat Join this group to extend the discussion of edchat topics! Saturday Ruth Herman Wells posted discussions Dec 6 Christine Hinkley updated an event ### OLC Innovate 2019 at Gaylord Rockies Resort and Convention Center April 3, 2019 at 8am to April 5, 2019 at 5pm At OLC Innovate, we will conduct a series of engaging hands-on sessions aimed at fostering organic interactions and collaborative cross-disciplinary problem-solving. Together we will build new foundations for stronger, better higher education environments. And because innovation scales best when ideas are shared, our work sessions will explore digital technologies and adapted teaching behaviors aimed at informing policy, inspiring leadership, and evolving practice at all levels impacting…See More Dec 6 katelyn brown and Janine Leiterman joined The Educator's PLN Dec 6 Kyle Kresge joined Child's Work's group ### Special Education A group based on Special Education needs. Where teachers, parents, and caregivers can share and distribute information for the classroom and beyond. See More Dec 5
MEDIUM DATA STRUCTURES AND ALGORITHMS # How to solve the Recover Binary Search Tree Problem TREESINORDER TRAVERSAL Written By Kenny Polyak Jai Pandya ## Introduction Recover Binary Search Tree The Recover Binary Search Tree problem asks us to restore a binary search tree to its original form after two of its nodes have been swapped. This problem requires a strong understanding of the structure and properties of a binary search tree, namely that all nodes to the left of the root are smaller than the root and all nodes to the right are larger. ## Problem Recover Binary Search Tree ### Example Inputs and Outputs #### Example 1 Input: `````` 1 / 3 \ 2 `````` Output: `````` 3 / 1 \ 2 `````` #### Example 2 Input: `````` 4 / \ 2 5 / \ \ 6 3 1 `````` Output: `````` 4 / \ 2 5 / \ \ 1 3 6 `````` Constraints • The number of nodes in this tree is in the range [2, 1000] • -5000 <= Node.val <= 5000 ## Solution Recover Binary Search Tree Watch Someone Solve Recover Binary Search Tree Netflix InterviewJava Interview Advance this person to the next round? Technical Skills: 3/4 Problem Solving Ability: 3/4 Communication Ability: 3/4 Show Transcript ### Recover Binary Search Tree Approaches #### Overview and Intuition In this question, we are provided with the root of a binary search tree, well, an almost binary search tree. Someone accidentally swapped two of its nodes, turning the BST into a regular binary tree. We are on a quest to find those two nodes. Once we find them, it shouldn't be difficult to swap them again to restore the original BST. Let's first understand the characteristics of this special kind of tree called a "binary search tree" or BST. The "special" part comes from the order in which it stores its nodes. As with a binary tree, each node of a BST can have at most two children, one left and one right. But each node maintains a special relationship with its children and parent. ``````class TreeNode: def __init__(self, data): self.data = data self.left = None self.right = None `````` In a BST, the value of any node is greater than the value of all nodes in its left subtree. In contrast, the value of any node is smaller than the value of all nodes in its right subtree. All nodes are related in this way to their respective left and right subtrees. Now that we have some clarity about BST, let's return to our original problem. It says that two of the nodes of the given BST have been swapped. Let's look at what a BST might look like if that were the case. Here we can see that the original tree satisfies all the properties of a BST. As soon as the nodes with values `1` and `6` are swapped, they no longer satisfy these properties. `6` is larger than `2` and `4`, but it is located in the left subtree of the two nodes after the swap. Similarly, `1` is smaller than `5` and `4`, but it is in the right subtree of `5` and `4`. We can take advantage of another interesting property of a binary search tree, which relates to its in-order 'depth first search' traversal. In in-order traversal, the left subtree is processed first, then the current node, and then the right subtree. We continue this recursively for each node. We already know that in a BST, the left child is always smaller, while the right child is always larger than the current node. The in-order traversal thus ensures that the nodes with smaller values are processed before those with higher values, i.e., in ascending sort order. For the given example - in-order traversal of the original configuration would produce the following array - `[1, 2, 3, 4, 5, 6]`. After swapping the two nodes, it changes to - `[6, 2, 3, 4, 5, 1]` Starting with this basic idea, we will examine three approaches to this problem. ### Approach 1: Export to an Array - Two Pass Solution #### Intuition In the previous section, we learned about the idea of in-order traversal. We take it further and implement it in this approach. We traverse the given tree in an in-order fashion. When we process a node, we push its reference into a list. If this were a perfect BST, the list or array would be perfectly sorted, and all nodes would be in ascending order of their values. In our case, however, there are two defectors. The problem has now changed slightly. Given a list of almost sorted nodes, we need to find the two nodes that are not in the correct order. Eventually, we will interchange their values. There are two possibilities - 1. When the defectors are next to each other in the resultant array. Example: `[1, 2, 4, 3, 5, 6]` We can see that `4` and `3`, which are right next to each other, have swapped positions. The ascending order is broken only at one place, namely at the pair `(4, 3)`, which are also the two nodes we are trying to find. 2. When the defectors are at least one element apart in the output list. Example: `[6, 2, 3, 4, 5, 1]` In this case, the perfect sort is broken for two pairs of values - `(6, 2)` and `(5, 1)`. It turns out that `5` and `1` are the errant nodes. #### Algorithm 1. We traverse the given tree in the in-order sequence, and push the processed nodes into a new list `almostSorted`. In the solution here, we use the recursive DFS algorithm to simplify the implementation. We could also use the iterative version. 2. If there are `N` elements in the list, iterate the index `i` through `almostSorted` from `0` till `N - 2`. 3. For each pair `almostSorted[i]` and `almostSorted[i + 1]`, if they are not in ascending order, we note them in a variable `swapped`. If `swapped` already contains two nodes, it means that we have already encountered a pair out of order. This means that `almostSorted[i + 1]` must be the second defector. 4. Now that we have identified the two swapped elements, we just need to swap their values again. 5. Return the `root` of the binary tree, which has now become a perfect BST. #### Code and Implementation ``````def recover_bst(self, root: Optional[TreeNode]) -> None: almost_sorted = [] swapped = None # traverse the given tree in in-order fashion # and populate `almostSorted` array def in_order(node: Optional[TreeNode]): if node is None: return # Notice the order here # 1. left child 2. current node 3. right child in_order(node.left) almost_sorted.append(node) in_order(node.right) in_order(root) # `almost_sorted` is a sorted array except for the two # nodes which have been swapped with each other. Find those # nodes for i in range(len(almost_sorted) - 1): if almost_sorted[i].val > almost_sorted[i + 1].val: if swapped is None: swapped = [almost_sorted[i], almost_sorted[i + 1]] else: swapped[1] = almost_sorted[i + 1] # Exchange the value of the swapped nodes swapped[0].val, swapped[1].val = (swapped[1].val, swapped[0].val) return root `````` #### Time/Space Complexity • Time Complexity - `O(n)`, where `n` is the number of nodes in the tree. During in-order traversal, we iterate through every node in the tree to build the almost sorted list. Then once again, we iterate through all the elements. So this solution takes `O(n)` time. In other words, the time taken grows linearly with the size of the tree. • Space Complexity - `O(n)`, where `n` is the number of nodes in the tree. We use an auxiliary data structure, a list, to store all processed nodes. It requires `O(n)` of memory. Recursion also requires space on the implicit stack, which can be as large as the height of the recursion tree - `O(n)` in the worst case. So this solution requires `O(n)` space on memory. ### Approach 2: In-order Recursive Traversal - Single Pass #### Intuition A tiny optimization over the naive approach can save us a pass through all nodes. In the previous solution, we create a list of nodes resulting from in-order traversal. But do we really need it? The only purpose of the list is to give us easy access to the pairs of nodes `almostSorted[i]` and `almostSorted[i + 1]]` in order. However, we could also access them during the traversal. When we traverse the tree in in-order fashion, we can treat the current node as `almostSorted[i + 1]`, and its predecessor (the last node processed) represents `almostSorted[i]`. It turns out that it makes no sense to create an auxiliary list. In a valid BST, the in-order predecessor should contain a smaller value than the current node. If we encounter a node that does not follow this rule, we note the predecessor and the current node. There can be, at most, two such nodes. #### Algorithm Traverse the tree in order. We use a recursive DFS algorithm here. Each node compares its value to its in-order predecessor. The node processed last is the in-order predecessor of a node. We store it in a global variable `lastProcessed`. Let's make a recursive function `inOrder` and pass `root` as the current `node`. 1. Termination condition / base case - when the current `node` is `null`, the function should simply return. 2. Recursively call `inOrder` with the left child as the current `node`. 3. If `lastProcessed` is empty, it indicates that the current node is the leftmost child of the original tree, so it has no predecessor. As a result, no comparisons as well. 4. If `lastProcessed` is not empty - compare its value to the current node. If `lastProcessed.value > node.value`, then do the following: We store the pair `lastProcessed` and `node` in a two cell array `swapped`. If `swapped` already contains an element, it means the current pair is the second pair in our search. So we store `node` at `swapped[1]`. 5. Update `lastProcessed` to the current node. 6. Make a recursive call to `inOrder` with the right child as the current `node`. 7. At the end, we swap the values of the nodes contained in `swapped`. 8. Return the `root` of the tree. #### Code and Implementation ``````def recover_bst(self, root: Optional[TreeNode]) -> None: swapped = last_processed = None # traverse the given tree recursively in # in-order fashion. left => current => right # and find the pair with swapped nodes def find_swapped_pair(current: Optional[TreeNode]): nonlocal swapped, last_processed if current is None: return find_swapped_pair(current.left) # compare the current nodes with its predecessor # there are two such nodes, so we find the condition # to be true at most two times if last_processed is not None: if last_processed.val > current.val: if swapped is None: # encounter an out of order node for the first time swapped = [last_processed, current] else: # encounter out of order node second time swapped[1] = current last_processed = current find_swapped_pair(current.right) find_swapped_pair(root) # exchange the values of both the nodes swapped[0].val, swapped[1].val = swapped[1].val, swapped[0].val return root `````` #### Time/Space Complexity • Time Complexity - `O(n)`, where `n` is the number of nodes in the tree. We traverse through all nodes in the tree. So, time taken in processing the given tree increases linearly with the tree size. The time complexity is `O(n)`. • Space Complexity - `O(n)`, where `n` is the number of nodes in the tree. We do not create a list of processed nodes, but the recursive implementation requires memory on the implicit recursion stack. The size of the recursion stack can grow to the height of the tree. When the tree is balanced, the height is `log n`. In the worst case, when each node has only one child, `O(n)` memory is required on the recursion stack. Thus, the space complexity is `O(n)`. ### Approach 3: In-order Iterative Traversal - Single Pass #### Intuition We have already seen a recursive implementation of the solution. By using an explicit stack in place of an implicit recursion stack, we can convert the solution to use iteration instead of recursion. We try to mimic what happens in the recursive implementation. During in-order traversal we process the left children before the current node. So we push the current node into a stack and move to the left child. In other words, we update the current pointer to point to the left child. We repeat this process until we reach a node without a left child. At this point because the left child is empty, we can process the current element. After processing the current element, the order needs to move to the right side. So we update the current pointer to point to the right side node. Now, the current node cant be processed before its left children. So we push it into the stack and update the current pointer to point to its left child. We should see a pattern repeating here, which can be easily implemented using a loop. In order to compare an element to its predecessor, we also keep track of the last processed node. The comparison happens right before the current pointer moves to the right side of a node. #### Algorithm Traverse the tree in in-order fashion. We use iteration here. Each node compares its value to its in-order predecessor. The node processed last is the in-order predecessor of a node. We store it in a global variable, `lastProcessed`. Initialize an empty stack `stack`, and a pointer `current` pointing at the root node. Initialize `lastProcessed` and `swapped` to contain `null` values. Do the following until the stack is not empty or `current` points at a valid node: 1. While `current` points at a valid node a. Push the node pointed by `current` into the stack b. Update `current` to point to its left child 2. `current` should be pointing at `null` at this moment. So we update `current` by popping the last stored node from the stack. 3. If we have processed a node earlier, compare the current node's value with the last processed node. If we find the last processed node to have a value greater than the current node, we have found a node out of order. We put both `lastProcessed` and `current` into the array `swapped`. If `swapped` already contains a node, it means the current node must be the second of the swapped pair. So we update `swapped[1]` with the current node and end the traversal here. At the end, we update `lastProcessed` to point to the current node. After traversal, we exchange the values of the nodes contained in `swapped` and return the `root` of the tree. #### Code and Implementation ``````def recover_bst(self, root: Optional[TreeNode]) -> None: stack = [] current = root last_processed = swapped = None # process until the stack contains something # or the current pointer points to a node while stack or current: # reach as far left as possible and put all # the nodes encountered into the stack while current: stack.append(current) current = current.left # current is None, repopulate it from the # stack's top current = stack.pop() # critical step - compare the current node with the # last processed node. If the node doesn't follow the # expected order, note the pair of nodes if last_processed and last_processed.val > current.val: if swapped: swapped[1] = current break else: swapped = [last_processed, current] # update last_processed node last_processed = current # move to the right side current = current.right # swap the values of the two nodes swapped[0].val, swapped[1].val = swapped[1].val, swapped[0].val return root `````` #### Time/Space Complexity • Time Complexity - `O(n)`, where `n` is the number of nodes in the tree. We traverse through all nodes in the tree. So, the time taken to process the given tree increases linearly with the size of the tree. The time complexity is `O(n)`. • Space Complexity - `O(n)`, where `n` is the number of nodes in the tree. The size of the stack used can grow to the height of the binary tree. When the tree is balanced, its height is `log n`. In the worst case, when each node has only one child, the tree height becomes `n`. Thus, `O(n)` memory is needed to accommodate the stack. Thus, the space complexity is `O(n)`. #### Bonus - Morris Traversal This approach may not be very relevant from the perspective of an interview. But it could be an interesting exercise if you want to try it out. We will not go into the details of this approach but will give you a basic idea here. In the recursive and iterative versions described above, we consume `O(n)` of space on the implicit or explicit stack. We use a stack to temporarily store the root nodes of different subtrees and the order in which they are processed. We put a node in the stack when we go down to the left child and remove it from the stack when we go back up. Could there be a way to skip the stack and still be able to go back up once we have examined the children of a node? This is precisely what we can do with Morris traversal. As we traverse a binary tree in order, the predecessor makes a temporary connection to the next node. Once we have explored all the nodes of the left subtree, the connection between the predecessor and its successor helps us move back up (or down) the tree. After that, the predecessor is returned to its original shape. This particular type of tree, where the predecessor nodes maintain a connection to their successor nodes, is also called a "threaded binary tree". ## Practice Questions Similar to Recover Binary Search Tree MEDIUM Data Structures and Algorithms Build a Max Heap Given an array of integers, transform the array in-place to a max heap. Watch 1 interview MEDIUM Data Structures and Algorithms Three Sum Given an array of integers, return an array of triplets such that i != j != k and nums[i] + nums[j] + nums[k] = 0. Watch 1 interview EASY Data Structures and Algorithms
## 187 – The P vs NP problem This note is based on lecture notes for the Caltech course Math 6c, prepared with A. Kechris and M. Shulman. 1. Decision problems Consider a finite alphabet ${A}$, and “words” on that alphabet (the “alphabet” may consist of digits, of abstract symbols, of actual letters, etc). We use the notation ${A^}$ to indicate the set of all “words” from the alphabet ${A}$. Here, a word is simply a finite sequence of symbols from ${A}$. For example, if ${A}$ is the usual alphabet, then $\displaystyle awwweeeedddfDDkH$ would be a word. We are also given a set ${V}$ of words, and we say that the words in ${V}$ are valid. (${V}$ may be infinite.) In the decision problem associated to ${V}$, we are given as input a word in this alphabet. As output we say yes or no, depending on whether the word is in ${V}$ or not (i.e., whether it is “valid”). We are interested in whether there is an algorithm that allows us to decide the right answer. For example: 1. ${V}$ may be the set of prime numbers. As input we are given a positive integer ${n}$. An algorithm to see whether ${n}$ is prime consists on dividing ${n}$ by ${2,3,\dots,n-1}$ and checking whether the answer is ever exact. If it is (i.e., if we find some number ${k}$ that divides ${n}$), then ${n}$ is not prime, and we return No. If we do not find such number ${k}$, then ${n}$ is prime and we return Yes. 2. ${V}$ may be the set of propositional tautologies. As input we are given a propositional formula ${A}$. An algorithm to see if ${A}$ is a tautology consists in writing down the truth table for ${A}$ and seeing if we always get truth value ${T}$. If so, we return Yes. 3. The “alphabet” could consists of the names for American cities. ${V}$ may be a set of pairs ${city_1city_2}$ for which we can travel by plane from ${city_1}$ to ${city_2}$ spending under $300. Websites like Expedia try to implement algorithms that allow us to solve the associated decision problem. 4. Using the usual alphabet together with the usual keyboard symbols, let ${V}$ be the set of computer programs in language ${C^{++}}$ that eventually stop if executed with input ${1}$. In this case there is no algorithm to see whether a given computer program is in ${V}$. In computer science this is called the halting problem. 5. If ${V}$ is the set of mathematical statements that are true, there is no algorithm either. This means that there is no mechanical way of checking, given a mathematical statement ${S}$, whether it holds. (${S}$ could be “${3>5}$” or Fermat’s last theorem or “There are integers ${x}$ and ${y}$ such that ${x^2-17y^2=-1}$” or …) Of course, one way of verifying the truth of ${S}$ could be to find a proof of ${S}$ or a proof of ${\lnot S}$. One could simply have a program that writes down all possible proofs, and if it ever writes down a proof of either ${S}$ or ${\lnot S}$, then we stop and have our answer. However, this method does not work in general. A remarkable result of Gödel from 1930 says that, no matter what axioms we begin with, there are mathematical truths that can neither be proved nor disproved. This is his famous Incompleteness theorem. If an algorithm exists for the decision problem corresponding to a set ${V}$, we say that ${V}$ is decidable. Otherwise we call it undecidable. Thus, modulo our rather vague definition of “algorithm,” we have the first main classification of (decision) problems as decidable or undecidable. Our next goal will be to classify further the decidable problems according to the (time) complexity of their algorithms and distinguish between those for which an efficient (feasible) algorithm is possible, the tractable problems, and those for which it is not, the intractable problems. There is a mathematical way of making precise the notion of algorithm, using “Turing machines.” We will not need the precise definition in what follows. For information on Alan Turing, please visit http://www.mathcomp.leeds.ac.uk/turing2012/ Other approaches are possible, and go by the common name of “models of computation.” It is a remarkable fact that the very different models that have been considered are all actually equivalent. 2. The Class ${P}$ The study of efficiency of algorithms is also called computational complexity. We will concentrate here on time complexity (how long it takes to solve the problem), but one can also discuss, for example, space complexity (how much “memory” storage is used in solving it). A natural measure of time complexity for us is the number of steps an algorithm requires to solve a given problem. Since different problems require different amounts of input, in order to compare the complexities of different algorithms and problems, we must measure complexity as a function of the input size. In addition, with computer speeds increasing rapidly, small instances of a problem can usually be solved no matter how difficult the problem is. For this reason, and also to eliminate the effect of “overhead” time, we will consider only the asymptotic complexity as the size of the input increases. First we recall the “big-${O}$ notation,” which gives a rigorous way of comparing asymptotic growth rates. Definition 1 Given two functions ${f,g:{\mathbb N}\rightarrow {\mathbb N}}$ we define $\displaystyle f=O(g)\text{ iff }\exists n_0\,\exists C\,\forall n\geq n_0(f(n)\leq Cg(n)).$ For example, if ${p(n)}$ is a polynomial, then ${p(n)=O(n^d)}$ for large enough ${d}$. Definition 2 Given any ${T:{\mathbb N}\rightarrow{\mathbb N}}$, we let ${time(T)}$ consist of all decision problems ${V}$ which can be decided by an algorithm in time ${t}$ for some ${t=O(T)}$. More precisely, a decision problem ${V}$ is in ${time(T)}$ if there is ${t=O(T)}$ and an algorithm ${M}$ on some alphabet ${B\supseteq A\cup\{Y,N\}}$ such that for ${w\in A^}$: 1. ${w\in P\Longrightarrow}$ on input ${w}$, ${M}$ halts in ${\leq t(\vert w\vert )}$ steps with output ${Y.}$ 2. ${w\not\in P\Longrightarrow}$ on input ${w}$, ${M}$ halts in ${\leq t(\vert w\vert )}$ steps with output ${N}$ (here ${\vert w\vert}$ = length of the word ${w}$). It is important to remark here that, even though it is standard convention, the notation ${f=O(g)}$ is somewhat imprecise, because there are many different functions ${f}$, all of which are ${O(g)}$ for the same ${g}$, so saying ${f=O(g)}$ and ${h=O(g)}$ does not mean that ${f=h}$, the ${=}$ symbol in “${f=O(g)}$” is just a convention, and should not be confused with any kind of actual equality. What sort of growth rate should we require of an algorithm to consider it manageable? After all, we must expect the time to increase somewhat with the input size. It turns out that the major distinctions in complexity are between “polynomial time” algorithms and faster-growing ones, such as exponentials. Definition 3 A decision problem is in the class ${P}$ (or it is polynomially decidable) if it is ${time(n^d)}$ for some ${d\geq 0}$, i.e. it can be decided in polynomial time. Problems in the class ${P}$ are considered tractable (efficiently decidable) and the others intractable. It is clear that the class ${P}$ provides an upper limit for problems that can be algorithmically solved in realistic terms. If a problem is in ${P}$, however, it does not necessarily mean that an algorithm for it can be practically implemented, for example, it could have time complexity of the order of ${n^{1,000,000}}$ or of the order ${n^3}$ but with enormous coefficients. However, most natural problems that have been shown to be polynomially decidable have been found to have efficient (e.g., very low degree) algorithms. Moreover, the class of problems in ${P}$ behaves well mathematically, and is independent of the model of computation, since any two formal models of computation can be mutually simulated within polynomial time. It is important to remark that time complexity, as explained here, is a worst case analysis. If a problem ${P}$ is intractable, then there is no polynomial time algorithm which for all ${n}$ and all inputs of length ${n}$ will decide ${P}$. But one can still search for approximate algorithms that work well on the average or for most practical (e.g., small) instances of the problem or give the correct answer with high probability, etc. Example 1 LINEAR PROGRAMMING is the decision problem given by: An input is an integer matrix ${(v_{ij})}$ for ${1\leq i\leq m}$ and ${1\leq j\leq n}$, along with integer vectors ${D=(d_i)^m_{i=1}}$ and ${C=(c_j)^n_{j=1}}$, and an integer ${B}$. The question we want to decide is whether there a rational vector ${X=(x_j)^n_{j=1}}$ such that ${\sum^n_{j=1} v_{ij}x_j\leq d_i}$, for ${1\leq i\leq m}$, and ${\sum^n_{j=1}c_jx_j\geq B}$? LINEAR PROGRAMMING turns out to be in ${P}$. This is a difficult result. It was proved by Khachian, in 1979. 3. The Class ${NP}$ and the ${P =NP}$ Problem Although a large class of problems have been classified as tractable or intractable, there is a vast collection of decision problems, many of them of great practical importance, that are widely assumed to be intractable but no one until now has been able to demonstrate it. These are the so-called ${NP}$-complete problems. In order to introduce these problems, we must first define the class ${NP}$ of non-deterministic polynomial decision problems. Definition 4 Let ${V}$ be a decision problem. We say that ${V}$ is in the class ${NP}$ if there is an algorithm ${M}$ on an alphabet ${B\supseteq A\cup\{Y,N\}}$ and a polynomial ${p(n)}$ such that for any ${w\in A^}$: ${w\in P\Longleftrightarrow \exists v\in B^}$ such that ${\vert v\vert\leq p(\vert w\vert)}$ and on input ${wv}$, ${M}$ stops with output ${Y}$ after at most ${p(\vert w\vert )}$ many steps. In other words, ${w\in P}$ iff there is a “guess” ${v}$, of length bounded by a polynomial in the length of ${w}$, such that ${w}$ together with ${v}$ pass a polynomial acceptance test. (This can be also viewed as a non-deterministic polynomial time algorithm.) Example 2 SATISFIABILITY, the decision problem of whether a propositional sentence is satisfiable, is in ${NP}$, since if we can guess a truth assignment, we can verify that it satisfies the given set of clauses in polynomial time. Example 3 TRAVELING SALESMAN is the following decision problem: We are given a finite collection of cities ${c_1,\dots,c_n}$, together with information about the distance between them, ${d(c_i,c_j)}$ being the distance between ${c_i}$ and ${c_j}$. We are also given a bound ${B}$. We want to decide whether here is a way to travel through all the cities and returning to the original point, without actually traveling more than distance ${B}$, i.e., whether there is a way to order the cities, say ${c_2,c_7,c_5,\dots,c_3}$ so that we have $\displaystyle d(c_2,c_7)+d(c_7,c_5)+d(c_5,c_3)+\dots+d(c_3,c_2)\le B.$ It turns out that TRAVELING SALESMAN is also in ${NP}$, since if we guess the order of the set of cities, we can calculate whether the length of the tour is ${\leq B}$ in polynomial time. In fact a vast number of problems like these, which involve some kind of search, are in ${NP}$. Clearly ${P\subseteq NP\subseteq\bigcup_d\text{TIME}(2^{n^d})}$. The most famous problem in theoretical computer science is whether $\displaystyle P =NP,$ that is, whether any problem with an efficient nondeterministic algorithm also has an efficient deterministic algorithm. This is known, unsurprisingly, as the ${P =NP}$ Problem. Recently the Clay Mathematics Institute included the ${P =NP}$ Problem as one of its seven Millenium Prize Problems, offering$1,000,000 for its solution. The prevailing assumption today is that ${P\neq NP}$. There are many excellent resources on the web with detailed information on this problem. I highly recommend Scott Aaronson’s blog, Shtetl-Optimized, http://www.scottaaronson.com/blog/ See for example his lectures on “Great Ideas in Theoretical Computer Science.” 4. ${NP}$-Complete Problems We can get a better understanding of the ${P =NP}$ problem by discussing the notion of an ${NP}$-complete problem. The ${NP}$-complete problems form sort of a “core” of the “most difficult” problems in ${NP}$. The key notion is the following: Definition 5 A (total) function ${f:A^\rightarrow B^}$, where ${A,B}$ are finite alphabets, is polynomial-time computable if there is an algorithm ${M}$ on a finite alphabet ${C\supseteq A\cup B}$, and a polynomial ${p}$, such that for every input ${w\in A^}$, ${M}$ terminates on at most ${p(\vert w\vert)}$ steps with output ${f(w)}$. Unsurprisingly, a polynomial-time reduction is a computable reduction which is in addition polynomial-time computable. Definition 6 A decision problem ${Q}$ in some alphabet ${B}$ is ${NP}$-complete if it is in ${NP}$, and for every ${NP}$ problem ${P}$ (in some alphabet ${A}$) there is a polynomial-time reduction of ${P}$ to ${Q}$. (That is, there is a polynomial-time computable function ${f:A^\rightarrow B^}$ such that ${w\in P}$ if and only if ${f(w)\in Q}$.) An ${NP}$-complete problem is in some sense a hardest possible problem in ${NP}$. For example, it is clear that if ${Q}$ is in ${P}$ and ${R}$ can be polynomial-time reduced to ${Q}$, then also ${R}$ is in ${P}$. It therefore follows that if any ${NP}$-complete problem is in ${P}$, then ${P = NP}$. Thus the ${P= NP}$ question is equivalent to the question of whether any given ${NP}$-complete problem is in ${P}$. It is clear from the definition that all ${NP}$-complete problems are “equivalent” in some sense, since each one can be reduced to the other by a polynomial-time computable function. The first example of an ${NP}$-complete problem is due to Cook and Levin in 1971. Theorem 7 (Cook, Levin) SATISFIABILITY is ${NP}$-complete. In a sense, this suggests that there may not be any way of solving SATISFIABILITY than by means of the very slow algorithm of writing down truth tables. However, there are polynomial-time algorithms that solve SATISFIABILITY in many (but not all) situations. This is a problem with many practical applications, and therefore “SAT-solving engines” that are “in general” fast are of great interest. Karp in 1972 has shown that TRAVELING SALESMAN and many other combinatorial problems are ${NP}$-complete, and since that time hundreds of others have been discovered in many areas of mathematics and computer science. Typeset using LaTeX2WP. Here is a printable version of this post. [Edit (Feb. 28, 2012): See also Search and Destroy (in Spanish), by Javier Moreno.]
NCERT Solutions: Mensuration (Exercise - 9.1, 9.2, 9.3) # NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration - 1 (Exercise 9.1) ### Exercise 9.1 Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. Ans: Area of the trapezium 1/2 × (a + b) × h 1/2 × (1.2 + 1) × 0.8 1/2 × 2.2 × 0.8 = 0.88 m2 Hence, the required area = 0.88 m2 Q2. The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel sides. Ans: Given: Area of trapezium = 34 cm2 Length of one of the parallel sides a = 10 cm height h = 4 cm Area of the trapezium = 1/2 × (a + b) × h 34 = 1/2 × (10 + b) × 4 ⇒ 34 = (10 + b) × 2 ⇒ 17 = 10 + b ⇒ b = 17 – 10 = 7 cm Hence, the required length = 7 cm. Q3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC. Ans: Length of the fence = Perimeter ∴ AB + BC + CD + DA = 120 or AB + 48 + 17 + 40 = 120 or AB + 105 = 120 or AB = 120 – 105 = 15 m ∵ Area of a trapezium = 1/2 * [Sum of parallel sides] * Height = 1/2 × (AD + BC) × AB = 1/2 × (40 + 48) × 15 M2 = 1/2 × 88 × 15 M2 = 44 × 15 M2 = 660 M2 Q4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field. Ans: Area of the field = area of ∆ABD + area of ∆BCD = 1/2 × b × h + 1/2 × b × h = 1/2 × 24 × 13 + 1/2 × 24 × 8 = 12 × 13 + 12 × 8 = 12 × (13 + 8) = 12 × 21 = 252 m2 Hence, the required area of the field = 252 m2. Q5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Ans: Given: d1 =7.5 cm and d2 = 12 cm Area of the rhombus = 1/2 × d1 × d2 = 1/2 × 7.5 × 12 = 7.5 × 6 = 45 cm2 Hence, area of the rhombus = 45 cm2. Q6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. Ans: Given: Side = 5 cm Altitude = 4.8 cm Length of one diagonal = 8 cm Area of the rhombus = Side × Altitude = 5 × 4.8 = 24 cm2 Area of the rhombus = 1/2 × d1 × d2 24 = 1/2 × d1 × d 24 = 4d2 d2 = 6 cm Hence, the length of other diagonal = 6 cm. Q7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4. Ans: Given: Number of tiles = 3000 Length of the two diagonals of a tile = 45 cm and 30 cm Area of one tile = 1/2 × d1 × d2 = 1/2 × 45 × 30 = 45 × 15 = 675 cm2 Area covered by 3000 tiles = 3000 × 675 cm2 = 2025000 cm2 = 202.5 m2 Cost of polishing the floor = 202.5 × 4 = ₹ 810 Hence, the required cost = ₹ 810. Q8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. Ans: Perpendicular distance (h) = 100 m (Given) Area of the trapezium-shaped field = 10500 m2 (Given) Let the side along the road be ‘x’ m and the side along the river = 2x m Area of the trapezium field = (1/2)×(a + b) × h 10500 = (1/2) × (x + 2x) × 100 10500 = 3x × 50 After simplifying, we have x = 70, which means the side along the river is 70 m Hence, the side along the river = 2x = 2(70) = 140 m. Q9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface. Ans: This regular octagon can be into two trapeziums (each having height 4 m and parallel sides as 11 m and 5 m) and a rectangular part with length 11 m and breadth 5 m. Area of a trapezium = 1/2 [11m + 5m] × 4m = 1/2 × 16 × 4m2 = 32m2 ∴ Area of both the trapeziums = 2 × 32 m2 = 64 m2 Now, Area of the rectangular part = l * b = 11 m × 5 m = 55 m2 ∴ Area of the regular octagon = 64 m2 + 55 m2 = 119 m2 Q10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area? Ans: Jyoti’s diagram: The given shape is splitted into two congruent trapeziums. Area of one trapezium = 1/2 × [15 + 30] × 15/2 m2 = 1/2 × 45 × 15/2 m2 = 675/4 m2 ∴ Area of the pentagonal shape = 2 × 675/4 m2 = 675/2 m2 = 337.5 m2 Kavita’s diagram: The given shape is splitted into a square and a triangle. Area of the square = 15 m × 15 m = 225 m2 Area of the triangle = 1/2 × 15 × 15 =225/2 m2 = 112.5 m2 ∴ Area of the pentagonal shape = 112.5 + 225 = 337.5 m2 Another way of finding the area of the pentagonal shape: By splitting the pentagonal shape into 3 triangles, we have: Area of triangle I = 1/2 × 15 × 15m= 225/2 m2 Area of triangle II = 1/2 × 15 × 15m= 225/2 m2 Area of triangle III = 1/2 × 15 × 15m= 225/2 m2 ∴ Area of the pentagonal shape = 225/2 m2 + 225/2 m2 + 225/2 m2 = 337.5 m2 Q11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same. Ans: Divide given figure into 4 parts, as shown below: Here two of the given figures (I) and (II) are similar in dimensions.And also, figures (III) and (IV) are similar in dimensions. Area of figure (I) = Area of trapezium = 1/2 × (a + b) × h = 1/2 × (28 + 20) × 4 = 1/2 × 48 × 4 = 96 Area of figure (I) = 96 cm2 Also, Area of figure (II) = 96 cm2 Now, Area of figure (III) = Area of trapezium = 1/2 × (a + b) × h = 1/2 × (24 + 16)4 = 1/2 × 40 × 4 = 80 Area of figure (III) is 80 cm2 Also, Area of figure (IV) = 80 cm2 ## Exercise - 9.2 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? Solution :- To determine which box requires less material, we are going to compare their surface areas. For the first cuboidal box, given – Length(l) = 60 cm Height (h) = 50 cm Total surface area of a cuboid = 2(lb + bh + hl) 2(60 × 40 + 40 × 50 + 50 × 60) (putting the value of l, b, h in the formula) = 2(2400 + 2000 + 3000) = 2 × 7400 = 14800 cm² For the second cuboidal box, given – Length(l) = 50 cm Height (h) = 50 cm Total surface area of a cuboid = 2(lb + bh + hl) 2(50 × 50 + 50 × 50 + 50 × 50) (putting the value of l, b, h in the formula) = 2(2500 + 2500 + 2500) = 2 × 7500 = 15000 cm² The total surface area of the cuboidal box (a) is less, cuboidal box (a) would require less material to make. (Box with the smaller surface area requires less material.) 2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases? Solution :- For the suitcase, given – Length(l) = 80 cm Height (h) = 24 cm Total surface area of a cuboid = 2(lb + bh + hl) Surface area of one suitcase = 2(80 × 48 + 48 × 24 + 24 × 80) cm² = 2(3840 + 1152 + 1920) = 2 × 6912 = 13824 cm² To cover 1 suitcase, area of tarpaulin cloth required will be = 13824 cm² Given, width of the tarpaulin cloth = 96 cm Area of tarpaulin cloth required to cover 1 suitcase = Total surface area of the suitcase Area of tarpaulin cloth required to cover 1 suitcase = 13824 cm² =) Length × Width = 13824 (since tarpaulin is in the form of rectange. Area of Rectangle = Length × Width) =) Length × 96 = 13824 =) Length = 13824/96 = 144 cm Length of tarpaulin cloth required to cover 100 suitcases = 100 × 144 = 144000 cm = 144000/100 (converting centimeter to meter by dividing with 100) = 144 m A total of 144 m of tarpaulin cloth with width of 96 cm is required to cover 100 suitcases. 3. Find the side of a cube whose surface area is 600 cm². Solution :- Total surface area of a cube = 6 × side². Given, total surface area of the cube is 600 cm² 600 cm² = 6 × side². side² = 600/6 Solving for side, side = √(600/6) = √100 = 10 cm 4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.Solution :- Given the dimensions of the cabinet – Length(l) = 1 m Height (h) = 1.5 m The surface area of the painted cabinet = Total surface area of the cabinet – Area of the bottom = 2(lb + bh + hl) – lb = lb + 2bh + 2hl = 2(1 × 2 + 2 × 2 × 1.5 + 2 × 1.5 × 1) = (2 + 6 + 3) m² = 11 m² Area of cabinet she covers is 11 m² 5. Daniel is painting the walls and ceiling of a cuboidal hall with length breadth and height of 15 m, 10 m, and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will he need to paint the room? Solution :- Given the dimensions of the wall – Length(l) = 15 m Height (h) = 7 m Surface area of walls and ceiling to be painted = 2h(l + b) + l × b = 2 × 7(15 + 10) + 15 × 10 = 2 × 7 × 25 + 150 = 350 + 150 = 500 m² Each can can paint and area of 100 m² Number of cans required for painting = Area to be painted/Area painted by 1 can = 500 m²/100 m² = 5 Daniel require 5 cans of paint to paint the hall 6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?Solution :- Similarities: – Both figures have the same height of 7 cm. – The diameter of the cylinder and the side length of the cube are both 7 cm. Differences: – The cylinder has a circular base while the cube has a square base. – The cylinder’s lateral surface is curved while the cube’s lateral surface consists of four flat squares. For the cylinder – Diameter = 7 cm = 7/2 cm Height = 7 cm Lateral Surface Area of the cube = 4 × side² = 4 × (7 cm × 7 cm) = 196 cm². Curbed lateral Surface Area of the cylinder = 2πr × h = 2 × 22/7 × 7/2 cm × 7 cm = 154 cm². The cube(2nd figure) has a larger lateral surface area than the cylinder. 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required? Solution :- The surface area of a cylinder includes the areas of its two circular ends and its lateral surface. Surface area = 2πr² (ends) + 2πrh (lateral surface). Given, for radius of the cylinder(r) = 7 m and Height (h) = 3 m: Surface area = 2 × 22/7 × 7² + 2 × 22/7 × 7 × 3 = 2 × 22 × 7 + 2 × 22 × 3 = 308 + 132 = 440 m² 440 m² sheet of metal is required to build a closed cylindrical tank 8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet? Solution :- When we cut the hollow cylinder along its height, we get a rectangular sheet of width 33 cm. This is height of the cylinder. Here we can assume radius of the hollow cylinder be r Lateral surface area of the cylinder = length of the rectangle × width of the rectangle. For a lateral surface area of 4224 cm² and a width of 33 cm: Lateral surface area of the cylinder = 2πrh 4224 = 2πrh (since, Lateral surface area = 4224) 2 × 22/7 × r × 33 = 4224 r = 4224 × 7/(2 × 22 × 33) = (7 × 96)/33 = (7 × 33)/11 = 21 cm On cutting the hollow cylinder, the base’s perimeter is equal to length of rectangular sheet Length of the rectangle = Perimeter of circular base Perimeter of circular base = 2πr = 2 × 22/7 × 21 = 128 cm. Perimeter of the rectangle = 2 × (length + width) = 2 × (128 cm + 33 cm) = 2 × 161 cm = 322 cm The perimeter of rectangular sheet is 322 cm 9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m. Hint: The road roller is in the shape of a cylinder. The area covered by the roller will be equal to curved surface area of the roller. Solution :- Given – Length of the roller = 1 m = 100 cm Diameter of road roller = 84 cm Curved surface area of the cylinder = 2πrh = 2 × 22/7 × 42 × 100 = 26400 cm² Area covered by the road roller in one revolution = Curved surface area of the roller = 26400 cm² (lets conver it to m²) = 26400/(100 × 100) = 2.64 m² Area covered by road roller in 750 revolutions = 750 × 2.64 = 1980 m² is the required area of the road 10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom what is the area of the label. Solution :- The label covers the lateral surface area of the cylinder, minus the top and bottom 2 cm. Radius of the cylinder, r = 14/2 cm = 7 cm. Effective height of the label, h = 20 cm – 2 cm – 2 cm = 16 cm. Lateral surface area for the label = 2πrh = 2 × π × 7 cm × 16 cm = 2 × 3.14 × 7 × 16 cm² = 704 cm². Therefore, the area of the label is 704 cm². ## Exercise - 9. 3 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To find how much it can hold. Volume (because volume measures the capacity of the tank). (b) Number of cement bags required to plaster it. Surface Area (because plastering involves covering the surface). (c) To find the number of smaller tanks that can be filled with water from it. Volume (as it’s a matter of the tank’s capacity). 2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area? Since the diameter of cylinder B is greater(double) than that of A, we can say that the volume of the cylinder B will be greater without performing any calculations. Let’s verify this by calculating the volume of both cylinders: For cylinder A: Radius r = diameter / 2 = 7 cm / 2 = 3.5 cm Height of cylinder A: h = 14 cm Volume V = π * r² * h = (22/7) * (3.5 cm)² * 14 cm Volume V = (22/7) * 12.25 cm² * 14 cm Volume V = 22 * 12.25 cm² * 2 cm (dividing numerator 14 by denominator 7) Volume V = 539 cm³ For cylinder B: Radius r = diameter / 2 = 14 cm / 2 = 7 cm Height of cylinder B: h = 7 cm Volume V = π * r² * h = (22/7) * (7 cm)² * 7 cm Volume V = (22/7) * 49 cm² * 7 cm Volume V = 22 * 7 cm * 7 cm Volume V = 1078 cm³ From the calculation, we have verified that the volume of the cylinder B is greater. Surface area comparison: For cylinder A: Surface area A = 2 * π * r (h + r) A = 2 * (22/7) * 3.5 cm (14 + 3.5) A = 22 (28 + 7)/2 A = (22 35)/2 A = 385 cm² For cylinder B: Surface area A = 2 * π * r (h + r) A = 2 * (22/7) * 7 cm (7 + 7) A = 2 * (22/7) * 7 cm (14) A = 44 * 14 A = 616 cm² Cylinder B has a greater surface area of 616 cm² compared to cylinder A’s surface area of 385 cm². Question 3: Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³. Formula for volume of a cuboid: Volume = Base Area × Height Given: Volume = 900 cm³, Base Area = 180 cm² Volume = Base Area × Height Height = Volume / Base Area Height = 900 cm³ / 180 cm² = 5 cm The height of cuboid is 5 cm Question 4: A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid? Given, dimensions of cuboid = 60 cm × 54 cm × 30 cm Length = 60 cm Width = 54 cm Height = 30 cm Formula for volume of the cuboid = Length × Width × Height = 60 cm × 54 cm × 30 cm = 97200 cm³ Given side of small cube is 6 cm Volume of a small cube = Side³ = 6 cm × 6 cm × 6 cm = 216 cm³ Since small cubes are placed in the given cuboid – Number of small cubes = Volume of the cuboid / Volume of a cube = 97200 cm³ / 216 cm³ = 450 Question 5: Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm. Radius of the base = Diameter / 2 = 140 cm / 2 = 70 cm = 0.7 m (since 1m = 100cm) Formula for volume of a cylinder: Volume = π × r² × Height Given: Volume = 1.54 m³, r = 0.7 m Height = Volume / (π × r²) = 1.54 m³ / (π × 0.7 m × 0.7 m) Height = 1 m (using π = 3.14) Question 6: A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank. Radius of the cylindrical tank = 1.5 m (given) Height of the cylindrical tank = 7m Formula for volume of a cylinder: Volume = π × r² × Height Given: r = 1.5 m, Height (length of the tank) = 7 m Volume = π × 1.5 m × 1.5 m × 7 m Volume = (22/7) × 1.5 m × 1.5 m × 7 m Volume = 49.5 m³ (using π = 22/7) Since 1 m³ = 1000 L Volume = 49.5 * 1000 = 49500 L Question 7: If each edge of a cube is doubled, (i) how many times will its surface area increase? (ii) how many times will its volume increase? (i) Increase in Surface Area:The formula for the surface area of a cube is 6 × (edge length)². Let’s say the original edge length is ‘a’. So, the original surface area is 6a². When the edge length is doubled, the new edge length becomes 2a. The new surface area is then 6 × (2a)² = 6 × 4a² = 24a². To find how many times the surface area increases, we divide the new surface area by the original surface area. So, it increases by a factor of 24a² / 6a² = 4 times. (ii) Increase in Volume: The formula for the volume of a cube is (edge length)³. The original volume is a³. When the edge length is doubled to 2a, the new volume becomes (2a)³ = 8a³. To find how many times the volume increases, we divide the new volume by the original volume. So, it increases by a factor of 8a³ / a³ = 8 times. Question 8: Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.Finding the Time to Fill the Reservoir: We are given the rate of water flow as 60 liters per minute and the volume of the reservoir as 108 m³. First, we need to convert the volume of the reservoir from cubic meters to liters, because the flow rate is in liters. 1 m³ = 1000 liters. So, the volume of the reservoir is 108 × 1000 = 108,000 liters. Now, we find the time taken to fill the reservoir. Time = Volume / Rate. So, the time taken to fill the reservoir is 108,000 liters / 60 liters per minute = 1800 minutes. To convert minutes into hours, we divide by 60. Thus, 1800 minutes / 60 = 30 hours. So, it will take 30 hours to fill the reservoir. The document NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration - 1 (Exercise 9.1) is a part of the Class 8 Course Mathematics (Maths) Class 8. All you need of Class 8 at this link: Class 8 ## Mathematics (Maths) Class 8 79 videos\|411 docs\|31 tests ## FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration - 1 (Exercise 9.1) 1. What are the basic concepts of mensuration? Ans. Mensuration involves the study of measurement of geometric figures such as length, area, volume, and surface area. It includes concepts like perimeter, circumference, and diagonal. 2. How is mensuration used in real life? Ans. Mensuration is used in various real-life situations such as calculating the area of a field, volume of a container, or surface area of a room. It is also used in construction, architecture, and engineering. 3. What is the difference between perimeter and area? Ans. Perimeter is the distance around a two-dimensional shape, while area is the measure of the space enclosed by the shape. Perimeter is measured in linear units, while area is measured in square units. 4. How do you calculate the volume of a three-dimensional shape? Ans. The volume of a three-dimensional shape is calculated by multiplying the area of the base by the height of the shape. Different shapes have different formulas for calculating volume, such as V = lwh for a rectangular prism. 5. Why is mensuration important in mathematics? Ans. Mensuration is important in mathematics as it helps in understanding and solving real-world problems involving measurement. It also forms the basis for more advanced mathematical concepts in geometry and calculus. ## Mathematics (Maths) Class 8 79 videos\|411 docs\|31 tests ### Up next Explore Courses for Class 8 exam ### Top Courses for Class 8 Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;
# Unit Fractions - Number Line Videos and solutions to help Grade 3 students learn how to place unit fractions on a number line with endpoints 0 and 1. Common Core Standards: 3.NF.2a, 3.NF.2b, 3.NF.3c, 3.NF.3d, 3.MD.4 Related Topics: Lesson Plans and Worksheets for Grade 3 Lesson Plans and Worksheets for all Grades New York State Common Core Math Grade 3, Module 5, Lesson 14 Lesson 14 Application Problem Mr. Ray is knitting a scarf. He says that he has completed 1 fifth of the total length of the scarf. Draw a picture of the final scarf. Label what he has finished and what he still has to make. Draw a number bond with 2 parts to show the fraction he has made and the fraction he has not made. Lesson 14 Concept Development Making Fraction Strips 1) Draw a horizontal line with your ruler that is a bit longer than 1 of your fraction strips. 2) Place a whole fraction strip just above the line you drew. 3) Make a small mark on the left end of your strip. 4) Label that mark 0 above the line. This is where we start measuring the length of the strip. 5) Make a small mark on the right end of your strip. 6) Label that mark 1 above the line. If we start at 0, the 1 tells us when we’ve travelled 1 whole length of the strip. Measure the Unit Fractions: 1. Place your fraction strip with halves above the line. 2. Make a mark on the number line at the right end of 1 half. This is the length of 1 half of the fraction strip. 3. Label that mark ½. Label 0 halves and 2 halves. 4. Repeat the process to measure and make other unit fractions on a number line. Lesson 14 Problem Set 1. Write number bonds. Partition the fraction strip to show the unit fractions of the number bond. Use the fraction strip to help you label the unit fractions on the number line. Include 0 unit fractions. 2. Trevor needs to let his puppy outside every quarter (1 fourth) hour to potty train him. Draw and label a number line from 0 hours to 1 hour to show every 1 fourth hour. Include 0 fourths and 4 fourths hour. Label 0 hours and 1 hour, too. 3. A ribbon is one meter long. Mrs. Lee wants to sew a bead every 1/5 m. The first bead is at 1/5 m. The last bead is at the 1 m. Draw and label a number line from 0 m to 1 m to show where Mrs. Lee will sew in a bead. Label all the fractions including 0 fifths and 5 fifths. Label 0 meters and 1 meter, too. Lesson 14 Homework 1. Write number bonds. Partition the fraction strip to show the unit fractions of the number bond. Use the fraction strip to help you label the unit fractions on the number line. Include 0 unit fractions. c. fifths 2. Carter needs to wrap 7 presents. He lays the ribbon out flat and says, "If I make 6 equally spaced cuts, IӬl have just enough pieces. I can use 1 piece for each package, and I wonӴ have any pieces left over." Does he have enough pieces to wrap all the presents? 3. Mrs. Rivera is planting flowers in her 1 meter long rectangular plant box. She divides the plant box into sections 1/9 m in length, and plants 1 seed in each section. Draw and label a fraction strip representing the plant box from 0m to 1m. Represent each section where Mrs. Rivera will plant a seed. Label all the fractions. a. How many seeds will she be able to plant in 1 plant box? b. How many seeds will she be able to plant in 4 plant boxes? c. Draw a number line below your fraction strip and mark all the fractions. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.3 \| Set 2 • Last Updated : 18 Mar, 2021 ### Question 21. (16/x) – 1 = 15/(x + 1), x ≠ 0, -1. Solution: We have equation, (16/x) – 1 = 15/(x + 1) (16 – x)/x = 15/(x + 1) 15x = (x + 1) (16 – x) 15x = 16x – x2 + 16 – x 15x – 16x + x2 -16 + x = 0 x2 – 16 = 0 (x – 4) (x + 4) = 0 Therefore, roots of the equation are 4 or -4. Question 22. (x + 3)/(x + 2) = (3x – 7)/(2x – 3), x ≠ -2, 3/2. Solution: We have equation, (x + 3) (2x – 3) = (3x – 7) (x + 2) 2x2 – 3x + 6x – 9 = 3x2 + 6x -7x -14 2x2 + 3x -9 = 3x2 -x -14 x2 -4x -5 = 0 We can factorize this equation as: x2 – 5x + x -5 = 0 x (x – 5) + 1 (x – 5) = 0 (x + 1) (x – 5) = 0 Therefore, roots of the equation are 5 or -1. ### Question 23. (2x/(x – 4)) + ((2x – 5)/(x – 3)) = 25/3, x ≠ 3, 4 Solution: We have equation, (2x/(x – 4)) + ((2x – 5)/(x – 3)) = 25/3 ((2x)(x – 3) + (2x – 5) (x – 4))/((x – 4) (x – 3)) = 25/3 25x2 – 175x + 300 = 12x2 – 57x + 60 13x2 – 118x + 240 = 0 We can factorize this equation as: 13x2 – 78x – 40x + 240 = 0 13x (x – 6) – 40 (x – 6) = 0 (13x – 40) (x – 6) = 0 Therefore, roots of the equation are 6 or 40/13. ### Question 24. ((x + 3)/(x – 2)) – ((1 – x)/(x)) = 17/4, x ≠ 0, 2 Solution: We have equation, ((x + 3)/(x – 2)) + ((1 – x)/(x)) = 17/4 8x2 + 8 = 17x2 – 34x 9x2 – 34x – 8 = 0 We can factorize this equation as: 9x2 – 36x + 2x – 8 = 0 9x(x – 4) + 2(x – 4) = 0 (9x + 2) (x – 4) = 0 Therefore, roots of the equation are 4 or -2/9. ### Question 25. ((x – 3)/(x + 3)) – ((x + 3)/(x – 3)) = 48/7, x ≠ 3, -3 Solution: We have equation, ((x – 3)/(x + 3)) – ((x + 3)/(x – 3)) = 48/7 -84x = 48x2 – 432 48x2 + 84x – 432 = 0 4x2 + 7x -36 = 0 We can factorize this equation as: 4x2 + 16x – 9x – 36 = 0 4x (x + 4) – 9 (x + 4) = 0 (4x – 9) (x + 4) = 0 Therefore, roots of the equation are -4 or 9/4. ### Question 26. (1/(x – 2)) + (2/(x – 1)) = 6/x, x ≠ 0 Solution: We have equation, (1/(x – 2)) + (2/(x – 1)) = 6/x 3x2 – 5x = 6x2 – 18x + 12 3x2 – 13x + 12 = 0 We can factorize this equation as: 3x2 – 9x -4x + 12 = 0 3x (x – 3) – 4 (x – 3) = 0 (3x – 4) (x – 3) = 0 Therefore, roots of the equation are 3 or 4/3. ### Question 27. ((x + 1)/(x – 1)) – ((x – 1)/(x + 1)) = 5/6, x ≠ 1, -1 Solution: We have equation, ((x + 1)/(x – 1)) – ((x – 1)/(x + 1)) = 5/6 5x2 – 5 = 24x 5x2 – 24x – 5 = 0 We can factorize this equation as: 5x2 – 25x + x – 5 = 0 5x (x – 5) + 1 (x – 5) = 0 (5x + 1) (x – 5) = 0 Therefore, roots of the equation are 5 or -1/5. ### Question 28. ((x – 1)/(2x + 1)) + ((2x + 1)/(x – 1)) = 5/2, x ≠ 1, -1/2 Solution: We have equation, ((x – 1)/(2x + 1)) + ((2x + 1)/(x – 1)) = 5/2 2 (5x2 + 2x + 2) = 5 (2x2 – x – 1) 9x + 9 = 0 9 (x + 1) = 0 Therefore, roots of the equation are -1. ### Question 29. (4/x) – 3 = 5/(2x + 3), x ≠ 0, -3/2 Solution: We have equation, (4/x) – 3) = 5/(2x + 3) 5x = (2x + 3) (4 – 3x) 5x = 8x – 6x2 + 12 – 9x 6x2 – 6x -12 = 0 x2 – x -2 = 0 We can factorize this equation as: x2 + 2x – x – 2 = 0 x (x + 2) -1 (x + 2) = 0 (x – 1) (x + 2) = 0 Therefore, roots of the equation are 1 or -2. ### Question 30. ((x – 4)/(x – 5)) + ((x – 6)/(x – 7)) = 10/3, x ≠ 5, 7 Solution: We have equation, ((x – 4)/(x – 5)) + ((x – 6)/(x – 7)) = 10/3 4x2 – 54x + 176 = 0 2x2 – 27x + 88 = 0 We can factorize this equation as: 2x2 – 16x -11x + 88 = 0 2x (x – 8) – 11 (x – 8) = 0 (2x – 11) (x – 8) = 0 Therefore, roots of the equation are 8 or 11/2. ### Question 31. ((x – 2)/(x – 3)) + ((x – 4)/(x – 5)) = 10/3, x ≠ 3, 5. Solution: We have equation, ((x – 2)/(x – 3)) + ((x – 4)/(x – 5)) = 10/3 We can rewrite it as : ((x – 3 + 1)/(x – 3)) + ((x – 5 + 1)/(x – 5)) = 10/3 1 + 1 + (1/(x – 3)) + (1/( x – 5)) = 10 3 4 (x2 – 8x + 15) = 6x – 24 4x2 – 38x + 84 = 0 2x2 -19x + 42 = 0 We can factorize this equation as: 2x2 – 12x – 7x + 42 = 0 2x (x – 6) – 7 (x – 6) = 0 (2x – 7) (x – 6) = 0 Therefore, roots of the equation are 6 or 7/2. ### Question 32. (( 5 + x)/(5 – x)) – ((5 – x)/(5 + x)) = 15/4, x ≠ 5, -5. Solution: We have equation, (( 5 + x)/(5 – x)) – ((5 – x)/(5 + x)) = 15/4 80x = 375 – 15x2 15x2 +80x -375 = 0 3x2 +16x – 75 = 0 We can factorize this equation as: 3x2 + 25x – 9x – 75 = 0 x (3x + 25) – 3 (3x + 25) = 0 (3x + 25) (x – 3) = 0 Therefore, roots of the equation are 3 or -25/3. ### Question 33. (3/(x + 1)) – (1/2) = 2/(3x – 1), x ≠ -1, 1/3 Solution: We have equation, (3/(x + 1)) – (1/2) = 2/(3x – 1) 2 (2x + 2) = (5 – x) (3x – 1) 4x + 4 = 15x – 5 – 3x2 + x 3x2 – 12x + 9 = 0 x2 – 4x + 3 = 0 We can factorize this equation as: x2 – 3x – x + 3 = 0 x (x – 3) – 1 (x – 3) = 0 (x – 3) (x – 1) = 0 Therefore, roots of the equation are 1 or 3. ### Question 34. (3/(x + 1)) + (4/(x – 1)) = 29/(4x – 1), x ≠ -1, 1, 1/4 Solution: We have equation, (3/(x + 1)) + (4/(x – 1)) = 29/(4x – 1) (7x + 1) (4x – 1) = 29 (x2 – 1) 28x2 – 7x + 4x – 1 = 29x2 – 29 x2 + 3x -28 = 0 We can factorize this equation as: x2 + 7x – 4x – 28 = 0 x (x + 7) – 4 (x + 7) = 0 (x – 4) (x + 7) = 0 Therefore, roots of the equation are 4 or -7. ### Question 35. (2/(x + 1)) + (3/(2(x – 2))) = 23/5x, x ≠ 0, -1, 2 Solution: We have equation, (2/(x + 1)) + (3/(2(x – 2))) = 23/5x 35x2 – 25x = 46 (x2 – x – 2) 11x2 – 21x – 92 = 0 We can factorize this equation as: 11x2 – 44x + 23x – 92 = 0 11x (x – 4) + 23 (x – 4) = 0 (11x + 23) ( x – 4) = 0 Therefore, roots of the equation are 4 or -23/11. ### Question 36. x2 – (√3 + 1) x + √3 = 0 Solution: We have equation, x2 – (√3 + 1) x + √3 = 0 x2 – √3x – x + √3 = 0 We can factorize this equation as: x (x – √3) – 1 (x – √3) = 0 (x -1) (x – √3) = 0 Therefore, roots of the equation are 1 or √3. ### Question 37. 3√5 x2 + 25x – 10√5 = 0 Solution: We have equation, 3√5 x2 + 25x – 10√5 = 0 √5 (3x2 + (25/√5) x – (10√5/√5)) = 0 √5 (3x2 + 5√5x – 10) = 0 We can factorize this equation as: 3x2 – √5x + 6√5x – 10 = 0 x (3x – √5) + 2√5 (3x – √5) = 0 (x + 2√5) (3x – √5) = 0 Therefore, roots of the equation are -2√5 or √5/3. ### Question 38. √3x2 – 2√2 x – 2√3 = 0 Solution: We have equation, √3x2 – 2√2 x – 2√3 = 0 Here a = √3, b = -2√2 and c = -2√3 Since, Discriminant D = b2 – 4ac and x = (-b ± √D)/2a Therefore, D = 8 + 24 = 32, and x = (-(-2√2) ± √32)/2√3 x = ( 2√2 ± 4√2)/2√3 Therefore, roots of the equations are (2√2 + 4√2)/2√3 or (2√2 – 4√2)/2√3. ### Question 39. 4√3x2 + 5x – 2√3 = 0 Solution: We have equation, 4√3x2 + 5x – 2√3 = 0 Here a = 4√3, b = 5 and c = -2√3 Since, Discriminant D = b2 – 4ac and x = (-b ± √D)/2a Therefore, D = 25 + 96 = 121, and x = (-(5) ± √121)/8√3 x = (-5 ± 11)/8√3 Therefore, roots of the equations are -2 /√3 or √3/ 4. ### Question 40. √2x2 – 3x – 2√2 = 0 Solution: We have equation, √2x2 – 3x – 2√2 = 0 Here a = √2, b = -3 and c = -2√2 Since, Discriminant D = b2 – 4ac and x = (-b ± √D)/2a Therefore, D = 9 + 16 = 25, and x = (-(-3) ± √25)/2√2 x = (3 ± 5)/2√2 Therefore, roots of the equations are 2√2 or -1/ √2. 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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # s5_4 - 4 MATRICES 170 4 Matrices 4.1 Denitions Definition... This preview shows pages 1–3. Sign up to view the full content. 4. MATRICES 170 4. Matrices 4.1. Definitions. Definition 4.1.1 . A matrix is a rectangular array of numbers. A matrix with m rows and n columns is said to have dimension m × n and may be represented as follows: A = a 11 a 12 · · · a 1 n a 21 a 22 · · · a 2 n . . . . . . . . . . . . a m 1 a m 2 · · · a mn = [ a ij ] Definition 4.1.2 . Matrices A and B are equal , A = B , if A and B have the same dimensions and each entry of A is equal to the corresponding entry of B . Discussion Matrices have many applications in discrete mathematics. You have probably encountered them in a precalculus course. We present the basic definitions associated with matrices and matrix operations here as well as a few additional operations with which you might not be familiar. We often use capital letters to represent matrices and enclose the array of numbers with brackets or parenthesis; e.g., A = a b c d or A = a b c d . We do not use simply straight lines in place of brackets when writing matrices because the notation a b c d has a special meaning in linear algebra. A = [ a ij ] is a shorthand notation often used when one wishes to specify how the elements are to be represented, where the first subscript i denotes the row number and the subscript j denotes the column number of the entry a ij . Thus, if one writes a 34 , one is referring to the element in the 3rd row and 4th column. This notation, however, does not indicate the dimensions of the matrix. Using this notation, we can say that two m × n matrices A = [ a ij ] and B = [ b ij ] are equal if and only if a ij = b ij for all i and j . Example 4.1.1 . The following matrix is a 1 × 3 matrix with a 11 = 2 , a 12 = 3 , and a 13 = - 2 . h 2 3 - 2 i This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 4. MATRICES 171 Example 4.1.2 . The following matrix is a 2 × 3 matrix. 0 π - 2 2 5 0 4.2. Matrix Arithmetic. Let α be a scalar, A = [ a ij ] and B = [ b ij ] be m × n matrices, and C = [ c ij ] a n × p matrix. (1) Addition: A + B = [ a ij + b ij ] (2) Subtraction: A - B = [ a ij - b ij ] (3) Scalar Multiplication: αA = [ αa ij ] (4) Matrix Multiplication: AC = " n X k =1 a ik c kj # Discussion Matrices may be added, subtracted, and multiplied, provided their dimensions satisfy certain restrictions. To add or subtract two matrices, the matrices must have the same dimensions. This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 8 s5_4 - 4 MATRICES 170 4 Matrices 4.1 Denitions Definition... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# Dividing Numbers that Result in Decimals Warmup Activity #1 Analyze a Base ten Diagram in Division. In this unit, we have introduced three techniques to divide whole numbers that result in a terminating decimal: base-ten diagrams, partial quotients, and long division. • Examine the method below of base-ten blocks used by Mai, and answer the questions that follow. Mai used base-ten diagrams to calculate 62 ÷ 5. She started by representing 62. She then made 5 groups, each with 1 ten. There was 1 ten left. She unbundled it into 10 ones and distributed the ones across the 5 groups. Here below is Mai’s diagram from 62 ÷ 5. (1.) Mai should have a total of 12 ones, but her diagram shows only 10. Why? (2.) She did not originally have tenths, but in her diagram each group has 4 tenths. Why? (3.) What value has Mai found for 62 ÷ 5? Explain your reasoning. Activity #2 Explore Long Division Method. • Examine here below, the long division method used by Lin to divide the same numbers Mai used above, and answer the questions that follow. Here is how Lin calculated 62 ÷ 5. (1.) Lin put a 0 after the remainder of 2. Why? Why does this 0 not change the value of the quotient? (2.) Lin subtracted 5 groups of 4 from 20. What value does the 4 in the quotient represent? (3.) What value did Lin find for 62 ÷ 5? • Use long division in the applet below to find the value of each expression on the work sheet below. (a.) 126 ÷ 8 (b.) 90 ÷ 12 • Use long division in the applet below to show that: (a.) 5 ÷ 4, or is 1.25 (b.) 4 ÷ 5, or is 0.8 (c.) 1 ÷ 8, or is 0.125 Activity #3 Practice Base ten Diagrams in Division. • Find the quotient of 511 ÷ 5 by drawing base-ten diagrams or by using the partial quotients method. Four students share a \$271 prize from a science competition. How much does each student get if the prize is shared equally? • Show your reasoning using the applet below. Challenge #1 Complete the calculations so that each shows the correct difference. Challenge #2 Use long division to show that the fraction and decimal in each pair are equal. Challenge #3 Noah said we cannot use long division to calculate 10 ÷ 3 because there will always be a remainder. (1.) What do you think Noah meant by “there will always be a remainder”? (2.) Do you agree with him? Explain your reasoning. Quiz Time
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 11.4: Theoretical and Experimental Probability Difficulty Level: At Grade Created by: CK-12 ## Introduction Weekend Woes “I don’t want to work on the weekend,” Carey said to Telly at lunch one day. “But that was part of the deal. We both have to work one day out of the weekend,” Telly said. “Well, which day do you want?” Carey asked. “I don’t know. I haven’t really thought about it,” Telly said. “But we could make it really random.” Telly took two pieces of paper and wrote Saturday on one and Sunday on the other. “Now we can figure out the probability of you getting Saturday or Sunday,” she said. We can stop there. This lesson is all about probability. Telly’s experiment is an example of experimental probability. Let’s talk more about this at the end of the lesson. What You Will Learn In this lesson, you will learn how to demonstrate the following skills. • Recognize the theoretical probability of an event as the ratio of favorable outcomes to possible outcomes. • Recognize the experimental probability of an event as the ratio of successful outcomes to trials attempted. • Write and compare probabilities as fractions, decimals and percents. • Make and compare predictions based on theoretical and experimental probabilities, justifying the use of either. Teaching Time I. Recognize the Theoretical Probability of an Event as the Ratio of Favorable Outcomes to Possible Outcomes In this chapter we have explored different outcomes for events, but not probability itself. Probability is defined as a mathematical way of calculating how likely an event is to occur. The probability of an event occurring is defined as the ratio of favorable outcomes to the number of possible equally likely total outcomes in a given situation. In ratio form, the probability of an event is: P(event)=favorable outcomes:total outcomes\begin{align}P \text{(event)}=\text{favorable outcomes} : \text{total outcomes}\end{align} Theoretical probability is probability that is based on an ideal situation. For example, since a flipped coin has two sides and each side is equally likely to land up, the theoretical probability of landing heads (or tails) is exactly 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world does not affect theoretical probability. The theoretical probability of an event remains the same no matter how events turn out in the real world. Example Find the probability of tossing a number cube and having it come up “4”. Step 1: Find the total number of outcomes Total outcomes= 1 2 3 4 5 6=6 total outcomes\begin{align}\text{Total outcomes} &= \bullet \ 1 \ \bullet \bullet \ 2 \ \bullet \bullet \bullet \ 3 \ \bullet \bullet \bullet \bullet \ 4 \ \bullet \bullet \bullet \bullet \bullet \ 5 \ \bullet \bullet \bullet \bullet \bullet \bullet \ 6 \\ &= 6 \ \text{total outcomes}\end{align} Step 2: Find the number of favorable outcomes. favorable outcomes= 1 2 3 4 5 6=1 favorable outcome\begin{align}\text{favorable outcomes} &= \bullet \ 1 \ \bullet \bullet \ 2 \ \bullet \bullet \bullet \ 3 \ \ {\color{red}\bullet \bullet \bullet \bullet \ 4} \ \bullet \bullet \bullet \bullet \bullet \ 5 \ \bullet \bullet \bullet \bullet \bullet \bullet \ 6 \\ &= 1 \ \text{favorable outcome}\end{align} Step 3: Find the ratio of favorable outcomes to total outcomes. Favorable:Total=1:6\begin{align}\text{Favorable}:\text{Total}= 1:6\end{align} While theoretical probability is based on the ideal, we can also figure out experimental probability. II. Recognize the Experimental Probability of an Event as the Ratio of Successful Outcomes to Trials Attempted Theoretical probability is based on an ideal situation. Since a flipped coin seems equally likely to land up or down, the theoretical probability of landing heads (or tails) is 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world is something you must determine with experimental probability. Experimental probability is probability based on doing actual experiments – flipping coins, spinning spinners, picking ping pong balls out of a jar, and so on. To compute the experimental probability of the number cube landing on 3 you would need to conduct an experiment. Suppose you were to toss the number cube 60 times. Favorable outcomes: Total outcomes: 60 tosses Experimental probability: \begin{align}P(3) =\frac{favorable \ outcomes}{total \ outcomes}=\frac{Number \ of \ 3's}{Total \ Number \ of \ tosses}\end{align} Write this comparison down in your notebooks. Example What is the experimental probability of having the number cube land on 3? trial 1 2 3 4 5 6 Total raw data:3s \begin{align}{\|}\end{align} \begin{align}{\|\|\|}\end{align} \begin{align}{\|}\end{align} \begin{align}{\|\|}\end{align} \begin{align}{\|\|}\end{align} favorable outcomes:3s 1 3 0 1 2 2 9 total tosses total outcomes 10 10 10 10 10 10 60 experimental probability: favorable outcomes to total outcomes x x x x x x \begin{align}9:60=3:20\end{align} The data from the experiment shows that 3 turned up on the number cube 9 out of 60 times. Simplified, this ratio becomes: \begin{align}\text{Favorable outcomes}:\text{total outcomes}= 3:20\end{align} You can see that it is only possible to calculate the experimental probability when you are actually doing experiments and counting results. Example A spinner was spun in a probability experiment 48 times. The results are shown in the table. Compute the experimental probability of the spinner landing on yellow. color red green yellow Total spins raw data \begin{align}\cancel{\|\|\|\|} \ \cancel{\|\|\|\|} \ \cancel{\|\|\|\|} \ {\|}\end{align} \begin{align}\cancel{\|\|\|\|} \ \cancel{\|\|\|\|} \ {\|\|\|\|}\end{align} \begin{align}\cancel{\|\|\|\|} \ \cancel{\|\|\|\|} \ \cancel{\|\|\|\|} \ {\|\|\|}\end{align} 48 total from tally 16 14 18 48 favorable outcomes:yellow x x 18 x experimental probability: x x \begin{align}\colorbox{yellow}{\color{red}18:48}\end{align} x The data from the experiment above shows that the arrow landed on yellow 18 out of 48 times. Simplified, this ratio becomes: \begin{align}\text{Favorable outcomes} : \text{total outcomes} &= 18:48\\ &= 3:8\end{align} Notice that the ratio 18 out of 48 was simplified to 3 out of 8. We can simplify probabilities because they are written in ratio form. III. Write and Compare Probabilities as Fractions, Decimals and Percents You’ve seen how to compute probabilities in terms of ratios. Since any ratio can be turned into a fraction, decimal, or percent, you can also turn any probability into a fraction, decimal, or percent. For example, when you toss a number cube, the probability of rolling a “3” is: \begin{align}P (3) &=\text{favorable outcomes} : \text{total outcomes}\\ P (3) &= 1 : 6\end{align} You can write the same probability as a fraction simply by rewriting the two numbers in the ratio as the numerator and denominator of a fraction. \begin{align}P(3)=\frac{1}{6}\end{align} This is the answer in fraction form. How can we turn this fraction into a decimal? You can turn a fraction into a decimal by dividing the numerator by the denominator. \begin{align}\frac{{\color{blue}1}}{{\color{red}6}}=\overset{ \ \ 0.167}{{\color{red}6} \overline{ ) {{\color{blue}1}.000 \;}}}\end{align} How can we turn the decimal into a percent? We can turn a decimal into a percent by multiply the decimal by 100 since a percent is out of 100. Then we can move the decimal point two decimal places to show the percent. \begin{align}0.167 = 0.167 \times 100 = 16.7\%\end{align} To summarize, the probability of rolling a 3 with a number cube is 1 out of 6, or: ratio fraction decimal percent 1:6 \begin{align}\frac{1}{6}\end{align} 0.167 16.7% IV. Make and Compare Predictions Based on Theoretical and Experimental Probabilities, Justifying the Use of Either A prediction is a reasonable guess about what will happen in the future. Good predictions should be based on facts and probability. There are two main types of predictions. Type 1: Predictions based on theoretical probability: These are the most reliable types of predictions, based on physical relationships that are easy to see and measure and that do not change over time. They include such things as: • coin flips • spinners • number cubes Type 2: Predictions based on data and experimental probability: These predictions are often reliable, but subject to change depending on the situation. They include such activities as: • batting averages, shooting percentages, and similar data from sports • predicting the weather • sales figures from such things as movies, TV shows, products • polls and surveys that measure opinion • historical data that measures past events The difference between the two types of prediction is best illustrated by the following examples. • Type 1: Coin flip prediction: of 100 flips, 50 are predicted to turn up heads • Type 2: Weather prediction: a 50 percent chance of rain tomorrow Note that both predictions are about the future – and you never know what might happen in the future. Though it is HIGHLY UNLIKELY to occur, a coin could land on heads 10, or 20, or even 50 times in a row. Similarly, a weather data might predict a zero percent chance of rain. But on a given day, the unexpected could happen. The winds could change clouds could unexpectedly move out or move in. That’s why predictions based on experimental probability are always less reliable than those based on theoretical probability. In general, the greater the number of outcomes you have, the closer a prediction based on probability is likely to be. Write this statement down in your notebooks. One hundred coin flips should turn out to be close to 50 percent heads; 1000 flips should be even closer to 50 percent; 10,000 flips should be closer yet. The same thing goes with weather forecasts. Over a year, forecasts are a lot more likely to be accurate than over a single day or a single week. To make predictions, use the following formulas. \begin{align}\mathbf{Type \ 1 \ Prediction}= \text{theoretical probability} \cdot \text{number of trials}\end{align} Or, for type 2 situations in which you do not have theoretical probability: \begin{align}\mathbf{Type \ 2 \ Prediction}= \text{experimental probability} \cdot \text{number of trials}\end{align} Problem: Out of 150 number cube rolls, predict how many will turn up greater than 4. Step 1: Find the theoretical or experimental probability. \begin{align}P(> 4) =\frac{favorable \ outcomes}{total \ outcomes}=\frac{2}{6}=\frac{1}{3}\end{align} Step 2: Multiply the theoretical probability by the number of total trials \begin{align}\text{Prediction} &=\text{theoretical probability} \cdot \text{number of trials}\\ &=\frac{1}{3} \cdot 150 \\ &= 50\end{align} Based on these numbers, you would predict that 50 out of 150 rolls would land on numbers greater than 4. Now let’s apply what you have learned about predictions from the problem in the introduction. ## Real-Life Example Completed Weekend Woes Here is the original problem from the introduction. Think about how this is an example of experimental probability and then figure out the probability of Carey working Saturday in fraction, decimal and percentage form. “I don’t want to work on the weekend,” Carey said to Telly at lunch one day. “But that was part of the deal. We both have to work one day out of the weekend,” Telly said. “Well, which day do you want?” Carey asked. “I don’t know. I haven’t really thought about it,” Telly said. “But we could make it really random.” Telly took two pieces of paper and wrote Saturday on one and Sunday on the other. “Now we can figure out the probability of you getting Saturday or Sunday,” she said. Solution to Real – Life Example Carey has a chance of working Saturday or Sunday. There are two possible outcomes. She has a one out of 2 chance of working on Saturday and a one out of two chance of working on Sunday. \begin{align}\frac{1}{2}\end{align} .50 50% chance or probability for each outcome. ## Vocabulary Here are the vocabulary words that are found in this lesson. Probability a mathematical way of calculating how likely an event is to occur. Favorable Outcome the outcome that you are looking for Total Outcomes all of the outcomes both favorable and unfavorable. Theoretical Probability probability based on an ideal situation relating favorable to total outcomes Experimental Probability probability based on doing actual experiments. Prediction a reasonable guess based on probability ## Time to Practice Directions: Solve each problem. A spinner has five sections: purple, yellow, green, blue and red. 1. Find the probability for the arrow landing on blue on the spinner: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 2. Find the probability for the arrow landing on red or green on the spinner: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 3. Find the probability for the arrow NOT landing on yellow on the spinner: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 4. Find the probability for rolling 6 on the number cube: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 5. Find the probability for rolling greater than 2 on the number cube: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 6. Find the probability for rolling less than 4 on the number cube: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 7. Find the probability for rolling 1 or 6 on the number cube: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. 8. A box contains 12 slips of paper numbered 1 to 12. Find the probability for randomly choosing a slip with a number less than 4 on it: 1. List each favorable outcome. 2. Count the number of favorable outcomes. 3. Write the total number of outcomes. 4. Write the probability. Directions: Use the table to answer the questions. Express all ratios in simplest form. Use the table to compute the experimental probability of flipping a coin and having it land on heads. trial 1 2 3 4 5 6 Total raw data(heads) \begin{align}\cancel{\|\|\|\|}\end{align} \begin{align}\cancel{\|\|\|\|} \ {\|}\end{align} \begin{align}\cancel{\|\|\|\|} \ {\|}\end{align} \begin{align}{\|\|\|}\end{align} \begin{align}\cancel{\|\|\|\|} \ {\|}\end{align} \begin{align}\cancel{\|\|\|\|}\end{align} number of heads 5 6 6 3 6 5 31 total number of flips 10 10 10 10 10 10 60 experimental probability x x x x x x 31:60 1. How many favorable outcomes were there in the experiment? 2. How many total outcomes were there in the experiment? 3. What was the experimental probability of the coin landing on heads? 31:60 Use the table to compute the experimental probability of a number cube landing on 6. trial 1 2 3 4 5 Total raw data \begin{align}{\|\|\|\|}\end{align} \begin{align}{\|}\end{align} \begin{align}{\|}\end{align} \begin{align}{\|\|}\end{align} \begin{align}{\|}\end{align} x number of 6's 4 1 1 2 1 9 total tosses 10 10 10 10 10 50 experimental probability x x x x x 9:50 1. How many favorable outcomes were there in the experiment? 2. How many total outcomes were there in the experiment? 3. What is the experimental probability of the arrow landing on yellow? Directions: Use what you have learned about probability, fractions, decimals and percents to answer each question. A bag has 6 red marbles, 5 blue marbles, 7 green marbles, 2 white marbles, and 5 yellow marbles. Find the probability of randomly picking out one of the following. 1. What is the probability in fraction form of choosing red marble? 2. What is the probability in decimal form of choosing green marble? 3. What is the probability in percent form of choosing a green or blue marble? 4. Which 3 marbles together have a 60 percent chance of being chosen? 5. Which 2 marbles together have a 40 percent chance of being chosen? 6. Which 3 marbles together have a 0.72 chance of being chosen? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
## Average Deviation 1429 Statistics & Excel In the world of statistics and data analysis, understanding how to measure the spread or dispersion of data is crucial. One way to do this is by calculating the average deviation. In this blog, we will explore the concept of average deviation, how to calculate it using Excel, and its importance in statistical analysis. Calculating the Mean (Average): Before diving into average deviation, let’s start with the basics by calculating the mean or average of a dataset. The mean represents the central point of a dataset. In Excel, you can easily calculate the mean using the formula `=AVERAGE(range)`, where “range” refers to the dataset. For example, if we have a simple dataset containing -6, 4, 4, and 6, the mean is calculated as: `=AVERAGE(-6, 4, 4, 6) = 0` You can also calculate the mean manually by summing all the data points and dividing by the number of data points. Introducing Average Deviation: Now that we have our mean, let’s explore the concept of average deviation. Average deviation measures how far individual data points are from the mean, without considering whether they are above or below the mean. Here’s the intuitive process: 1. Find the difference between each data point (x) and the mean (μ). • For our example: • (6 – 0) = 6 • (4 – 0) = 4 • (-4 – 0) = -4 • (-6 – 0) = -6 2. Take the absolute value of these differences to ensure all values are positive. • Absolute values: 6, 4, 4, 6 3. Sum up these absolute differences. • Sum = 6 + 4 + 4 + 6 = 20 4. Finally, divide the sum by the number of data points to calculate the average deviation. • Average Deviation = Sum / Number of Data Points = 20 / 4 = 5 Why Average Deviation Matters: Average deviation provides a numerical representation of the spread of data around the mean. It gives us an idea of how dispersed the data points are without considering their direction from the mean. While average deviation might seem less intuitive compared to the mean, it becomes valuable when comparing different datasets. In comparative settings, analyzing average deviations can help you understand how similar or dissimilar two datasets are in terms of data spread. However, as we delve further into statistics, we will introduce more advanced measures like standard deviation and variance, which offer deeper insights into data variability. Conclusion: Average deviation is a fundamental concept in statistics that measures the average distance of data points from the mean. It provides a basic understanding of data spread, making it a useful tool in comparative analysis. In future blogs, we will explore more advanced statistical concepts, including standard deviation and variance, to enhance our data analysis skills. Stay tuned for deeper insights into statistical analysis using Excel and other tools.
# Basic Algebra Terms Related Topics: More Algebra Lessons Basic algebra terms you need to know are constants, variables, coefficients, terms, expressions, equations and quadratic equations. These are some algebra vocabulary that will be useful. The following diagram gives an example to illustrate the following algebra vocabulary that you will need to know: constants, variables, coefficients, terms, expressions, and equations. Scroll down the page for more examples and explanations. ### Constants A fixed quantity that does not change. For example: 3, –6, π, ### Variables A variable is a symbol that we assign to an unknown value. It is usually represented by letters such as x, y, or t. For example, we might say that l stands for the length of a rectangle and w stands for the width of the rectangle. We use variables when we need to indicate how objects are related even though we may not know the exact values of the objects. For example, if we want to say that the length of a rectangle is 3 times the length of its width then we can write l = 3 × w ### Coefficients The coefficient of a variable is the number that is placed in front of a variable. For example, 3 × w can be written as 3w and 3 is the coefficient. Coefficient ### Terms A term can be any of the following: • a constant: e.g. 3, 10, π, • the product of a number (coefficient) and a variable: e.g. –3x, 11y, • the product of two or more variables: e.g. x2, xy, 2y2, 7xy Like terms are terms that differ only in their numerical coefficients. For example: 3a, 22a, are like terms. ### Expressions An expression is made up of one or more terms. For example: 3w + 4xy + 5 ### Equations An equation consists of two expressions separated by an equal sign. The expression on one side of the equal sign has the same value as the expression on the other side. For example: 4 + 6 = 5 × 2 l = 3 × w 3w + 4xy + 5 = 2w + 3 A Quadratic Equation is an equation of the form: ax2 + bx + c = 0, where a, b and c are numbers and a ≠ 0 For example: x2 + 2x + 3 = 0 2x2 + 5x – 7 = 0 2x2 + 5x = 8 is a quadratic equation because it can be changed to 2x2 + 5x – 8 = 0 x2 + x = 0 is a quadratic with c = 0 2x27 = 0 is a quadratic with b = 0 2x + 3 = 0 is not a quadratic because a cannot be 0 ### Algebraic Fraction An algebraic fraction is a fraction that contains an algebraic expression in its numerator and/or denominator. For example: $$\frac{4}{{2x - 3}},\frac{{3x - 5}}{{x + 3}}$$ #### How to Understand Algebraic Variables This video will show you how to understand algebraic variables. In algebra, variables are placeholder letters (capitalized and lowercase) that represent the unknown, or what you're solving for. This video shows you what variables can look like and what they mean. Understanding variables help make algebra easier for you. #### How to Understand the Vocabulary Of Algebra In this video you will learn how to understand the vocabulary of algebra. Knowing the symbols and expressions used in algebra makes understanding algebra easier. This video explains some common algebra symbols and phrases, such equations, operations, variables, and constants Expression, term, equation, operation, variable, constant, exponent, simplify, factor, solve Differences between an algebraic expression and an algebraic equation It also explains terms, coefficients and constants. This video gives the following algebra vocabulary. Operation, term, variable, coefficient, expression, equation Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Video: KS2-M16 • Paper 3 • Question 1 The numbers in this sequence increase by 14 each time. ＿ 82 96 ＿ 124 138 ＿. Write the missing numbers. 04:19 ### Video Transcript The numbers in this sequence increase by 14 each time. And then the sequence shows a missing number, 82, 96, another missing number, 124, 138, and then a third missing number. Write the missing numbers. In this question, we’ve been given a sequence of seven numbers. But three of the numbers are missing. So we need to apply the rule to find out what they are. What is the rule in this sequence? Well, often a question will ask us to work out what the rule is. But in this problem, we’re given the rule as part of the question. The numbers in the sequence increase by 14. So as we move from each number to the next in the sequence, we add 14 each time. If we look at our three missing numbers, one of them is a little bit more tricky to work out than the others. Which is it? Well, it’s the first missing number in the sequence. Because it’s the first number, we don’t have a number to add 14 to, to give us the missing number. With the other two missing numbers, we do have the number that comes before in the sequence. So how are we going to work out our first missing number? Let’s use the number 82 to help. At the moment, we know that our missing number plus 14 equals 82. And so we can start with 82 and take away 14 to give us our missing number. Let’s change the arrow on our sequence to show how to calculate the first missing number. Start with 82, take away 14 to give us our missing number. 82 take away 10 is 72 and then take away four gives us 68. Let’s just check that works. 68 plus 10 equals 78, and then plus four equals 82. So there is a difference of 14. The next missing number in the sequence is a little bit easier because we know the number that comes before. What is 96 plus 14? Let’s add 10 and four, as before. 96 plus 10 equals 106. And we know that six and four make the nearest 10. So 106 plus four equals 110. Again, we can quickly check that the number is correct by adding 14 to 110. And we can see that it does equal the next number in the sequence which is 124. Finally, let’s work out the last missing number. Again, we know the number before it. So we just need to add 14 to 138. 138 plus 10 equals 148, and then plus the four equals 152. How did we find our three missing numbers? Well, we didn’t have to work out the rule. We were told it in the question. We knew that the numbers increased by 14 each time. We knew that this meant there was a difference of 14 between each number. For two of the missing numbers in our sequence, we knew the number that came before. And so we just added 14 to the number that came before to find the answer. One of our numbers came right at the start. And so we didn’t have a number to add 14 to. So we used our knowledge that there’s a difference of 14. We started with 82 and we subtracted 14 to find the number that came before. And so the missing numbers are 68, 110, and 152.
# Quick Answer: How Many Three Digits Numbers Can Be Formed Using 2 3 4 And 5 With The Digit Being Repeated? ## How many three digit numbers can be formed? 648 differentIf each digit must be distinct, then no repetition is allowed and the number of possibilities is: nine for the first digit [1-9], times nine (including zero) for the second digit, times eight for the third digit, for a total of 648 different 3-digit numbers.. ## How many different combinations of 3 digit numbers can be formed using the numbers 1 2 3 4 and 5 if repetitions are not allowed? Explanation: The first digit of the 3-digits can take 7 distinct values: 1, 2, 3, 4, 5, 7, 9. As repetition is allowed, the second digit can also take 7 distinct values, and the third can take 7 distinct values aswell, giving a total of 7⋅7⋅7=343 distinct combinations of numbers. ## What is the sum of all 4 digit numbers formed using the digits 2 3 4 and 5 without repetition? Answer and Explanation: The sum of all the 4-digit numbers formed using the digits 2, 3, 4, and 5 (without repetition) is 93, 324. ## How many 4 digits number has sum of their digits is four? Four digit numbers =4⋅3⋅2⋅1=24 ways we can form a four digit number. Since it’s a 4 digit number, each digit will appear 6=24/4 times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is 6(1+2+5+6)=84. ## How many combinations are there of 4 numbers without repeating? 15Answer and Explanation: The number of possible combinations with 4 numbers without repetition is 15. ## How many permutations of 3 digits are there from the numbers 0 9? There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. ## What are all the combinations for a 3 Number Lock? A standard 3-digit lock has 999 possible combinations, which is why the 4- digit locks cost a bit of a step more money, as they have 9999 possibles. ## How many combinations of 3 numbers can 5 numbers make? 10 possible combinationsSo 5 choose 3 = 10 possible combinations. ## How many 4 digit combinations are there? 10,000 possibleThere are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code. ## How many 3 digit even numbers can be formed from the digits 0 1 2 3 4 and 5 when none of the digits are repeated? Therefore total three digit even numbers formed from the digits 0 ,1, 2, 3, 4, 5 and 6 such that no repetition of digit is allowed is 105. There are nine possible first digits, because numbers beginning in 0 drop the 0, so 012 is really just 12, a two digit number. … Here, we get to count 0.12 as a three digit number, a leading zero before the decimal is retained. So there are 10 possible first digits, 9 possible second digits, and… ## How many three digit numbers can be formed using the digits 2 3 4 5 6 if digits can be repeated? Using the same argument we can fill up the Units digit in 4 ways. Hence we can form (6)(5)(4) = 120 numbers of three digits using the digits 1,2,3,4,5,6 without repetition. What is the largest 7-digit number that has only the digits 4, 8, 7, 9, and 5? ## How many different three digit numbers can be formed by 0 3 and 5 if none of them is repeated in a number? It’s quite easy to reason this out. You start with 35 and you form all possible combinations adding all digits (except 3 and 5 as per the requirements) to the right: 351 352 354 356 357 358 359 350 = 8 possible; than you repeat putting the extra digit in the middle 315 325 … ## How many 4 digit numbers can be formed from the digits 1 2 3 4 5 6 and 7 which are divisible by 5 when none of the digits are repeated? Answer is 840! This is a question of permutation and combination. We have to make 4 digit number without repetition using 1,2,3,4,5,6,7. ## How many 2 digit numbers are there? 90 twoHow many two-digit numbers are there? There are 90 two-digit numbers in all.
2019 AIME I Problems/Problem 7 Problem 7 There are positive integers $x$ and $y$ that satisfy the system of equations $$\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60$$$$\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.$$ Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$. Solution Add the two equations to get that log x+log y+2(log(gcd(x,y))+log(lcm(x,y)))=630. Then, use the theorem log a+log b=log ab to get the equation log xy+2(log(gcd(x,y))+log(lcm(x,y)))=630. Use the theorem that the product of the gcd and lcm of two numbers equals to the product of the number along with the log a+log b=log ab theorem to get the equation 3log xy=630. This can easily be simplified to log xy=210, or xy = 10^210. 10^210 can be factored into 2^210 * 5^210, and m+n equals to the sum of the exponents of 2 and 5, which is 210+210 = 420. Multiply by two to get 2m +2n, which is 840. Then, use the first equation, which is log x + 2log(gcd(x,y)) = 60, to realize that x has to have a lower degree of 2 and 5 than y, therefore making the gcd x. Then, turn the equation into 3log x = 60, yielding log x = 20, or x = 10^20. Factor this into 2^20 * 5^20, and add the two 20's, resulting in m, which is 40. Add m to 2m + 2n (which is 840) to get 40+840 = 880. Solution 2 (Almost same as above but cleaner) First simplifying the first and second equations, we get that $$\log_{10}(x\cdot\text{gcd}(x,y)^2)=60$$ $$\log_{10}(y\cdot\text{lcm}(x,y)^2)=570$$ Thus, when the two equations are added, we have that $$\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630$$ When simplified, this equals $$\log_{10}(x^3y^3)=630$$ so this means that $$x^3y^3=10^{630}$$ so $$xy=10^{210}.$$ Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x and $x$ and $y$ are both powers of $10$. This means the first equation would simplify to $$x^3=10^{60}$$ and $$y^3=10^{570}.$$ Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so $$3\cdot 40 + 2\cdot 380 = \boxed{880}.$$ Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution. See Also 2019 AIME I (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
# How do you write sqrt (x^5) as an exponential form? Jan 31, 2016 The square root is expressed as an exponent of $\frac{1}{2}$, so $\sqrt{{x}^{5}}$ can be expressed as ${x}^{\frac{5}{2}}$. #### Explanation: Roots are expressed as fractional exponents: $\sqrt[2]{x} = {x}^{\frac{1}{2}}$ $\sqrt[3]{x} = {x}^{\frac{1}{3}}$ and so on. This makes sense, because when we multiply we add exponents: $\sqrt{x}$ x $\sqrt{x}$ = $x$ ${x}^{\frac{1}{2}}$ x ${x}^{\frac{1}{2}}$ = ${x}^{\left(\frac{1}{2} + \frac{1}{2}\right)}$ = ${x}^{1}$ = $x$ When an exponent is raised to another exponent, the exponents are multiplied: $\sqrt{{x}^{5}} = {\left({x}^{5}\right)}^{\frac{1}{2}} = {x}^{5 \cdot \frac{1}{2}} = {x}^{\frac{5}{2}}$
Question # If a square is inscribed in a circle, find the ratio of the areas of the circle and the square. Hint: - Area of circle is $\pi {r^2}{\text{ sq}}{\text{.units}}$, where r is the radius of the circle. Let, we assume ABCD is a square inscribed in a circle with center O. Let the side of the square be $b$. $\Rightarrow AB = BC = CD = DA = b$. Let the radius of the circle be $r$. $\Rightarrow OB = OD = r$ As we know area of the circle$= \pi {r^2}{\text{ sq}}{\text{.units}}$ And the diagonal of the square $\left( {DB} \right)$is passing through the center O. Which is the diameter of the circle. $\Rightarrow BD = 2r$ And we know that the side of a square is perpendicular to each other. Therefore from figure $\Delta BCD{\text{ }}$makes a right angle triangle so, apply Pythagoras Theorem $\begin{gathered} \Rightarrow {\left( {BD} \right)^2} = {\left( {BC} \right)^2} + {\left( {CD} \right)^2} \\ \Rightarrow {\left( {2r} \right)^2} = {b^2} + {b^2} = 2{b^2} \\ \Rightarrow {b^2} = \dfrac{{4{r^2}}}{2} \Rightarrow 2{r^2} \\ \end{gathered}$ As we know area of square is $= {\left( {{\text{side}}} \right)^2} = {b^2} = 2{r^2}{\text{ sq}}{\text{.units}}$ Now we have to find out the ratio of the areas of the circle and the square. Therefore required ratio $= \dfrac{{{\text{Area of circle}}}}{{{\text{Area of square}}}} = \dfrac{{\pi {r^2}{\text{ sq}}{\text{.units}}}}{{{\text{2}}{{\text{r}}^2}{\text{ sq}}{\text{.units}}}} = \dfrac{\pi }{2}$ So, $\dfrac{\pi }{2}$ is the required ratio of the areas. Note: -In such types of questions the key concept we have to remember is that the diagonal of the square is always passing through the center of the circle, and the sides of square is perpendicular to each other so triangle BCD makes a right angle triangle from this we easily calculate the side of the square, then find out the area of the circle and area of the square, then divide them we will get the required answer.
# Video: Ordering Rational Numbers Which of the following choices shows the amounts in order from the least to greatest? [A] 1/2, 65% of 1, 0.329 [B] 2/5, 0.31, 49% of 1 [C] 45% of 1, 0.5, 2/3 [D] 80% of 1, 3/4, 0.492 04:19 ### Video Transcript Which of the following choices shows the amounts in order from the least to greatest? Is it (A) one-half, 65 percent of one, and 0.329; (B) two-fifths, 0.31, 49 percent of one; (C) 45 percent of one, 0.5, two-thirds, (D) 80 percent of one, three-quarters, and 0.492? To work out whether the choices are in order from least to greatest, we need to convert all the values to either fractions, decimals, or percentages. In this question, we will convert them to decimals. Let’s begin by considering how we convert the percentages to decimals. The word “percent” means out of 100 and the word “of” in mathematics means multiply. Therefore, 65 percent of one is the same as 65 over 100 multiplied by one. Any number multiplied by one remains the same. Therefore, this is equal to 65 over 100. As the line in a fraction means divide, we need to divide 65 by 100. We move all our digits two places to the right, giving us a decimal value 0.65. 65 percent of one is equal to 0.65. We could repeat this with 49 percent of one such that this is equal to 0.49. 45 percent of one is equal to 0.45. 80 percent of one is equal to 0.8 or 0.80. We also need to convert the four fractions to decimals. Most of these are common fractions that we should know the conversions of. One-half is equal to 0.5. We know that one-fifth is equivalent to two-tenths. Therefore, it is equal to 0.2. This means that two-fifths is equal to 0.4. One-third is equal to 0.3 recurring. This can be denoted with a dot or bar above the three. Two-thirds is, therefore, equal to 0.6 recurring. Finally, we know that one-quarter is equal to 0.25 as this is a half of a half. Three-quarters is, therefore, three times this and it’s equal to 0.75. We have now converted all of our values to decimals. We need to find the list that is in order from least to greatest. Option (A) is incorrect as 0.329 is less than 0.65 and 0.5. The same is true for option (B) as 0.31 is less than 0.4. In option (D), the numbers are in descending order. They’re going from greatest to least. 0.8 is greater than 0.75, which is also greater than 0.492. The correct answer is, therefore, option (C). 0.45 is less than 0.5, which is less than 0.6 recurring. Option (C) shows the amounts in order from least to greatest, 45 percent of one, 0.5, and then two-thirds.
# Bar Graph ## Bar Graph for Class 5 Maths The bar graph is a visual representation of the data. Here students will learn about what is a bar graph. Bar graph images and how to draw a bar graph. In this learning concept, the students will also learn to • Classify the types of bar graphs. • Evaluate the horizontal bar graph and vertical bar graph. • Identify the advantage of bar graphs. Each concept is explained to class 5 maths students using illustrations, examples, and mind maps. Students can assess their learning by solving the two printable worksheets given at the page’s end. Download the bar graph worksheet for class 5 and check the solutions to the bar graph questions for class 5 provided in PDF format. #### What Is a Bar Graph? A bar graph is a graphical representation of data, by rectangular bars. • Data can be represented by bars. • In a bar graph each bar represents a number. • The length of bars represents numerical value. • In a bar graph, bars can be drawn vertically or horizontally. Example:In a school, there are 15, 20, 45, and 39 students with yellow, red, blue, and green dresses, respectively. We can represent this data using a bar graph: #### To make a bar graph, we need: Data categories:Type of things in the data. Data value:Numerical value of each category. Scale:To draw the bar with numbers by scaling. Title:We need to give an appropriate title to the graph. • In a bar graph, there should be equal spacing between the bars. #### Vertical Bar Graph • In vertical bar graphs, the bars are drawn vertically to represent the data. Example: The number of children in five different batches of an educational institute is given below by a vertical bar graph. Here, 1 unit length = 10 children. From this graph, we can find Question: What is the number of students in Batch 3? From the graph, the number of students in batch 3 is 40. Question: The maximum number of students present in which batch? The maximum number of students present in batch 2, the number of students is 50. #### Horizontal Bar Graph • In the horizontal bar graph represents the data by the horizontal bars. Example: The data for the baking of cakes in a bakery from Monday to Saturday is shown below by a horizontal bar graph. 1 unit = 10 cakes Question 1: Maximum numbers of cakes baked on which day? Question 2: Minimum numbers of cakes baked on which day? Minimum numbers of cakes baked on Monday. #### How to Draw a Bar Graph? Let us consider, We have four different types of animals, such as cat, dog, rabbit, and hamster and the corresponding numbers are 40, 30, 10, and 70 respectively. #### The bar graph becomes as given below: Note: Scale: 1 unit=10 animals. Transformation of a horizontal bar graph to a vertical bar graph: From this graph, we can make the table The following table shows number of visitors to park for the months January to March Month January February March Month 150 300 250 Also, from the graph we have • The greatest number of visitors come to the park in February. • Also, the maximum number of visitors in one month is 300. • The number of visitors that come in March is 250. • The minimum number of visitors comes in January. • The number of visitors that come in January is 150. Also, we can make the vertical graph by the similar data: • Bar graph summaries the large set of data in simple visual form. • Bar graph displays each category of data in the frequency distribution. • Bar graph displays each category of data in the frequency distribution. • The bar graph clarifies the trend of data better than the table. • Bar graph helps in estimating the key values at a glance. #### The disadvantage of bar graph • Sometimes, the bar graph fails to reveal the patterns, cause, effects, etc. • It can be easily manipulated to yield fake information. Did you know? Bar graph use to show the increase of daily cases of coivd-19
Absolute Value Function Worksheet - Page 2 \| Problems & Solutions # Absolute Value Function Worksheet - Page 2 Absolute Value Function Worksheet • Page 2 11. Find the values of $x$, if \|4$x$ + 1\| = $\frac{4}{3}$. a. $\frac{-1}{12}$, $\frac{7}{12}$ b. 1, -7 c. 1, $\frac{1}{12}$ d. $\frac{1}{12}$, $\frac{-7}{12}$ #### Solution: \|4x + 1\| = 4 / 3 4x + 1 = 4 / 3 or 4x + 1 = -4 / 3 [The expression 4x + 1 is equal to -4 / 3 or 4 / 3.] 4x + 1 - 1 = 4 / 3 - 1 or 4x + 1 - 1 = -4 / 3 - 1 [Subtracting 1 from the two sides of the equation.] 4x = 1 / 3 or 4x = -7 / 3 [Simplify.] x = 1 / 12 or x = -7 / 12 [Divide throughout by 4.] The two solutions of the expression are 1 / 12 and -7 / 12. 12. Find the values of $x$, if \|4$x$ - 16\| - 19 = 12. a. $\frac{47}{4}$, $\frac{47}{4}$ b. $\frac{47}{4}$, $\frac{-15}{4}$ c. $\frac{47}{4}$, $\frac{1}{4}$ d. $\frac{-47}{4}$, $\frac{-15}{4}$ #### Solution: \|4x - 16\| - 19 = 12 \|4x - 16\| - 19 + 19 = 12 + 19 [Add 19 to both sides of the equation.] \|4x - 16\| = 31 [Simplify.] \|4x - 16\| = 31, 4x - 16 equals to 31 or - 31. 4x - 16 = 31 or 4x - 16 = - 31. 4x - 16 + 16 = 31 + 16 or 4x - 16 + 16 = - 31 + 16 [Add 16 to both sides of the equation.] 4x = 47 or 4x = -15 [Simplify.] x = 47 / 4 or x = -15 / 4 [Divide throughout by 4.] The equation has two solutions: 47 / 4 and -15 / 4. 13. Find the values of $x$, if \|11 - 6$x$\| - $\frac{1}{4}$ = 5.6. a. -0.86, 2.81 b. 0.86, 2.81 c. 0.86, -2.81 d. -0.86, -2.81 #### Solution: \|11 - 6x\| - 1 / 4 = 5.6 \|11 - 6x\| - 1 / 4 + 1 / 4 = 5.6 + 1 / 4 [Add 1 / 4 to both sides of the equation.] \|11 - 6x\| = 5.6 + 0.25 = 5.85 [Simplify.] \|11 - 6x\| = 5.85, 11 - 6x equals 5.85 or - 5.85. 11 - 6x = 5.85 or 11 - 6x = - 5.85 11 - 6x - 11 = 5.85 - 11 or 11 - 6x - 11 = - 5.85 - 11 [Subtracting 11 from the two sides of the equation.] x = 0.86 or x = 2.81 [Simplify.] The equation has two solutions: 0.86 and 2.81. 14. Which of the following is an absolute value function? a. $f$($x$) = \|8$x$ - 10\| b. $f$($x$) = 10$x$3 - 8$x$ - 5 c. $f$($x$) = d. $f$($x$) = 10${e}^{8x+5}$ #### Solution: f(x) = 10x3 - 8x - 5 is a cubic function. f(x) = 10x - 5x+8 is a rational function. f(x) = 10e8x+5 is an exponential function. f(x) = \|8x - 10\| an absolute value function. 15. 0 a. $b$ = - 9 or $b$ = - 14 b. $b$ = 4 or $b$ = 9 c. $b$ = - 4 or $b$ = - 14 d. $b$ = 4 or $b$ = - 14 #### Solution: \|b + 5\| = 9 b + 5 = 9 or b + 5 = - 9 b = 4 or b = - 14 16. 0 a. $a$ = 20 b. $a$ = 20 or $a$ = - 30 c. $a$ = 105 or $a$ = -151 d. $a$ = 29.2 or $a$ = -19.2 #### Solution: \|5a + 25\| - 2 = 123 \|5a + 25\| = 125 [Add 2 to both sides of the equation.] 5a + 25 = 125 or 5a + 25 = - 125 [Express as a disjunction.] 5a = 100 or 5a = - 150 [Subtracting 25 from the two sides of the equation.] a = 20 or a = - 30 [Divide throughout by 5 .] 17. 0 a. $z$ = $\frac{6}{5}$ or $z$ = 6 b. $z$ = $\frac{33}{5}$or $z$ = - 6 c. $z$ = - $\frac{33}{5}$ or $z$ = - 44 d. $z$ = - $\frac{33}{5}$ or $z$ = 11 #### Solution: \|11 - 5z\| = 44 11 - 5z = 44 or 11 - 5z = - 44 - 5z = 33 or - 5z = - 55 [Subtracting 11 from the two sides of the equation.] z = - 33 / 5 or z = 11 [Divide throughout by 5.] 18. Determine whether the statement is true or False: If $y$ < 0, then \|6 - $y$\| = 6 + $y$ a. True b. False #### Solution: let y = - a, where a is any positive real number. [y < 0.] \|6 - (- a)\| = 6 - a [Replace y = - a in the given statement.] \|6 + a\| = 6 - a 6 + a = 6 - a, which is always false because a > 0. [\|x\| = x, if x is positive.] 19. Determine whether the given statement is true or false: If $x$ ≥ 0, then \|$x$ - 4\| = 4 - $x$ a. True b. False #### Solution: If x ≥ 0, then \|x - 4\| = 4 - x Take one example x = 8, which is greater than zero. \|8 - 4\| = 4 - 8 4 = - 4, which is not true. Hence, the given statement is not true. 20. Determine whether the given statement is True or false: If $k$ ≤ 0, then \|$k$ - 3\| = \|3 - $k$\| a. False b. True #### Solution: Since k ≤ 0, let k = - b, where ' b ' is any positive real number. \|- b - 3\| = \|3 - (- b)\| [Substitute the values.] \|- (b + 3)\| = \|3 + b\| \|b + 3\| = \|b + 3\|, which is true. [\|- k\| = \|k\|.] So, the given statement is true.
## 1. Mathematical Induction Principle of Mathematical Induction: Let P be a property of positive integers such that: 1. Basis step: P(1) is true, and 2. Inductive step: if P(k) is true for all 1 ≤ k ≤ n then P(n + 1) is true. Then P(n) is true for all positive integers. The premise P(n) in the inductive step is called Induction Hypothesis. The validity of the Principle of Mathematical Induction is obvious. The basis step states that P(1) is true. Then the inductive step implies that P(2) is also true. By the inductive step again we see that P(3) is true, and so on. Consequently the property is true for all positive integers. In the basis step we may replace 1 with some other integer m. Then the conclusion is that the property is true for every integer n greater than or equal to m. Example: Prove that the sum of the n first odd positive integers is n2, i.e., 1 + 3 + 5 + · · · + (2n − 1) = n2 Proof: n=1, S(1) = 12 = 1 (1); n=k, S(k) = (k)2 ⇒ n=2k+1, S(k+1) = S(k) + 2k + 1 = k2 + 2k + 1 = (k+1)2 (2); ## 2. Recursiveness A definition such that the object defined occurs in the definition is called a recursive definition. For instance, consider the Fibonacci sequence 0, 1, 1, 2, 3, 5, 8, 13, . . . It can be defined as a sequence whose two first terms are F0 = 0, F1 = 1 and each subsequent term is the sum of the two previous ones: Fn = Fn−1 + Fn−2 (for n ≥ 2). Other examples: Factorial: 1. 0! = 1 2. n! = n · (n − 1)! (n ≥ 1) Power: 1. a0 = 1 2. an = an−1 a (n ≥ 1) In all these examples we have: 1. A basis, where the function is explicitly evaluated for one or more values of its argument. 2. A recursive step, stating how to compute the function from its previous values. ## 3. Divide and Conquer Many useful algorithms are recursive in structure: to solve a given problem, they call themselves recursively one or more times to deal with closely related subproblems. These algorithms typically follow a divide-and-conquer approach: they break the problem into several subproblems that are similar to the original problem but smaller in size, solve the subproblems recursively, and then combine these solutions to create a solution to the original problem. The divide-and-conquer paradigm involves three steps at each level of the recursion: • Divide the problem into a number of subproblems that are smaller instances of the same problem. • Conquer the subproblems by solving them recursively. If the subproblem sizes are small enough, however, just solve the subproblems in a straightforward manner. • Combine the solutions to the subproblems into the solution for the original problem. ``````MERGE-SORT(A, p, r) 1 if p < r 2 q = ⌊(p + r) / 2 ⌋ 3 MERGE-SORT(A, p, q) 4 MERGE-SORT(A, q, r) 5 MERGE(A, p, q, r) MERGE(A, p, q, r) 1 n1 = q - p + 1 2 n2 = r - q 3 let L[1..n1 + 1] and R[1..n2] be new arrays 4 for i = 1 to n1 5 L[i] = A[p + i - 1] 6 for j = 1 to n2 7 R[j] = A[q + j] 8 L[n1 + 1] = ∞ 9 R[n2 + 1] = ∞ 10 i = 1 11 j = 1 12 for k = p to r 13 if L[i] ≤ R[j] 14 A[k] = L[i] 15 i = i + 1 16 else A[k] = R[j] 17 j = j + 1`````` ### 3.1. Analyzing divide-and-conquer algorithms When an algorithm contains a recursive call to itself, we can often describe its running time by a recurrence equation or recurrence, which describes the overall running time on a problem of size n in terms of the running time on smaller inputs. We can then use mathematical tools to solve the recurrence and provide bounds on the performance of the algorithm. A recurrence for the running time of a divide-and-conquer algorithm falls out from the three steps of the basic paradigm. • We let T(n) be the running time on a problem of size n. If the problem size is small enough, say n ≤ c for some constant c, the straightforward solution takes constant time, which we write as Θ(1). • Suppose that our division of the problem yields a subproblems, each of which is 1/b the size of the original. For merge sort, both a and b are 2, but we shall see many divide-and-conquer algorithms in which a ≠ b. It takes time T(n/b) to solve one subproblem of size n/b, and so it takes time a.T(n/b) to solve a of them. • If we take D(n) time to divide the problem into subproblems and _C(n) time to combine the solutions to the subproblems into the solution to the original problem, we get the recurrence T(n) = O(1) if n ≤ c, a.T(n/b) + D(n) + C(n) otherwise. ### 3.2. Proof of the master theorem The master method provides a “cookbook” method for solving recurrences of the form T(n) = a.T(n/b) + f(n) where a ≥ 1 and b > 1 are constants and f(n) is an asymptotically positive function. For merge sort, we see the T(n) that roughly: T(n) = 2T(n/2) + n Replacing n with n/2 we have T(n/2) = 2T(n/4) + n/2, hence: T(n) = 2T(n/2) + n = 2(2T(n/4) + n/2) + n = 4T(n/4) + 2n Repeating k times we get: T(n) = 2kT(n/2k) + k.n So for k = log2n we have: T(n) = nT(1) + nlog2n = Θ(n.lgn) ### 3.3. The maximum-subarray problem ``````// 53. Maximum Subarray // Medium // // Given an integer array nums, find the contiguous subarray (containing at least one number) which // has the largest sum and return its sum. // // A subarray is a contiguous part of an array. // // Example 1: // // Input: nums = [-2,1,-3,4,-1,2,1,-5,4] // Output: 6 // Explanation: [4,-1,2,1] has the largest sum = 6. // // Example 2: // // Input: nums = [1] // Output: 1 // // Example 3: // // Input: nums = [5,4,-1,7,8] // Output: 23 // // // // Constraints: // // 1 <= nums.length <= 10^5 // -10^4 <= nums[i] <= 10^4 // // // // Follow up: If you have figured out the O(n) solution, try coding another solution // using the divide and conquer approach, which is more subtle. package maxSubArray import ( "math" ) // divide-and-conquer // nums[low, mid, high] // nums[low,...,mid], nums[low,...,mid,...,high], nums[mid,...,high] func maxSubArray(nums []int) int { var findMaxCrossSubArray func(nums []int, low, mid, high int) int findMaxCrossSubArray = func(nums []int, low, mid, high int) int { leftMax := math.MinInt sum := 0 for i := mid - 1; i >= low; i-- { sum += nums[i] if leftMax < sum { leftMax = sum } } sum = 0 rightMax := math.MinInt for i := mid; i < high; i++ { sum += nums[i] if rightMax < sum { rightMax = sum } } return leftMax + rightMax } var findMaxSubArray func(nums []int, low, high int) int findMaxSubArray = func(nums []int, low, high int) int { if high-low <= 1 { // bottom-out, base-case, only one number, O(1) return nums[low] } mid := (low + high) / 2 left := findMaxSubArray(nums, low, mid) // T(n/2) cross := findMaxCrossSubArray(nums, low, mid, high) // O(n), n = high - low right := findMaxSubArray(nums, mid, high) // T(n/2) // fmt.Println(left, cross, right) if left >= right && left >= cross { // O(1) return left } else if right >= left && right >= cross { return right } return cross } // T(n) = O(1) + 2T(n/2) + O(n) + O(1) = 2T(n/2) + O(n) => O(nlgn), n > 1 // T(n) = O(1), n == 1 return findMaxSubArray(nums, 0, len(nums)) } // brute-force O(n^2) // func maxSubArray(nums []int) int { // max := nums[0] // for i := 0; i < len(nums); i++ { // sum := 0 // for j := i; j < len(nums); j++ { // sum += nums[j] // if max < sum { // max = sum // } // } // } // return max // }``````
# 1819 C2X §5.Refresh What are these graphs called? 1 / 12 Slide 1: Woordweb WiskundeMiddelbare schoolhavo, vwoLeerjaar 2 In deze les zitten 12 slides, met interactieve quiz en tekstslides. Lesduur is: 50 min ## Onderdelen in deze les What are these graphs called? #### Slide 1 -Woordweb Quadratic formulas are the type of formulas you will use most often in Mathematics. Working with quadratic formulas requires you to work precisely and neatly and you need the skills you've learned in previous chapters. #### Slide 2 -Tekstslide Recap: Order of Operations When doing calculations, you have to follow the order of operations: 1. First calculate what is given between brackets. 2. Then calculate powers and roots. 3. Then multiply and divide from left to right. 4. Then add and subtract from left to right. #### Slide 4 -Tekstslide Recap: Square roots The square root of 25 is 5, because the square of 5 is 25. The square root of 6.25 is 2.5, because the square of 2.5 is 6.25 There is no square root of -16 (or any other negative number), because there is no number you can square to get a negative result #### Slide 5 -Tekstslide Recap: Writing formulas shorter You can write formulas shorter by: - multiplying factors - combining like terms - not writing down the multiplication sign between a number and a variable #### Slide 7 -Tekstslide If a formula contains a the square of a variable (and not a higher exponent), it is called a quadratic formula. The corresponding graph is called a parabola. When filling in negative numbers for the variable in a quadratic formula you should use brackets, because you want to square the negative number, not take the opposite of a positive square. #### Slide 10 -Tekstslide Pen&Paper or Digital
Lesson Video: Simplifying Algebraic Fractions \| Nagwa Lesson Video: Simplifying Algebraic Fractions \| Nagwa # Lesson Video: Simplifying Algebraic Fractions Mathematics In this video, we will learn how to factor algebraic expressions and simplify algebraic fractions. 14:55 ### Video Transcript In this video, we’ll learn how to factor algebraic expressions and simplify algebraic fractions. Let’s begin by recalling the general method for simplifying a fraction made up of purely numerical values, such as 27 over 63. Our job is to find the greatest common factor of the numerator and denominator of our fraction, then divide through by this value. So take the fraction 27 over 63. Let’s find the greatest common factor of these numbers. The greatest common factor of 27 and 63 is nine. So we divide the numerator and denominator of our fraction by nine, giving us an answer of three-sevenths. We say that three and seven are coprime. They share no factors other than one. So we can say that our fraction is now fully simplified. And believe it or not, the same process goes for algebraic fractions. Let’s take five 𝑥 squared over 10𝑥 to the fifth power. This time, we need to work out the greatest common factor of five 𝑥 squared and 10𝑥 to the fifth power. Well, the greatest common factor of the numerical part that is the greatest common factor of five and 10 is five and, of our algebraic part, it’s 𝑥 squared. So the greatest common factor of our numerator and denominator is five 𝑥 squared. We then divide both parts of this fraction by five 𝑥 squared. Five 𝑥 squared divided by five 𝑥 squared is one. And 10𝑥 to the fifth power divided by five 𝑥 squared is two 𝑥 cubed. So in its simplest form, our fraction is one over two 𝑥 cubed. But what if our algebraic fraction is more complicated? Let’s see what that might look like. Fully simplify 𝑥 plus four times 𝑥 plus three over 𝑥 plus two times 𝑥 plus four. Remember, to simplify fractions, we divide the numerator and denominator by the greatest common factor of each. Let’s look carefully at the numerator and denominator of our fraction. The factors of the numerator are 𝑥 plus four and 𝑥 plus three. Then the factors of the denominator are 𝑥 plus two and 𝑥 plus four. We can therefore say that the greatest common factor is 𝑥 plus four. It’s the only factor other than one that these two expressions have in common. So to simplify this algebraic fraction, we’re going to divide through by 𝑥 plus four on both parts. 𝑥 plus four times 𝑥 plus three divided by 𝑥 plus four leaves us with 𝑥 plus three on our numerator. Similarly, 𝑥 plus two times 𝑥 plus four divided by 𝑥 plus four leaves us with 𝑥 plus two on our denominator. That leaves us with 𝑥 plus three over 𝑥 plus two. 𝑥 plus three and 𝑥 plus two are now coprime. They share no other factors other than one. So we’re finished. In its simplest form, our fraction is 𝑥 plus three over 𝑥 plus two. Now, this fraction was quite easy to simplify because it was fully factored on both its numerator and denominator. We’ll now have a look at an example where that’s not necessarily the case. Fully simplify 𝑥 minus three times 𝑥 squared minus six 𝑥 plus nine over 𝑥 minus three cubed. Remember, to simplify fractions, we divide the numerator and denominator by the greatest common factor of each. Now, we can see that we have a common factor of 𝑥 minus three. There’s an 𝑥 minus three in the numerator and the denominator. But is that the greatest common factor? Well, to check, we factor any nonfactored expressions. So we’re going to fully factor the expression 𝑥 squared minus six 𝑥 plus nine from the numerator of our fraction. This is a quadratic expression, but there are no common factors in 𝑥 squared, negative six 𝑥, and nine. So that tells us we factor into two parentheses. We know that, in order to achieve an 𝑥 squared, we need an 𝑥 here and an 𝑥 here. And we need to find two numbers whose product is nine and whose sum is negative six. Well, that must be negative three and negative three, since negative three times negative three is positive nine. But negative three plus negative three is negative six. If we replace 𝑥 squared minus six 𝑥 plus nine with its factored form, our fraction becomes 𝑥 minus three times 𝑥 minus three times 𝑥 minus three over 𝑥 minus three cubed. But of course, it should be quite clear that 𝑥 minus three times 𝑥 minus three times 𝑥 minus three is actually 𝑥 minus three cubed. Notice now that our numerator and denominator are actually equal. We’re dividing a number by itself. And when we divide a number by itself, we get one. So in simplified form, this fraction is simply one. Now, note that we could’ve approached this slightly differently. Instead of factoring at the start 𝑥 squared minus six 𝑥 plus nine, we could have divided through by a factor of 𝑥 minus three. We eventually saw that this isn’t the greatest common factor, but it’s a good start. We divide the numerator and the denominator by 𝑥 minus three, noting that 𝑥 minus three cubed divided by 𝑥 minus three is 𝑥 minus three squared. Then we could’ve factored and spotted that we had further common factors of 𝑥 minus three squared. Either method ultimately results in us dividing both the numerator and denominator by the greatest common factor of 𝑥 minus three cubed. And either method results in an answer of one. Let’s now have a look at an example that involves factoring more than one expression. Fully simplify 𝑥 plus two times 𝑥 squared plus seven 𝑥 plus 12 over 𝑥 plus seven times 𝑥 squared plus 10𝑥 plus 21. Remember, to simplify fractions, we divide the numerator and denominator by their greatest common factor. Now, looking at our fraction, it’s not instantly obvious what the greatest common factor is. And so we look for any unfactored expressions and we factor them. In fact, there are two. On the numerator, we have 𝑥 squared plus seven 𝑥 plus 12 and, on the denominator, 𝑥 squared plus 10𝑥 plus 21. Note that 12 and 21 share a common factor of three. That’s an indication to us that we might end up canceling by a factor of, say, 𝑥 plus three. That’s not hugely helpful just yet, but certainly something to bear in mind. We begin by fully factoring the expression 𝑥 squared plus seven 𝑥 plus 12. It’s a quadratic expression. And there are no common factors throughout. So we have two brackets with an 𝑥 at front of each bracket. We’re looking for two numbers that multiply to make 12 and add to make seven. Well, that’s three and four. So 𝑥 squared plus seven 𝑥 plus 12 can be written as 𝑥 plus three times 𝑥 plus four. Let’s repeat this process for the expression 𝑥 squared plus 10𝑥 plus 21. Once again, it’s two pairs of parentheses with the 𝑥 at the front of each one. This time, though, we want two numbers whose product is 21 and whose sum is 10. That’s seven and three. So this expression becomes 𝑥 plus seven times 𝑥 plus three. Let’s now go back to our original fraction and replace each expression with its factored form. In doing so, it becomes 𝑥 plus two times 𝑥 plus three times 𝑥 plus four over 𝑥 plus seven times 𝑥 plus seven times 𝑥 plus three. We’re now ready to look for the greatest common factor of both parts of our fraction. If we look carefully, we see that each part shares a factor of 𝑥 plus three. So we’re going to divide both the numerator and the denominator by 𝑥 plus three. Dividing the numerator by 𝑥 plus three and we leave ourselves with 𝑥 plus two times 𝑥 plus four. Similarly, dividing the denominator by 𝑥 plus three leaves us with 𝑥 plus seven times 𝑥 plus seven. Noting, of course, that we can write 𝑥 plus seven times itself as 𝑥 plus seven squared, we see that this simplifies to 𝑥 plus two times 𝑥 plus four over 𝑥 plus seven squared. Remember also that multiplication is commutative. We can write our numerator alternatively as 𝑥 plus four times 𝑥 plus two. And we’re still getting the same result. We’ve fully simplified our original fraction. In our next example, we’ll look at a fraction whose numerator and denominator are both fully unfactored. Fully simplify 𝑥 squared plus five 𝑥 minus 24 over 𝑥 squared plus 15𝑥 plus 56. We begin by recalling that, to simplify fractions, we divide the numerator and denominator by their greatest common factor. Now, it’s not instantly obvious what the greatest common factor of 𝑥 squared plus five 𝑥 minus 24 and 𝑥 squared plus 15𝑥 plus 56 is. And so we look for any unfactored factorable expressions, and we fully factor them. So let’s begin by factoring 𝑥 squared plus five 𝑥 minus 24. We have a quadratic expression whose three terms are coprime. That is, their greatest common factor is one. That tells us we’re going to factor the expression into two pairs of parentheses, at the front of which will be 𝑥. Then to find the numerical parts, we need two numbers whose product is negative 24 and whose sum is five. Those are negative three and eight. So 𝑥 squared plus five 𝑥 minus 24 can be written as 𝑥 minus three times 𝑥 plus eight. Our next job is to factor the denominator, 𝑥 squared plus five 𝑥 plus 56. We might notice that 56 has a factor of eight. That’s an indication to us that we might end up dividing through by a factor of 𝑥 plus eight at the end. Let’s factor this expression. Once again, we have two pairs of parentheses. This time we want two numbers whose product is positive 56 and whose sum is positive 15. Well, that’s seven and eight. So this factors to 𝑥 plus seven times 𝑥 plus eight. And so we rewrite our fraction completely. We write it as 𝑥 minus three times 𝑥 plus eight over 𝑥 plus seven times 𝑥 plus eight. And now, we see that the greatest common factor of 𝑥 minus three times 𝑥 plus eight and 𝑥 plus seven times 𝑥 plus eight is indeed 𝑥 plus eight. And so let’s divide both the numerator and denominator by this value. Dividing our numerator by 𝑥 plus eight leaves us with 𝑥 minus three. Similarly, dividing our denominator by 𝑥 plus eight leaves us with 𝑥 plus seven. And so we fully simplified our fraction. It’s 𝑥 minus three over 𝑥 plus seven. In our final example, we’ll look at how we can fully simplify an algebraic fraction where either the numerator or the denominator’s coefficient of 𝑥 squared is not equal to one. Fully simplify 𝑥 squared minus 𝑥 minus 20 over two 𝑥 squared plus nine 𝑥 plus four. We know that, to simplify any fraction, we need to divide both the numerator and the denominator by their greatest common factor. The problem is, with algebraic fractions, it’s not instantly obvious what that is. And so before we can divide through by the greatest common factor, we look to fully factor any unfactorized factorable expressions. We actually have two. We have the numerator, 𝑥 squared minus 𝑥 minus 20, and then the denominator, two 𝑥 squared plus nine 𝑥 plus four. Let’s factor 𝑥 squared minus 𝑥 minus 20. This is a quadratic expression whose three terms are coprime. They share no other factors than one. This tells us we factor into two pairs of parentheses with an 𝑥 at the front of each. Now, we need to find two numbers whose product is negative 20 and whose sum is negative one. That’s negative five and plus four. So we fully factored the numerator. But what about the denominator? This one is a little trickier since the coefficient of 𝑥 squared is no longer one, it’s two. This tells us that, in one of our parentheses, we’re going to have two 𝑥 at the front. And so we can’t use the standard tricks. We do know we need two numbers whose product is four. Well, this could be one and four, which means that we could have four and one in our parentheses. Alternatively, we could have two and two. Now, we could use a bit of trial and error, but let’s look carefully at our numerator. One of the factors of our numerator is 𝑥 plus four. That’s a good indication to us that our denominator must share that factor, meaning that it could factor to be two 𝑥 plus one times 𝑥 plus four. Let’s check by redistributing our parentheses. We multiply two 𝑥 by 𝑥 to give us two 𝑥 squared. We then multiply two 𝑥 by four to get eight 𝑥. We multiply one by 𝑥 to get 𝑥. And then we multiply one by four to get four. Collecting like terms, and we see that we do indeed get two 𝑥 squared plus nine 𝑥 plus four as required. And so we see we can write our fraction as 𝑥 minus five times 𝑥 plus four over two 𝑥 plus one times 𝑥 plus four. The greatest common factor here then must be 𝑥 plus four. So we’ll divide both the numerator and the denominator by 𝑥 plus four. Dividing our numerator leaves us with 𝑥 minus five and dividing our denominator leaves us with two 𝑥 plus one. And so we fully simplified our algebraic fraction. It’s 𝑥 minus five over two 𝑥 plus one. Now, it is worth going back to this step here, where we factorized two 𝑥 squared plus nine 𝑥 plus four. We made an assumption that our algebraic fraction was going to simplify and that there must be a common fact of 𝑥 plus four. It might have been, however, but this was not how two 𝑥 square plus nine 𝑥 plus four should have been factored. If that had been the case and we hadn’t ended up with some common factor, then we could’ve deduced that our fraction was already in its simplest form. In this video, we learned that we can simplify algebraic fractions the same way we do numerical fractions, by dividing both the numerator and the denominator by their greatest common factor. We also saw that there will be occasions where the greatest common factor isn’t instantly obvious. In those occasions, we’ll need to factor any unfactored expressions before it becomes apparent. 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# How To Find Square Root Of A Number In Three Easy Steps A very important topic in Quantitative Aptitude of IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC CGL, SSC CHSL and other competitive exams is squares and cubes. In our previous post, we learnt what squares are and how to find the square of any number in your head, without any pen and paper. Here, we are going to learn how to find the square root of a given number. Conventionally, we have been finding square root of a number through the ‘long division method’ that was taught to us in school. Here, we are going to discuss the shortcut to finding square root of a number. ### Importance Of How To Find Square Root Of A Number In Less Time? Finding square roots of numbers can help you save the much-needed time in competitive exams, such as SSC CGL 2017, SSC CHSL 2017, etc, allowing you to attempt more questions in the limited given amount of time. Finding whether the given number is a perfect square of not, is a lengthy process. If you have a question from simplifications, where you need to find square root in one of the steps, consider the given number to be a perfect square. ONLY when the question is given under the topic approximations, it may not be a perfect square. ### How To Find Square Root Of A Number In Three Easy Steps Firstly, you must memorise the squares of numbers from 1 to 30. We have already learnt that there is a certain pattern seen when it comes to square of any number: SMART METHOD ### How to find square root of a number in a smart method is a 3-step process Example 1: Find the √2401 Solution: Step1: Decide the unit’s place of the result. 2401 ends with 1 = Square root will end with 1 or 9 _ 1 or _ 9 Step2: Leave the last 2 digits and find the immediate perfect square before the remaining number. The immediate perfect square before 24 = 16, which is 42 That is the number in ten’s place 41 or 49 Step 3: To determine whether the square root of 2401 is 41 or 49 Method 1: Take the number that lies exactly between 41 and 49 and square it. 452 = 2025 (We have learnt earlier how to find square of numbers ending with 5) 2025 < 2401 45 < Square root of 2401 And since 45 < 49 Square root of 2401 is 49 Example 2: Find the √6084 Solution: Step1: Decide the unit’s place of the result. 6084 ends with 4 = Square root will end with 2 or 8 _2 or _8 Step 2: Leave the last 2 digits and find the immediate perfect square before the remaining number. The immediate perfect square before 60 = 49, which is 72 72 or 78 Step 3: To determine whether the square root of 6084 is 72 or 78 Method 2: Take squares of numbers 70 and 80, since 72 and 78 lie within the range. 702 = 4900 802 = 6400 Since, 6084 is closer to 6400, the square root must also be closer to 80, which is 78 Square root of 6084 is 78 Example 3: Find the √12769 Solution: Step 1: Decide the unit’s place of the result. 12769 ends with 9 = Square root will end with 3 or 7 _3 or _7 Step 2: Leave the last 2 digits and find the immediate perfect square before the remaining number. The immediate perfect square before 127 = 121, which is 112 113 or 117 Step 3: To determine whether the square root of 12769 is 113 or 117 Method 1: Take the number that lies exactly between 113 and 117 and square it. 1152 = 13225 12769 < 13225 And since 113 < 115 Square root of 12769 is 113 Method 2: Take squares of numbers 110 and 120, since 113 and 117 lie within the range. 1102 = 12100 1202 = 14400 Since 12100 is closer to 12769, the square root must also be closer to 110, which is 113 Square root of 12769 is 113 Practice this technique well and ensure that the next time you are asked to find the square root of a perfect square, you don’t spend more than 3 seconds. 0 Followers Most reacted comment 18 Comment authors Recent comment authors Subscribe Notify of Guest Ahmed saleem1314 We need the saqrue root of number 7584516 plz Guest SENDIL Last number maybe 4(or)6 Guest Meera That number’s square root is 2754. Guest Navya How do we find square roots of non perfect square numbers? Guest gfjhgjk They can take the nearest no. i think Guest sabari that is the number in ten’s place which means? plz tell me Guest karz how to find square root of 6 Guest Vineeshkumar We have already learnt that there is a certain pattern seen when it comes to a square of any number: How To Find Square Root the 2nd table on this table perfect square ends with 9 then square root ends with 3 or 7, not 3 or 6 Guest sunil soir pls tell me what is the root for 1935 Guest Meera I don’t think so that becomes a square because it is ending with a lot of decimals Guest Jj =16 which is 42. Why is it 42? Guest Jj The immediate perfect square before 24 = 16 which is 42. Why is it 42? Guest fds root of 16 = 4, last digit of root of numbre could be 2 Guest arun it’s four square ! not 42. Guest dev perfect step to find square root thank oyu Guest Aarushi How to find difficult roots? Like 4root 15 Guest Rahul R Sekhar How to find 11449 sq root Guest MayBe The immediate perfect square before 24 = 16 which is 42. Why is it 42? Please explain in detail Guest Priyasha How to find √7 Guest P.N.John What do u mean by immediate perfect square.?
# Solving polynomial Equations in Factored Form MM1A2f: Goal: solve polynomial equations Factor trinomials of the form x2 +bx +c. ## Presentation on theme: "Solving polynomial Equations in Factored Form MM1A2f: Goal: solve polynomial equations Factor trinomials of the form x2 +bx +c."— Presentation transcript: Solving polynomial Equations in Factored Form MM1A2f: Goal: solve polynomial equations Factor trinomials of the form x2 +bx +c Zero Product Property The zero-product property is used to solve an equation when one side is zero and the other side is a product of polynomial factors The zero-product property is used to solve an equation when one side is zero and the other side is a product of polynomial factors Solutions are also called roots Solutions are also called roots Zero Product Property Solve: Solve: (x – 3)(x + 6) = 0 (x – 3)(x + 6) = 0 x – 3 = 0 or x + 6 = 0 zero prod. Prop x – 3 = 0 or x + 6 = 0 zero prod. Prop x = 3 or x = -6 solve x = 3 or x = -6 solve Zero Product Property Solve: Solve: (4x + 5)(4x – 5) = 0 (4x + 5)(4x – 5) = 0 4x+5 = 0 or 4x – 5 =0 zero prod. prop 4x+5 = 0 or 4x – 5 =0 zero prod. prop 4x = -5 or 4x = 5 solve 4x = -5 or 4x = 5 solve x = -5/4 or x = 5/4 x = -5/4 or x = 5/4 Solve by factoring Solve: 6x 2 + 12x = 0 Solve: 6x 2 + 12x = 0 6x (x + 2) = 0 factor 6x (x + 2) = 0 factor 6x = 0 or x + 2 = 0 zero prod. Prop 6x = 0 or x + 2 = 0 zero prod. Prop x = 0 or x = -2 solve x = 0 or x = -2 solve Solve by Factoring Solve: Solve: -10b 2 + 25b = 0 -10b 2 + 25b = 0 -5b (2b + 5) = 0 factor -5b (2b + 5) = 0 factor -5b = 0 or 2b + 5 = 0 zero prod. Prop -5b = 0 or 2b + 5 = 0 zero prod. Prop b = 0 or b = -5/2 b = 0 or b = -5/2 Factor trinomials x 2 +bx+c Use guess and check Use guess and check Use x-factor Use x-factor Factor k 2 + 6k -7 k 2 + 6k -7 x factor x factor or guess and check or guess and check (x -1)(x +7) (x -1)(x +7) x – 1 = 0 x + 7 = 0 x – 1 = 0 x + 7 = 0 + 1 + 1 -7 -7 + 1 + 1 -7 -7 x = 1 x = -7 x = 1 x = -7 6 -7 Factor d 2 + 5d – 50 d 2 + 5d – 50 x factor or guess and check x factor or guess and check (x + 10) or (x – 5 ) (x + 10) or (x – 5 ) x + 10 = 0 x – 5 = 0 x + 10 = 0 x – 5 = 0 -10 -10 +5 +5 -10 -10 +5 +5 x = -10 x = 5 x = -10 x = 5 5 -50 Factor -2b 2 + 14b – 20 -2b 2 + 14b – 20 -2(b 2 – 7b + 10) Factor out -2 -2(b 2 – 7b + 10) Factor out -2 x factor or guess and check x factor or guess and check (x - 2) or (x – 5 ) (x - 2) or (x – 5 ) x - 2 = 0 x – 5 = 0 x - 2 = 0 x – 5 = 0 +2 +2 +5 +5 +2 +2 +5 +5 x = 2 x = 5 x = 2 x = 5 -7 10 Download ppt "Solving polynomial Equations in Factored Form MM1A2f: Goal: solve polynomial equations Factor trinomials of the form x2 +bx +c." Similar presentations
# Prime Factorization A number in general has two diverging aspects. A number can be viewed as a factor of a larger number and at the same it is formed as product of some smaller numbers.The concepts of factor decomposition and product composition are applied at any level of Mathematical problem solving. Prime Factorization of a number is finding the prime factors of the number. Every natural number greater than 1, is either a prime or a product of primes in a unique way. ## Prime Factorization Definition Prime Factorization means breaking up a composite number into prime factors. The concept of factors and the idea of factorization originate from the divisibility of numbers. If a number divides another number with a remainder "0", then the second number is said to be divisible by the first number. Solved Example Question: Find the prime factors of 12. Solution: When 12 is divided by 3, the quotient is 4 and the remainder is 0. So we can say that 12 is divisible by 3. Hence prime factors of 12, 12 = 2 x 2 x 3 = 22 x 3. ## What is Prime Factorization? Prime factorization is the process of writing a composite number as a product of its prime factors. A composite number written as a product of prime numbers is known as the prime factorization of the number. In prime factorization the order of numbers does not matter. This makes there is only one prime factorization for a number. Solved Example Question: Prime factorization of 18. Solution: Prime factors of 18, 18 = 2 x 3 x 3 = 2 x 32 or 3 x 3 x 2 = 32 x 2 or 3 x 2 x 3 = 32 x 2. ## How to do a Prime Factorization of a Number? The divisibility test can be used to identify the prime factors of a number. Divisibility test help to find the factors of a number. The following table details the divisibility tests for some of the prime numbers. Divisible by Divisibility Test 2 The ones digit is 0, 2, 4, 6 or 8. In other words the number should be an even number. 3 The sum of the digits is divisible by 3.Example: 231231 = 2 + 3 + 1 = 6, 6 is divisible by 3.So 231 is divisible by 3. 5 The ones digit is 0 or 5Example: 120Here ones digit is zeroSo 120 is divisible by 5. 7 Double the ones digit and subtract that from the sum of the rest of the digits.Repeat the procedure till a two digit number is arrived. If the two digit number got is divisible by 7, then the given number is divisible by 7. 11 Subtract the sum of the digits in odd places from the sum of the digits in the even places. If the difference obtained is 0 or divisible by 11, then the given number is divisible by 11. Solved Example Question: Write - ### Prime Factorization of 72 Solution: By divisibility test: Since 72 is an even number, is divisible by 2 72 = 2 x 36. Continuing the division with 2, 36 = 2 x 18 and 18 = 2 x 9. 9 = 3 x 3 Hence prime factorization of 72, 72 = 2 x 2 x 2 x 3 x 3 = 23x 32 ## Prime Factorization Tree Another method of doing prime factorization is using a prime factorization tree also commonly called a factor tree. The factor tree is formed by branching out the factors at each level of prime division. Solved Example Question: Use factor tree to find - ### Prime Factorization of 245 Solution: The ones digit is 5, so the number is divisible by 5. Dividing 245 by 5 we get the quotient as 49. again 49 is divisible by 7 and 49 = 7 x 7. The factor tree can be extended and completed as follows: The process is stopped at that level when the quotient on division by a prime is also a prime. Hence the prime factorization is 245, 245 = 5 x 7 x 7 = 5 x 72. ## Prime Factorization in Finding the GCF Prime factorization is used in finding the greatest common factor of two or more numbers. The greatest common factor of two or more numbers is the product of all common factors of the numbers. Solved Example Question: Find the greatest common factor of the numbers 24 and 60. Solution: The Prime factorization of the two numbers 24 and 60 is shown here:: Prime factors of 24 = 2 x 2 x 2 x 3 Prime factors of 60 = 2 x 2 x 3 x 5 Hence GCF of 24 and 60 = 2 x 2 x 3 = 12 ## Prime Factorization Examples Below you could see some example of prime factorizations: ## Solved Examples Question 1: Write - ### Prime Factorization of 36 Solution: By divisibility test: 36 is even number, a number with 6 as ones digit. So 2 is a prime factor of 36. On division by 2 the quotient is 18. => 36 = 2 x 18 Again 18 is an even number and can be written as 18 = 2 x 9. Now we have 36 = 2 x 2 x 9, 9 is divisible by 3 and 9 = 3 x 3 Hence the prime factorization of 36, 36 = 2 x 2 x 3 x 3 If the repeated factors are replaced by powers, then 36 = 22 x 32. Question 2: Write - ### Prime Factorization of 48 Solution: By divisibility test: 48 being an even number is divisible by 2. 48 = 2 x 24 Continuing the division with 2, 24 = 2 x 12 and 12 = 2 x 6 and 6 = 2 x 3. Hence the prime factorization for 48 is 48 = 2 x 2 x 2 x 2 x 3 = 24 x 3. Question 3: Write - ### Prime Factorization of 75 Solution: By divisibility test: As the ones digits is 5, this number is divisible by 5. Dividing 75 by 5 the quotient is 15, 75 = 5 x 15. 15 can be factored in a similar manner as 15 = 3 x 5. Hence the prime factorization of 75 is 75 = 5 x 5 x 3 = 52x 3
Find the area of the shaded region in the figure. http://i575.photobucket.com/albums/ss196/felixduhc... And how do you get the answer? Explain how you get the answer please. This question kinda confuses me. Relevance Do it in bits. First find the area of the whole circle... πr²=4π Then find the area of the sector of the circle you're interested in ... The angle at the centre is 120, and this is 1/3 of the whole angle [As 120/360 = 1/3] So area sector=4π/3 Then take off the triangle portion, to leave just the shaded segment... Area triangle=(1/2) x (one side) x (another side) x (sine of angle between) = 0.5 x 2 x 2 x sin(120) = 2 x √3 /2 =√3 Area of segment = (4π/3) - √3 People often leave answers like that, but you can work it out as a decimal, if you want... 2.46 Source(s): Maths teacher OK so this is how I'm going to solve the question. FIRST, i recognize that 120deg. is one third of 360deg., a whole circle which means i can find a third of the area of the whole circle simply by dividing the area by 3. SECOND, I will find the area of the 120deg triangle and subtract that from 1/3 of the area of the circle. Two steps. Simple, right? Applying the first step. A=πr^2 A=π(2^2) A=4π A=12.56 One third of this is 12.56 / 3 = 4.18 Applying the second step. The are of the triangle will equal two times the area of half the triangle, and finding the area of half of that triangle is easy, so that's what we'll do. We need first to find the shortest distance between the centre of the circle and the line bounding the shaded area. This distance is 2cos(60deg.) = 1, meaning that, from Pythagoras' theorem, the other side is sqrt(3). Now we simply apply the area formula: A = 1/2 bh A = 0.5 sqrt(3) A = 0.866 Multiply this by two and we get the area of the whole triangle. =1.73 units^2 This we now subtract from 4.18 to get the final answer. = 4.18 - 1.73 = 2.44 square units. So just recapping...I first found 1/3 of the whole area of the circle, then subtracted the area of the triangle. I found the area of the triangle by splitting it into to two triangles, through the middle so it makes a 90deg angle with the line binding the shaded area. I then used the cosine rule to find the adjacent side, then Pythagoras to the find the other side, then found the area and multiplied it by two. Source(s): High School Maths.
0 Upcoming SlideShare × # Math practice 744 Published on Published in: Technology, Education 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment • Be the first to like this Views Total Views 744 On Slideshare 0 From Embeds 0 Number of Embeds 1 Actions Shares 0 7 0 Likes 0 Embeds 0 No embeds No notes for slide ### Transcript of "Math practice" 1. 1. MATH REVIEW ACTIVITES QUIZ 1 2. 2. PLACE VALUE: http://www.sheppardsoftware.com/math.htm• Look for DECIMALS and practice with all the games there: 1) Decimal models tenth and hundreth 2) Compare and order decimals 3) Decimal place value 4) Decimal rounding and comparing• Look for PLACE VALUE and practice games 1) Place value tutorial 2) Scooter quest 3) Fruit shoot place value 4) Underline place value 5) Expanded form 6) How to read numbers 7) Compare and order numbers 3. 3. MATH PRACTICE• PLACE VALUE - LESSONS: 1.1 – 1.7 http://www.eduplace.com/kids/mw/practice/5/ep5_01.html•POINTS, LINES RAYS, ETC. ( lesson 15.1)http://www.eduplace.com/kids/mw/practice/5/ep5_06.html 4. 4. Digit Values What is the value of the underlined digit?632,814 - The value of the digit 6 is 6 hundred-thousands, or 600,000.632,814 - The value of the digit 3 is 3 ten-thousands, or 30,000.632,814 - The value of the digit 2 is 2 thousands, or 2,000.632,814 - The value of the digit 8 is 8 hundreds, or 800.632,814 - The value of the digit 1 is 1 tens, or 10.632,814- The value of the digit 4 is 4 ones, or 4.Write the value of the underlined digit.a. 198,752 - __________________ b. 956,726 - __________________c. 472,861 - __________________ d. 764,509- __________________e. 896,804 - __________________ f. 601,099 - __________________g. 467,530 - __________________ h. 50,402 - __________________ 5. 5. CONTINUE PLACE VALUE4 5 6, 8 0 21. In the number above, which digit has the greatest value? ______________2. In the number above, which digit has the least value? ______________3. What is the value of the digit in the thousands place of the number above? ______________4. What is the value of the digit in the ten-thousands place of the number above? ______________5. Write in word form and short word form: _______________________________ _________________________________________________________________ 6. 6. ROUNDING NUMBERS1. 690 rounded to the nearest hundred is... a. 600 b. 650 c . 7002. 2,829 rounded to the nearest hundred is... a. 2,800 b. 2,700 c. 3,0003. 9,567 rounded to the nearest hundred is... a. 10,000 b. 9,000 c. 9,6004. 2,159 rounded to the nearest hundred is... a. 2,000 b. 2,200 c. 2,300 7. 7. COMPARING NUMBERSRead and answer the questions. 1. There are 686,923 people living in Alaska. There are 873,092 people living in Delaware. Which state has the greater population? __________________________________ 2. The size of Texas is 268,581 square miles. Minnesota is 86,939 square miles. Which state has a smaller area? __________________________________ 3. The distance around the Earths equator is 24,901miles. The distance around Saturns equator is 236,672miles. Which planet has the shorter distance aroundits equator? ________________________________ 8. 8. TENTH, HUNDRETH, THOUSANTH a. three and six tenths _________ b. thirty and six tenths _________ c. three and five tenths _________ d) twenty six hundredth _________ e) four tenth _________ f ) four hundreth WORD FORM WRITE IN _________ g) two hundred six thousandth _________ 4.5 ___________________________3.68 ___________________________.9 ___________________________.09 ___________________________ 9. 9. COMPAREa) 3.4 ____ 3.40b) .9 ____ .09c) 4.09 ____ 4.90d) 7 ____ .7e) .67 ____ .670 10. 10. ORDER FROM LEAST TO GREATa) 7.89 - .78 - 7.98 - .56b) 567,895 - 567,859 - 576,589 11. 11. EXPONENTS SIMPLIFY ( add an exponent)a) 5 _______________________b) 4 _______________________c) 3 _______________________d) 10 _______________________e) 10 _______________________ 12. 12. WORD FORM AND EXPANDED FORM• Write in word form , short word form, and expanded form.• 1,234,789 ____________________________________________________________________________________________________• 34.567,890 ______________________________________________________________________________________________________ 13. 13. LINES• DRAW A :• POINT,• LINE,• RAY,• LINE SEGMENT,• PARALLEL LINES• INTERSECTING LINES• PERPENDICULAR LINES 14. 14. IDENTIFY LINES 15. 15. lines 16. 16. CONTESTA CORRECTAMENTE.• cómo se leen las siguientes cantidades:a) 345,678 __________________________b) 89,008 __________________________c) 800,109 __________________________• Que número se forma con:a) Cinco unidades de millar, 8 centenas, 3 unidades ____________________b) 3 centenas de millar, 2 decenas de millar, 5 unidades de millar, 7 decenas, 1unidad _____________________________________________________________c) 4 decenas de millar, 3 centenas, 3 decenas, 1 unidadd) _____________________________________________________________ 17. 17. Contesta• A) 456, 789 b) 1, 896,721 + 12,456 + 4,003,567 c) 345, 632 d) 5,000 - 34,567 - 1,544 18. 18. 4,322 5,678 X 23 x 64 364 5 357 1. #### A particular slide catching your eye? Clipping is a handy way to collect important slides you want to go back to later.
### Win or Lose? A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with? ### Fixing the Odds You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two bags so as to make the probability of choosing a red ball as small as possible and what will the probability be in that case? ### Scratch Cards To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize? # In a Box ### Why do this problem? This problem offers opportunties to consider different methods of listing systematically. It can be used to introduce or revisit sample space diagrams, and with some students, tree diagrams. ### Possible approach This printable worksheet may be useful: In a Box. Play the game a few times for real. "Is this a fair game? How can we be sure?" Class work in pairs trying to decide and to develop an argument to justify their conjectures. After about ten minutes, stop to discuss the merits of different arguments and representations. This may be an appropriate point to highlight the benefits of different systematic methods for listing all possibilities, using sample space diagrams and, if pupils have met them before, tree diagrams. Finding a fair game can become a class activity: Students help to create a class list of all distinct starting points for the game (for example, four ribbons can be either $1R$ and $3B$ or $2R$ and $2B$). These are written on the board for $3, 4, 5, \ldots$ribbons. Distribute the task of checking which combinations are fair and record them on the board as pairs of pupils decide. There are not many solutions that work and if pupils are to notice a pattern amongst the combinations that are fair they may need to consider up to a total of $16$ ribbons. Spend some time conjecturing about more than $16$ ribbons and test. ### Key questions How can you decide if the game is fair? How many goes do you think we need to be confident of the likelihood of winning? Are there efficient systems for recording the different possible combinations?
# 4. The Derivative as an Instantaneous Rate of Change The derivative tells us the rate of change of one quantity compared to another at a particular instant or point (so we call it "instantaneous rate of change"). This concept has many applications in electricity, dynamics, economics, fluid flow, population modelling, queuing theory and so on. Wherever a quantity is always changing in value, we can use calculus (differentiation and integration) to model its behaviour. In this section, we will be talking about events at certain times, so we will be using Δt instead of the Δx that we saw in the last section Derivative from First Principles. Note: This section is part of the introduction to differentiation. We learn some (much easier) rules for differentiating in the next section, Derivatives of Polynomials. ## Velocity We learned before that velocity is distance divided by time. But this only works if the velocity is constant. We need a new method if the velocity is changing all the time. If we have an expression for s (displacement) in terms of t (time), then the velocity at any particular instant t is given by: v=lim_(Deltat->0)(Deltas)/(Deltat To make the algebra simple, we will use h for Δt and write: v=lim_(h->0)(f(t+h)-f(t))/h ### Example An object falling from rest has displacement s in cm given by s = 490t2, where t is in seconds (s). What is the velocity when t = 10 s? Note: In the time given above, t = 10 s, the "s" (non-italic) is the official metric symbol for "seconds". Don't confuse it with s (using italics), which is the variable commonly used for displacement (as used in the first sentence of this Example, s = 490t2). The derivative tells us: • the rate of change of one quantity compared to another • the slope of a tangent to a curve at any point • the velocity if we know the expression s, for displacement: v=(ds)/(dt) • the acceleration if we know the expression v, for velocity: a=(dv)/(dt) "Yes, but what does dy/dt really mean?" In summary, dy/dt means "change in y compared to change in t at a precise value of t." It is used where the quantity "y" is undergoing constant change. Let's use the example of temperature. Say you are in Melbourne, Australia (where daily extremes of temperature are common :-), and we want to know how fast the temperature is increasing right now. In winter, at night, the temperature might typically be 2°"C". In summer (6 months later) at night, it may be 26°"C". The average rate of change is (26 - 2)/6 = 24/6 = 4^@ per month This is a long term average change. It is not dy/dt. But now let's think of one day in summer. At 6:00 am the temperature might be 13° and by 1:00 pm it is (say) 27^@. The average change now is (27 - 13)/7 = 14/7 = 2^@ per hour. We still do not have dy/dt. Now let's consider at 9:00 am it is 20° and at 10:00 am it is 22.4°. So the average change is (22.4 - 20)/60 = 2.4/60 = 0.04^@ per min (equivalent to 2.4° per hour) We could keep going for smaller and smaller time intervals (like second, then millisecond, then nanosecond and so on) to get a precise change in temperature at 9:00 am. This precise change is represented by the concept of dy/dt. Historically, what I have described in The Slope of a Curve (Numerical) was what they had to do before Newton and Leibniz gave us differential calculus. In Derivative from First Principles we saw the algebraic approach that Newton and Leibniz developed. Now we can find precise values of dy/dt using a mathematical process based on a function, without having to substitute numbers all over the place. ### Coming up... In the next section, we will see some (much simpler) rules for differentiation. We won't use "differentiation from first principles" very often from here on, but it is good to have an understanding of where differentiation comes from and what it can do for us. ### Online Algebra Solver This algebra solver can solve a wide range of math problems. (Please be patient while it loads.) ### Calculus Lessons on DVD Easy to understand calculus lessons on DVD. See samples before you commit.
Largest number that divides x and is co-prime with y Given two positive numbers x and y. Find the maximum valued integer a such that: 1. a divides x i.e. x % a = 0 2. a and y are co-prime i.e. gcd(a, y) = 1 Examples : Input : x = 15 y = 3 Output : a = 5 Explanation: 5 is the max integer which satisfies both the conditions. 15 % 5 =0 gcd(5, 3) = 1 Hence, output is 5. Input : x = 14 y = 28 Output : a = 1 Explanation: 14 % 1 =0 gcd(1, 28) = 1 Hence, output is 1. Approach: Here, first we will remove the common factors of x and y from x by finding the greatest common divisor (gcd) of x and y and dividing x with that gcd. Mathematically: x = x / gcd(x, y) â€”â€” STEP1 Now, we repeat STEP1 till we get gcd(x, y) = 1. At last, we return a = x Algorithm: Step 1: Define a function named gcd to find gcd of two numbers a and b. Step 2: If a or b is equal to 0, return 0. If a is equal to b, return a. Step 3: If a is greater than b, return gcd(a-b, b). Step 4: If b is greater than a return gcd(a b-a). Step 5: Define a function named cpFact to find the largest coprime divisor of two numbers x and y. Step 6: While gcd(x, y) is not equal to 1, divide x by gcd(x,y). Step 7: Return x as the largest coprime divisor. below is the code implementation of the above approach: C++ // CPP program to find the // Largest Coprime Divisor #include using namespace std; // Recursive function to return gcd // of a and b int gcd(int a, int b) { // Everything divides 0 if (a == 0 \|\| b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return gcd(a - b, b); return gcd(a, b - a); } // function to find largest // coprime divisor int cpFact(int x, int y) { while (gcd(x, y) != 1) { x = x / gcd(x, y); } return x; } // divisor code int main() { int x = 15; int y = 3; cout << cpFact(x, y) << endl; x = 14; y = 28; cout << cpFact(x, y) << endl; x = 7; y = 3; cout << cpFact(x, y); return 0; } Java // java program to find the // Largest Coprime Divisor import java.io.*; class GFG { // Recursive function to return gcd // of a and b static int gcd(int a, int b) { // Everything divides 0 if (a == 0 \|\| b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return gcd(a - b, b); return gcd(a, b - a); } // function to find largest // coprime divisor static int cpFact(int x, int y) { while (gcd(x, y) != 1) { x = x / gcd(x, y); } return x; } // divisor code public static void main(String[] args) { int x = 15; int y = 3; System.out.println(cpFact(x, y)); x = 14; y = 28; System.out.println(cpFact(x, y)); x = 7; y = 3; System.out.println(cpFact(x, y)); } } // Python3 # Python3 code to find the # Largest Coprime Divisor # Recursive function to return # gcd of a and b def gcd (a, b): # Everything divides 0 if a == 0 or b == 0: return 0 # base case if a == b: return a # a is greater if a > b: return gcd(a - b, b) return gcd(a, b - a) # function to find largest # coprime divisor def cpFact(x, y): while gcd(x, y) != 1: x = x / gcd(x, y) return int(x) # divisor code x = 15 y = 3 print(cpFact(x, y)) x = 14 y = 28 print(cpFact(x, y)) x = 7 y = 3 print(cpFact(x, y)) # This code is contributed by "Sharad_Bhardwaj". C# // C# program to find the // Largest Coprime Divisor using System; class GFG { // Recursive function to return gcd // of a and b static int gcd(int a, int b) { // Everything divides 0 if (a == 0 \|\| b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return gcd(a - b, b); return gcd(a, b - a); } // function to find largest // coprime divisor static int cpFact(int x, int y) { while (gcd(x, y) != 1) { x = x / gcd(x, y); } return x; } // divisor code public static void Main() { int x = 15; int y = 3; Console.WriteLine(cpFact(x, y)); x = 14; y = 28; Console.WriteLine(cpFact(x, y)); x = 7; y = 3; Console.WriteLine(cpFact(x, y)); } } // This code is contributed by vt_m. PHP \$b) return gcd(\$a - \$b, \$b); return gcd(\$a, \$b - \$a); } // function to find largest // coprime divisor function cpFact( \$x, \$y) { while (gcd(\$x, \$y) != 1) { \$x = \$x / gcd(\$x, \$y); } return \$x; } // Driver Code \$x = 15; \$y = 3; echo cpFact(\$x, \$y), "\n"; \$x = 14; \$y = 28; echo cpFact(\$x, \$y), "\n"; \$x = 7; \$y = 3; echo cpFact(\$x, \$y); // This code is contributed by aj_36 ?> Javascript Output : 5 1 7 Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule. Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks! Previous Next
# A Mathematician for President [Image courtesy of the Images of American Political History.] In 1876, a politician made mathematical history. James Abram Garfield, the honorable Congressman from Ohio, published a brand new proof of the Pythagorean Theorem in The New England Journal of Education. He concluded, “We think it something on which the members of both houses can unite without distinction of party.” ## The Politician’s Proof To show that $c^2 = a^2 + b^2$, Garfield used a trapezoid: Given: Trapezoid ACED, constructed using congruent right triangles ABC and BDE. (1) Show that ABD is a right triangle. Alex showed this at the bottom of The Pythagorean Proof. (2) Find the total area of the three triangles. $Area = \frac12 a b + \frac12 a b + \frac12 c^2 = ab + \frac12 c^2$ (3) Find the area of the whole trapezoid. $Area = \frac12 \left( height \right) \left( sum\; of\; bases \right)$ $= \frac12 \left( a + b \right) \left( a + b \right)$ $= \frac12 \left( a^2 + 2ab + b^2 \right)$ (4) The whole trapezoid is equal to the sum of its parts. $ab + \frac12 c^2 = \frac12 \left( a^2 + 2ab + b^2 \right)$ (5) Simplify the equation. $ab + \frac12 c^2 = \frac12 \left( a^2 + 2ab + b^2 \right)$ $ab + \frac12 c^2 = \frac12 a^2 + ab + \frac12 b^2$ $\frac12 c^2 = \frac12 a^2 + \frac12 b^2$ $c^2 = a^2 + b^2$ Q.E.D. ## The Rest of the Story Garfield went on to become the 20th president of the United States. He won his election by the narrowest margin in U.S. history — less than 10,000 votes, or 1/10 of 1%. He was the last president born in a log cabin, and he appointed Abraham Lincoln’s son to his Cabinet. Then in 1881, just four months after taking office, President Garfield was shot. One bullet grazed his arm, but a second one lodged somewhere inside. Sixteen doctors dug around (with unwashed hands). No one could find the bullet. Inventor Alexander Graham Bell tried a new idea: a metal detector. He said he found the bullet, much deeper than they first thought. Doctors operated, without sterilization or success. It turned out that Bell had detected a metal mattress spring. More than three months after he was shot, President Garfield died. An autopsy showed that the bullet would not have killed him, if the doctors had left him alone. The gunman was hanged for assassination. The doctors sent a bill to the government — and got paid. ## To Be Continued… Read all the posts from the May/June 1999 issue of my Mathematical Adventures of Alexandria Jones newsletter. ## 9 thoughts on “A Mathematician for President” 1. I loved it, … But Garfield was not JUST a politician, he was a Professor of Math in Ohio before he became a senator… I copied the intro and linked your post at my blog with some additional comments.. you have a great site… We who teach the non-stay at home math kids, thank you for all the hard work.. Pat 2. I had heard that, too, or read it somewhere, but the sources I found as I revised this article from my old newsletter all said he taught “Classics” or ancient languages. 3. In order to understand this proof, you need to first master basic algebra and geometry. If you have passed algebra 1 and high school geometry, this should be easy to you. If you haven’t gotten that far in math, then don’t worry about understanding this proof — just keep working to learn the basics. 4. Beautiful pictures, If you go here http://www.pballew.net/arithm11.html#pythagor you will see a proof that is very visual, and very similar… Keep in mind that the white areas is not changing, so the square a^2 and b^2 added together, make up the same area as the square built on side c,with area c^2. The white area is equal to both a^2 + b^2, and also to c^2, so they must be equal. I hope that helps… it is better with the pictures, so give it a look. Pat This site uses Akismet to reduce spam. Learn how your comment data is processed.
# Finding inverse of a 3X3 Matrix • Aug 4th 2008, 02:47 PM Kitty216 Finding inverse of a 3X3 Matrix http://i146.photobucket.com/albums/r...6/problem4.jpg I've found inverses for 2X2 matrices in the past but have never done a 3X3. I looked at a few sites and they performed things such as row operations which I've never seen before. I tried to use the method myself but got more confused (Worried). Does anyone know a way that is easy to understand? • Aug 4th 2008, 04:00 PM o_O Have you learned how to reduce a matrix to row echelon form before? If so, then take your matrix and reduce it to row echelon form. For each step you do, you do it to the identity matrix. What is left after row reducing is your inverse matrix. For example, say that your first step to reducing B to the identity matrix is dividing the entire first row by 4, then you do the same step to the identity matrix: $\displaystyle \left[ \begin{array}{ccc} 4 & 5 & 8 \\ 7 & 6 & -1 \\ 0 & 9 & 2 \end{array} \Bigg\| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \: \: \Rightarrow$$\displaystyle {\color{white}.} \: \: \left[ \begin{array}{ccc} 1 & \frac{5}{4} & 2 \\ 7 & 6 & -1 \\ 0 & 9 & 2 \end{array} \Bigg\| \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$ See how what I did to the left side, I did to the right side? Now continue to row reduce the left matrix to the identity matrix and perform the same steps to the right matrix. Again, whatever your result is on the right side after row reducing the left matrix is your inverse. ----------------------- If not, you can always use this ghastly formula: $\displaystyle B^{-1} = \frac{adj(B)}{det(B)}$ $\displaystyle adj(B)$ is found by finding the cofactors of each entry of B and then taking its transpose. Here's a resource that you may find helpful: Paul's Online Notes: Linear Algebra • Aug 4th 2008, 08:45 PM Kitty216 oo I see what you did, but how did you know to divide by 4? Is there some way of telling what to multiply or divide by? Are we able to subtract and add too?
# Video: Integrating a Function by Recalling a Standard Trigonometric Derivative Calculate β« 3/(4 + π₯Β²) dπ₯. [A] (4/3) tanβ»ΒΉ (π₯) + πΆ [B] (3/2) tanβ»ΒΉ (π₯/2) + πΆ [C] (3/2) tanβ»ΒΉ (π₯) + πΆ [D] 3 tanβ»ΒΉ (π₯/2) + πΆ 04:26 ### Video Transcript Calculate the integral of three divided by four plus π₯ squared with respect to π₯. Weβre given four options. a) four-thirds multiplied by the inverse tangent of π₯ plus our constant of integration πΆ. b) three over two multiplied by the inverse tangent of π₯ divided by two plus our constant of integration πΆ. c) three over two multiplied by the inverse tangent of π₯ plus our constant of integration πΆ. And option d), three multiplied by the inverse tangent of π₯ divided by two plus the constant of integration πΆ. The question asks us to integrate three divided by four plus π₯ squared with respect to π₯. One thing we can do in any question asking us to integrate is to look at our integrand, that is the function inside of our integral, and to see if this reminds us of any standard derivatives that we might know. We can recall that the derivative with respect to π₯ of the inverse tangent of π₯ is equal to one divided by one plus π₯ squared. And the derivative with respect to π₯ of the inverse cotangent function of π₯ is equal to negative one divided by one plus π₯ squared. And we can see that these derivatives are very similar to the integrand given to us in the question. What we can then do is integrate both sides of our equation to give us the integral of the derivative of the inverse tangent of π₯ with respect to π₯ is equal to the integral of one divided by one plus π₯ squared with respect to π₯. We then notice that the antiderivative of our derivative function of the inverse tangent of π₯ is just equal to the inverse tangent of π₯. And we add our constant of integration πΆ. We could do the same process with the inverse cotangent of π₯ to get that the inverse cotangent of π₯ plus πΆ is equal to the integral of negative one divided by one plus π₯ squared with respect to π₯. At this point, we could continue using either of these identities and arrive at a correct answer. However, the options listed in the question are all in terms of the inverse tangent function. So we will continue with using the inverse tangent function. So letβs start with our integral of three divided by four plus π₯ squared with respect to π₯. Weβve already shown we know how to integrate one divided by one plus π₯ squared with respect π₯. So letβs try and rearrange this to look like this integral. Since the numerator of three is just a constant, we can take the constant factor of three outside of our integral. Next, we can take a factor of four outside of our denominator to get four multiplied by one plus π₯ squared divided by four. Then, we can notice that π₯ squared divided by four is actually equal to π₯ over two all squared. This gives us three multiplied by the integral of one divided by four multiplied by one plus π₯ over two all squared with respect to π₯. And similarly to how we did before, we could take the constant factor of four in our denominator outside of our integral. This gives us three-quarters multiplied by the integral of one divided by one plus π₯ over two all squared with respect to π₯. We can now see that our integrand is very similar to one divided by one plus π₯ squared. The only difference is, instead of an π₯ squared, we have an π₯ over two squared. To rectify this, letβs use the substitution π’ is equal to π₯ divided by two. This gives us that the derivative of π’ with respect to π₯ is equal to a half. Itβs worth noting here that dπ’ dπ₯ is not actually a fraction. However, when using integration by substitution, it does behave a little bit like a fraction. This gives us that two dπ’ is equal to dπ₯. Substituting this information into our integral gives us three quarters multiplied by the integral of one divided by one plus π’ squared multiplied by two with respect to π’. We could take the factor of two outside of our integral and then cancel it with the division by four. Weβre now ready to use our integral result. The integral of one divided by one plus π’ squared with respect to π’ is equal to the inverse tangent of π’ plus a constant of integration πΆ one. Finally, we distribute three over two over our parentheses and use our substitution π’ is equal to π₯ divided by two to get three over two multiplied by the inverse tangent of π₯ divided by two. And instead of adding three over two multiplied by πΆ one, weβll add a new constant of integration which we will call πΆ. Therefore, what we have shown is that the integral of three divided by four plus π₯ squared with respect to π₯ is equal to three over two multiplied by the inverse tangent of π₯ divided by two plus a constant of integration πΆ, which was our option b.
Courses Courses for Kids Free study material Offline Centres More Store # A convex lens of focal length $25\,cm$ and a concave lens of focal length $10\,cm$ are placed in contact with each other.(a) What is the power of this combination?(b) What is the focal length of this combination?(c) Is this combination converging or diverging? Last updated date: 12th Aug 2024 Total views: 347.7k Views today: 6.47k Verified 347.7k+ views Hint:In order to solve this question, we will use the general formula of finding the net focal length of combination of lenses and net power of combination of lenses. The focal length of a convex lens is positive since it’s a converging lens and the focal length of a concave lens is negative because concave lens is a diverging lens. Formula used: Power and focal length of a lens is related as $P = \dfrac{1}{f}$ . Complete step by step answer: According to the question, Focal length of convex lens is ${f_{convex}} = + 25cm = + 0.25m$ Power of this lens will be ${P_{convex}} = \dfrac{1}{{0.25}}$ ${P_{convex}} = 4D$ And, focal length of concave lens is ${f_{concave}} = - 10cm = - 0.1m$ Power of this lens will be ${P_{concave}} = - \dfrac{1}{{0.1}}$ ${P_{concave}} = - 10D$ (a) Net power of the combination can be calculated as ${P_{net}} = {P_{convex}} + {P_{concave}}$ Putting the values of parameters we get, ${P_{net}} = - 10 + 4$ $\therefore {P_{net}} = - 6\,D$ (b) Net focal length of the combination can be found as Since net power we have calculated is ${P_{net}} = - 6D$ Then, net focal length can be written as ${f_{net}} = \dfrac{1}{{{P_{net}}}}$ $\Rightarrow {f_{net}} = - \dfrac{1}{6}$ $\therefore {f_{net}} = - 0.1666\,m = - 16.66\,cm$ Hence, the focal length of a combination of such systems is $- 16.66\,cm$. (c) Since, the net focal length of the combination of the system is $- 16.66\,cm$ which is negative and the focal length of concave lens is negative which is a diverging lens hence, the system will behave as a diverging lens. Note: It should be remembered that, the unit of power is Dioptre and denoted by $D$.$1D$ is the ratio of lens having focal length of $1m$ and basic unit of conversions used are $1m = 100cm$ and if two lenses were kept at a distance of x meter then net focal length will be calculated by using the formula $\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{x}{{{f_1}{f_2}}}.$
# Intermediate Algebra Tutorial 25 Intermediate Algebra Tutorial 25: Polynomials and Polynomial Functions WTAMU > Virtual Math Lab > Intermediate Algebra Learning Objectives After completing this tutorial, you should be able to: 1. Identify a term, coefficient, constant term, and polynomial. 2. Tell the difference between a monomial, binomial, and trinomial. 3. Find the degree of a term and polynomial. 4. Evaluate a polynomial function. 5. Combine like terms. Introduction In this tutorial we will be looking at the different components of polynomials. Then we will move on to evaluating polynomial functions as well as adding and subtracting them. Some of these concepts are based on ideas that were covered in earlier tutorials. A lot of times in math you are using previous knowledge to learn new concepts. The trick is to not reinvent the wheel each time, but recognize what you have done before and draw on that knowledge to help you work through the problems. Tutorial Let’s start with defining some words before we get to our polynomial. Term A term is a number, variable or the product of a number and variable(s). Examples of terms are , z Coefficient A coefficient is the numeric factor of your term. Here are the coefficients of the terms listed above: Term Coefficient 3 5 2 z 1 Constant Term A constant term is a term that contains only a number. In other words, there is no variable in a constant term. Examples of constant terms are 4, 100, and -5. Standard Form of a Polynomial where n is a non-negative integer. is a constant. In other words, a polynomial is a finite sum of terms where the exponents on the variables are non-negative integers. Note that the terms are separated by +’s and -‘s. An example of a polynomial expression is . Degree of a Term The degree of a term is the sum of the exponents on the variables contained in the term. Degree of the Polynomial The degree of the polynomial is the largest degree of all its terms. Descending Order Note that the standard form of a polynomial that is shown above is written in descending order. This means that the term that has the highest degree is written first, the term with the next highest degree is written next, and so forth Also note that a polynomial can be “missing” terms. For example, the polynomial written above starts with a degree of 5, but notice there is not a term that has an exponent of 4. That means the coefficient on it is 0, so we do not write it. Some Types of Polynomials Type Definition Example Monomial A polynomial with one term 5x Binomial A polynomial with two terms 5x - 10 Trinomial A polynomial with three terms Let’s go through some examples that illustrate these different definitions. Example 1: Find the degree of the term . What do you think? Since the degree is the sum of the variable exponents and 5 is the only exponent, the degree would have to be 5. Example 2: Find the degree of the term 8. What do you think? This one is a little bit tricky. Where is the variable? When you have a constant term, it’s degree is always 0, because there is no variable there. Since this is a constant term, it’s degree is 0. Example 3: Find the degree of the term . What do you think? Since the degree is the sum of the variable exponents and it looks like we have a 1 and a 3 as our exponents, the degree would have to be 1 + 3 = 4. Example 4: Find the degree of the polynomial and indicate whether the polynomial is a monomial, binomial, trinomial, or none of these. Since the degree of the polynomial is the highest degree of all the terms, it looks like the degree is 2. Since there are three terms, this is a trinomial. Example 5: Find the degree of the polynomial and indicate whether the polynomial is a monomial, binomial, trinomial, or none of these. Since the degree of the polynomial is the highest degree of all the terms, it looks like the degree is 6. Make sure that you don’t fall into the trap of thinking it is always the degree of the first term. This polynomial is not written in standard form (descending order). So we had to actually go to the second term to get the highest degree. Since there are two terms, this is a binomial. Example 6: Find the degree of the polynomial and indicate whether the polynomial is a monomial, binomial, trinomial, or none of these. -20 Since the degree of the polynomial is the highest degree of all the terms, it looks like the degree is 0. Since there is one term, this is a monomial. Polynomial Function Since a polynomial does fit the definition of a function, which can be found in Tutorial 13: Introduction to Functions, we can write a polynomial using function notation. Evaluating a polynomial function is exactly the same concept as evaluating any function, which can be found in Tutorial 13: Introduction to Functions. Example 7: If find P(-2). Plugging -2 into the polynomial function we get: Replace x with -2 Exponent Multiplication Subtraction Combining Like Terms Recall that like terms are terms that have the exact same variables raised to the exact same exponents. One example of like terms is . Another example is . You can only combine terms that are like terms. You think of it as the reverse of the distributive property. It is like counting apples and oranges. You just count up how many variables you have the same and write the number in front of the common variable part. Example 8: Simplify by combining like terms: . First we need to identify the like terms. Let’s rewrite this so that we have the like terms next to each other. It looks like we have two terms that have an x squared that we can combine and we have two terms that have an x that we can combine. The poor 5 does not have anything it can combine with so it will have to stay 5. Combine the x squared terms together and then the x terms together Step 1: Remove the ( ) . If there is only a + sign in front of ( ), then the terms inside of ( ) remain the same when you remove the ( ). Step 2: Combine like terms. Example 9: Perform the indicated operation and simplify: Remove the ( ) Subtracting Polynomials Step 1: Remove the ( ) . If there is a - in front of the ( ) then distribute it by multiplying every term in the ( ) by a -1 . Or you can think of it as negating every term in the ( ). Step 2: Combine like terms. Example 10: Perform the indicated operation and simplify: Dist. the - through second ( ) Combine like terms Example 11: Perform the indicated operation and simplify: Dist. the - through second ( ) Combine like terms Practice Problems These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer. Practice Problems 1a - 1b: Find the degree of the term. 1a. -3 Practice Problems 2a - 2c: Find the degree of the polynomial and indicate whether the polynomial is a monomial, binomial, trinomial, or none of these. Practice Problem 3a: Evaluate the polynomial function. 3a. If , find P(-3) Practice Problems 4a - 4b: Perform the indicated operation and simplify. Need Extra Help on these Topics? The following are webpages that can assist you in the topics that were covered on this page: http://www.purplemath.com/modules/polydefs.htm This webpage helps you with the different parts of a polynomial. This webpage helps with adding and subtracting polynomials. Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions. Last revised on July 13, 2011 by Kim Seward.
Lesson 17 Solutions Exercise 1) 1) Let $A=\{1, 2\}, B=\{a, b\}$, and $C=\{3,4\}$. Then the Cartesian product $A\times B$ of A and B is a perfectly good set, and so we can take its Cartesian product with C, denoted by $(A\times B)\times C$. How many elements are in $(A\times B)\times C$?. Solution: The quick answer is that there are 8 elements in $(A\times B)\times C$. The longer answer would involve an explanation as to why this is so, so let me provide that now. We already know that the Cartesian product of a set with N elements and one with M elements is a set with $N\times M$ elements. Accordingly, we can simply view $(A\times B)\times C$ as the Cartesian product of a 4-element set with a 2-element set. This is because $(A\times B)\times C$ is the Cartesian product of $A\times B$, which has 4 elements (since it is itself the Cartesian product of two 2-element sets), with C, which has 2 elements. To be even more explicit, and to check that we’re right, we can simply write out all of the elements in $(A\times B)\times C$. To do this, we first need all of the elements in $A\times B$. These are simply $(1, a), (1, b), (2,a), \mathrm{\ and\ }(2,b)$. Thus, simply inserting these 4 elements into the “first slot” of the pairs in the set $(A\times B)\times C$, we have the 8 elements $((1,a), 3), ((1, a), 4), ((1,b),3), ((1,b),4), ((2,a),3), ((2,a),4), ((2,b),3), \mathrm{\ and\ }((2,b),4)$. Exercise 2) Let $A=\{1, 2, 3\}$. How many elements are in $A\times A$? Solution: After doing the above exercise, I think this one is rather clear, the main point of it being to point out that we can indeed take Cartesian products of sets with themselves (just as we can with unions and intersections). Thus, since A is a 3-element set, the Cartesian product of A with itself is a 9-element set. That’s the answer. To be more clear, however, we can list these 9 elements: $(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2),\mathrm{\ and\ } (3, 3)$. Note, however, that while we can take unions, intersections, and Cartesian products of sets with themselves, the results of doing so are vastly different. The union and intersection of any set with itself are both simply the original set (as you can check), whereas the Cartesian product of a set with itself is vastly different from the original set. The difference lies in the fact that the elements in the Cartesian product are pairs of elements in the original set, and not the elements themselves. Note also that $(1, 2)$ and $(2, 1)$ are different elements, since the pairs involve picking the “1” and “2” from different copies of A. We can think of this as picking the elements in a different order: one involves picking 1 first and 2 second, and the other involves picking 2 first and 1 second, thus giving rise to different pairs. Back to Lesson 17 On to Lesson 18
# A triangle has corners at (9 ,3 ), (2 ,7 ), and (3 ,4 ). How far is the triangle's centroid from the origin? May 25, 2016 Triangle's centroid is $6.6$ units away from the origin. #### Explanation: Centroid of a triangle, whose corners are $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$, is given by $\left(\frac{1}{3} \left({x}_{1} + {x}_{2} + {x}_{3}\right) , \frac{1}{3} \left({y}_{1} + {y}_{2} + {y}_{3}\right)\right)$ Hence centroid of the triangle whose corners are $\left(9 , 3\right)$, $\left(2.7\right)$ and $\left(3 , 4\right)$ is $\left(\frac{1}{3} \left(9 + 2 + 3\right) , \frac{1}{3} \left(3 + 7 + 4\right)\right)$ or $\left(\frac{14}{3} , \frac{14}{3}\right)$ And its distance from origin $\left(0 , 0\right)$ is sqrt((14/3-0)^2+(14/3-0)^2))=sqrt((14/3)^2+(14/3)^2) = $\frac{14}{3} \times \sqrt{2} = \frac{14}{3} \times 1.4142 = 6.6$
## Intermediate Algebra (12th Edition) $\bf{\text{Solution Outline:}}$ Complete the table by substituting the given value of the variable in the given equation, $3x+2y=10 .$ $\bf{\text{Solution Details:}}$ If $x=0 ,$ then \begin{array}{l}\require{cancel} 3x+2y=10 \\\\ 3(0)+2y=10 \\\\ 0+2y=10 \\\\ 2y=10 \\\\ y=\dfrac{10}{2} \\\\ y=5 .\end{array} If $y=0 ,$ then \begin{array}{l}\require{cancel} 3x+2y=10 \\\\ 3x+2(0)=10 \\\\ 3x+0=10 \\\\ 3x=10 \\\\ x=\dfrac{10}{3} .\end{array} If $x=2 ,$ then \begin{array}{l}\require{cancel} 3x+2y=10 \\\\ 3(2)+2y=10 \\\\ 6+2y=10 \\\\ 2y=10-6 \\\\ 2y=4 \\\\ y=\dfrac{4}{2} \\\\ y=2 .\end{array} If $y=-2 ,$ then \begin{array}{l}\require{cancel} 3x+2y=10 \\\\ 3x+2(-2)=10 \\\\ 3x-4=10 \\\\ 3x=10+4 \\\\ 3x=14 \\\\ x=\dfrac{14}{3} .\end{array}
Find the image of the point (2, -1, 5) in the line Find the image of the point (2, -1, 5) in the line Find the image of the point in the line $\stackrel{\to }{\mathrm{r}}=\left(11\stackrel{^}{i}–2\stackrel{^}{j}–8\stackrel{^}{\mathrm{k}}\right)+\lambda \left(10\stackrel{^}{i}–4\stackrel{^}{j}–11\stackrel{^}{k}\right)$ \overrightarrow{\mathrm{r}}=(11 \hat{i}-2 \hat{j}-8 \hat{\mathrm{k}})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k}) Given: Point Equation of line The equation of line can be re-arranged as The general point on this line is $\left(10r+11,–4r–2,–11r–8\right)$ (10 r+11,-4 r-2,-11 r-8) Let be the foot of the perpendicular drawn from the point on the given line. Then, this point is for some fixed value of . D.r.’s of PN are D.r.’s of the given line is Since, PN is perpendicular to the given line, we have, $10\left(10r+9\right)–4\left(–4r–3\right)–11\left(–11r–3\right)=0$ 10(10 r+9)-4(-4 r-3)-11(-11 r-3)=0 $\begin{array}{l}100r+90+16r+12+121r+33=0\\ 237r=135\\ r=\frac{135}{237}\end{array}$ \begin{array}{l} 100 r+90+16 r+12+121 r+33=0 \\ 237 r=135 \\ r=\frac{135}{237} \end{array} Then, the image of the point is $\frac{\alpha –11}{–1.1}=0,\frac{\beta +2}{\overline{I}}=1,\frac{\gamma +8}{9}=1$ \frac{\alpha-11}{-1.1}=0, \frac{\beta+2}{\bar{I}}=1, \frac{\gamma+8}{9}=1 Therefore, the image is .
Question # Solve and give the correct answer with using second derivative of the function as follows displaystyle f{{left({x}right)}}={left({x}+{9}right)}^{2} Modeling data distributions Solve and give the correct answer with using second derivative of the function as follows $$\displaystyle f{{\left({x}\right)}}={\left({x}+{9}\right)}^{2}$$ 2021-02-13 The given function is $$\displaystyle f{{\left({x}\right)}}{i}{s}{\left({x}+{9}\right)}^{2}.$$ Obtain the first and second derivative of the function as follows. $$\displaystyle f{{\left({x}\right)}}={\left({x}+{9}\right)}^{2}$$ $$\displaystyle{f}'{\left({x}\right)}={2}{\left({x}+{9}\right)}$$ $$\displaystyle{f}'{\left({x}\right)}={2}$$ Find the critical points as follows $$\displaystyle{f}'{\left({x}\right)}={0}$$ $$\displaystyle{2}{\left({x}+{9}\right)}={0}$$ $$\displaystyle\frac{{{2}{\left({x}+{9}\right)}}}{{2}}=\frac{0}{{2}}$$ $$\displaystyle{x}=-{9}$$ Thus, the function has critical point at $$\displaystyle{x}=-{9}$$ Since the second derivative is a constant function and is greater than zero for all values of x, the function has only local minima. Thus, the function has local minima or minimum at $$\displaystyle{\left(–{9},{0}\right)}.$$
# Free Ordering Rational Numbers Worksheet A Logical Numbers Worksheet might help your kids become a little more informed about the methods powering this percentage of integers. In this worksheet, students will be able to remedy 12 different issues linked to logical expression. They will figure out how to multiply two or more phone numbers, group them in sets, and determine their goods. They will also exercise simplifying rational expressions. After they have learned these ideas, this worksheet might be a important device for advancing their reports. Free Ordering Rational Numbers Worksheet. ## Rational Amounts can be a percentage of integers The two main varieties of figures: irrational and rational. Logical numbers are understood to be total phone numbers, whereas irrational phone numbers will not repeat, and possess an limitless number of numbers. Irrational amounts are non-absolutely nothing, non-terminating decimals, and sq . roots that are not excellent squares. They are often used in math applications, even though these types of numbers are not used often in everyday life. To outline a logical number, you must know just what a realistic number is. An integer can be a complete variety, as well as a logical variety is a rate of two integers. The rate of two integers will be the number on the top split by the variety on the bottom. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction. ## They may be produced in a fraction A reasonable amount has a denominator and numerator that are not absolutely nothing. This means that they are often conveyed as being a small percentage. Along with their integer numerators and denominators, rational figures can also have a bad value. The bad value needs to be placed to the left of as well as its absolute importance is its length from zero. To streamline this example, we will say that .0333333 is actually a fraction that could be published being a 1/3. As well as adverse integers, a logical quantity can even be created in a fraction. For instance, /18,572 is actually a rational number, when -1/ is not really. Any portion made up of integers is reasonable, so long as the denominator will not have a and will be published for an integer. Likewise, a decimal that ends in a level can be another logical number. ## They are sensation Regardless of their brand, logical numbers don’t make a lot sensation. In math, these are solitary organizations having a exclusive duration in the amount series. This means that if we count up one thing, we could order the shape by its ratio to the unique quantity. This holds accurate regardless if there are limitless reasonable numbers between two certain numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer. If we want to know the length of a string of pearls, we can use a rational number, in real life. To obtain the duration of a pearl, for example, we could count up its width. One particular pearl is 10 kilograms, which is a logical amount. Moreover, a pound’s weight equates to twenty kilograms. As a result, we will be able to divide a lb by ten, without the need of concern yourself with the length of a single pearl. ## They can be conveyed as a decimal You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount might be published like a several of two integers, so 4 times 5 various is equivalent to 8. The same issue involves the repetitive small percentage 2/1, and each side should be divided up by 99 to have the appropriate solution. But how would you have the conversion process? Here are a few examples. A realistic number may also be designed in various forms, which include fractions along with a decimal. One method to signify a logical variety in a decimal is usually to split it into its fractional equal. You can find 3 ways to break down a reasonable amount, and all these approaches yields its decimal equivalent. One of these simple ways is usually to split it into its fractional equivalent, and that’s what’s called a terminating decimal.
## Sunday, November 7, 2021 ### Solving Equations with Radicals and Extraneous Solutions Solving Equations with Radicals and Extraneous Solutions Introduction For equations such as: x^2 = a Solving for x requires us to take the root of both sides with the solutions: x = +a and x = -a. However when have something in the form of: √(f(x)) = a we can take the square of both sides and solve for x. f(x) = a^2 Sometimes our solutions of x may not work out. This is where we run into the case of extraneous solutions. What is an extraneous solution? This is a solution that may appear to be a valid solution by solving an equation through valid methods, but the catch is that this solution does not solve the equation. Think of an extraneous solution as a "false solution". We are going to see some examples in the next section. Principal Square Root When we refer to the square root of a number, most of the time we are referring to the principal square root. That is, the principal square root only refers to the non-negative square root. Example: The principal square root of 256 is: √256 = 16 Calculators use the principal square root for their square root function. Example Problems where Extraneous Solutions May Exist In all the examples, √ is used for the principal square root. In fact, unless when specified, the √ symbol will always refer to the principal square root. Other root operators, such as cube root and quartic root, will operating the same way. Example 1: x - 3 = √(3x - 9) Start by squaring both sides: x^2 - 6x + 9 = 3x - 9 Subtract (3x - 9) from both sides: x^2 - 9x + 18 = 0 The left side can be factored into: (x - 6)(x - 3) = 0 Giving potential solutions of: x = 6 and x = 3. Checking both solutions: x = 6: 6 - 3 = 3; √(63 - 9) = 3 x = 3: 3 - 3 = 0; √(33 - 9) = 0 In this first case, both solutions, x = 3 and x = 6, are valid. Example 2: √(5x - 1) + 7 = 3 An attempt to solve for x: √(5x - 1) = -4 5x - 1 = 16 5x = 17 x = 17/5 But checking for when x = 17/5: √(5 * 17/5 - 1) + 7 = √(16) + 7 = 11 ≠ 3 In this case, x = 17/5 is an extraneous solution, and does not solve the equation. Example 3: x + √(4x + 1) = 5 An attempt to solve for x: √(4x + 1) = 5 - x 4x + 1 = 25 - 10x + x^2 0 = 24 -14x + x^2 0 = (x - 2)(x - 12) Possible solutions: x = 2, x = 12 Checking: x = 2: 2 + √(42 + 1) = 2 + √(9) = 5 x = 12: 12 + √(412 + 1) = 12 + √(49) = 19 ≠ 5 Hence only x = 2 is a true solution, as x = 12 is extraneous. The bottom line: when solving equations involving radicals and roots, it is best to check your answers. This is not a good practice in school, but in life. Eddie All original content copyright, © 2011-2021. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
# Exploring the Properties of Polygons with an Interior Angle Sum of 1080 ## Introduction to Understanding the Interior Angle Sum of 1080: What is a Pentagon? A pentagon is a polygon with five sides and five angles. It is one of the most recognized shapes in geometry, often appearing in architecture, artwork, and other design endeavors. Each side of a pentagon has an equal length and all the angles are equal to 108 degrees. The sum of the angles of any polygon is equal to the number of sides minus two multiplied by 180. Therefore, when dealing with a pentagon like any other n-sided polygon, the interior angle sum equals 1080 degrees (5 – 2) x 180 = 1080° . This means that each interior angle measures 108° . For example if you were to draw three lines from one vertex (the point where two or more lines meet) to all its adjacent vertices, then each line would measure 108 degrees at that shared vertex. The concept applies for any regular polygon (a polygon whose sides are all equal in length and whose interior angles are all equal). To calculate for the interior angle sum for any given n-sided regular polygon simply use this equation: (n – 2) x 180° = Interior Angle Sum In this equation n represents your total number of sides while ‘Interior Angle Sum’ means your total amount measured in degrees. As an example if shown a hexagon you could use this formula to calculate its interior angle sum by inserting 6 into your equation thus: (6 – 2) x 180° = 720° That being said applying this concept is very useful when attempting problem solving involving finding missing/unknown measures or angles in related or similar polygons or shapes such as triangles and quadrilaterals. ## How Do We Determine Which Polygon Has an Interior Angle Sum of 1080? Determining the interior angle sum of a polygon is a simple mathematical calculation that can be used to identify different types of polygons. To determine which polygon has an interior angle sum of 1080, we first need to understand the formula for calculating the interior angle sum of any polygon. The formula for the interior angle sum of any polygon is (n-2) * 180, where n represents the number of sides on the polygon. This means that if we know how many sides are in a certain type of polygon, we can use this formula to determine its interior angle sum. For example, if a triangle has 3 sides then its interior angle sum would be (3-2) * 180 = 180 degrees. Similarly, if a pentagon has 5 sides then its interior angle sum would be (5-2) * 180 = 540 degrees. Using this same formula for any given number of sides will always yield an interior angle sum that is a multiple of 180 degrees – specifically the degree equivalent to one less than double the number of sides present in the polygon (e.g. 3 sides yields 180 degrees). In turn, this means that only multiples of 180 degrees will ever appear as possible solutions when working out an unknown polygon’s interior angle sum and therefore it’s easier to think about solving such problems as finding which multiple produces or nearest approximates our desired solution figure – in this case 1080. Taking into account our required solution figure (1080), we know that it must fit within some range – either 9 times 180 degrees between 0 and 1620, or 10 times 180 degrees between 0 and 1800 depending on whether you round up or down from 1080 respectively. Starting with 9 times being closer to 1080 than 10 times we can ascertain that our solution must have at least 10 sides but no more than 11 sides – producing 1801 / 1215 respectively as limiting angles sums either side of our target ## Step by Step Guide to Exploring the Interior Angle Sum of 1080 in a Pentagon A pentagon is a polygon with five sides and can be an excellent foundation for understanding higher level mathematics. Here, we will dive into finding the interior angle sum of a regular pentagon. This means discovering the total degrees in all ten of its internal angles. To start off, let’s refresh our understanding of what an interior angle is. An interior angle is an angle formed by two sides of a shape that meet at one vertex inside the shape. In this instance, the interior angles are all measured within the pentagon itself, so none of their measurements extend to outside points or lines in the world around them. With this definition established, let’s begin exploring how to find their sum. First things first – what is needed to calculate the sum? We will want to use properties from geometry and trigonometry including: • The formula for measuring angles in a regular polygon • Formulas for finding angles using Pythagorean’s theorem • Properties about parallel lines related to interior angles In order to begin our math journey, let’s consider some basics about the pentagon: 1. All five sides are equal lengths (a regular pentagon). 2. When connected together side-to-side they form 360° combined in total (a line across any two vertices). 3. Each of these line pieces have equal length 180° inside angles when measuring them outwards (versus inside) each corner point at opposite sides forming an “X” as pictured below: Using these basic facts we will now solve for each interior angle measurement using formulas from geometry and trigonometry listed above along with properties related to parallel lines referenced below: • For any regular polygon, like this 5 sided figure we are studying here (pentagon), the formula for determining its internal angle measurement per vertex specifies that it ## FAQs About Exploring the Interior Angle Sum of 1080 in a Pentagon Q: What is the interior angle sum of a pentagon? A: The interior angle sum of a pentagon is 1080 degrees. This is true for any regular, convex pentagon (i.e. one with sides that are all the same length and angles that all point outward from the center). Q: What does it mean when an angle sum is said to be “1080”? A: In geometry, angles can be measured in degrees (°), which is a unit of angular measurement. An angle measure of “1080” means that the total internal angles formed by five straight lines add up to 1080 degrees. Q: How do you find the individual angles in a pentagon? A: To find the individual angles in a Pentagon you need to divide the total angle sum by 5. So if you have an interior angle sum of 1080, then each internal angle would equal 1080/5 or 216 degrees. ## Top 5 Fascinating Facts about the Interior Angle Sum of 1080 and Its Relation to the Pentagon The Interior Angle Sum of 1080 is one of the most interesting facts in mathematics. It has been studied by mathematicians since the time of Euclid and its relation to the Pentagon has remained a source of fascinating speculation ever since. Here are some of the most notable facts and theories regarding this number: 1. Euclid Proved That The Interior Angle Sum Of A Pentagon Equals 1080 Degrees: In his famous treatise Elements, Euclid was the first to prove that the interior angle sum of any pentagon is exactly 1080 degrees. This proved to be an important stepping stone for many future mathematical breakthroughs related to pentagons and other regular polygons. 2. There’s A Geometric Way Of Deriving The Number 1080: Besides being able to prove it through arithmetic expressions (like 108 * 10 = 1080), in modern times it can also be derived geometrically by constructing various triangles with specific angles inside a pentagon. This technique involves dissecting and rearranging specific parts of a pentagon which then results in an interior angle sum of 1080 degrees (a fun exercise for anyone interested!). 3. The Pentagon Has Other Properties Related To The Number 1080: Not only does its interior angle sum equal this number, but if you were to draw three lines connecting all 5 corners together you would find out that these three line segments form a unique triangle whose sides have lengths that relate to the number 1080! 4. This Number Symbolizes Both Simple Symmetry And Complexity: For simple shapes like a pentagon, having such high-order symmetry (1080 being divisible by many small numbers) reflects how simple life is on earth; it’s amazing how many complex phenomena follow from simple mathematical descriptions! At the same time, this information fuels our desire for exploring even further — what other symmetries do polygons have and where else are they found in nature? 5. Humans Have Adopted The Number For Other ## Conclusion: A Comprehensive Look at Exploring the Interior Angle Sum of 1080 The interior angle sum of 1080 is most commonly explored in the realm of geometry and spatial awareness. While many will think of this as a simple problem to solve, the reality is that it has much more complexity than initially meets the eye. By exploring the interior angle sum of 1080, we can gain insight into several concepts related to geometry, including properties of circles and polygons, relationships within shapes, measuring angles, and basic trigonometry. When attempting to answer the question “What is the interior angle sum of 1080?” one must first understand what an interior angle actually is. As its name suggests, an interior angle refers to any portion of one or more angles inside a shape or object. Furthermore, an angle consists of two rays that meet at a common point – these are traditionally denoted by Greek letters such as α (alpha) and θ (theta). The size of each particular degree measure defines how large (or small) the resulting angles are after joining; 180 degrees would equal a full circle while two parallel lines joined together creates a zero-degree angle. Following this understanding, let’s explore how adding up all the interior angles in a polygon would yield our number – 1080! Taking shape into account helps us break down this calculation: triangles contain 3 angles across their 3 sides while rectangles contain 4 angles across their 4 sides etc… An octagon has 8 edges quads which gives us 360 degrees + 720 × 8 = 2880° shared amongst its 8 congruent internal angles . So when evaluating each individual corner/angle on its own , we obtain 2880 / 8 = 360° for every single measurement! To summarize , if you round out ∠ A + ∠ B + ∠ C + …etc for all possible internal-facing radii then you ultimately arrive at 1080° as your total answer . By now hopefully some lightbulbs have gone off; after exploring this topic further it becomes clear that
In this video, we will be learning how to find a percent of a number using proportions. After you finish this lesson, view all of our Pre-Algebra and Algebra lessons and practice problems. In math, “of” means multiply Percent Formula: $\frac{Percent}{100}=\frac{Part (is)}{Whole(of)}$ ## Example of Finding the Percent of a Number What is 8% of 40? $\dfrac{8}{100}=\dfrac{x}{40}\leftarrow$ First we cross-multiply $\dfrac{320}{100}=\dfrac{100x}{100}\leftarrow$ Then we divide by 100 on both sides to isolate x $3.2=x$ ## Example 1 What is $17\%$ of $30$? First, we cross-multiply $\dfrac{17}{100}=\dfrac{x}{30}$ Then, we divide by $100$ on both sides to isolate $x$ $\dfrac{510}{100}=\dfrac{100x}{100}$ $5.1=x$ The answer is $5.1$ ## Example 2 What percent of $50$ is $4.8$? We have the “of” and the “is”. Our formula should look like this: $\dfrac{x}{100}=\dfrac{4.8}{50}$ Let’s cross multiply $x\times50$ is $50x$ $4.8 \times 100$ is $480$ We have: $50x=480$ Now, let’s divide both sides by $50$ $\dfrac{50x}{50}=\dfrac{480}{50}$ $x=9.6$ Therefore, the answer is $9.6\%$ ## Video-Lesson Transcript In this video, we will be learning how to find a percent of a number using proportions. In math, ‘of’ means ‘multiply‘. So if you’re asked $8\%$ of $40$, it means $8\% \times 40$. We can’t get the answer right away. We have to change $8\%$ into a decimal. So we have to move the decimal two spaces from the right to left. $8\%$ will become $0.08$ then we can multiply $0.08 \times 40 = 3.2$ Another way to find the percent of a number is the percent formula. Percent formula is $\dfrac{Percent}{100} = \dfrac{Part (is)}{Whole (of)}$ What is $8\%$ of $40$? Let’s use the formula $\dfrac{8}{100} = \dfrac{x}{40}$ Let’s cross-multiply $8 \times 40$ is $320$ $x \times 100$ is $100x$ Now we have $320 = 100x$ To isolate $x$, let’s do the inverse operation. We just have to divide both sides by $100$ And we get $3.2 = x$ Which is the same answer using a different method. To sum up, you can multiply or use the percent formula to get the answer if the missing value is the “is” or the part. But if the missing value is the percent or the whole, you have to use the percent formula. To give an example, let’s answer this question: What percent of $40$ is $3.2$? Here, we’re looking for the percent. We have the “of” and the “is”. The formula is $\dfrac{Percent}{100} = \dfrac{Part (is)}{Whole (of)}$ Since we’re missing the percent, our formula should look like this $\dfrac{xt}{100} = \dfrac{3.2}{40}$ Then let’s cross multiply $x \times 40$ is $40x$ $3.2 \times 100$ is $320$ Now, let’s do the inverse operation of multiplication which is division. So, let’s divide both sides by $40$ $\dfrac{40x}{40} = \dfrac{320}{40}$ The answer is $x = 8$ So the answer to what percent of $40$ is $3.2$? The answer is $8\%$
Factors # What Are the Factors of 30? \| Factor Tree of 30 \| Prime Factors of 30 Written by Prerit Jain Updated on: 12 Aug 2023 ### What Are the Factors of 30? \| Factor Tree of 30 \| Prime Factors of 30 Factors of 30: A list of integers that can be divided by 30 with 0 as a remainder is known as 30 factors. The 30 factor has both negative and positive numbers. A number that is a factor of 30 with a negative sign is known as a negative factor of 30. Also, the numbers which are factors of 30 with positive signs are known as positive factors of 30. In this blog, let us understand how to find factors of 30 along with solved examples. Scroll down to find out more. ## What Are the Factors of 30? Factors are the numbers that divide the original number without leaving any remainder. In the case of 30, the numbers 1, 2, 3, 5, 6, 10, 15, and 30 are the numbers that divide the number 30 without leaving any remainder. Therefore, factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Since 30 is a composite number, it has more than two factors, which are both positive and negative. The positive and negative factors of 30 are given below: • 1, 2, 3, 5, 6, 10, 15, and 30 are positive factors of 30. • -1, -2, -3, -5, -6, -10, -15, and -30 are negative factors of 30. ## Write Factors of 30 in Pairs? The factors of 30 can be written in two pairs, namely positive pairs and negative pairs. The positive and negative pairs of 30 are given below: ### Positive Pair Factors of 30 Below are the positive pair factors of 30: ### Negative Pair Factors of 30 Below are the negative pair factors of 30: ## How To Find Factors of 30 Through Division Method? You can find factors of 30 through the division method. For that, you will have to divide the number 30 by the sequence of natural numbers up to 30. A number that is divided by 30, and doesn’t leave any remainder, is the number that is considered a factor of 30. The details on how to find the factors of number 30 are given below: • 30 ÷ 1 = 30 • 30 ÷ 2 = 15 • 30 ÷ 3 = 10 • 30 ÷ 5 = 6 • 30 ÷ 6 = 5 • 30 ÷ 10 = 3 • 30 ÷ 15 = 2 • 30 ÷ 30 = 1 As we can see, only the above numbers don’t leave a remainder after division, so that means 1, 2, 3, 5, 6, 10, 15, and 30 are the factors of 30. ## What Is the Prime Factorisation of 30? We can also find prime factors of 30 through prime factorization. Let’s see the steps below for the prime factorization of 30. • Step 1: Divide the number 30 by the least prime number. Hence, the least prime number that is divisible by 30 is 2. 30/2 = 15 • Step 2: When 30 is divided by 2, we will get the number 15. Now, we will divide it again by the smallest prime number that 15 can be divided by, i.e 3. 15/3 = 5 • Step 3: Since 5 is a prime number, we can only divide it by 5. 5/5 = 1 Therefore, 2 X 3 X 5 = 30. So, 2, 3, and 5 are the prime factors of 30. ## Write the Factor Tree of 30. We can also find the prime factors of the number 30 using the factor tree method. The steps to finding prime factors of 30 using the factor tree method are given below: • Step 1: Place the number 30 at the top of the factor tree. • Step 2: Now, as the tree branches, we will write down the pair factors of 30 we can divide 30 by Here we are using the pair factor 3, 15 as the first branch of 30. • Step 3: Since 3 is a prime number, we can no longer divide it. But, we can divide 15 by 2 and 5. • Step 4: As 2 and 5 are prime numbers, we cannot further separate them. Hence, circling all the branches gives us factors 2, 3, and 5. Therefore, the prime factors of 30 using a factor tree are 2 X 3 X 5 = 30. The factor tree of 30 is given below: ## List All the Factors of 30 From Least to Greatest The factors of 30 are 3, 5, 10, 15, 30, 1, 2, and 6. Now, let’s arrange the factors of 30 from least to great or in increasing order. Hence, factors of 30 in increasing order will be: 1, 2, 3, 5, 6, 10, 15, and 30. ## Solved Examples on Factors of 30 Some of the solved examples on factors of 30 are given below: What is the sum of all the Factors of 30? The factors number 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Hence adding 1 + 2 + 3 + 5 + 6 + 10 + 15 +30 = 72 The sum of factors of 30 is 72. What are the factors and prime factors of 30? Factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The prime factors of 30 are 2, 3, and 5. What’s the highest common factor of 45 and 30? Factors of 45: 1, 3, 5, 9, 15, and 45. Factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30. Hence the common factors of 45 and 30 are 1, 3, 5, and 15. Therefore the highest common factor of 45 and 30 is 15. What’s the highest common factor of 50 and 30? Factors of 50: 1, 2, 5, 10, 25, and 50. Factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30. From the above, the common factors of 50 and 30 are 1, 2, 5, and 10. Hence the highest common factor of 50 and 30 is 10. ## FAQs on Factors of 30 The frequently asked questions on 30 factors are given below: List all the factors of 30. 1, 2, 3, 5, 6, 10, 15, and 30 are factors of 30. What are the composite factors of 30? 6, 10, 15, and 30 are the only composite factors of 30. What are the factors of 30 in pairs? (1, 30), (2, 15), (3, 10), and (5, 6) are the pairs of factors of 30. What are the positive factors of 30? 1, 2, 3, 5, 6, 10, 15, and 30 are the positive factors of 30. What are the common factors of 30 and 100? Factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30 Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, and 10 Hence, the common factors of 30 and 100 are 1, 2, and 5. Also, the least common factor of 30 and 100 is 1. What are the common factors of 30 and 42? Factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30 Factors of 42: 1, 2, 3, 6, 7, 14, 21, and 42 Hence, the common factors of 30 and 42 are 1, 2, 3, and 6. Also, the least common factor of 30 and 42 is 1. That’s all about factors of 30 for now. We hope this blog has been useful for you and you were able to learn a lot. If you still have any doubts popping on your mind, feel free to comment below and we will reply with answers as soon as possible. Calculate Factors of The Factors are https://wiingy.com/learn/math/factors-of-30/ Written by by Prerit Jain Share article on
# Problem of the Week Problem E and Solution Shape Building ## Problem Sina drew square $$ABCD$$ with side length $$6$$ cm on a piece of paper and passed the paper to Theo. Theo drew a circle on top of the square so that the circle passes through $$A$$ and $$D$$, and the circle is tangent to side $$BC$$ at point $$P$$. Determine the radius of the circle. Note: You may find the following known result about circles useful: If a line is tangent to a circle, then the perpendicular to that line at the point of tangency passes through the centre of the circle. ## Solution Let $$O$$ be the centre of the circle and $$r$$ be the radius. Construct line segment $$PQ$$ perpendicular to $$CB$$ with $$Q$$ on side $$AD$$ of the square. Since $$CB$$ is tangent to the circle with point of tangency $$P$$, $$PQ$$ must pass through the centre of the circle, $$O$$. Therefore, $$PO=r$$. Since $$PQ\perp BC$$, $$PQ\parallel AB,$$ and $$PQ=AB=6$$, then $$QO=PQ-PO=6-r$$. Since $$A$$ and $$D$$ are on the circle, $$AO=DO=r$$. Using the Pythagorean Theorem, $$AQ^2=AO^2-QO^2=r^2-(6-r)^2$$ and $$DQ^2=DO^2-QO^2=r^2-(6-r)^2$$. Therefore, $$AQ^2=DQ^2$$ and $$AQ=DQ$$ follows. Since $$AQ=DQ$$ and $$AQ+QD=AD=6$$, we can substitute to obtain $$AQ+AQ=2AQ=6$$ or $$AQ=3$$. Using the Pythagorean Theorem in $$\triangle AQO$$, \begin{aligned} AO^2&=AQ^2+QO^2\\ r^2&=3^2+(6-r)^2\\ r^2&=9+36-12r+r^2\\ 12r&=45\\ r&=\frac{45}{12}\\ r&=3.75 \end{aligned} Therefore, the radius of the circle is $$3.75$$ cm.
Hong Kong Stage 1 - Stage 3 # Algebraic Equations Lesson We have already started to look at how to turn written sentences into algebraic equations. Let's continue now by looking at some more complex examples, involving more than one operation. Remember! • Addition "$+$+" can be expressed by words such as "more than", "sum", "plus", "add" and "increased by". • Subtraction "$-$" can be expressed by words such as "less than", "difference", "minus", "subtract" and "decreased by". • Multiplication "$\times$×" can be expressed by words such as "groups of", "times", "product" and "multiply". • Division "$\div$÷" can be expressed by words such as "quotient" and "divided by". We usually represent division using fractions instead of using the "$\div$÷" operator. • Equality "$=$=" can be expressed by words such as "is", "equal to" and "the same as". A number sentence needs one of these symbols to be an equation! #### Example Write down an equation in simplest form to represent "$v$v is $5$5 less than $3$3 lots of $u$u". Think: What symbol, number or variable can we use to represent each part of the sentence? Do: "$v$v is" means that $v$v will be on one side of the "$=$=" sign and everything else will be on the other side. "$3$3 lots of $u$u" means $3\times u$3×u, and "$5$5 less than" means we are going to subtract $5$5 from this amount (using the "$-$" operator). So we have $v=3\times u-5$v=3×u5, which we can write more simply as $v=3u-5$v=3u5. Careful! The order of the numbers in the sentence is not necessarily the same as the order in the equation! In the example above, "$5$5 less than" meant that $5$5 was to be subtracted from the following term "$3$3 lots of $u$u". So the equation was written as $v=3u-5$v=3u5. Let's watch some worked video examples: ##### Question 1 Write an equation in simplest form for: $y$y is $x$x divided by $3$3 plus $12$12 ##### Question 2 Write an equation for: $y$y equals $5$5 times the sum of $x$x and $10$10
### Learning Outcomes By the end of this section, you will be able to: • Factor a quadratic equation to solve it. • Use the square root property to solve a quadratic equation. • Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle. • Complete the square to solve a quadratic equation. • Use the discriminant to determine the number and type of solutions to a quadratic equation. The left computer monitor in the image below is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods. ## Factoring and the Square Root Property An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as $2{x}^{2}+3x - 1=0$ and ${x}^{2}-4=0$ are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics. Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if $a\cdot b=0$, then $a=0$ or $b=0$, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero. Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression $\left(x - 2\right)\left(x+3\right)$ by multiplying the two factors together. $\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&={x}^{2}+3x - 2x - 6\hfill \\ \hfill&={x}^{2}+x - 6\hfill \end{array}$ The product is a quadratic expression. Set equal to zero, ${x}^{2}+x - 6=0$ is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied. The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, $a{x}^{2}+bx+c=0$, where a, b, and c are real numbers and $a\ne 0$. The equation ${x}^{2}+x - 6=0$ is in standard form. We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section. ### A General Note: The Zero-Product Property and Quadratic Equations The zero-product property states $\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0$, where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example $a{x}^{2}+bx+c=0$ where a, b, and c are real numbers, and $a\ne 0$. It is in standard form. In the quadratic equation ${x}^{2}+x - 6=0$, the leading coefficient, or the coefficient of ${x}^{2}$, is 1. We have one method of factoring quadratic equations in this form. ### How To: Given a quadratic equation with the leading coefficient of 1, factor it 1. Find two numbers whose product equals c and whose sum equals b. 2. Use those numbers to write two factors of the form $\left(x+k\right)\text{ or }\left(x-k\right)$, where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and $-2$, the factors are $\left(x+1\right)\left(x - 2\right)$. 3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable. ### Example: Factoring and Solving a Quadratic with Leading Coefficient of 1 Factor and solve the equation: ${x}^{2}+x - 6=0$. ### Try It Factor and solve the quadratic equation: ${x}^{2}-5x - 6=0$. ### Using the Square Root Property When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the ${x}^{2}$ term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the ${x}^{2}$ term so that the square root property can be used. ### A General Note: The Square Root Property With the ${x}^{2}$ term isolated, the square root propty states that: $\text{if }{x}^{2}=k,\text{then }x=\pm \sqrt{k}$ where k is a nonzero real number. ### How To: Given a quadratic equation with an ${x}^{2}$ term but no $x$ term, use the square root property to solve it 1. Isolate the ${x}^{2}$ term on one side of the equal sign. 2. Take the square root of both sides of the equation, putting a $\pm$ sign before the expression on the side opposite the squared term. 3. Simplify the numbers on the side with the $\pm$ sign. ### Example: Solving a Simple Quadratic Equation Using the Square Root Property Solve the quadratic using the square root property: ${x}^{2}=8$. ### Example: Solving a Quadratic Equation Using the Square Root Property Solve the quadratic equation: $4{x}^{2}+1=7$ ### Try It Solve the quadratic equation using the square root property: $3{\left(x - 4\right)}^{2}=15$. ### Using the Pythagorean Theorem One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle and states the relationship among the lengths of the sides as ${a}^{2}+{b}^{2}={c}^{2}$, where $a$ and $b$ refer to the legs of a right triangle adjacent to the $90^\circ$ angle, and $c$ refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side. The Pythagorean Theorem is given as ${a}^{2}+{b}^{2}={c}^{2}$ where $a$ and $b$ refer to the legs of a right triangle adjacent to the ${90}^{\circ }$ angle, and $c$ refers to the hypotenuse. ### Example: Finding the Length of the Missing Side of a Right Triangle Find the length of the missing side of the right triangle. ### Try It Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse. ## Completing the Square and the Quadratic Formula Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example ${x}^{2}+4x+1=0$ to illustrate each step. 1. Given a quadratic equation that cannot be factored and with $a=1$, first add or subtract the constant term to the right sign of the equal sign. ${x}^{2}+4x=-1$ 2. Multiply the b term by $\frac{1}{2}$ and square it. $\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}$ 3. Add ${\left(\frac{1}{2}b\right)}^{2}$ to both sides of the equal sign and simplify the right side. We have $\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}$ 4. The left side of the equation can now be factored as a perfect square. $\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}$ 5. Use the square root property and solve. $\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}$ 6. The solutions are $x=-2+\sqrt{3}$, $x=-2-\sqrt{3}$. ### Example: Solving a Quadratic by Completing the Square Solve the quadratic equation by completing the square: ${x}^{2}-3x - 5=0$. ### Try It Solve by completing the square: ${x}^{2}-6x=13$. The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by $-1$ and obtain a positive a. Given $a{x}^{2}+bx+c=0$, $a\ne 0$, we will complete the square as follows: 1. First, move the constant term to the right side of the equal sign: $a{x}^{2}+bx=-c$ 2. As we want the leading coefficient to equal 1, divide through by a: ${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$ 3. Then, find $\frac{1}{2}$ of the middle term, and add ${\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}$ to both sides of the equal sign: ${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$ 4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: ${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$ 5. Now, use the square root property, which gives $\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$ 6. Finally, add $-\frac{b}{2a}$ to both sides of the equation and combine the terms on the right side. Thus, $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$ ### A General Note: The Quadratic Formula Written in standard form, $a{x}^{2}+bx+c=0$, any quadratic equation can be solved using the quadratic formula: $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$ where a, b, and c are real numbers and $a\ne 0$. ### How To: Given a quadratic equation, solve it using the quadratic formula 1. Make sure the equation is in standard form: $a{x}^{2}+bx+c=0$. 2. Make note of the values of the coefficients and constant term, $a,b$, and $c$. 3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula. 4. Calculate and solve. Solve the quadratic equation: ${x}^{2}+5x+1=0$. Use the quadratic formula to solve ${x}^{2}+x+2=0$. . Notice they are written in standard form of a complex number. When a solution is a complex number, you must separate the real part from the imaginary part and write it in standard form. ### Try It Solve the quadratic equation using the quadratic formula: $9{x}^{2}+3x - 2=0$. ### The Discriminant The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, ${b}^{2}-4ac$. The discriminant tells us whether the solutions are real numbers or complex numbers as well as how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation. Value of Discriminant Results ${b}^{2}-4ac=0$ One rational solution (double solution) ${b}^{2}-4ac>0$, perfect square Two rational solutions ${b}^{2}-4ac>0$, not a perfect square Two irrational solutions ${b}^{2}-4ac<0$ Two complex solutions ### A General Note: The Discriminant For $a{x}^{2}+bx+c=0$, where $a$, $b$, and $c$ are real numbers, the discriminant is the expression under the radical in the quadratic formula: ${b}^{2}-4ac$. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect. ### Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation Use the discriminant to find the nature of the solutions to the following quadratic equations: 1. ${x}^{2}+4x+4=0$ 2. $8{x}^{2}+14x+3=0$ 3. $3{x}^{2}-5x - 2=0$ 4. $3{x}^{2}-10x+15=0$ ## Key Concepts • Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions. • Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. • Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. • Completing the square is a method of solving quadratic equations when the equation cannot be factored. • A highly dependable method for solving quadratic equations is the quadratic formula based on the coefficients and the constant term in the equation. • The discriminant is used to indicate the nature of the solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. • The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. ## Glossary completing the square a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square discriminant the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots. Pythagorean Theorem a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems one of the methods used to solve a quadratic equation in which the ${x}^{2}$ term is isolated so that the square root of both sides of the equation can be taken to solve for x
## Solution We have $$\frac{1}{(1-x)^2} = \sum_{k=0}^\infty (k+1)x^k$$ Multiplying by $x^2$ on both sides we get \begin{aligned} \frac{x^2}{(1-x)^2} &= \sum_{k=0}^\infty (k+1)x^{k+2} \\ \end{aligned} Substituting $x= \frac{1}{2}$ in the above identity, we get \begin{aligned} \sum_{k=0}^\infty \frac{k}{2^{k+1}} = 1 \end{aligned} Differentiating and mutliplying the equation below by $x^2$ $$\frac{x}{(1-x)^2} = \sum_{k=0}^\infty (k+1)x^{k+1}$$ we get $$\sum_{k=0}^\infty (k+1)^2 x^{k+2} = \frac{x^2(x+1)}{(1-x)^3}$$ Substituting $x= \frac{1}{2}$ in the above identity, we get \begin{aligned} \sum_{k=0}^\infty \frac{k^2}{2^{k+1}} = 3 \end{aligned} The last two identities can be obtained using the same approach as above. All the identities below are a generalization of the above identities
# The sum of the digits of a two-digit numeral is 8. If the digits are reversed, the new number is 18 greater than the original number. How do you find the original numeral? Apr 16, 2016 Solve equations in the digits to find the original number was $35$ #### Explanation: Suppose the original digits are $a$ and $b$. Then we are given: $\left\{\begin{matrix}a + b = 8 \\ \left(10 b + a\right) - \left(10 a + b\right) = 18\end{matrix}\right.$ The second equation simplifies to: $9 \left(b - a\right) = 18$ Hence: $b = a + 2$ Substituting this in the first equation we get: $a + a + 2 = 8$ Hence $a = 3$, $b = 5$ and the original number was $35$.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 3.1: Discrete Random Variables Difficulty Level: At Grade Created by: CK-12 Learning Objectives • Demonstrate an understanding of the notion of discrete random variables by using them to solve for the probabilities of outcomes, such as the probability of the occurrence of five heads in 14 coin tosses. You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins. (a) Determine the theoretical probability for the teacher. (b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials. Work through Chapter 3 and then revisit this problem to find the solution. Whenever you run and experiment, flip a coin, roll a die, pick a card, you assign a number to represent the value to the outcome that you get. This number that you assign is called a random variable. For example, if you were to roll two dice and asked what the sum of the two dice might be, you would design the following table of numerical values. These numerical values represent the possible outcomes of the rolling of two dice and summing of the result. In other words, rolling one die and seeing a \begin{align}{\color{red}6}\end{align} while rolling a second die and seeing a \begin{align}{\color{blue}4}\end{align}. Adding these values gives you a ten. The rolling of a die is interesting because there are only a certain number of possible outcomes that you can get when you roll a typical die. In other words, a typical die has the numbers 1, 2, 3, 4, 5, and 6 on it and nothing else. A discrete random variable can only have a specific (or finite) number of numerical values. A random variable is simply the rule that assigns the number to the outcome. For our example above, there are 36 possible combinations of the two dice being rolled. The discrete random variables (or values) in our sample are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, as you can see in the table below. We can have infinite discrete random variables if we think about things that we know have an estimated number. Think about the number of stars in the universe. We know that there are not a specific number that we have a way to count so this is an example of an infinite discrete random variable. Another example would be with investments. If you were to invest \$1000 at the start of this year, you could only estimate the amount you would have at the end of this year. Well, how does this relate to probability? Example 1: Looking at the previous table, what is the probability that the sum of the two dice rolled would be 4? Solution: \begin{align}P(4) & =\frac{3}{36} \\ P(4) & =\frac{1}{12}\end{align} Example 2: A coin is tossed 3 times. What are the possible outcomes? What is the probability of getting one head? Solution: If our first toss were a heads... If our first toss were a tails... Therefore the possible outcomes are: \begin{align}&{\color{blue}HHH}, \ {\color{blue}HH}T, \ {\color{blue}H}T{\color{blue}H}, \ {\color{blue}H}TT, \ T{\color{blue}HH}, \ T{\color{blue}H}T, \ TT{\color{blue}H}, TTT\\ & P(1\ {\color{blue}\text{head}}) = \frac{3}{8}\end{align} Alternate Solution: We have one coin and want to find the probability of getting one head in three tosses. We need to calculate two parts to solve the probability problem. Numerator (Top) In our example, we want to have 1 H and 2Ts. Our favorable outcomes would be any combination of HTT. The number of favorable choices would be: \begin{align}\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\ \# \ of\ favorable\ choices & = \frac{3\ letters!}{1\ head! \times 2\ tails!} \\ \# \ of\ favorable\ choices & = \frac{3 \times 2 \times 1}{1 \times (2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{6}{2} = 3\end{align} Denominator (Bottom) The number of possible outcomes \begin{align}= 2 \times 2 \times 2 = 8\end{align} We now want to find the number of possible times we could get one head when we do these three tosses. We call these favorable outcomes. Why? Because these are the outcomes that we want to happen, therefore they are favorable. Now we just divide the numerator by the denominator. \begin{align}P(1\ head) = \frac{3}{8}\end{align} Remember: Possible outcomes \begin{align}= 2^n\end{align} where \begin{align}n =\end{align} number of tosses. Here we have \begin{align}{\color{red}3}\end{align} tosses. Therefore, Possible outcomes \begin{align}= 2^{\color{red}n}\end{align} Possible outcomes \begin{align}= 2^{\color{red}3}\end{align} Possible outcomes \begin{align}= 2 \times 2 \times 2\end{align} Possible outcomes \begin{align}= 8\end{align} Note: The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples: \begin{align}{\color{blue}4!} & {\color{blue}\ = 4 \times 3 \times 2 \times 1 = 24}\\ {\color{blue}7!} & {\color{blue}\ = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040} \\ {\color{blue}1!} & {\color{blue}\ = 1}\end{align} Note: It is generally agreed that \begin{align}0! = 1\end{align}. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations. Example 3: A coin is tossed 4 times. What are the possible outcomes? What is the probability of getting one head? Solution: If our first toss were a heads... If our first toss were a tails... Therefore there are 16 possible outcomes: \begin{align}&{\color{blue}HHHH}, \ {\color{blue}HHH}T, \ {\color{blue}HH}T{\color{blue}H}, \ {\color{blue}HH}TT, \ {\color{blue}H}T{\color{blue}HH}, \ {\color{blue}H}T{\color{blue}H}T, \ {\color{blue}H}TT{\color{blue}H}, \ {\color{blue}H}TTT, \ T{\color{blue}HHH},\\ & \ T{\color{blue}HH}T, \ T{\color{blue}H}T{\color{blue}H}, \ T{\color{blue}H}TT, \ TT{\color{blue}HH}, \ TT{\color{blue}H}T, \ TTT{\color{blue}H}, \ TTTT \\ & P(1\ {\color{blue}\text{head}}) = \frac{4}{16} \\ & P(1\ {\color{blue}\text{head}}) = \frac{1}{4}\end{align} Alternate Solution: We have one coin and want to find the probability of getting one head in four tosses. We need to calculate two parts to solve the probability problem. Numerator (Top) In our example, we want to have 1 H and 3 Ts. Our favorable outcomes would be any combination of HTTT. The number of favorable choices would be: \begin{align}\#\ of\ favorable\ choices & = \frac{\# \ possible\ letters\ in\ combination!}{letter\ X! \times letter \ Y!} \\ \# \ of\ favorable\ choices & = \frac{4\ letters!}{1\ head! \times 3\ tails!} \\ \# \ of\ favorable\ choices & = \frac{4 \times 3 \times 2 \times 1}{1 \times (3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{24}{6} \\ \# \ of\ favorable\ choices & = 4\end{align} Denominator (Bottom) The number of possible outcomes \begin{align}= 2 \times 2 \times 2 \times 2 = 16\end{align} We now want to find the number of possible times we could get one head when we do these four tosses (or our favorable outcomes). Remember: Possible outcomes \begin{align}= 2^n\end{align} where \begin{align}n =\end{align} number of tosses. Here we have \begin{align}{\color{red}4}\end{align} tosses. Therefore, Possible outcomes \begin{align}= 2^{{\color{red}n}}\end{align} Possible outcomes \begin{align}= 2^{{\color{red}4}}\end{align} Possible outcomes \begin{align}= 2 \times 2 \times 2 \times 2\end{align} Possible outcomes \begin{align}= 16\end{align} Now we just divide the numerator by the denominator. \begin{align}P(1\ head) & = \frac{4}{16} \\ P(1\ head) & = \frac{1}{4}\end{align} Technology Note: Let’s take a look at how we can do this using the TI-84 calculators. There is an application on the TI calculators called the coin toss. Among others (including the dice roll, spinners, and picking random numbers), the coin toss is an excellent application for when you what to find the probabilities for a coin tossed more than 4 times or more than one coin being tossed multiple times. Let’s say you want to see one coin being tossed one time. Here is what the calculator will show and the key strokes to get to this toss. Let’s say you want to see one coin being tossed ten times. Here is what the calculator will show and the key strokes to get to this sequence. Try it on your own. We can actually see how many heads and tails occurred in the tossing of the 10 coins. If you click on the right arrow (>) the frequency label will show you how many of the tosses came up heads. We could also use randBin to simulate the tossing of a coin. Follow the keystrokes below. This list contains the count of heads resulting from each set of 10 coin tosses. If you use the right arrow (>) you can see how many times from the 20 trials you actually had 4 heads. Now let’s go back to our original chapter problem and see if we have gained enough knowledge to answer it. You are in statistics class. Your teacher asks what the probability is of obtaining five heads if you were to toss 14 coins. (a) Determine the theoretical probability for the teacher. (b) Use the TI calculator to determine the actual probability for a trial experiment for 20 trials. Solution: (a) Let’s calculate the theoretical probability of getting 5 heads for the 14 tosses. Numerator (Top) In our example, we want to have 5 H and 9 Ts. Our favorable outcomes would be any combination of HHHHHTTTTTTTTT. The number of favorable choices would be: \begin{align}\#\ of\ favorable\ choices & = \frac{\#\ possible\ letters\ in\ combination!}{letter\ X! \times letter\ Y!} \\ \# \ of\ favorable\ choices & = \frac{14\ letters!}{5\ head! \times 9\ tails!} \\ \# \ of\ favorable\ choices & = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)\times (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(120) \times (362880)} \\ \# \ of\ favorable\ choices & = \frac{8.72 \times 10^{10}}{(43545600)} \\ \# \ of\ favorable\ choices & = 2002\end{align} Denominator (Bottom) The number of possible outcomes \begin{align}= 2^{14}\end{align} The number of possible outcomes \begin{align}= 16384\end{align} Now we just divide the numerator by the denominator. \begin{align}P(5\ heads) & = \frac{2002}{16384} \\ P(5\ heads) & = 0.1222\end{align} The probability would be 12% of the tosses would have 5 heads. b) Looking at the data that resulted in this trial, there were 4 times of 20 that 5 heads appeared. \begin{align}P(5\ \text{heads}) = \frac{4}{20}\end{align} or 20%. Lesson Summary Probability in this chapter focused on experiments with random variables or the numbers that you assign to the probability of events. If we have a discrete random variable, then there are only a specific number of variables we can choose from. For example, tossing a fair coin has a probability of success for heads = probability of success for tails \begin{align}= 0.50\end{align}. Using tree diagrams or the formula \begin{align}P = \frac{\#\ of\ favorable\ outcomes}{total\ \#\ of\ outcomes}\end{align}, we can calculate the probabilities of these events. Using the formula requires the use of the factorial function where numbers are multiplied in descending order. Points to Consider • How is the calculator a useful tool for calculating probability in discrete random variable experiments? • Are TREE Diagrams useful in interpreting the probability of simple events? Vocabulary Discrete Random Variables Only have a specific (or finite) number of numerical values. Random Variable A variable that takes on numerical values governed by a chance experiment. Factorial Function (symbol: !) – The function of multiplying a series of descending natural numbers. Theoretical Probability A probability calculated by analyzing a situation, rather than performing an experiment, given by the ratio of the number of different ways an event can occur to the total number of equally likely outcomes possible. The numerical measure of the likelihood that an event, \begin{align}E\end{align}, will happen. \begin{align}P(E) = \frac{number\ of\ favorable\ outcomes}{total\ number\ of\ possible\ outcomes}\end{align} Tree Diagram A branching diagram used to list all the possible outcomes of a compound event. Review Questions 1. Define and give three examples of discrete random variables. 2. Draw a tree diagram to represent the tossing of two coins and determine the probability of getting at least one head. 3. Draw a tree diagram to represent the tossing of one coin three times and determine the probability of getting at least one head. 4. Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at least one red. 5. Draw a tree diagram to represent the drawing two marbles from a bag containing blue, green, and red marbles and determine the probability of getting at two blue marbles. \begin{align}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\\ & P(\text{two blue marbles}) =\frac{1}{9}\end{align} 6. Draw a diagram to represent the rolling two dice and determine the probability of getting at least one 5. 7. Draw a diagram to represent the rolling two dice and determine the probability of getting two 5s. 8. Use randBin to simulate the 6 tosses of a coin 20 times to determine the probability of getting two tails. 9. Use randBin to simulate the 15 tosses of a coin 25 times to determine the probability of getting two heads. 10. Calculate the theoretical probability of getting 4 heads for the 12 tosses. 11. Calculate the theoretical probability of getting 8 heads for the 10 tosses. \begin{align}P(8 \ heads) = \frac{45}{1024}\end{align} 12. Calculate the theoretical probability of getting 8 heads for the 15 tosses. 2. \begin{align}P(\text{at least}\ I \ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\ P(\text{at least}\ I \ H) & = \frac{3}{4}\end{align} 3. \begin{align}P(\text{at least}\ I\ H) & = \frac{HHH,HHT,HTH,HTT,THH,THT,TTH}{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} \\ P(\text{at least}\ I\ H) & = \frac{7}{8}\end{align} 4. \begin{align}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\end{align} \begin{align}P(\text{at least one red}) &= \frac{3}{9} \\ P(\text{at least one red}) &= \frac{1}{3}\end{align} 5. . 6. \begin{align}P (\text{at least one}\ 5) = \frac{11}{36}\end{align} 7. \begin{align}P\end{align}(two 5's) \begin{align}=\frac{1}{36}\end{align} 8. \begin{align}P (2 \ \text{heads}) = \frac{4}{20} = 20\%\end{align} 9. \begin{align}P\end{align}(4 heads) \begin{align}= \frac{6}{25} = 24\%\end{align} 10. \begin{align}\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{967680} \\ \# \ of\ favorable\ choices & = 495\end{align} The number of possible outcomes \begin{align}= 2^{12}\end{align} The number of possible outcomes \begin{align}= 4096\end{align} Now we just divide the numerator by the denominator. \begin{align}P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%\end{align} 11. \begin{align}P(8\ \text{heads}) = 4.39\%\end{align} 12. \begin{align}\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\ \# \ of\ favorable\ choices & = 6446\end{align} The number of possible outcomes \begin{align}= 2^{15}\end{align} The number of possible outcomes \begin{align}= 32768\end{align} Now we just divide the numerator by the denominator. \begin{align}P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%\end{align} Answer Key for Review Questions (even numbers) 2. \begin{align}P(\text{at least}\ I\ H) & = \frac{HH,HT,TH}{HH,HT,TH,TT} \\ P(\text{at least}\ I\ H) & = \frac{3}{4}\end{align} 4. \begin{align}&\text{Possible Outcomes}:\\ & BB, BG, BR, GB, GG, GR, RB, RG, RR\\ & P(\text{at least one red}) = \frac{3}{9} \\ & P(\text{at least one red}) = \frac{1}{3}\end{align} 6. \begin{align}P\end{align} (at least one 5) \begin{align}= \frac{11}{36}\end{align} 8. \begin{align}P\end{align} (2 heads) \begin{align}= \frac{4}{20} = 20\%\end{align} 10. \begin{align}\# \ of\ favorable\ choices & = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{(24) \times (40320)} \\ \# \ of\ favorable\ choices & = \frac{479001600}{967680} \\ \# \ of\ favorable\ choices & = 495\end{align} The number of possible outcomes \begin{align}= 2^{12}\end{align} The number of possible outcomes \begin{align}= 4096\end{align} Now we just divide the numerator by the denominator. \begin{align}P(4\ heads) & = \frac{495}{4096} \\ P(4\ heads) & = 0.121 \\ P(8\ \text{heads}) & = 19.7\%\end{align} 12. \begin{align}\# \ of\ favorable\ choices & = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{(40320) \times (5040)} \\ \# \ of\ favorable\ choices & = \frac{1.31 \times 10^{12}}{203212800} \\ \# \ of\ favorable\ choices & = 6446\end{align} The number of possible outcomes \begin{align}= 2^{15}\end{align} The number of possible outcomes \begin{align}= 32768\end{align} Now we just divide the numerator by the denominator. \begin{align}P(8\ heads) & = \frac{6446}{32768} \\ P(8\ heads) & = 0.197 \\ P(8\ \text{heads}) & = 19.7\%\end{align} ### Notes/Highlights Having trouble? 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Major and Minor Axes of the Ellipse We will discuss about the major and minor axes of the ellipse along with the examples. Definition of the major axis of the ellipse: The line-segment joining the vertices of an ellipse is called its Major Axis. The major axis is the longest diameter of an ellipse. Suppose the equation of the ellipse be $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 then, from the above figure we observe that the line-segment AA’ is the major axis along the x-axis of the ellipse and it’s length = 2a. Therefore, the distance AA' = 2a. Definition of the minor axis of the ellipse: The shortest diameter of an ellipse is the minor axis. Suppose the equation of the ellipse be $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 then, putting x = 0 in equation we get, y = ± b. Therefore, from the above figure we observe that the ellipse intersects y-axis at B (0, b) and B’ (0, - b). The line segment BB’ is called the minor Axis of the ellipse. The minor axis of the ellipse $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 is along the y-axis and its length = 2b. Therefore, the distance BB' = 2b. Solved examples to find the major and minor axes of an ellipse: 1. Find the lengths of the major and minor axes of the ellipse 3x^2 + 2y^2 = 6. Solution: The given equation of the ellipse is 3x$$^{2}$$ + 2y$$^{2}$$ = 6. Now dividing both sides by 6, of the above equation we get, $$\frac{x^{2}}{2}$$ + $$\frac{y^{2}}{3}$$ = 1 ………….. (i) This equation is of the form $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 (a$$^{2}$$ > b$$^{2}$$), where a$$^{2}$$ = 2 i.e., a = √2 and b$$^{2}$$ = 3 i.e., b = √3. Clearly, a < b, so the major axis = 2b = 2√3 and the minor axis = 2a = 2√2. 2. Find the lengths of the major and minor axes of the ellipse 9x$$^{2}$$ + 25y$$^{2}$$ - 225 = 0. Solution: The given equation of the ellipse is 9x$$^{2}$$ + 25y$$^{2}$$ - 225 = 0. Now form the above equation we get, 3x$$^{2}$$ + 2y$$^{2}$$ = 225 Now dividing both sides by 225, we get $$\frac{x^{2}}{25}$$ + $$\frac{y^{2}}{9}$$ = 1 ………….. (i) Comparing the above equation $$\frac{x^{2}}{25}$$ + $$\frac{y^{2}}{9}$$ = 1 with the standard equation of ellipse $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 (a$$^{2}$$ > b$$^{2}$$) we get, a$$^{2}$$ = 25 ⇒ a = 5 and b$$^{2}$$ = 9 ⇒ b = 3. Clearly, the centre of the ellipse (i) is at the origin and its major and minor axes are along x and y-axes respectively. Therefore, the length of its major axis = 2a = 25 = 10 units and the length of minor axis = 2b = 23 = 6 units. ● The Ellipse 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. Types of Fractions \|Proper Fraction \|Improper Fraction \|Mixed Fraction Jul 12, 24 03:08 PM The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato… 2. Worksheet on Fractions \| Questions on Fractions \| Representation \| Ans Jul 12, 24 02:11 PM In worksheet on fractions, all grade students can practice the questions on fractions on a whole number and also on representation of a fraction. This exercise sheet on fractions can be practiced 3. Fraction in Lowest Terms \|Reducing Fractions\|Fraction in Simplest Form Jul 12, 24 03:21 AM There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1… 4. Conversion of Improper Fractions into Mixed Fractions \|Solved Examples Jul 12, 24 12:59 AM To convert an improper fraction into a mixed number, divide the numerator of the given improper fraction by its denominator. The quotient will represent the whole number and the remainder so obtained…
• 570 # Pythagoras Pythagoras was born in 570BC and lived until 495 BC. He is credited with finding, and the proof of, the Pythagorean Theorom. • 570 # The Equation The Pythagorean Theorom definition: In any right-angled triangle, the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares whose sides are the two legs. From this definition we can form the equation, a^2+b^2=c^2 where c is the hypotenuse. • 570 # Solving the equation step 1 Find one leg of your right triangle, not the hypotenuse. This will be a. • 570 # Solving the Equation Step 2 After you have found a, sqaure it. (Multiply it by itsself). This is your value for a. • 570 # Solving the Equation Step 3 Find the second leg on the right triangle, not the hypotenuse. This will be b. • 570 # Solving the Equation Step 4 After you have found side b, square it (multiply it by itsself). This is your value for b. • 570 # Solving the Equation Step 5 Take your values for a and b, then add them together. This value is your c^2 value. • 570 # Solving the Equation Step 6 Take the square root of your c ^2value to find the value of c. You have found your hypotenuse! • 570 # Example of a Solved Equation a^2+b^2=c^2 a=5 b=8 c=? a= 5x5 b=8x8 25+64=c^2 89=c^2 √89 = 9.43 <-- This is the length of they hypotenuse. • 570 # Real Life Example of Pythagorean Theorem Imagine that you're a firefighter. There is a woman in a burning building on the 2nd floor. There is also a hedge that pertrudes about 5 feet from the base of the building. Your ladder is 25 feet long. Will your ladder reach the lady?
#  Calculate the perimeter & area for each figure.. ## Presentation on theme: " Calculate the perimeter & area for each figure.."— Presentation transcript:  Calculate the perimeter & area for each figure. Friday May 10 th  Slope  Distance  How can you determine the triangle’s classification on a graph?  We use the distance formula to measure the length of each side. Two pairs of parallel sides  Is this a PARALLELOGRAM?  How did you determine this? Find the slope of each side, if opposite sides have the same slope, then the quadrilateral is a PARALLELOGRAM  A (  6, 3), B (  2, 0), C (  2,  5), D (  6,  2) Four right angles Is this a RECTANGLE? How did you determine this? Find the slope of each side, if consecutive sides have the perpendicular slopes, then the quadrilateral is a RECTANGLE  J (  1, 8), K (1, 6), L (  5, 0), M (  7, 2) Four congruent sides  Is this a RHOMBUS?  How did you determine this? Find the length of each side, if each sides has the same length, then the quadrilateral is a RHOMBUS  A (1, 8), B (4, 6), C (1,  2), D (  2, 0) Four congruent sides and four right angles  Is this a SQUARE?  How did you determine this? Find the slope of each side and the length of each side, if consecutive sides have the perpendicular slopes and each side has the same length, then the quadrilateral is a SQUARE  A (  5, 14), B (  2, 11), C (  5, 8), D (  8, 11) Exactly one pair of parallel sides  Is this a TRAPEZOID?  How did you determine this? Find the slope of each side, if one pair of opposite sides have the same slope and the other pair does not, then the quadrilateral is a TRAPEZOID  A (  3, 4), B (0, -1), C (2, -1), D (7, 4) Exactly one pair of parallel sides and congruent legs  Is this an ISOSCELES TRAPEZOID?  How did you determine this? Find the slope of each side and the length of the legs, if one pair of opposite sides have the same slope and the other pair does not and the legs have the same length, then the quadrilateral is a ISOSCELES TRAPEZOID  A (  4, 2), B (  1, 5), C (3, 5), D (6, 2) 1. Q (  5, 1), R (  1,  2), S (  1,  7), T (  5,  4) 2. A(2, 0), B(  1, 3), C(2, 6), D(5, 3)  Triangle/Rhombus ◦ On coordinate plane: Length of each side, if all the same, it is a rhombus  Parallelogram/Rectangle ◦ On coordinate plane: Slope of each side. If opposite sides have same slope, it is a parallelogram. If consecutive sides have perpendicular slopes, it is a rectangle  Trapezoid ◦ On coordinate plane: Slopes of each side. If only one pair of opposite sides have same slope, it is a trapezoid. ◦ Find length of two legs, if legs are congruent, then it is isosceles.  Worksheet Monday May 13 th  On Friday, we classified figures on the coordinate plane.  Today we will use those graphs to calculate area and perimeter of figures.  For area we will leave out some figures because we are not going to find altitudes that are necessary for the calculations.  How do you calculate perimeter for any figure?  How do you do this on a coordinate plane? ◦ We will use the DISTANCE FORMULA to find each side length then add each side. Do you remember the area formulas?  Triangle =  Parallelogram = (This includes rectangle, rhombus, & square)  Trapezoid =  Worksheet
# When You Add Two Negative Numbers Do You Get A Positive? It is possible to produce a positive result by multiplying two negatives. This is because the two negative signs cancel each other out. Why don’t the good indicators cancel each other out? When you have two negative signs, one of them rolls over, and the two negative signs add together to form a positive. There is one dash left over when you have a positive and a negative. The answer is negative in this situation. ## Can you add two negative numbers together? When we combine a negative number with a positive number, or when we combine two negative numbers, it might be difficult to calculate the result. There are, however, a few easy regulations to follow, which we will outline below for your convenience. Rule 1: Adding positive numbers to positive numbers is as simple as adding two numbers together. ## Can numbers be positive or negative? Numbers can have either a positive or negative value. The Number Line is as follows: Negative Numbers () are a kind of number. Numbers with a plus sign (+) The negative sign is represented by the symbol ‘-‘. The positive sign is represented by the letter ‘+’. ## What are the rules for adding and subtracting positive and negative numbers? Positive and negative integers are added and subtracted according to certain rules. Whenever two signs are written next to each other, the following are the rules for adding and subtracting numbers: When two signs are dissimilar, they combine to form a negative sign. When two signs are the same, they combine to form a positive sign. ## What is multiplication of two negative numbers positive? • Positive results from the multiplication of two negative integers. • Proof that is not mathematical in nature. • We are all aware that two negatives in a statement equals one positive. The use of two negatives changes the meaning of the statement from negative to positive.This diamond is not an unusual find.This indicates that the diamond is common.The experiment’s results are not inconclusive in nature. This indicates that the results are conclusive. ## Can the sum of two negative numbers ever be positive? No, the sum of two negative numbers can never be a positive number since they are both negative. It is possible to owe more money to someone if you owe them money and then borrow additional funds from them. A minus plus a minus is a minus, according to the rule. Adding two negative numbers indicates that you should proceed down the number line and then back down. You might be interested: What Is The Most Powerful Dinosaur In Jurassic World The Game? ## What is the rule for adding two negative numbers? Take note: What happens when you add two negative integers together? Simply add the absolute values of each number together, followed by a minus sign, and you will have your answer! ## When 2 negative integers are added we get? As a result, if we add two negative numbers, we always end up with a negative integer. ## How minus into minus is plus? Multiply minus one is known as minus multiplicand. The minus sign is plus. In other words, the product of two negative integers is always a positive number when multiplied by themselves. ## What do you call plus and minus signs? • A mathematical symbol that means ‘plus or minus’ and is used to indicate the precision of an approximation (as in ‘The result is 10 0.3, which means the result is anywhere in the inclusive range from 9.7 to 10.3), or as a convenient shorthand for a quantity that has two possible values of opposing sign and identical magnitude (as in ‘The result is 10 0.3’, which means the result is anywhere in the inclusive range from 9.7 to 10.3). ## What happens when you add two positive numbers? Rule 1: Adding positive numbers to positive numbers is as simple as adding two numbers together. For example, you may say, ″This is what you’ve been learning.″ Three plus two equals two positive numbers. You can solve these problems using the same formula you’ve been using: 3 + 2 = 5. ## What happens when you add a negative number? Each time you add a positive number to your total, you advance to the right on the number line. When you add a negative integer to your equation, you shift to the left side of the equation.
1 / 16 # Regents Review #2 Regents Review #2. Equations. What type of Equations do we need to solve?. Simple Equations Equations with Fractions Quadratic Equations Literal Equations ( solving for another variable ) Equations that help us solve word problems. Simple Equations. ½ (2x – 10) = 5x – (6x + 9) ## Regents Review #2 An Image/Link below is provided (as is) to download presentation Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. During download, if you can't get a presentation, the file might be deleted by the publisher. E N D ### Presentation Transcript 1. Regents Review #2 Equations 2. What type of Equations do we need to solve? Simple Equations Equations with Fractions Quadratic Equations Literal Equations (solving for another variable) Equations that help us solve word problems 3. Simple Equations ½ (2x – 10) = 5x – (6x + 9) x – 5 = 5x – 6x – 9 x – 5 = -x – 9 2x = -4 x = -2 Always check solution(s) to any equation ½ (2 -2 – 10) = 5(-2) – (6 -2 + 9) ½ (-14) = -10 – (-3) -7 = -7 It checks! 4. How do we solve Quadratic Equations? 1) x2 = a Example: x2 = 16 Take the square root of both sides x = x = 4 or x = {4,-4} • x2 + bx + c = 0 Example: x2 – 5x + 6 = 0 Set all terms equal to zero (x – 2)(x – 3)= 0 Factor x – 2 = 0 x – 3 = 0 Set each factor equal to zero x = 2 x = 3 Solve x = {2,3} 5. Equations with Fractions There are two ways to “clear” fractions in Equations… • Multiply both sides of the equation by the LCD (Least Common Denominator) 2) Cross Multiply (only works for proportions ) 6. Equations with Fractions Cross Multiplication (solving proportions) Check for solutions that will make the denominator equal to zero. x = {3,1} 7. Equations With Fractions Multiplying by the LCD CHECK 8. Literal Equations When solving literal equations, isolate the indicated variable using inverse operations 9. Word Problems We can use equations to solve many different types of word problems. When solving a word problem, remember to… • Define all unknowns (set up “Let” statements) • Write an equation relating all unknowns • Solve the equation • Determine the value of all unknowns • Answer the question (use appropriate units) 10. Word Problems Find two consecutive integers whose sum is -35. -18 -17 (-18 + 1) x: 1st consecutive integer x + 1: 2nd consecutive integer Remember: Consecutive integers count by 1’s Ex: x, x+1, x+2, x+3…. Consecutive odd or even integers count by 2’s Ex: x, x+2, x+4, x+6… 11. Word Problems Ticket sales for a music concert totaled \$2,160. Three times as many tickets were sold for the Saturday night concert than the Sunday afternoon concert. Two times as many tickets were sold for the Friday night concert than the Sunday afternoon concert. Tickets for all three concerts sold for \$2 each. Find the number of tickets sold for the Saturday night concert. x: number of tickets sold for the Sunday concert 3x: number of tickets sold for the Saturday concert 2x: number of tickets sold for the Friday concert 180 tickets 540 tickets 3(180) 360 tickets 2(180) 2x + 2(3x) + 2(2x) = 2160 There were 540 tickets sold for the Saturday night concert 2x + 6x + 4x = 2160 12x = 2160 x = 180 12. Word Problems A person has 23 coins made up of dimes and quarters worth \$3.35. How many coins of each type are there? 0.10x + 0.25(23 – x) = 3.35 16 dimes 7 quarters (23 – 16) 10x + 25(23 – x) = 335 10x + 575 – 25x = 335 -15x + 575 = 335 -15x = -240 x = 16 Check 7(25) + 16(10) = 335 175 + 160 = 335 335 = 335 13. Word Problems The area of a rectangle is 120 square inches. If the length is two more than the width, find the dimensions of the figure. A = lw 120 = (x + 2)(x) 120 = x2 + 2x 0 = x2 + 2x – 120 0 = (x + 12)(x – 10) x + 12 = 0 x – 10 = 0 x = -12 x = 10 Reject Width: x Length: x + 2 The width is 10 inches The length is 12 inches 14. Word Problems Distance, Rate, Time (D = RT) • Two Options • Finding Rate • 2) Finding Time D R T How do you know what to look for? Distance: How far? Rate: How fast? Time: How long? 15. Word Problems Hannah took a trip to her cousin’s house. She drove 120 miles to get there. If it took her 1.2 hours to get halfway to her cousin’s house, what was her average rate of speed? Important Information: Distance traveled: 60 miles (120/2) Time: 1.2 hours Rate: ? If Hannah decided to travel 10 mph faster on the way home, how long would it take her? 16. Now it’s your turn to review on your own! Using the information presented today and your review packet, complete the practice problems in the packet.Regents Review #3 is FRIDAY, May 17thBE THERE!!!! More Related
How to calculate maximums, minimums, and inflection points in a function # How to calculate maximums, minimums, and inflection points in a function In this lesson I am going to teach you how to calculate maximums, minimums and inflection points of a function when you don’t have its graph. The relative extremes of a function are maximums, minimums and inflection points (point where the function goes from concave to convex and vice versa). ## How to obtain maximums, minimums and inflection points with derivatives The relative extremes (maxima, minima and inflection points) can be the points that make the first derivative of the function equal to zero: These points will be the candidates to be a maximum, a minimum, an inflection point, but to do so, they must meet a second condition, which is what I indicate in the next section. ## How to know if a point is a maximum, a minimum or an inflection point Once we have obtained the points for which the first derivative of the function is equal to zero, for each point we must check the following: If the value of the second derivative at that point is greater than zero, then that point is minimum: If the value of the second derivative at that point is less than zero, then that point is maximum: If the second derivative at that point is equal to zero, then that point is an inflection point, as long as the third derivative at that point is other than zero: Let’s see with an example everything explained so far. ## Resolved exercise on how to calculate maximums, minimums and inflection points Let’s get the relative extremes of the next function: First, we are going to obtain the possible relative extremes, obtaining the first derivative of the function and equaling it to 0. The first derivative of the function is: We equal it to zero to get the points that meet that condition: To solve the equation, we previously simplified it: As it is a third degree equation, I break it down into factors by the Ruffini rule: Whose solutions are: That correspond to possible maximums, minimums or inflection points. Now we are going to check what each point corresponds to, studying the sign of the second derivative. For this we obtain the second derivative of the function: And we calculate the value of the second derivative for each one of the values that we have just calculated and that make the first derivative zero (x=-2, x=-1 and x=1). We start by calculating the value of the second derivative for x=-2: The result is greater than zero, so in x=-2 there is a minimum: We calculated the value of f”(x) for x=-1: The result is less than zero, so in x=-1 there is a maximum And finally, we calculate the value of the second derivative for x=1: Whose value is greater than zero, so in x=1 there is a minimum: With the values of x obtained from equalizing the first derivative to zero, we have had no value of f”(x) equal to zero, that is, we have not found any inflection point. Therefore, we are going to calculate the points that make the second derivative equal to 0: We equate the second derivative to 0: And we solve the equation, whose results are: These two values are possible inflection points, as long as they meet the third derivative for those points is other than zero. We calculate the third derivative of the function: And we find the value of the third derivative for x=0.21: Which is different from 0, so at x=0.21 there is an inflection point: We do the same with x=1.24: The result is also non-zero, so at x=-1,24 there is an inflection point. In short, the relative extremes we have found are: • A minimum in x=-2 • A maximum in x=-1 • A minimum in x=1 • An inflection point in x=0.21 • An inflection point in x=-1,24
# Quick Answer: Does And In Probability Mean Multiply? ## What does or stand for in probability? Probability theory indicates the probability of either event A or event B occurring (“or” in this case means one or the other or both).. ## 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 are the four rules of maths? Teaching the ‘4 Rules of Number’The ‘4 rules’ (addition, subtraction, multiplication and division) are at the heart of calculation and problem solving.Over the years a range of teaching methods has been adopted by schools and it is sometimes the case that parents’ experiences are not the same as those of their children.More items… ## Does and mean multiply? In probability, the words “AND” and “OR” are kind of special, and they usually mean multiply the probabilities (for AND) and add the probabilities (for OR), respectively. – ## What is the and/or rule in probability? Sometimes we want to know the probability of getting one result or another. When events are mutually exclusive and we want to know the probability of getting one event OR another, then we can use the OR rule. … P(A or B) = P(A) + P(B) for mutually exclusive events. ## What does and mean in math terms? & This symbol is called an ampersand. It almost always means “and,” both in and outside of mathematics. * This symbol is called an asterisk. In mathematics, we sometimes use it to mean multiplication, particularly with computers. For example, 53 = 5 times 3 = 15. ## What does it mean to multiply probabilities? If A and B are two independent events in a probability experiment, then the probability that both events occur simultaneously is: P(A and B)=P(A)⋅P(B) In case of dependent events , the probability that both events occur simultaneously is: P(A and B)=P(A)⋅P(B \| A) ## What is the general multiplication rule? The multiplication rule is a way to find the probability of two events happening at the same time (this is also one of the AP Statistics formulas). There are two multiplication rules. The general multiplication rule formula is: P(A ∩ B) = P(A) P(B\|A) and the specific multiplication rule is P(A and B) = P(A) P(B). ## How do I know if I add or multiply probabilities? But the probability that either event will occur (A or B) is typically found by adding: When you’re looking for the probability that two events, A andB, will BOTH occur, the probability of this coincidence is small, and you multiply the separate probabilities of A and B to get a smaller number. ## What is both in probability? both occur. Rule of Multiplication The probability that Events A and B both occur is equal to the probability that Event A occurs times the probability that Event B occurs, given that A has occurred. P(A ∩ B) = P(A) P(B\|A) ## Why is there an order of operations? The order of operations is a rule that tells you the right order in which to solve different parts of a math problem. … Subtraction, multiplication, and division are all examples of operations.) The order of operations is important because it guarantees that people can all read and solve a problem in the same way. ## Is and plus or multiply? If all the events happen (an “and question”) Multiply the probabilities together. If only one of the events happens (an “or question”) Add the probabilities together. ## Why do you multiply in probability? When you want to learn about the probability of two events occurring together, you’re multiplying because it means “expanding the possibilities.” Because: Now, the possibilities are four, not two. … It’s multiplication because you’re trying to find the probability inside another probability. ## What are the 3 rules of probability? There are three main rules associated with basic probability: the addition rule, the multiplication rule, and the complement rule. You can think of the complement rule as the ‘subtraction rule’ if it helps you to remember it. ## What does P AUB mean? P(A U B) is the probability of the sum of all sample points in A U B. Now P(A) + P(B) is the sum of probabilities of sample points in A and in B. Since we added up the sample points in (A ∩ B) twice, we need to subtract once to obtain the sum of probabilities in (A U B), which is P(A U B).
Angle Between Two Vectors Home > Lessons > Angle Between Vectors Search \| Updated April 9th, 2019 Introduction This lesson page will explain how to calculate the angle between two vectors. Here are the sections within this lesson page: Prerequisites In order to understand how to calculate the angle between two vectors, you must have learned several other concepts and skills. Make sure you are familiar with the content below before proceeding with this lesson. esson: Vectors esson: Vector Magnitude esson: Dot Product of Vectors esson: Inverse Trigonometric Functions As a side note, this lesson uses two notations for vectors. If a letter is bold or it has an arrow ( → ) above it, it is a vector. For instance, these two symbols both mean vector w. Why Do I Need to Know the Angle? We need to know how to calculate the angle between two vectors for numerous reasons. Vectors are all around us. Vectors are the forces that are acting on beams and other supports within structures. Vectors are used to represent wind, pressure, humidity, and many conditions for predicting weather patterns and climates. The air that flows around an aircraft’s wing, the fluid that flows within a pipe, and several other situations are modeled using vectors. These vectors help researchers create aircraft that is fuel efficient and pipes that contain extreme pressures. When two forces interact, the angle between those forces is important for determining the resulting force. Angle Equation The equation for finding the angle between two vectors, u and v, is this. This equation is not a simple one. It involves a trigonometric function, the dot product of two vectors, and the magnitude of two vectors. The examples below will explain, in detail, how to use the equation to find theta, the angle between two vectors. Example 1 Let us start with two vectors, u and v, so that we can determine the angle (in degrees) between the two vectors. The equation requires us to determine the dot product of the two vectors. The equation requires us to calculate the magnitudes of the vectors; so, here are their magnitudes. Now, we can place this information into the equation, like so. To solve for the angle, theta, we must use the inverse cosine or arccosine, like so. If your calculator is set for degrees, the angle rounded to the nearest thousandth is… Use these resources to help yourself learn more about the angle between two vectors. Example 2 Here is another example. So, here are two new vectors, u and v. We will again determine the angle (in degrees) between the two vectors. Here is the dot product between the vectors. Here are the magnitudes of the two vectors. We can plug in the values into their appropriate places within the equation. This problem does not require the use of a calculator, for two reasons. The first is that knowledge of the unit circle should tell us that the cosine of 90 degrees is equal to zero. The second reason is that the dot product being equal to zero means that the vectors are orthogonal. This also means the angle between them is 90 degrees. See our lesson on Orthogonal Vectors. Use these resources to help yourself learn more about the angle between two vectors. Instructional Videos This video reveals what the formula is, provides two examples, and explains the graphical representation. Interactive Quizmasters Try this quiz to see if you understand how to find the angle between two vectors. Related Lessons Try these lessons, which contain more information on vectors. esson: Vectors esson: Engineering Notation (Vectors) esson: Unit Vectors esson: Vector Projection
# Proportions to Find Percent, P ## Use proportions to cross - multiply and find percents. % Progress MEMORY METER This indicates how strong in your memory this concept is Progress % Use Proportions to Find Percents Have you ever thought about voting and percents? Take a look at this dilemma. A senator wants to start a program to encourage more people to vote in his state. In County A, 32,100 people voted. In neighboring County B, 57,800 people voted. Which county needs the program more? Well, it depends on how many people are in each of the counties. We can’t compare the polling rates unless you use a percent. If we know that the first county has a population of 39,150 people and the second county has a population of 81,400 people, we can now find what percent of the people voted. We are comparing the number of people who voted with the population of each county. We are actually going to find two percents here. Do you know how to do this? Pay attention and you will understand how to complete this task by the end of the Concept. ### Guidance A percent is a part of a whole that represents a quantity out of 100. Fractions and decimals are also parts of a whole. Sometimes, you will be given information, but not a percent. You will need to know how to figure out the percent. Percents, fractions, decimals and proportions can all help to you solve problems and figure out percents. You began using proportions to figure out a percent when writing fractions as percents. Remember that proportions involve comparing quantities. A proportion is a comparison between two equal ratios. Percents are also written to compare a quantity to 100. Because both of these are comparing, we can use proportions to help us figure out a percent. That is a great question. First, we write the proportion using \begin{align}a\end{align} over \begin{align}b\end{align}. \begin{align}\frac{a}{b}\end{align} This is equal to the percent which is out of 100. \begin{align}\frac{p}{100}\end{align} Here is the proportion: \begin{align}\frac{a}{b} = \frac{p}{100}\end{align} Now let's apply this proportion. Take a look at this dilemma. 15 out of 30 is what percent? To work on this problem, first, we write a ratio comparing our given values to the missing percent. \begin{align}\frac{15}{30} = \frac{p}{100}\end{align} We know that fifteen is half of thirty, and 50 is half of 100. Write each as a percent. #### Example A 18 out of 50 Solution: \begin{align}36\%\end{align} #### Example B 22 out of 40 Solution: \begin{align}55\%\end{align} #### Example C 78 out of 80 Solution: \begin{align}97.5\%\end{align} Now let's go back to the dilemma from the beginning of the Concept. For each county, we will use the proportion \begin{align}\frac{a}{b} = \frac{p}{100}\end{align} where \begin{align}a\end{align} is the number of people that voted and \begin{align}b\end{align} is the total population. \begin{align}& \text{County A} \qquad \qquad \qquad \text{County B}\\ & \ \frac{32100}{39150} = \frac{p}{100} \qquad \qquad \ \frac{57800}{81400} = \frac{p}{100}\\ & 39150p = 32100 \cdot 100 \quad 81400p = 57800 \cdot 100\\ & 39150p = 3210000 \qquad \ 81400p = 5780000\\ & \qquad \ \ p = 82 \% \qquad \qquad \qquad \ p=71 \%\end{align} In order to find the percent in each case, we used cross products as we would for any proportion. Now we can see that in County A, 82% of the people voted, while in County B only 71% of the people voted. The senator should push the program more in County B. ### Guided Practice Here is one for you to try on your own. John ran 8 out of 9 miles. What percent of the total miles did he run? Solution To figure this out, let's first write a proportion so that we can figure out the percent. \begin{align}\frac{8}{9} = \frac{p}{100}\end{align} Now we can cross-multiply and divide. \begin{align}9p &= 800 \\ p &= \frac{800}{9} \\ p &= 88.8\end{align} We can round up for our answer. Our answer is 89%. John ran 89% of the total miles. ### Explore More Directions: Find \begin{align}p\end{align} in the given problems using cross products. Round to the nearest tenths place. 1. \begin{align}\frac{7}{15}=\frac{p}{100}\end{align} 2. \begin{align}\frac{52}{3810}=\frac{p}{100}\end{align} 3. \begin{align}\frac{16}{17}=\frac{p}{100}\end{align} 4. \begin{align}\frac{3}{4}=\frac{p}{100}\end{align} 5. \begin{align}\frac{3}{5}=\frac{p}{100}\end{align} 6. \begin{align}\frac{1}{5}=\frac{p}{100}\end{align} 7. A dentist filled cavities in 8 of his 30 patients on Tuesday. What percent had cavities filled? 8. A florist delivered 18 out of 25 bouquets. What percent was delivered? 9. The baker sold 3 out of 4 dozen rolls. What percent was sold? 10. What percent is 85 out of 5000? 11. What percent is 15 out of 30? 12. What percent is 88 out of 1200? 13. What percent is 99 out of 200? 14. What percent is 100 out of 330? 15. What percent is 224 out of 5400? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes
7th Grade Math Circles Unit Test 13 questions #1 of 13:Mild Circles <p>In the circle shown below, what is the length of the radius?</p><newline></newline><imageui data-src="https://d3nx71g9uvfl9m.cloudfront.net/image_db/4eadc2e2-acc3-11ec-be3c-0242c0a84003.png" data-alt="circle" data-caption="" data-width="196px"></imageui> #2 of 13:Medium Circles <p>In the circle shown below, what is the length of the radius?</p><newline></newline><imageui data-src="https://d3nx71g9uvfl9m.cloudfront.net/image_db/f79e1140-acc3-11ec-be3c-0242c0a84003.png" data-alt="circle" data-caption="" data-width="196px"></imageui> #3 of 13:Mild Circles <p>Calculate the length of the diameter in the circle shown below.</p> <imageui data-src="https://d3nx71g9uvfl9m.cloudfront.net/image_db/c45146e2-ac0d-11ec-be3c-0242c0a84003.png" data-alt="circle" data-caption="" data-width="196px"></imageui> #4 of 13:Spicy Circles <p>In the circle shown below, what is the length of the radius?</p><newline></newline><imageui data-src="https://d3nx71g9uvfl9m.cloudfront.net/image_db/1567b30c-acc4-11ec-be3c-0242c0a84003.png" data-alt="circle" data-caption="" data-width="196px"></imageui> #5 of 13:Spicy Circles <p>A circle has a diameter of `45.52` feet. Calculate the circumference of the circle. Give your answer in terms of `π` and rounded to the nearest hundredth.</p> #6 of 13:Mild Circles <p>A circle has a radius of `22` centimeters. Calculate the circumference of the circle. Give your answer in terms of `π` and rounded to the nearest hundredth.</p> #7 of 13:Medium Circles <p>A circle has a radius of `23.7` feet. Calculate the circumference of the circle. Give your answer in terms of `π` and rounded to the nearest hundredth.</p> #8 of 13:Medium Circles <p>Victoria always starts her day by jogging around Morris Playground, a circular-shaped park. The radius of the park is `30` meters. </p><p>If Victoria does one lap around the park, approximately how much distance does she cover? </br><highlight data-color="#666" data-style="italic">Round your answer to the nearest hundredth.</highlight></p> #9 of 13:Medium Circles <p>A circle has a diameter of `14.1` inches. Calculate the circumference of the circle. Give your answer in terms of `π` and rounded to the nearest hundredth.</p> #10 of 13:Mild Circles <p>Samantha recently earned second place in her swimming competition. She received a silver medal for her prize. The medal had a radius of `4` in. </p><p>What was the area of the medal? <br><highlight data-color="#666" data-style="italic">Round your answer to the nearest hundredth of a square inch. </highlight></p> #11 of 13:Medium Circles <p>Calculate the area of the circle that has a diameter of `26` inches. Give your answer in terms of `π` and rounded to the nearest hundredth.</p> #12 of 13:Spicy Circles <p>The circumference of a circle is 9.4π centimeters. What is the diameter, radius, and area of the circle?</p> #13 of 13:Spicy Circles <p>Find the area of the shape.<br><highlight data-color="#666" data-style="italic">Enter your answer in terms of `π`.</highlight></p><imageui data-src="https://d3nx71g9uvfl9m.cloudfront.net/image_db/c856406e-ab66-11ec-be3c-0242c0a84003.png" data-alt="circle" data-caption="" data-width="196px"></imageui> A circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called “center”. Any straight line joining a point on the circle to the center is called a radius. Taking circles u... Show all Circles What teachers are saying about BytelearnWhat teachers are saying Stephen Abate 19-year math teacher Carmel, CA Any math teacher that I know would love to have access to ByteLearn. Jennifer Maschino 4-year math teacher Summerville, SC “I love that ByteLearn helps reduce a teacher’s workload and engages students through an interactive digital interface.” Rodolpho Loureiro Dean, math program manager, principal Miami, FL “ByteLearn provides instant, customized feedback for students—a game-changer to the educational landscape.” What teachers are saying about BytelearnWhat teachers are saying Stephen Abate 19-year math teacher Carmel, CA Any math teacher that I know would love to have access to ByteLearn. Jennifer Maschino 4-year math teacher Summerville, SC “I love that ByteLearn helps reduce a teacher’s workload and engages students through an interactive digital interface.” Rodolpho Loureiro Dean, math program manager, principal Miami, FL “ByteLearn provides instant, customized feedback for students—a game-changer to the educational landscape.”
# Solving for x problems In algebra, one of the most important concepts is Solving for x problems. So let's get started! ## Solve for x problems We will also give you a few tips on how to choose the right app for Solving for x problems. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane. The most common type of function is the linear function. A linear function is a function in which the input and output are related by a straight line. College algebra is the study of linear functions and their properties. It investigates how these functions can be used to model real-world situations. In addition, college algebra also covers topics such as graphing, solving equations, and manipulating algebraic expressions. As a result, college algebra is an important course for any student who plans on pursuing a career in mathematics or another field that uses mathematics. Partial fractions is a method for decomposing a fraction into a sum of simpler fractions. The process involves breaking up the original fraction into smaller pieces, each of which can be more easily simplified. While partial fractions can be used to decompose any fraction, it is particularly useful for dealing with rational expressions that contain variables. In order to solve a partial fraction, one must first determine the factors of the denominator. Once the factors have been determined, the numerator can be factored as well. The next step is to identify the terms in the numerator and denominator that share common factors. These terms can then be combined, and the resulting expression can be simplified. Finally, the remaining terms in the numerator and denominator can be solve for using basic algebraic principles. By following these steps, one can solve any partial fraction problem. Quadratic equations are a common type of algebraic equation that can be difficult to solve. However, there are a number of Quadratic equation solvers that can help to make the process easier. These solvers will typically provide a step-by-step solution, making it easy to see how to solve the equation. In addition, some Quadratic equation solvers will also provide a visual representation of the solution, which can be helpful in understanding the concept. There are a number of Quadratic equation solvers available online, and many of them are free to use. ## Help with math It a wonderful app to solve your math problems and it shows all the steps nicely and one by one with detailed explanation. It's very useful for students as well as teachers and parents. Marlee Gonzales If you're struggling with math, get this app. I only have the free version but it's a great tool to check your answers and provide steps to solve problems. Even though the explanations are limited without paying for pro it's still enough. Maci Roberts How to do algebra equations Pre algebra math help Escribir problemas de matematicas y resolverlos Purple math problem solver Solve for angle in right triangle
# Find the function y = mx + n if the points (1;2) and (-1;-1) are on the graph of the function. neela \| Student y = mx+n, the points (1,2) and (-1,-1) are on the graph. Since points are on the graph, the coordinates of the two given points should satisfy the graph. Point (1,2) is on y = mx+n. So m1+n= 2. Or m+n = 2....(1) Point (-1, -1) is on the graph y mx+n . So m(-1) +n = -1, Or -m+n = -1....(2). Adding eq(1) and eq (2) , we get: m+n-m+n = 2-1 = 1. 2n = 1. n = 1/2. Eq(1) -eq(2) gives: m+n - (m+n) = 2- (-1) = 3. 2m = 3. m = 3/2. Therefore the value of m = 1/2 and n = 3/2. giorgiana1976 \| Student The function that has to be determined is a linear function. A linear function is determined when it's coefficients are determined. y = f(x) = mx + n So, in order to determine y, we'll have to calculate the coefficients m and n. Since the function is determined by the points (1,2) and (-1,-1), that means that if we'll substitute the coordinates of the points into the expression of the function, we'll get the relations: f(1) = 2 f(1) = m1 + n m + n = 2 (1) f(-1) = -1 f(-1) = m(-1) + n -m + n = -1 (2) m + n -m + n = 2 - 1 We'll combine and eliminate like terms: 2n = 1 n = 1/2 We'll substitute n in (1): m + n = 2 m + 1/2 = 2 m = 2 - 1/2 m = 3/2 The expression of the linear function is: f(x) = 3x/2 + 1/2
Question Video: Differentiating a Combination of Logarithm Arithmic and Polynomial Functions Using the Quotient Rule \| Nagwa Question Video: Differentiating a Combination of Logarithm Arithmic and Polynomial Functions Using the Quotient Rule \| Nagwa # Question Video: Differentiating a Combination of Logarithm Arithmic and Polynomial Functions Using the Quotient Rule Mathematics • Third Year of Secondary School ## Join Nagwa Classes Find dπ¦/dπ₯, given that π¦ = 9π₯/ln 9π₯. 02:37 ### Video Transcript Find dπ¦ by dπ₯ given that π¦ is equal to nine π₯ divided by the natural logarithm of nine π₯. The question wants us to find dπ¦ by dπ₯. Thatβs the first derivative of π¦ with respect to π₯. And we can see that π¦ is the quotient of two functions. Itβs the quotient of nine π₯ and the natural logarithm of nine π₯. So weβll find this derivative by using the quotient rule. We recall the quotient rule tells us if π¦ is the quotient of two functions π’ over π£, then dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all divided by π£ squared. So to use the quotient rule, weβll start by setting π’ of π₯ to be the function in our numerator, thatβs nine π₯, and π£ of π₯ to be the function in our denominator, thatβs the natural logarithm of nine π₯. And to apply the quotient rule, weβre going to need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with finding dπ’ by dπ₯. Thatβs the derivative of nine π₯ with respect to π₯. And nine π₯ is just a linear function. So its derivative is the coefficient of π₯, which, in this case, is nine. Letβs now find an expression for dπ£ by dπ₯. Thatβs the derivative of the natural logarithm of nine π₯ with respect to π₯. And we can do this by using one of our standard derivative results for logarithmic functions. For any positive constant π, the derivative of the natural logarithm of ππ₯ with respect to π₯ is equal to one divided by π₯. So in our case, the derivative of the natural logarithm of nine π₯ with respect to π₯ is equal to one divided by π₯. So weβre now ready to find the dπ¦ by dπ₯ by using the quotient rule. The quotient rule tells us dπ¦ by dπ₯ will be equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ divided by π£ squared. Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get dπ¦ by dπ₯ is equal to the natural logarithm of nine π₯ multiplied by nine minus nine π₯ times one over π₯ all divided by the natural logarithm of nine π₯ squared. And we can simplify this expression. First, weβll cancel π₯ multiplied by one over π₯. Next, we want to take out a factor of nine in our numerator. And this gives us nine times the natural logarithm of nine π₯ minus one all divided by the natural logarithm of nine π₯ squared. And this is our final answer. Therefore, weβve shown if π¦ is equal to nine π₯ divided by the natural logarithm of nine π₯, then dπ¦ by dπ₯ is equal to nine times the natural logarithm of nine π₯ minus one all divided by the natural logarithm of nine π₯ squared. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
math fraction-to-decimal conversions The more familiar you are with some simple fraction-to-decimal conversions, the quicker you'll be able to solve math problems within the 70-minute time limit. You should definitely use your calculator too, but memorizing these conversions will be a huge advantage for you on test day. 0.01 = 1/100 = 1% 0.1 = 1/10 = 10% 0.25 = 1/4 = 25% 0.5 = 1/2 = 50% 0.2 = 1/5 = 20% 0.75 = 3/4 = 75% math terms to know The math problems on the SAT will use words that you need to be familiar with in order to solve them. Here are some key terms that The Princeton Review encourages everyone to memorize. 1. An integer is a non-decimal, non-fractional number. Integers can be positive, negative, or zero. Examples: -10, 0, 1, 5 2. A prime number is a number that can be divided only by itself and 1. The first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The number 1 is NOT prime. The number 2 is the only even prime. 3. Zero is an integer, it's neither positive nor negative, and it's an even number. 4. Absolute value is always expressed as a positive number. It is the numerical value of any real number regardless of its sign. Examples: \|-4\| = 4 \|5\| = 5 5. To find the union of two sets, combine all the elements into a new set of distinct elements. To find the intersection of two sets, combine only the elements that belong to both sets into a new set of distinct elements. 6. Congruent means having the same size and shape — figures are congruent if their corresponding angles and sides are equal. 7. The mode is the number that appears the most often. Example: the mode of {2, 2, 3, 4, 5, 5, 5, 6, 7} is 5. (The SAT won't ask you for the mode of a set of numbers like this — 2, 2, 5, 5 — or this — 1, 2, 3, 4.) 8. MADSPM reminds you when you Multiply exponents, you Add them; when you Divide exponents you Subtract them; and when there are Parentheses you Multiply the exponents. 9. FOIL stands for "first, inner, outer, last." FOIL describes the steps you use to multiply factors to form a quadratic equation. 10. A line that's tangent to a circle touches the circle at exactly one point and is perpendicular to the radius. Visit the next page for reading tips! Connect with Seventeen Today's Video Daily Freebies 5 Winners! Enter for a chance to win a light blue bikini in a cool halter style. More Freebies Most Popular Celebs Connect with Seventeen Seventeen Network: Seventeen CosmoGirl TeenMag MisQuinceMag DonateMyDress Welcome!
Monday, September 25, 2023 # What Is Standard Form In Math 3rd Grade ## Standard Form Of A Decimal Number Writing in the Standard Form \| Mathematics Grade 3 \| Periwinkle The standard form of a decimal number in Britain is known as Scientific notation, where the number is written in the following way: 4527.7 = 4.5277 x 10³ A Number In Scientific Notation In this example, 4527.7 can be written as 4.5277 × 10³ in scientific notation Because, 4527.7 = 4.5277 × 1000 = 4.5277 × 10³ The standard form of decimal numbers in the United States, and in other countries using the US conventions is written in expanded form. Example: 4.327 in expanded form can be written as: 4.327=4×1+3×110+2×1100+7×11000 4.327=4×1+3×110+2×1100+7×11000 4.327 = 4 + 0.3 + 0.03 + 0.007 Therefore, The expanded form of 4.327 is 4 + 0.3 + 0.03 + 0.007. ## How To Write Decimal Numbers In Standard Form Now, let us discuss how to write the decimal numbers in standard form. Step 1: Write the first non-zero digit from the given number Step 2: Add the decimal point after the first non-zero digit Step 3: Find the number of the decimal point that shifts from the given number and write it to the power of 10. For example, 0.0000017 in standard form is as follows: Step 1: In the given number, the first non-zero digit is 1 Step 2: Add decimal point after the number 1. Hence, it becomes 1. Step 3: Here, the decimal point shifts 6 places to the right. Hence, the standard form of a number 0.0000017 is 1.7 ×10-6 In this section, we shall learn about the representation of numbers in the expanded form. ## What Is The Standard Form Of Equation In the standard form of an equation, 0 is usually present on the right side, whereas the rest of the expressions are on the left. Also, the terms are arranged in the descending order of their exponents. To convert an equation into stadard form, just apply arithmetic operations on both sides to turn the right side to be 0. Recommended Reading: College Algebra Chapter 1 Study Guide ## Numbers In Expanded Form Let us consider a 6 digit number 657891. In the expanded form this number can be written as: 657891 = 6 × 100000 + 5 × 10000 + 7 × 1000 + 8 × 100 + 9 × 10 + 1 We can express this number in the exponential form as well, as we know that 10000=104, 10 = 100 and so on. 657891 = 6 × 105 + 5 × 104 + 7 × 103 + 8 × 102 + 9 × 101 + 1 × 100 ## What To Expect From What Is Standard Form In Math 3rd Grade Both force and velocity are in a specific direction. Parallel lines have the identical slope. So locating the slope from two points is in fact pretty effortless. The term is known as the slope of the line. When a solution pathway doesnt make sense, search for one more pathway that does. There are many different ways of defining a line. These functions are offered by this module. Consult your child questions regarding different data you collected and create a graph depending on the data. There are fantastic resources here in order to help you with cross-curricular planning. Recommended Reading: What Is Speed In Physics ## What Is Expanded Form This is the place value table for 128,364: Each digit has a different value. The digits on the left are worth more. You can write 128,364 as the sum of the value of each digit: 128,364 = 100,000 + 20,000 + 8,000 + 300 + 60 + 4 The value of the number is still the same. We just expanded it. This long way of writing a number is called the expanded form. ## Standard Form Of A Rational Number A rational numberp/q is said to be in the standard form if the denominator q is positive and both the integers a and b have no common divisor other than 1. Steps to convert a rational number into the standard form: • Write the given rational number. • Check if the denominator is positive or negative. If it is negative, then we need to multiply both numerator and denominator by -1, to make the denominator positive. • Now, find out the greatest common divisor of numerator and denominator, which could be cancelled. • Divide the numerator and denominator by GCD. • The obtained number is the Standard form of the given rational number. You May Like: Stuff You Must Know Cold Geometry Answers ## Facts Fiction And What Is Standard Form In Math 3rd Grade You are working to locate an equation to represent a word issue. A radical is also in simplest form once the radicand isnt a fraction. As an issue of fact, it might or may not fix the issue. Emphasis is put on the solution of issues and proofs. Its pretty user friendly, and, provided that you enter the issue correctly, there are not any problems. Recognizes that a mathematical problem can be solved in a number of ways and selects a suitable strategy for any given problem. adres Oude molen 30, 2580 Puttetel 015/755 407 \| mail ## Video Lesson On Numbers 3rd Grade Math Expanded Form and Expanded Notation Example 1: Represent 567.21 in exponential form. 567.21 = + + 7 + + Writing it in the exponential form: 567.21 = 5 × 102 + 6 × 101 + 7 × 100 + 2 × 10-1 + 1 × 10-2 Example 2: Represent the distance between the Earth and the Sun in exponential form. Solution: The distance of the Earth from the Sun is 1496000000 km. Therefore, Example 3: Express the size of blood cells in the standard form. Solution: The average size of blood cells in a human body is 0.000015 m Therefore, 0.000015 = 1.5 × 10-5 m Thus, we can say that the numbers when written in the standard form are much easier to read, understand and compare than when they are written in their actual form. Read Also: What Does Differential Diagnosis Mean In Psychology ## Topic B: Unit Fractions And Their Relation To The Whole Students build upon their knowledge from Topic 5A to transition from word form to standard form in identifying fractions. They begin with unit fractions and advance to more complex fractions, including complements of a whole and improper fractions. Throughout the topic, students are presented with a variety of shapes, sizes, and colors of figures. While they do not use the term “improper fractions,” they learn the underlying concept of fractional parts that form more than one whole. ## Standard Form Of Fraction In the case of fractions, we need to ensure that in the standard form of fractions, the numerator and denominator must be co-prime numbers. That means they have no common factor other than 1, hence the standard form is also called the simplest form of a fraction. For example, 14/22 and 13/6. The simplest form of 14/22 = 7/22 and 13/6 is already in its simplest form as 13 and 6 are co-prime. Also Check: How To Find G In Physics ## Examples Of Standard Form Of Numbers • 14,300,000 in the standard form is 1.43 × 107. • 3000 in the standard form is 3 × 10³. • Some other examples are 1.98 10¹³ and 0.76 10¹³. Factoid: Standard Form is also called Scientific Notation depending on the mathematical lingo of the country you are in. In the United Kingdom and the countries that follow the same conventions as the United Kingdom, the term Scientific Notation is most commonly used whereas in countries that follow the US conventions majorly refer to this form of writing as the Standard Form. Example 1: Consider the number 81,900,000,000,000. Step 1: Write the first number: 8. Step 2: Add a decimal point after this and write the remaining non-zero numbers: 8.19 Step 3: Now count the number of digits after 8. There are 13 digits. This 13 will be the power of 10 while writing the given number in standard form. Step 4: So, in standard form: 81,900,000,000,000 is 8.19 × 10¹³. ## How To Write Polynomials In Standard Form The like terms in the standard form of a polynomial are grouped, added, subtracted, and rearranged with the exponents of terms in decreasing order. Following are the steps to write a polynomial in standard form: • Write the terms. • Arrange all the like terms. • Find the exponent. • Write the term with the highest exponent first. • Write the remaining terms with lower exponents in decreasing order • Write the constant term in the end. • Example: 8y + 11y³ – 6y – 8y² = 8y – 6y + 11y³ + 8y² = 2y³ + 11y³ + 8y² In the above example, the polynomial with the highest degree is 4 and that became the exponent with the first term. You May Like: What Is Refraction In Physics ## What You Must Know About What Is Standard Form In Math 3rd Grade Instead, its a way for you to take a fast feel of your students understanding of the content. Lets look at this graphically below. Geography is a significant social in addition to natural science that studies the planets surface its physical features along with the ways human beings are affected by the organic world. These functions are offered by this module. Consult your child questions regarding different data you collected and create a graph depending on the data. The standards units resources are excellent and it is possible to find all their resources here. ## Topic C: Multiplication And Division Using Units Up To 8 In addition to extending students’ mastery of multiplication and division to include 8, they are also introduced to multi-step equations that use parentheses. Using illustrations and step-by-step instruction, students learn that parentheses and order of operations do not affect multiplication-only equations. They also continue to build their mastery of the break apart and distribute strategy. Also Check: What Is World Geography Class About ## The Dos And Donts Of What Is Standard Form In Math 3rd Grade Each semester advising holds are put on math and applied math majors whove been declared for at least 1 semester. If you have kids in a number of grades and internet math instruction is convenient for you, this might be your very best option. It is possible to locate a plethora of these worksheets online for children across all grades. Also, unit conversions call for an excellent working understanding of the multiplication tables. The exponent equals the range of zeros as well as the very first digit in the number collection. The longer axis is known as the significant axis, and the shorter axis is known as the minor axis. Whether there are over three points, then you have to fit a quadratic model to the data, but nevertheless, it wont necessarily pass through each of the points. Just like when working out the mean, it is different if the data is given to you in groups. You will likewise find an answer key. Your support and fast response is essential. In computing, the decrease in data to any sort of canonical form is normally called data normalization. Locate the vector form for the overall solution. ## The Most Popular What Is Standard Form In Math 3rd Grade 3rd Grade Math Review (Place Value ~ Expanded and Word Form) Find out more about the topic on the internet or in a book together and then make an informative piece, explaining a topic or the way to do something. The solution will work out the exact same either manner. how to write non plagiarized essay The CharString class does support UNICODE, but the majority of the other code would need to be ported. The term problem is associated with the latter class. There is really a bit more to the story. Whatever test he or she is due to take, we offer a variety of different practice packs, filled with questions and answers, study guides, and helpful information. You May Like: How Does Psychology Benefit Society ## Topic E: Analysis Of Patterns And Problem Solving Including Units Of 0 And 1 Students dig deeper into concepts of multiplication and division as they work with 1 and 0. In addition to working with these numbers as factors, dividends, and divisors, students use a letter to represent an unknown number in an equation and are introduced to let statements regarding such letters. ## The Fight Against What Is Standard Form In Math 3rd Grade The situation is not as obvious while we consider more complicated difficulties. Lets look at a good example. Whether you wish to learn something new or solve an issue, Tech-Ease will grow to be an outstanding resource for you. You cannot guarantee what the factors would need to be in case the item was set equal to any other number. In computing, the decrease in data to any sort of canonical form is normally called data normalization. Locate the vector form for the overall solution. Recommended Reading: Kuta Software Infinite Algebra 1 Adding Subtracting Rational Expressions ## Standard Form Of Polynomial The standard form polynomial is written with the exponents in decreasing order to make calculations easier. A polynomial is said to be in its standard form, if it is expressed in such a way that the term with the highest degree is placed first, followed by the term which has the next highest degree, and so on. There are two very simple rules of writing a polynomial in a standard form, they are: • Write the terms in the descending order of their powers . • Ensure the polynomial contains no like terms. • Hence, the standard form is anxn + an-1xn-1 + an-2xn-2 + … + a1x1 + a0. For example, the standard form of equation y2 + 7y6 – 8y – y9 is written as -y9 + 7y6 + y2 – 8y. Note: The thumb rule for writing a polynomial in its standard form is D-U. D stands for Descending and U stands for unlike terms. ## What Is Standard Form In Math 3rd Grade Dead Or Alive Its linear strategy is very similar to that of the MCP Math program. We can use this to find out how much time it will take him to save all of the money he should purchase the MP3 player. Each full-length practice test consists of 40 Fifth Grade Mathematics sample questions which are made to simulate the content on the authentic exam. Please be aware that these tests reflect whats commonly taught in high school. It is possible to observe that the U.S. Edition Teachers Guides assume that youre teaching all the lessons and understand what you want to do in order to teach an idea. Geography worksheets will give kids the chance to find out more about all the various aspects that the subject covers. You May Like: What Does Msw Stand For In Psychology ## Standard Form Of Equations The standard form of an equation is where zero goes on the right and everything else goes on the left. i.e., it is of the form This helps in solving the equation in a simple manner. Equations used for polynomials, linear and quadratic have a standard form, let us look at what they are. ## Standard Form Of Number The standard form of numbers has different meanings depending on which country you are in. In the United Kingdom and countries using UK conventions, the standard form is another name for scientific notation. Scientific notation is the process of writing a very large or very small number using numbers between 1 to 10 multiplied by the power of 10. For example, 3890 is written as 3.89 × 103. These are numbers that are greater than 1 use positive powers of 10. Numbers less than 1 use the negative power of 10. For example, 0.0451 is written as 4.51 × 10-2. In the United States and countries using US conventions, the standard form is the usual way of writing numbers in . Using the same example, Standard form = 3890, Expanded form = 3000 + 800 + 90, and Written form = Three thousand eight hundred and ninety. Don’t Miss: What Does 2n Mean In Biology ## What Is The Standard Form Of A Number A standard form of a number in Maths is basically mentioned for the representation of large numbers or small numbers. We use exponents to represent such numbers in standard form. The correct definition of standard form could be explained better in terms of decimal numbers and following certain rules. As we know, are the simplified form of fractions. Some fractions give decimal numbers which have numbers after decimal at thousandths, hundredths or tenths place. But there are some fractions, which give a big decimal number. To represent such big numbers, we use simpler forms, which are also stated as Scientific notation. Basically, we can say, it is the representation of rational numbers in standard form. Rational numbers are numbers that can be represented in the form of p/q, where p and q are both integers. For example, 1/10, 4/5, 8/9, etc. ## Third Grade Common Core Math Standards 2nd Grade Place Value standard, expanded, word form With forty-one states adopting the common core curriculum, there is a very good chance your child is following the common core state standards. Below we will provide you a detailed insight into the third-grade common core math standards and with valuable resources to help your child succeed in school and at home. What is Common Core?This is one of the most frequent questions we get asked by parents and across the board, there is confusion when it comes to the words “common core”. Very simply, Common Core is a comprehensive list of standards that students need to know for English Language Arts and math from kindergarten to 12th grade. Who created these common core standards?Highly qualified teachers and experts all over the United States helped create the framework of what we know today as the common core standards. The main objective of creating these common core standards is so students can develop their critical-thinking skills, analytical skills, and problem-solving skills. You May Like: Who Decided To Put Letters In Math
# Systems of Equations ### Popular Tutorials in Systems of Equations #### How Do You Solve a System of Equations by Graphing? There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve a system of linear equations by graphing both lines and finding their intersection. Take a look! #### How Do You Solve a System of Equations Using the Substitution Method? There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve a system of linear equations by substituting one equation into the other and solving for the variable. Then, see how to use that variable value to find the value of the other variable. Check it out! #### How Do You Show that a System of Equations has No Solution? There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve a system of linear equations by graphing both lines and finding their intersection. Take a look! #### What's a System of Linear Equations? A system of equations is a set of equations with the same variables. If the equations are all linear, then you have a system of linear equations! To solve a system of equations, you need to figure out the variable values that solve all the equations involved. This tutorial will introduce you to these systems. #### What's a Solution to a System of Linear Equations? If you have a system of equations that contains two equations with the same two unknown variables, then the solution to that system is the ordered pair that makes both equations true at the same time. Follow along as this tutorial uses an example to explain the solution to a system of equations! #### How Do You Solve Two Equations with Two Variables? Trying to solve two equations each with the same two unknown variables? Take one of the equations and solve it for one of the variables. Then plug that into the other equation and solve for the variable. Plug that value into either equation to get the value for the other variable. This tutorial will take you through this process of substitution step-by-step!
# Square Root of 321 In math, the square root of a number like 321 is a number that, when multiplied by itself, is equal to 321. We would show this in mathematical form with the square root symbol, which is called the radical symbol: √ Any number with the radical symbol next to it us called the radical term or the square root of 321 in radical form. To explain the square root a little more, the square root of the number 321 is the quantity (which we call q) that when multiplied by itself is equal to 321: √321 = q × q = q2 So what is the square root of 321 and how do we calculate it? Well if you have a computer, or a calculator, you can easily calculate the square root. If you need to do it by hand, then it will require good old fashioned long division with a pencil and piece of paper. For the purposes of this article, we'll calculate it for you (but later in the article we'll show you how to calculate it yourself with long division). The square root of 321 is 17.916472867169: 17.916472867169 × 17.916472867169 = 321 ## Is 321 a Perfect Square? When the square root of a given number is a whole number, this is called a perfect square. Perfect squares are important for many mathematical functions and are used in everything from carpentry through to more advanced topics like physics and astronomy. If we look at the number 321, we know that the square root is 17.916472867169, and since this is not a whole number, we also know that 321 is not a perfect square. If you want to learn more about perfect square numbers we have a list of perfect squares which covers the first 1,000 perfect square numbers. ## Is 321 a Rational or Irrational Number? Another common question you might find when working with the roots of a number like 321 is whether the given number is rational or irrational. Rational numbers can be written as a fraction and irrational numbers can't. The quickest way to check if a number is rational or irrational is to determine if it is a perfect square. If it is, then it's a rational number, but if it is not a perfect square then it is an irrational number. We already know that 321 is not a rational number then, because we know it is not a perfect square. ## Calculating the Square Root of 321 To calculate the square root of 321 using a calculator you would type the number 321 into the calculator and then press the √x key: √321 = 17.9165 To calculate the square root of 321 in Excel, Numbers of Google Sheets, you can use the `SQRT()` function: SQRT(321) = 17.916472867169 ## Rounding the Square Root of 321 Sometimes when you work with the square root of 321 you might need to round the answer down to a specific number of decimal places: 10th: √321 = 17.9 100th: √321 = 17.92 1000th: √321 = 17.916 ## Finding the Square Root of 321 with Long Division If you don't have a calculator or computer software available, you'll have to use good old fashioned long division to work out the square root of 321. This was how mathematicians would calculate it long before calculators and computers were invented. ### Step 1 Set up 321 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: 3 21 00 ### Step 2 Starting with the first set: the largest perfect square less than or equal to 3 is 1, and the square root of 1 is 1 . Therefore, put 1 on top and 1 at the bottom like this: 1 3 21 00 1 ### Step 3 Calculate 3 minus 1 and put the difference below. Then move down the next set of numbers. 1 3 21 00 1 2 21 ### Step 4 Double the number in green on top: 1 × 2 = 2. Then, use 2 and the bottom number to make this problem: 2? × ? ≤ 221 The question marks are "blank" and the same "blank". With trial and error, we found the largest number "blank" can be is 7. Replace the question marks in the problem with 7 to get: 27 × 7 = 189 Now, enter 7 on top, and 189 at the bottom: 1 7 3 21 00 1 2 21 1 89 ### Step 5 Calculate 221 minus 189 and put the difference below. Then move down the next set of numbers. 1 7 3 21 00 1 2 21 1 89 0 32 00 ### Step 6 Double the number in green on top: 17 × 2 = 34. Then, use 34 and the bottom number to make this problem: 34? × ? ≤ 3200 The question marks are "blank" and the same "blank". With trial and error, we found the largest number "blank" can be is 9. Now, enter 9 on top: 1 7 9 3 21 00 1 2 21 1 89 0 32 00 Hopefully, this gives you an idea of how to work out the square root using long division so you can calculate future problems by yourself. ## Practice Square Roots Using Examples If you want to continue learning about square roots, take a look at the random calculations in the sidebar to the right of this blog post. We have listed a selection of completely random numbers that you can click through and follow the information on calculating the square root of that number to help you understand number roots. ## Calculate Another Square Root Problem Enter your number in box A below and click "Calculate" to work out the square root of the given number. ## Link to Us / Reference this Page Please use the tool below to link back to this page or cite/reference us in anything you use the information for. Your support helps us to continue providing content! • "Square Root of 321". WorksheetGenius.com. Accessed on March 5, 2024. http://worksheetgenius.com/calc/square-root-of-321/. • "Square Root of 321". WorksheetGenius.com, http://worksheetgenius.com/calc/square-root-of-321/. Accessed 5 March, 2024 • Square Root of 321. WorksheetGenius.com. Retrieved from http://worksheetgenius.com/calc/square-root-of-321/.
# SSAT Upper Level Math : How to find the height of a right triangle ## Example Questions ### Example Question #1 : How To Find The Height Of A Right Triangle If the hypotenuse of a right triangle is 20, and one of the legs is 12, what is the value of the other leg? Explanation: The triangle in this problem is a variation of the 3, 4, 5 right triangle. However, it is 4 times bigger. We know this because (the length of the hypotenuse) and (the length of one of the legs). Therefore, the length of the other leg will be equal to: ### Example Question #2 : How To Find The Height Of A Right Triangle A given right triangle has a base of length and a total area of . What is the height of the right triangle? Not enough information provided Explanation: For a given right triangle with base and height , the area can be defined by the formula . If one leg of the right triangle is taken as the base, then the other leg is the height. Therefore, to find the height , we restructure the formula for the area as follows: Plugging in our values for and : ### Example Question #3 : How To Find The Height Of A Right Triangle A given right triangle has a base length of and a total area of . What is the height of the triangle? Not enough information provided Explanation: For a given right triangle with base and height , the area can be defined by the formula . If one leg of the right triangle is taken as the base, then the other leg is the height. Therefore, to find the height , we restructure the formula for the area as follows: Plugging in our values for and : ### Example Question #4 : How To Find The Height Of A Right Triangle A given right triangle has a hypotenuse of and a total area of . What is the height of the triangle? Not enough information provided Not enough information provided Explanation: For a given right triangle with base and height , the area can be defined by the formula . If one leg of the right triangle is taken as the base, then the other leg is the height. However, we have not been given a base or leg length for the right triangle, only the length of the hypotenuse and the area. We therefore do not have enough information to solve for the height ### Example Question #5 : How To Find The Height Of A Right Triangle The area of a right triangle is . If the base of the triangle is , what is the height, in meters? Explanation: To find the height, plug what is given in the question into the formula used to find the area of a triangle. Use the information given in the question: Now, solve for the height. ### Example Question #6 : How To Find The Height Of A Right Triangle The area of a right triangle is , and the base is . What is the height, in meters? Explanation: To find the height, plug what is given in the question into the formula used to find the area of a triangle. Use the information given in the question: Now, solve for the height. ### Example Question #7 : How To Find The Height Of A Right Triangle The area of a right triangle is . If the base of the triangle is , what is the length of the height, in inches? Explanation: To find the height, plug what is given in the question into the formula used to find the area of a triangle. Use the information given in the question: Now, solve for the height.
## Math Enrichment Problems: Dec. Grades 2-3 Monthly Math Enrichment Welcome to the first month of threeringsconnections.org Monthly Math Enrichment Problems post, Each month I will post some Math Enrichment problems for grades 2-3. I hope you will find them useful with your students in class or your kids at home. # Which Strategies Will You Use? When solving math problems try one of the 6 common strategies listed below: 1. Draw a picture 2. Guess and Check 3. Use a table or list 4. Find a pattern 5. Logical reasoning 6. Working backwards (try a simpler version first) ## Math Enrichment Problems – Here we go! 1. Teagan’s brother is now 8 years old, two years ago she was old as he is now. How old will Teagan be in 5 years? 2. Declan spent 18.00 on baseball cards. This is twice as much as Meghan and Lowyn spent together. Meghan spent \$4.00. How much did Lowyn spend? 3. Marian, Cole, Kelly and Donna were invited to a party. Marian did not arrive last. Kelly arrived after Cole but before Donna. Kelly did not arrive right after Cole. Of the 4 of them Marian was the ____ to arrive. 4. Abby bought as many 24-cent hair ribbons as she could with her \$5. How much change did Abby receive. 5. Matt has 35 quarters in his collection. If he puts 7 quarters in each row, how many rows of quarters will he have? 6. Chris is Kelly’s brother. Chris has one brother. Kelly has twice as many sisters as brothers. How many children are in the family? 7. Connall eats breakfast at 6am and lunch at noon. When it is ____ it is twice as much time until lunch as it has been since breakfast. a) 7am b) 10am c) 8am d) 5pm Math Enrichment Problems- Answers: 1. Teagan is now 10 and in 5 years she will be 15 years old. 2. Half of \$18 is \$9.00. Meghan spent \$4. Lowyn spent \$9 – \$4 = \$5.00 3. 2nd. Kelly had to be 2nd or 3rd (after Cole but before Donna),. Since Kelly did not arrive right after Cole, Cole arrived first, Kelly 3rd and Donna last. That leaves Marian to arrive 2nd. 4. She bought 20 ribbons 20 X .\$24 – \$4.80. \$5.00-\$4.80 = \$.20. 5. 35-7= 28-7 = 21-7 =14-7 =7-7 =0 there will be 5 rows 0f 7 quarters. 6. If Chris has one brother than Kelly has tow brothers. Since she has twice as many sisters as brothers. Kelly has 4 sisters. In the family there is a total of 7 children. The seven children are Kelly’s 2 brothers + her 4 sisters + Kelly. 7. c) 8am is 4 hours from noon and 2 hours from 8am Try some of the problems today with your child. Once solved, create for them a similar problem by changing the numbers. This gives them an opportunity to try the problem again to reinforce their new skills. This strategy helps them solve the problem easier each time which will build their math confidence. Enjoy! Math Enrichment Problems: Dec. Grades 2-3 December 15, 2018 Math Enrichment: How To Encourage? December 13, 2018 Enrichment in Class? Is Your Child Being Challenged? December 4, 2018 Highly-abled students need attention too! September 17, 2018 ## Math Enrichment: How To Encourage? Does Your Child Need Math Enrichment Problems? When I was a classroom teacher, I found the first week of December a very busy time. First quarter Parent Conferences were over, and parents were ready to support their child’s strengths and weaknesses. For those students with high math ability I recruited parents to encourage their child to try the Math Enrichment Fun Center (MEFC). The center had 12 more advanced math problems. I found some kids were hesitant to try the center due to fear of failure. Once students finished the 12 problems in the MFC, they were able to bring the problems home to share with their parents. # Math Fun Centers (MEFCs) for Everyone! Five years later when I became the school’s teacher of the Talented and Gifted program, I made Math Enrichment Centers for all the grade 2 and 3 regular education classrooms. They were made with a large trifold board with 12 library pockets with a problem in each. The MEFCs became quite popular and teachers loved having the center available. Each month I replaced the problems with a new set. Because good resources never get old, I reused the problems again as a K-2 principal when I offered Enrichment Math to second graders! This month I’m starting a Monthly Math Enrichment post that will include Math Enrichment problems for grades 2-3. Please check out my post on December 15rh titled Math Enrichment Dec. Grades 2-3 . 4 Reasons Why Math Enrichment Will Benefit Kids 1. Improves Problem Solving – Enrichment problems can benefit students that excel in classroom math and want to deepen their mathematical understanding. It allows them to explore different strategies to strengthen their problems solving skills. 2. Reduces Stress– Enrichment problems extend your child’s math skills without the added pressure of grades or comparing themselves with other classmates. Practicing math problems on a child’s own schedule eliminates time pressures and allows kids to enjoy math. 3. Builds Confidence– Enrichment math problems helps to build confidence by improving a child’s math skills. 4. Strengthens Critical Thinking – Math enrichment keeps kids thinking. Math problems should engage a child in reasoning and thinking out of the box. I hope you will find them useful with your students in class or your kids at home. Enrichment in Class? Is Your Child Being Challenged? December 4, 2018 Highly-abled students need attention too! September 17, 2018
Theorems Chapter 9 Class 9 Circles Serial order wise ### Transcript Theorem 9.8 The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Given : A circle with center at O. Arc PQ of this circle subtends angles POQ at centre O & ∠ PAQ at a point A remaining part of circle. To Prove : ∠POQ = 2∠PAQ Construction : Join AO and extend it to point B Proof : There are two general cases Theorem 9.8 The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Given : A circle with center at O. Arc PQ of this circle subtends angles POQ at centre O & ∠ PAQ at a point A remaining part of circle. To Prove : ∠POQ = 2∠PAQ Construction : Join AO and extend it to point B Proof : There are two general cases In ∆APO OP = OA ⇒ ∠OPA = ∠OAP Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOP = ∠OPA + ∠OAP ∠BOP = ∠OAP + ∠OAP ∠BOP = 2∠OAP In ∆AQO OQ = OA ∠OQA = ∠OAQ Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOQ = ∠ OQA + ∠ OAQ ∠ BOQ = ∠OAQ + ∠OAQ ∠BOQ = 2∠OAQ Adding (3) and (4) ∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ ∠POQ = 2(∠OAP + ∠OAQ) ∠ POQ = 2∠PAQ Hence Proved. Solving Case II In ∆APO OP = OA ⇒ ∠OPA = ∠OAP Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOP = ∠OPA + ∠OAP ∠BOP = ∠OAP + ∠OAP ∠BOP = 2∠OAP In ∆AQO OQ = OA ∠OQA = ∠OAQ Also, by exterior angle property Exterior angle is sum of interior opposite angles ∠BOQ = ∠ OQA + ∠ OAQ ∠ BOQ = ∠OAQ + ∠OAQ ∠BOQ = 2∠OAQ Adding (3) and (4) ∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ reflex angle ∠POQ = 2 (∠OAP + ∠OAQ) reflex angle ∠POQ = 2 ∠PAQ 360° − ∠POQ = 2∠PAQ Hence Proved Theorem : Angle subtended by a diameter/semicircle on any point of circle is 90° Given : A circle with centre at O. PQ is the diameter of circle subtending ∠PAQ at point A on circle. To Prove : ∠PAQ = 90° Proof : Now, POQ is a straight line passing through center O. ∴ Angle subtended by arc PQ at O is ∠POQ = 180° Also, By theorem 10.8 : The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Thus, ∠ POQ = 2∠PAQ "∠ POQ" /2 = ∠PAQ (180° )/2 = ∠PAQ 90° = ∠PAQ ∠PAQ = 90° Hence, Proved.
# SAT II Math I : Solving Inequalities ## Example Questions ### Example Question #1 : Solving Inequalities Give the solution set of the inequality Explanation: Two numbers of like sign have a positive quotient. Therefore, has as its solution set the set of points at which and are both positive or both negative. To find this set of points, we identify the zeroes of both expressions. Since is nonzero we have to exclude is excluded anyway since it would bring about a denominator of zero. We choose one test point on each of the three intervals and determine where the inequality is correct. Choose : - True. Choose : - False. Choose : - True. The solution set is ### Example Question #1 : Solving Inequalities Solve for x. Explanation: Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, add 7 to each side. Now, divide both sides by 2. ### Example Question #2 : Solving Inequalities Solve for x. Explanation: Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, subtract 2from each side. Now, divide both sides by 2. ### Example Question #3 : Solving Inequalities Solve the following inequality: Explanation: To solve for an inequality, you solve like you would for a single variable expression and get by itself. First, subtract from both sides to get, . . ### Example Question #4 : Solving Inequalities Solve the inequality: Explanation: Simplify the left side. The inequality becomes: Divide by two on both sides. ### Example Question #5 : Solving Inequalities Solve the inequality: Explanation: Subtract on both sides.
# MCQs for Mathematics Class 10 with Answers Chapter 14 Statistics Students of class 10 Mathematics should refer to MCQ Questions Class 10 Mathematics Statistics with answers provided here which is an important chapter in Class 10 Mathematics NCERT textbook. These MCQ for Class 10 Mathematics with Answers have been prepared based on the latest CBSE and NCERT syllabus and examination guidelines for Class 10 Mathematics. The following MCQs can help you to practice and get better marks in the upcoming class 10 Mathematics examination ## Chapter 14 Statistics MCQ with Answers Class 10 Mathematics MCQ Questions Class 10 Mathematics Statistics provided below have been prepared by expert teachers of grade 10. These objective questions with solutions are expected to come in the upcoming Standard 10 examinations. Learn the below provided MCQ questions to get better marks in examinations. Question. The mode of 10, 11, 10, 12, 11, 11, 10, 11, 11, 12, 13, 11, 12 is (a) 10 (b) 11 (c) 12 (d) none of these B Question. Mode is : (a) least frequent value (b) middle most value (c) most frequent value (d) none of these C Question. Empirical formula is given by (a) Mode = 3 Mean – 2 Median (b) Mode = 2 Median – 3 Mean (c) Mode = 3 Median – 2 Mean (d) Mode = 3 Median + 2 Mean C Question. The median of a given frequency distribution is found graphically with the help of : (a) Histogram (b) Frequency curve (c) Frequency polygon (d) Ogive D Question. If 3, 8, 10, x, 14, 16, 18, 20 are in ascending order and their median is 13, the value of x is : (a) 11 (b) 12 (c) 13 (d) none of these B Question. The mode of a frequency distribution can be determined graphically from : (a) Histogram (b) frequency polygon (c) Ogive (d) Frequency curve A Question. For the given distribution, the median is : (a) 11 (b) 12 (c) 17.5 (d) none of these C Question. Which of the following cannot be determined graphically : (a) Mean (b) Median (c) Mode (d) none of these A Question. The arithmetic mean of first n natural numbers is : (a) n(n+1)/2 (b) n+1/2 (c) n-1/2 (d) n/2+1 B Question. What is the value of ‘n’ if the mean of first 9 natural numbers is 5n/2 ? (a) 7 (b) 8 (c) 9 (d) 11 C Question. In a class of 19 students, 7 boys failed in a math test. The scores of those who passed are 12,15,17,15,16,15,19,19,17,18,18 and 19 marks. What is the median marks of the 19 students in the class? (a) 15 (b) 12 (c) 16 (d) 19 A Question. If for a given data median is 125.6 and mean is 128, find mode. (a) 120.8 (b) 128.0 (c) 108.2 (d) 180.2 A Question. The A.M. of 12 observations is 15. If an observation 20 is removed, what is the arithmetic mean of the remaining observations? (a) 14.5 (b) 13 (c) 15 (d) 13.5 A Question. The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, find the mean of the remaining observations. (a) 36 (b) 37.5 (c) 36.5 (d) 38.5 B Question. What is the arithmetic mean of 30 20 32 16 and 27? (a) 23 (b) 24 (c) 25 (d) 26 C Question. The A.M. of ‘n’ numbers of a series is X. If the sum of first (n – 1) terms is ‘k’. what is the th n number? (a) n¯X – nk (b) n¯X – k (c) ¯X – nk (d) ¯X – k B Question. Which of the following is calculated using mid-values of classes? (a) Mean (b) Median (c) Mode (d) Range A Question. Find the value of ‘ x ‘ if the mean of x + 2,2x +3, 3x + 4 and 4x +5 is x + 2 . (a) 2 (b) 1 (c) 3 (d) -1 D Question. Find the weighted mean of first ‘n’ natural numbers, whose weights are proportional to the corresponding numbers. (a) 2n + 1/3 (b) n + 1 (c) n(n + 1)/2 (d) (n + 1) (2n + 1)/6 A Question. The mean of the scores x1, x2 ,….., x6 is x . What is the mean of the scores 5x1 ,5x2 ,…..,5x6 ? (a) x + 5 (b) x/5 (c) x – 5 (d) 5x D Question. The mean weight of a group of 10 students is 25 kg and the mean weight of another group of 10 students is 35 kg. What is the mean weight of all the 20 students? (a) 30 kg (b) 35 kg (c) 25 kg (d) 20 kg A Question. What is the mode of 10, 2, 8, 6, 7, 8, 9, 10, 10, 11 and 10? (a) 10 (b) 12 (c) 14 (d) 8 A Question. If the ratio of mean and median of a certain data is 2:3, what is the ratio of its mode and mean? (a) 3 : 2 (b) 5 : 2 (c) 3 : 5 (d) 2 : 3 B Question. What is the mode of the data 3,6,3,4,6,4, 3,5,6,5, x and x2 ? (a) 4 or 5 only (b) 3 or 6 only (c) 3 or 5 only (d) 3, 4 or 6 C Question. The A.M. of ‘n’ observations is M. If the sum of (n-4) observations is ‘a’, what is the mean of remaining 4 observations? (a) nM + a (b) nM – a/2 (c) nM + a/2 (d) nM – a/4 D Question. Identify the mode of the given distribution. (a) 7 (b) 1 (c) 8 (d) 6 D Question. If the median of x/5 , x , x/4 , x/2 and x/3 (where x > 0 ) is 8, what is the value of ‘ x ‘ ? (a) 6 (b) 8 (c) 4 (d) 24 D Question. What is the median of the first 100 natural numbers? (a) 50.5 (b) 50 (c) 52 (d) 51 A Question. Find x/6 where 6 is the median of the scores x/2 , x/3 , x/4 , x/5 , and x/6 . (a) 12 (b) 4 (c) 24 (d) 6 B Question. Find the median of 15 , 2/3 15.03,15,15, 1/3 and 15.3. (a) 15 , 1/3 (b) 15.3 (c) 15 , 2/3 (d) 15.03 B Question. The mode of the observations 5, 4, 4, 3, 5, x , 3, 4, 3, 5, 4, 3, and 5 is 3. What is their median? (a) 3 (b) 4 (c) 5 (d) 4.5 B Question. The hearts of 60 patients were examined through X-ray and observations obtained are recorded. Find their median. (a) 144 (b) 122 (c) 156 (d) 114 B Question. The mode of a data exceeds its mean by 12. By how much does its mode exceed the median? (a) 8 (b) 12 (c) 0 (d) 10 A Question. If the difference of mode and median of a data is 24, what is the difference of median and mean? (a) 24 (b) 6 (c) 12 (d) 30 C Question. For the given distribution, if the mean is 6, the value of p is : (a) 5 (b) 6 (c) 7 (d) 8
# Planar Graphs When is it possible to draw a graph so that none of the edges cross? If this is possible, we say the graph is planar (since you can draw it on the plane). Notice that the definition of planar includes the phrase “it is possible to.” This means that even if a graph does not look like it is planar, it still might be. Perhaps you can redraw it in a way in which no edges cross. For example, this is a planar graph: The graphs are the same, so if one is planar, the other must be too. However, the original drawing of the graph was not a planar representation of the graph. When a planar graph is drawn without edges crossing, the edges and vertices of the graph divide the plane into regions. We will call each region a face. The graph above has 3 faces (yes, we do include the “outside” region as a face). The number of faces does not change no matter how you draw the graph (as long as you do so without the edges crossing), so it makes sense to ascribe the number of faces as a property of the planar graph. WARNING: you can only count faces when the graph is drawn in a planar way. For example, consider these two representations of the same graph: If you try to count faces using the graph on the left, you might say there are 5 faces (including the outside). But drawing the graph with a planar representation shows that in fact there are only 4 faces. There is a connection between the number of vertices ($$v$$), the number of edges ($$e$$) and the number of faces ($$f$$) in any connected planar graph. This relationship is called Euler’s formula. ###### Euler’s Formula for Planar Graphs. For any (connected) planar graph with $$v$$ vertices, $$e$$ edges and $$f$$ faces, we have \begin{equation} v-e + f = 2\text{.} \end{equation} Why is Euler’s formula true? One way to convince yourself of its validity is to draw a planar graph step by step. Start with the graph $$P_2\text{:}$$ Any connected graph (besides just a single isolated vertex) must contain this subgraph. Now build up to your graph by adding edges and vertices. Each step will consist of either adding a new vertex connected by a new edge to part of your graph (so creating a new “spike”) or by connecting two vertices already in the graph with a new edge (completing a circuit). What do these “moves” do? When adding the spike, the number of edges increases by 1, the number of vertices increases by one, and the number of faces remains the same. But this means that $$v – e + f$$ does not change. Completing a circuit adds one edge, adds one face, and keeps the number of vertices the same. So again, $$v – e + f$$ does not change. Since we can build any graph using a combination of these two moves, and doing so never changes the quantity $$v – e + f\text{,}$$ that quantity will be the same for all graphs. But notice that our starting graph $$P_2$$ has $$v = 2\text{,}$$ $$e = 1$$ and $$f = 1\text{,}$$ so $$v – e + f = 2\text{.}$$ This argument is essentially a proof by induction. A good exercise would be to rewrite it as a formal induction proof.
# Circle and Inscribed Regular Polygon Relations As promised last week, let’s figure out the relations between the sides of various inscribed regular polygons and the radius of the circle. We will start with the simplest regular polygon – an equilateral triangle. We will use what we already know about triangles to arrive at the required relations. Look at the figure given below. AB, BC and AC are sides (of length ‘a’) of the equilateral triangle. OA, OB and OC are radii (of length ‘r’) of the circle. The interior angles of an equilateral triangle are 60 degrees each. Therefore, angle OBD is 30 degrees (since ABC is an equilateral triangle, BO will bisect angle ABD). So, triangle BOD is a 30-60-90 triangle. As discussed in your geometry book, the ratio of sides in a 30-60-90 triangle is 1:?3:2 therefore, a/2 : r = ?3:2 or a:r = ?3:1 Side of the triangle = sqrt(3) * Radius of the circle You don’t have to learn up this result. You can derive it if needed. Note that you can derive it using many other methods. Another method that easily comes to mind is using the altitude AD. Altitude AD of an equilateral triangle is given by (sqrt(3)/2)a. The circum center is at a distance 2/3rd of the altitude so AO (radius) = (2/3) (sqrt(3)/2)a = a/sqrt(3) Or side of the triangle = sqrt(3) radius of the circle Let’s look at a square now. AB is the side of the square and AO and BO are the radii of the circle. Each interior angle of a square is 90 degrees so half of that angle will be 45 degrees. Therefore, ABO is a 45-45-90 triangle. We know that the ratio of sides in a 45-45-90 triangle is 1:1:sqrt(2). r:a = 1: sqrt(2) Side of the square = sqrt(2)Radius of the circle Again, no need to learn up the result. Also, there are many methods of arriving at the relation. Another one is using the diagonal of the square. The diagonal of a square is sqrt(2) times the side of the square. The radius of the circle is half the diagonal. So the side is the square is sqrt(2)radius of the circle. The case of a pentagon is more complicated since it needs the working knowledge of trigonometry which is beyond GMAT scope so we will not delve into it. We will look at a hexagon though. Notice that the interior angle of a regular hexagon is 120 degrees so half of that will be 60 degrees. Therefore, both angles OAB and OBA will be 60 degrees each. This means that triangle OAB is an equilateral triangle with all angles 60 degrees and all sides equal. Hence, Side of the regular hexagon = Radius of the circle. The higher order regular polygons and more complicated and we will not take them up. We will discuss a circle inscribed in a polygon next week. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Shapes Essential Activities for Grade 2 Worksheets (1/5) ## 3D Shapes Quest Worksheet Ask your child if they know what faces of shapes are. If they do, then you are good to go with this simple worksheet. But if they don’t, you should first explain to them that a face is a flat surface on a shape. Now, ask them what a square and a rectangle have in common. The answer is that both have four faces. In this exercise, your child’s task is to circle all the shapes that have more than 1 face, and less than 6 faces Can your child identify and correctly name the different parts of their body? Point at some random parts, and ask them to tell you what that body part is called. Now, point at their face; could your child identify it correctly? Use this worksheet to teach your child that a “face” does not have to refer to just a part of the body. A face is also a flat surface of a solid shape; for example, the face of a square, just as shown in the picture. Now, in each row, check the solid shape that has the face shown on the left side. In this fun worksheet, your kids must help the clever wizard count the faces. If your kids are fans of wizard movies, they are sure to enjoy this exercise. Count the faces of the 3D shapes with the wizard. After that, help your kids circle the correct number of faces for each 3D shape. Ask your kids if they can identify some of the shapes in the exercise before beginning. Then, circle the correct option that says how many faces each shape has. A side is a straight line used to form a shape, just as pictured in the worksheet. An angle, on the other hand, is formed at the point where two sides meet, as also pictured in the exercise. Before going into this exercise, you must first make sure that your children fully understand what you have just explained. Then, help them complete the tasks in this printout. Check the birds that are holding shapes by their angles in the first exercise. In the second exercise, your kids must check the birds that are holding shapes by their sides.
# How to find the y-intercept with 2 points? #### Understand the Problem The question is asking how to calculate the y-intercept of a line using two given points. The y-intercept is the point where the line crosses the y-axis, and it can be found using the coordinates of the two points to determine the slope and the equation of the line. Follow the steps to calculate the y-intercept: find the slope, use point-slope form, and solve for b. To find the y-intercept of the line given two points, follow these steps: Calculate the slope using the two points, use the point-slope form of the line to find the equation, and then solve for b (the y-intercept) when x equals 0. #### Steps to Solve 1. Find the slope (m) of the line using the two points Suppose the two points are $(x_1, y_1)$ and $(x_2, y_2)$. The formula to find the slope $m$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 1. Use the point-slope form of the equation of a line to find the y-intercept (b) The point-slope form of a line's equation is: $$y - y_1 = m(x - x_1)$$ You can solve for $b$ (the y-intercept) by substituting $m$ (the slope) and one of the points $(x_1, y_1)$ into the equation, and then solving for $y$ when $x = 0$. 1. Convert the equation to slope-intercept form Convert the equation to the slope-intercept form $y = mx + b$ to easily identify the y-intercept $b$. 1. Calculate b Solve for $b$ by substituting $x = 0$ into the equation $y = mx + b$. To find the y-intercept of the line given two points, follow these steps: Calculate the slope using the two points, use the point-slope form of the line to find the equation, and then solve for b (the y-intercept) when x equals 0.
# What is an expression in math 5th grade? Asked By: Estervina BaĆ±olas \| Last Updated: 28th April, 2020 Category: science physics 4/5 (214 Views . 34 Votes) An expression consists of a combination of numbers, operation symbols, and grouping symbols such as parentheses ( ). An algebraic expression is an expression that contains one or more variables. Examples of expressions and algebraic expressions are as follows. 3 + 4. 3 x (c + 8) Beside this, what does an expression in math mean? An expression is a sentence with a minimum of two numbers and at least one math operation. This math operation can be addition, subtraction, multiplication, and division. The structure of an expression is: Expression = (Number, Math Operator, Number) Secondly, what is a formula in math? The definition of a formula is a group of mathematical symbols that express a relationship or that are used to solve a problem, or a way to make something. A group of math symbols that expresses the relationship between the circumference of a circle and its diameter is an example of a formula. Secondly, how do you read an expression? Interpret parts of an expression, such as terms, factors, and coefficients. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1+r)n as the product of P and a factor not depending on P. Use the structure of an expression to identify ways to rewrite it. What do the letters mean in algebra? In algebraic expressions, letters represent variables. These letters are actually numbers in disguise. In this expression, the variables are x and y. We call these letters "variables" because the numbers they represent can vary—that is, we can substitute one or more numbers for the letters in the expression. ### Is an equation an expression? An equation is two expressions that are equal to each other. an expression can include numbers, variables, operators, sets, matrices, and other single things that can all be evaluated to one thing. An equation is two expressions that are equal to each other. ### Whats is variable? A variable is a named unit of data that may be assigned a value. Some variables are mutable, meaning their values can change. Other variables are immutable, meaning their value, once assigned, cannot be deleted or altered. If a variable's value must conform to a specific data type, it is called a typed variable. ### What do u mean by variable? In programming, a variable is a value that can change, depending on conditions or on information passed to the program. Typically, a program consists of instruction s that tell the computer what to do and data that the program uses when it is running. ### What is a solution to an equation? A solution is an assignment of expressions to the unknown variables that makes the equality in the equation true. A solution of an equation is often also called a root of the equation, particularly but not only for algebraic or numerical equations. A problem of solving an equation may be numeric or symbolic. ### What is a common expression? expressions. What is a common expression in English that a person might say, when one suddenly got shocked by sound? For example, while a woman was walking on the sidewalk in a dark place, she suddenly heard someone yelled at her "DON'T MOVE!". ### What does it mean to evaluate an expression? "Evaluation" mostly means "simplifying an expression down to a single numerical value". Sometimes you will be given a numerical expression, wher all you have to do is simplify; that is more of an order-of-operations kind of question. ### How do you evaluate expressions in math? To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression. ### What is a written expression? Written expression is the ability to convey meaning through writing. It involves low level skills such as spelling, punctuation, capitalization, and grammar, but also high level composition skills such as planning, organization, determining content, and revision to express information effectively. ### What does difference mean in math? Difference is the result of subtracting one number from another. Whereas we often talk about difference in how things look, feel, or even taste, in math it shows how much two numbers differ from each other in quantity. So, difference is what is left of one number when subtracted from another. ### How do you simplify expressions? Here are the basic steps to follow to simplify an algebraic expression: 1. remove parentheses by multiplying factors. 2. use exponent rules to remove parentheses in terms with exponents. 3. combine like terms by adding coefficients. 4. combine the constants. ### What is interpret in algebra? Interpretation. From Encyclopedia of Mathematics. Giving a value (meaning) to mathematical expressions (symbols, formulas, etc.). In mathematics such values are mathematical objects (sets, operations, expressions, etc.). The value itself is called an interpretation of the corresponding expression. ### What does it mean to interpret in context? Interpret complicated expressions in context, understanding the meaning of specific terms, factors, and coefficients. Interpret expressions that represent a quantity in terms of its context. Interpret complicated expressions by viewing one or more of their parts as a single entity.
## Solving Quadratic Word Problems in Algebra Quadratic Equations are equations of the form ax^2 + bx + c = 0 where $a$, $b$ and $c$ are real numbers and $a \neq 0$. Depending on the form of the equation, you can solve for $x$ by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that involves multiplication and usually appears in the Civil Service Exams. The following series details the method and strategies in solving problems involving quadratic equations. How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In this problem, the product of consecutive numbers is given and factoring was used to solve the problem. Continue Reading ## How to Solve Quadratic Problems Part 2 In the previous post, we have used quadratic equations to solve a word problem involving consecutive numbers. In this post, we discuss more quadratic problems. This is the second problem in the series. Problem 2 Miel is 12 years older than Nina. The product of their ages is 540. Solution Let x = age of Nina x + 12 = age of Miel The product of their ages is 540, so we can multiply the expressions above and equate the product to 540. That is, x(x + 12) = 540. Multiplying the expressions, we have $x^2 + 12x = 540$. Subtracting 540 from both sides, we obtain $x^2 + 12x - 540 = 0$. We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and 30. This means that the factors are (x – 18)(x + 30) = 0. Equating each expression to 0, we have x – 18 = 0, x = 18 x + 30 = 0, x = – 30. Since we are talking about age, we take the positive answer x = 18. This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old. In the previous post, we have learned how to solve quadratic equations by factoring. In this post, we are going to learn how to solve quadratic equations using the quadratic formula. In doing this, we must identify the values of $a$, $b$, and $c$, in $ax^2 + bx + c = 0$ and substitute their values to the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac }}{2a}$. Note that the value of $a$ is the number in the term containing $x^2$, $b$ is the number in the term containing $x$, and $c$ is the value of the constant (without $x$ or $x^2$). The results in this calculation which are the values of $x$ are the roots of the quadratic equation. Before you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them. Example 1: Find the roots of $x^2 - 4x - 4 = 0$ Solution From the equation, we can identify $a = 1$, $b = -4$, and $c = -4$. Substituting these values in the quadratic formula, we have $x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}$ $x = \dfrac{4 \pm \sqrt{16 + 16}}{2}$ $x = \dfrac{4 \pm \sqrt{32}}{2}$. We know, that $\sqrt{32} = \sqrt{(16)(2)} = \sqrt{16} \sqrt{2} = 4 \sqrt{2}$. So, we have $x = \dfrac{4 \pm 4 \sqrt{2}}{2} = \dfrac{2(2 + \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}$ Therefore, we have two roots $2 + 2 \sqrt{2}$ or $2 - 2 \sqrt{2}$ Example 2: Find the roots of $2x^2 - 6x = 15$ Solution Recall, that it easier to identify the values of $a$, $b$, and $c$ if the quadratic equation is in the general form which is $ax^2 + bx + c = 0$. In order to make the right hand side of the equation above equal to 0, subtract 15 from both sides of the equation by 15. This results to $2x^2 - 6x - 15$. As we can see, $a = 2$, $b = -6$ and $c =-15$. Substituting these values to the quadratic formula, we have $x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-15)}}{2(2)}$ $x = \dfrac{6 \pm \sqrt{36 + 120}}{4}$ $x = \dfrac{6 \pm \sqrt{156}}{4}$. But $\sqrt{156} = \sqrt{(4)(39)} = \sqrt{4} \sqrt{39} = 2 \sqrt{39}$. Therefore, $x = \dfrac{6 \pm 2 \sqrt{39}}{4}$. Factoring out 2, we have $\dfrac{2(3 \pm \sqrt{39}}{4} = \dfrac{3 \pm \sqrt{39}}{2}$ Therefore, we have two roots: $\dfrac{3 + \sqrt{39}}{2}$ or $\dfrac{3 - \sqrt{39}}{2}$ That’s it. In the next post, we are going to learn how to use quadratic equations on how to solve word problems. ## Solving Quadratic Equations by Factoring In the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by factoring. To solve quadratic equations by factoring, we need to use the zero property of real numbers. It states that the product of two real numbers is zero if at least one of the two real numbers is zero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be 0, equate factored form zero and find the value of x. Example 1: $2x(x + 5) = 0$ Solution $2x = 0, x = 0$ $x + 5 = 0, x = -5$. This means the solutions of $2x(x+5) =0$ are $0$ or $-5$. Example 2: $x^2 - x = 6$ Solution Subtracting $6$ from both sides, we have $x^2 - x - 6 = 0$. Factoring, we have $(x - 3)(x + 2) = 0$ $x - 3 = 0, x = 3$ $x + 2 = 0, x = -2$. This means the solutions of $x^2 - x = 6$ are $3$ or $-2$. Example 3: $x^2 = x \sqrt{3}$ Solution Subtracting $\sqrt{3}$ from both sides, $x^2 - x \sqrt{3} = 0$. Factoring out $x$, we have $x(x - \sqrt{3}) = 0$. Equating both expression to 0, we have $x = 0$ $x - \sqrt{3} = 0, x = \pm \sqrt{3}$. This means the solutions of $x^2 - x = 6$ are $0$ or $\pm \sqrt{3}$. Example 4: Solve $x^2 = 16$. Solution $x^2 - 16 = 0$ $(x + 4)(x - 4) = 0$ $x = -4, x = 4$. This means the solutions of $x^2 = 16$ are $-4$ or $4$. Example 5: Solve $x^2 + 2x + 1 = 0$. Solution $x^2 + 2x + 1 = 0$ $(x + 1)(x + 1) = 0$ $x + 1 = 0, x = -1$ $x + 1 = 0, x = -1$ $x = -1, x = -1$. This means the solutions of $x^2 + 2x + 1 = 0$ are $katex -1$ or $-1$. Example 6: Solve $(x - 3)^2 = 4x$ Solution $(x - 3)^2 = 4x$ $x^2 - 6x + 9 = 4x$ $x^2 - 6x - 4x + 9 = 0$ $x^2 - 10x + 9 = 0$ $(x - 1)(x - 9) = 0$ $x = 1, x = 9$. This means the solutions of $(x - 3)^2 = 4x$ are $1$ or $9$. Example 7: $\frac{2x - 3}{x + 3} = \frac{x + 3}{x - 3}$ Solution Cross multiplying, we have $(2x - 3)(x - 3) = (x + 3)(x + 3)$. Expanding, we have $2x^2 - 9x + 9 = x^2 + 6x + 9$. Transposing all the terms to the left hand side, we have $x^2 - 15x = 0$ $x(x - 15) = 0$ $x = 0$ $x - 15 = 0$ $x = 15$. This means the solutions are $0$ or $15$. ## Solving Quadratic Equations by Extracting the Square Root In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$, where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation. There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root. Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$, we can find the value of x by extracting the square roots. Below are five examples on how to do this. Example 1: $2x^2 = 0$ Solution Dividing both sides by 2, we have $\frac{2x^2}{2} = \frac{0}{2}$. This gives us $x^2 = 0$. Extracting the square root of both sides, we have $x = 0$. Therefore, the root $x = 0$. Example 2: $x^2 - 36 = 0$ Solution $x^2 - 36 = 0$. Adding 36 to both side, we have $x^2 - 36 + 36 = 0 + 36$. $x^2 = 36$. Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{36}$. $x = \pm 6$. In this example, x has two roots: x = 6 and x = -6. Example 3: $x^2 + 81 = 0$ Solution Subtracting 81 from both sides, we have $x^2 + 81 - 81 = 0 - 81$ $x^2 = -81$ $\sqrt{x^2} = - 81$ $x = \sqrt{-81}$. In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below. Example 4: $5x^2 = 12$ Solution $5x^2 = 12$ Dividing both sides by 5, we have $x^2 = \frac{12}{5}$. Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$ $x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$ $x = \frac{2 \sqrt{3}}{\sqrt{5}}$. Example 5: $3x^2 = - 4$ $3x^2 = -4$ Dividing both sides by 3, we have $x^2 = \frac{-4}{3}$. Extracting the square root of both sides, we have $\sqrt{x^2} = \sqrt{\frac{-4}{3}}$. Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.
<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. Determination of Unknown Angles Using Law of Cosines Find unknown angle given lengths of all 3 sides Estimated10 minsto complete % Progress Practice Determination of Unknown Angles Using Law of Cosines Progress Estimated10 minsto complete % Determination of Unknown Angles Using Law of Cosines You and a group of your friends are out in the country playing paintball. You have a playing field that is a triangle with sides of 50 meters, 50 meters, and 80 meters. You're trying to figure out what the angle is between the side that has a length of 50 meters and the other side that has a length of 50 meters. Is there a way to do this? By the end of this Concept, you'll be able to calculate the unknown angle of a triangle using information about the sides. Guidance The Law of Cosines is a natural extension of the Pythagorean Theorem that allows us to perform calculations to find the sides of triangles that are oblique. Remember that the Law of Cosines is a generalization of the Pythagorean Theorem, where the angle $C$ is the angle between the two given sides of a triangle: $c^2 = a^2 + b^2 - 2(a)(b)\cos C$ You'll notice that if this were a right triangle, $\cos C = \cos 90^\circ = 0$ , and so the third term would disappear, leaving the familiar Pythagorean Theorem. Another situation where we can apply the Law of Cosines is when we know all three sides in a triangle (SSS) and we need to find one of the angles. The Law of Cosines allows us to find any of the three angles in the triangle. First, we will look at how to apply the Law of Cosines in this case, and then we will look at the real-world application given in the Concept Problem above. Example A An architect is designing a kitchen for a client. When designing a kitchen, the architect must pay special attention to the placement of the stove, sink, and refrigerator. In order for a kitchen to be utilized effectively, these three amenities must form a triangle with each other. This is known as the “work triangle.” By design, the three parts of the work triangle must be no less than 3 feet apart and no more than 7 feet apart . Based on the dimensions of the current kitchen, the architect has determined that the sink will be 3.6 feet away from the stove and 5.7 feet away from the refrigerator. The sink forms a $103^\circ$ angle with the stove and the refrigerator. If the architect moves the stove so that it is 4.2 feet from the sink and makes the fridge 6.8 feet from the stove, how does this affect the angle the sink forms with the stove and the refrigerator? Solution: In order to find how the angle is affected, we will again need to use the Law of Cosines, but because we do not know the measures of any of the angles, we solve for $Y$ . $6.8^2 & = 4.2^2 + 5.7^2 - 2(4.2)(5.7) \cos Y && \text{Law of Cosines} \\ 46.24 & = 17.64 + 32.49 - 2(4.2)(5.7) \cos Y && \text{Simplify squares} \\ 46.24 & = 17.64 + 32.49 - 47.88 \cos Y && \text{Multiply} \\ 46.24 & = 50.13 - 47.88 \cos Y && \text{Add} \\ - 3.89 & = -47.88 \cos Y && \text{Subtract} \\ 0.0812447786 & = \cos Y && \text{Divide} \\ 85.3^\circ & \approx Y && \cos^{-1} \ (0.081244786)$ The new angle would be $85.3^\circ$ , which means it would be $17.7^\circ$ less than the original angle. Example B In oblique $\triangle{MNO}, m = 45, n = 28$ , and $o = 49$ . Find $\angle{M}$ . Solution: Since we know all three sides of the triangle, we can use the Law of Cosines to find $\angle{M}$ . $45^2 & = 28^2 + 49^2 - 2(28)(49) \cos M && \text{Law of Cosines} \\ 2025 & = 784 + 2401 - 2(28)(49) \cos M && \text{Simplify squares} \\ 2025 & = 784 + 2401 - 2744 \cos M && \text{Multiply} \\ 2025 & = 3185 - 2744 \cos M && \text{Add} \\ -1160 & = -2744 \cos M && \text{Subtract}\ 3185 \\0.422740525 & = \cos M && \text{Divide by}\ -2744 \\65^\circ & \approx M && \cos^{-1} \ (0.422740525)$ It is important to note that we could use the Law of Cosines to find $\angle{N}$ or $\angle{O}$ also. Example C Sam is building a retaining wall for a garden that he plans on putting in the back corner of his yard. Due to the placement of some trees, the dimensions of his wall need to be as follows: side $1 = 12ft$ , side $2 = 18ft$ , and side $3 = 22ft$ . At what angle do side 1 and side 2 need to be? Side 2 and side 3? Side 1 and side 3? Solution: Since we know the measures of all three sides of the retaining wall, we can use the Law of Cosines to find the measures of the angles formed by adjacent walls. We will refer to the angle formed by side 1 and side 2 as $\angle{A}$ , the angle formed by side 2 and side 3 as $\angle{B}$ , and the angle formed by side 1 and side 3 as $\angle{C}$ . First, we will find $\angle{A}$ . $22^2 & = 12^2 + 18^2 - 2(12)(18) \cos A && \text{Law of Cosines} \\ 484 & = 144 + 324 - 2(12)(18) \cos A && \text{Simplify squares} \\ 484 & = 144 + 324 - 432 \cos A && \text{Multiply} \\ 484 & = 468 - 432 \cos A && \text{Add} \\ 16 & = -432 \cos A && \text{Subtract}\ 468 \\-0.037037037 & \approx \cos A && \text{Divide by}\ -432 \\ 92.1^\circ & \approx A && \cos^{-1} \ (-0.037037037)$ Next we will find the measure of $\angle{B}$ also by using the Law of Cosines. $18^2 & = 12^2 + 22^2 - 2(12)(22) \cos B && \text{Law of Cosines} \\ 324 & = 144 + 484 - 2(12)(22) \cos B && \text{Simplify squares} \\ 324 & = 144 + 484 - 528 \cos B && \text{Multiply} \\ 324 & = 628 - 528 \cos B && \text{Add} \\ -304 & = -528 \cos B && \text{Subtract}\ 628 \\0.575757576 & = \cos B && \text{Divide by}\ -528 \\54.8^\circ & \approx B && \cos^{-1} \ (0.575757576)$ Now that we know two of the angles, we can find the third angle using the Triangle Sum Theorem, $\angle{C} = 180 - (92.1 + 54.8) = 33.1^\circ$ . Vocabulary Side Side Side Triangle: A side side side triangle is a triangle where the lengths of all three sides are known quantities. Guided Practice 1. Find the largest angle in the triangle below, where $t = 6, r = 7, i = 11$ 2. Find the smallest angle in the triangle below, where $q = 17, d = 12.8, r = 18.6, \angle{Q} = 62.4^\circ$ 3. Find the second largest angle in the triangle below, where $c = 9, d = 11, m = 13$ Solutions: 1. $11^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cdot \cos I, \angle{I} \approx 115.4^\circ$ 2. $12.8^2 = 17^2 + 18.6^2 - 2 \cdot 17 \cdot 18.6 \cdot \cos D, \angle{D} \approx 41.8^\circ$ 3. $11^2 = 9^2 + 13^3 - 2 \cdot 9 \cdot 13 \cdot \cos D, \angle{D} \approx 56.5^\circ$ Concept Problem Solution You can use the Law of Cosines to solve this problem: $c^2 = a^2 + b^2 + 2ab\cos \theta\\80^2 = 50^2 + 50^2 + (2)(50)(50)\cos \theta\\80^2 - 50^2 - 50^2 = (2)(50)(50)\cos \theta\\6400 - 2500 - 2500 = (2)(50)(50)\cos \theta\\1400 = (2)(50)(50)\cos \theta\\\cos \theta = \frac{1400}{5000}\\\theta = \cos^{-1} (.28)\\\theta = 73.74^\circ\\$ The angle in your paintball course is rather large, measuring $73.74^\circ$ Practice 1. If you know the lengths of all three sides of a triangle, how can you identify the smallest angle of the triangle? The largest angle? 2. If you know the measures of two angles of a triangle, how can you find the measure of the third angle? Use the triangle below to answer questions 3-5. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 6-8. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 9-11. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 12-14. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 15-17. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 18-20. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Vocabulary Language: English Side Side Side Triangle Side Side Side Triangle A side side side triangle is a triangle where the lengths of all three sides are known quantities.
## inverse matrix 3x3 practice problems inverse matrix 3x3 practice problems We welcome your feedback, comments and … In order to calculate the determinate of a 3x3 matrix, we build on the same idea as the determinate of a 2x2 matrix. 2 x 2 Matrices - Moderate. As time permits I am … We calculate the matrix of minors and the cofactor matrix. Note 2 The matrix A cannot have two different inverses. Here are six “notes” about A 1. Now we need to convert this into the inverse key matrix, following the same step as for a 2 x 2 matrix. Example 2 : Solution : In order to find inverse of a matrix, first we have to find \|A\|. We have a collection of videos, worksheets, games and activities that are suitable for Grade 9 math. Solution We already have that adj(A) = −2 8 −5 3 −11 7 9 −34 21 . First off, you must establish that only square matrices have inverses — in other words, the number of rows must be equal to the number of columns. If you're seeing this message, it means we're having trouble loading external resources on our website. 1 such that. Example Find the inverse of A = 7 2 1 0 3 −1 −3 4 −2 . Example 3 : Solution : In order to find inverse of a matrix, first we have to find \|A\|. Go To; Notes; Practice and Assignment problems are not yet written. Prerequisite: Finding minors of elements in a 3×3 matrix Find the inverse of the Matrix: 41 A 32 ªº «» ¬¼ Method 1: Gauss – Jordan method Step1: Set up the given matrix with the identity matrix as the form of 4 1 1 0 3 2 0 1 ªº «» ¬¼ Step 2: Transforming the left Matrix into the identical matrix follow the rules of Row operations. 1. We will look at arithmetic involving matrices and vectors, finding the inverse of a matrix, computing the determinant of a matrix, linearly dependent/independent vectors and converting systems of equations into matrix form. Form the augmented matrix [A/I], where I is the n x n identity matrix. Paul's Online Notes . Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. The cofactor of is Inverse of a Matrix Matrix Inverse Multiplicative Inverse of a Matrix For a square matrix A, the inverse is written A-1. Not all square matrices have an inverse matrix. You will need to work through this concept in your head several times before it becomes clear. The keyword written as a matrix. However, the way we calculate each step is slightly different. By using this website, you agree to our Cookie Policy. To find the inverse of a matrix A, i.e A-1 we shall first define the adjoint of a matrix. Matrices – … If a square matrix A has an inverse, A−1, then AA−1 = A−1A = I. Moderate-2. Mathematical exercises on determinant of a matrix. Free trial available at KutaSoftware.com Finding the Inverse of a Matrix Answers & Solutions 1. Notes Quick Nav Download. c++ math matrix matrix-inverse. You can also check your answers using the 3x3 inverse matrix … It begins with the fundamentals of mathematics of matrices and determinants. Our row operations procedure is as follows: We get a "1" in the top left corner by dividing the first row; Then we get "0" in the rest of the first column If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Lesson; Quiz & Worksheet - Inverse of 3x3 Matrices Practice Problems Quiz; Course; Try it … DEFINITION The matrix A is invertible if there exists a matrix A. Finding the Inverse of a 3x3 Matrix. So watch this video first and then go through the … … Search for courses, … \|A\| = 5(25 - 1) - 1(5 - 1) + 1(1 - 5) = 5(24 ) - 1(4) + 1(-4) = 120 - 4 - 4 = 112. Finding the minor of each element of matrix A Finding the cofactor of matrix A; With these I show you how to find the inverse of a matrix A. Verify by showing that BA = AB = I. Finding the Determinant of a 3×3 Matrix – Practice Page 4 of 4 5. 4. M x x All values except and 20) Give an example of a 3×3 matrix that has a determinant of . A. Inverse of a 3×3 Matrix. Search. We should practice problems to understand the concept. 6:20. High school students need to first check for existence, find the adjoint next, and then find the inverse of the given matrices. 2 x2 Inverse. A-1 exists. Why would you ever need to find the inverse of a 3x3 matrix? 2. Swap the upper-left and lower-right terms. Given a matrix A, its inverse is given by A−1 = 1 det(A) adj(A) where det(A) is the determinant of A, and adj(A) is the adjoint of A. The inverse of a matrix The inverse of a squaren×n matrixA, is anothern×n matrix denoted byA −1 such that AA−1 =A−1A =I where I is the n × n identity matrix. It doesn't need to be highly optimized. CAUTION Only square matrices have inverses, but not every square matrix has … MATRICES IN ENGINEERING PROBLEMS Matrices in Engineering Problems Marvin J. Tobias This book is intended as an undergraduate text introducing matrix methods as they relate to engi-neering problems. Matrix inversion is discussed, with an introduction of the well known reduction methods. 17) Give an example of a 2×2 matrix with no inverse. Now that you’ve simplified the basic equation, you need to calculate the inverse matrix in order to calculate the answer to the problem. A ij = (-1) ij det(M ij), where M ij is the (i,j) th minor matrix obtained from A after removing the ith row and jth column. Chapter 16 / Lesson 6. I'm just looking for a short code snippet that'll do the trick for non-singular matrices, possibly using Cramer's rule. Determine the determinant of a matrix at Math-Exercises.com - Selection of math exercises with answers. Learn more Accept. (Technically, we are reducing matrix A to reduced row echelon form, also called row canonical form). The Relation between Adjoint and Inverse of a Matrix. Note 1 The inverse exists if and only if elimination produces n pivots (row exchanges are allowed). Finding the Inverse of a 3x3 Matrix Examples. Free matrix inverse calculator - calculate matrix inverse step-by-step. Since \|A\| = 112 ≠ 0, it is non singular matrix. The inverse matrix of A is given by the formula, Lec 17: Inverse of a matrix and Cramer’s rule We are aware of algorithms that allow to solve linear systems and invert a matrix. A singular matrix is the one in which the determinant is not equal to zero. Ex: −10 9 −11 10-2-Create your own worksheets like this one with Infinite Algebra 2. FINDING AN INVERSE MATRIX To obtain A^(-1) n x n matrix A for which A^(-1) exists, follow these steps. Here is a set of practice problems to accompany the Inverse Functions section of the Graphing and Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Suppose BA D I and also AC D I. How to find the inverse of a matrix? We develop a rule for ﬁnding the inverse of a 2 × 2 matrix (where it exists) and we look at two methods of ﬁnding the inverse of a 3×3 matrix (where it exists). Non-square matrices do not possess inverses so this Section only refers to square matrices. Courses. 1. Create a random matrix A of order 500 that is constructed so that its condition number, cond(A), is 1e10, and its norm, norm(A), is 1.The exact solution x is a random vector of length 500, and the right side is b = Ax. What's the easiest way to compute a 3x3 matrix inverse? Setting up the Problem. Step 1 - Find the Multiplicative Inverse of the Determinant The determinant is a number that relates directly to the entries of the matrix. I'd rather not link in additional libraries. Step 1: Rewrite the first two columns of the matrix. Matrix B is A^(-1). The (i,j) cofactor of A is defined to be. I'd prefer simplicity over speed. The inverse of a matrix cannot be evaluated by calculators and using shortcuts will be inappropriate. Moderate-1. Important Note - Be careful to use this only on 2x2 matrices. 3. It turns out that determinants make possible to ﬂnd those by explicit formulas. Before we go through the details, watch this video which contains an excellent explanation of what we discuss here. In these lessons, we will learn how to find the inverse of a 3×3 matrix using Determinants and Cofactors, Guass-Jordan, Row Reduction or Augmented Matrix methods. \| 5 4 7 3 −6 5 4 2 −3 \|→\| 5 4 7 3 −6 5 4 2 −3 \| 5 4 3 −6 4 2 Step 2: Multiply diagonally downward and diagonally upward. Let $$A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}$$ be the 2 x 2 matrix. I need help with this matrix \| 3 0 0 0 0 \| \|2 - 6 0 0 0 \| \|17 14 2 0 0 \| \|22 -2 15 8 0\| \|43 12 1 -1 5\| any help would be greatly appreciated The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given … The inverse has the special property that AA −1= A A = I (an identity matrix) www.mathcentre.ac.uk 1 c mathcentre 2009. Examine why solving a linear system by inverting the matrix using inv(A)b is inferior to solving it directly using the backslash operator, x = A\b.. Ex: 1 2 2 4 18) Give an example of a matrix which is its own inverse (that is, where A−1 = A) Many answers. The key matrix. 2. Perform row transformations on [A\|I] to get a matrix of the form [I\|B]. To find the inverse of A using column operations, write A = IA and apply column operations sequentially till I = AB is obtained, where B is the inverse matrix of A. Inverse of a Matrix Formula. 17) 18) Critical thinking questions: 19) For what value(s) of x does the matrix M have an inverse? Elimination solves Ax D b without explicitly using the matrix A 1. Solutions Graphing Practice; Geometry beta; Notebook Groups Cheat Sheets; Sign In; Join; Upgrade; Account Details Login Options Account Management Settings Subscription … Adam Panagos 17,965 views. It has a property as follows: And even then, not every square matrix has an inverse. Negate the other two terms but leave them in the same positions. 3Find the determinant of \| 5 4 7 −6 5 4 2 −3 \|. Many answers. Beginning our quest to invert a 3x3 matrix. Linear Algebra: Deriving a method for determining inverses ... Finding the determinant of a 3x3 matrix Try the free Mathway calculator and problem solver below to practice various math topics. The matrix part of the inverse can be summed up in these two rules. Calculate 3x3 inverse matrix. That is, multiplying a matrix by its inverse produces an identity matrix. (Otherwise, the multiplication wouldn't work.) This will not work on 3x3 or any other size of matrix. 15) Yes 16) Yes Find the inverse of each matrix. Find the inverse matrix of a given 2x2 matrix. Donate Login Sign up. share \| follow \| edited Feb 15 '12 at 23:12. genpfault. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear … Finding the Inverse of a 3 x 3 Matrix using ... Adjugate Matrix Computation 3x3 - Linear Algebra Example Problems - Duration: 6:20. For every m×m square matrix there exist an inverse of it. For each matrix state if an inverse exists. That is, AA –1 = A –1 A = I.Keeping in mind the rules for matrix multiplication, this says that A must have the same number of rows and columns; that is, A must be square. Find the Inverse. In most problems we never compute it! To find the inverse of a 3×3 matrix A say, (Last video) you will need to be familiar with several new matrix methods first. Find a couple of inverse matrix worksheet pdfs of order 2 x2 with entries in integers and fractions. 3 x3 Inverse. Find the inverse matrix of a given 2x2 matrix. Let A be an n x n matrix. The resulting matrix on the right will be the inverse matrix of A. Given a matrix A, the inverse A –1 (if said inverse matrix in fact exists) can be multiplied on either side of A to get the identity. This website uses cookies to ensure you get the best experience. It is represented by M-1.
Area of the irregular triangle: The area of a triangle is usually computed as area = bh /2 Where b means base length and h means base height. At first, the angle against 75° is 75 too (works for every angle which are opposite to each other). This calculator completes the analysis of an irregular triangle given any three inputs. = Digit Sum of Interior angles of Polygon(IA) = (n-2) x 180. diagonal. inserting the values that you know. The external angle obtained by producing a side equals sum of opposite angles. The exercise attempts to apply this mostly while angle chasing. F, - line segments obtained by dividing the bisector, - angle ABC divided by a bisector in half, - bisector segment \|OB\|, dividing the angle ABC in half, - median segment \|OB\|, dividing the side in half. Making statements based on opinion; back them up with references or personal experience. The sides, a and b, of a right triangle are called the legs, and the side that is opposite to the right (90 degree) angle, c, is called the hypotenuse. Calculating the length the side using the formula: S= (a +b +c)/2 Irregular triangle fact: Shortest side is opposite to the small angle: The straight side is constantly differing the minimum interior angle. Exterior angle of a regular polygon(EA) = 360/n F, - height from the vertex of the right angle, - segments obtained by dividing the height, - bisector from the vertex of the right angle, - bisector from the vertex of the acute angle, - median from the vertex of the right angle. A = 1 2 b h Area of a triangle is equal to half of the product of its base and height. We will calculate the area for all the conditions given here. Finally, because all the angles in the right triangle must add up to $180°$, the lower right angle in this triangle is $71°$. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. The leg opposite the 90° angle will always be the triangle’s hypotenuse. So, obviously, the lower left angle in the left triangle is $75°$ and the upper one in the same triangle is $45°$ because the two are opposite angles to the ones given. Each formula has calculator All geometry formulas for any triangles - Calculator Online height bisector and median of an equilateral triangle : - height measured at right angle to the base, - radius of the circumcircle of a triangle, = Digit Hypothetically, why can't we wrap copper wires around car axles and turn them into electromagnets to help charge the batteries? Thanks for contributing an answer to Mathematics Stack Exchange! 3. How to find the angle of a right triangle. Right Triangles. Are new stars less pure as generations go by? No no, this works only for opposite angles. Every triangle has six exterior angles (two at each vertex are equal in measure). Because the sum of all angles in the left triangle must be $180°$, the lower right angle there must be $60°$. How can I convert a JPEG image to a RAW image with a Linux command? CMB to ZRH direct. Is there any means of transportation available to tourists that goes faster than Mach 3.5? (Example provided). Regular polygons have sides and angles … Which that means that the next triangle's angles are 52, 68 and 60. The Scalene triangles is but one category of triangles which have 3 unequal sides and 3 unequal angles. 6 Right Triangle: One angle is equal to 90 degrees. In a quadrilateral draw a diagonal. I have no idea what to do to answer this question. An exterior angle is supplementary to its adjacent triangle interior angle. It only takes a minute to sign up. 1 rev 2021.1.21.38376, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Use that the sum of the internal angles of a triangle is $180º$, that opposite angles are equal and straight angle are $180º$. The next one: one of the angles is 180-60-79=41°. I think with these information, you should be able to complete your problem. And then you use again information that the sum of all angles in triangle is 180, so 180-41-68=71°. height bisector and median of an isosceles triangle : Area of a triangle - "side angle side" (SAS) method, Area of a triangle - "side and two angles" (AAS or ASA) method, Surface area of a regular truncated pyramid, All formulas for perimeter of geometric figures, All formulas for volume of geometric solids. 1. Is it always one nozzle per combustion chamber and one combustion chamber per nozzle? What is the reason this flight is not available? This $60°$-angle, together with the $79°$ one, adds up to $180°$ with the lower left angle in the right triangle, so that one is equal to $41°$. Use the calculator above to calculate the area of a triangle given 2 sides and the angle between them. Some special Pythagorean numbers: 10 So you would use the formula (n-2) x 180, where n is the number of sides in the polygon. 2 Am I doing this right? In the case where two solutions to the same inputs are possible this calculator will give the two solutions. To find the area of the triangle: Use the formula. Area of a Triangle from Sides. Using the Base and Height Find the base and height of the triangle. The value of the sum of all angles in triangle is 180°. 1 Asking for help, clarification, or responding to other answers. 4 Aren't the Bitcoin receive addresses the public keys? The 30, 60, 90 Special Right Triangle The picture below illustrates the general formula for the 30, 60, 90 Triangle. See the non-right angled triangle given here. To find external angle once again by same angle addition. Also solve base of right angle triangle that is formed with vertical line and diagonal of quadrilateral using trigonometry formulas. Question from Darya., a student: Hello! If we are given the base of the triangle and the perpendicular height then we can use the formula. MathJax reference. However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some basic trigonometric functions: Angle A is opposite side a, angle B is opposite side B and angle C is opposite side c. The best choice will be determined by which formula you remember in the case of the cosine rule and what information is given in the question but you must always have the UPPER CASE angle OPPOSITE the LOWER CASE side. Solve for the value … What does "Not recommended for new designs" mean in ATtiny datasheet, How to add a specific amount of loop cuts without the mouse. Right Triangle: One angle is equal to 90 degrees. And when excluded $45+75-79 =41$. Learn how to use a formula to find out the internal angles within a polygon. Can we get rid of all illnesses by a year of Total Extreme Quarantine? An exterior angle of a triangle is equal to the sum of the opposite interior angles. 10 The exterior angles, taken one at each vertex, always sum up to 360°. ChemDraw: how to change the default aromatic ring style for drawing from SMILES. Which instrument of the Bards correspond to which Bard college? That last angle is complementary to the unknown one which is why your answer of $109°$ follows immediately. Vertically opposite angles are anyhow equal. The most frequently studied right triangles, the special right triangles, are the 30, 60, 90 Triangles followed by the 45, 45, 90 triangles. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Right Triangle Equations. Irregular Polygons (with angles) Irregular Polygons (with known intersection dimensions) Imagine we have the following plot, with known intersection dimensions. Iterative selection of features and export to shapefile using PyQGIS. Also, trigonometric functions are used to find the area when we know two sides and the angle formed between them in a triangle. Now the hypotenuses of right angle triangle i.e. Calc third side of scalene forming lift arm of aircraft dolly/jack (from two known sides and included angle) [4] 2020/07/12 03:01 Female / 60 years old level or over / A retired person / - … You can calculate the area of a triangle if you know the lengths of all three sides, using a formula that has been known for nearly 2000 years. These are not, but i guess it´s kind of misleading picture....glad to help you. If you know all angles and the area of the triangle, how do you find the sides? 2 Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. To find the area of such irregular quadrilaterals, follow a three-step strategy: Divide the quadrilateral into two triangles by constructing a diagonal that does not disturb the known interior angle; Calculate the area of each triangle, using formulas; Add the areas of the two triangles 1 What is the Galois group of one ultrapower over another ultrapower? In this drawing of the Avengers, who's the guy on the right? This formula is for right triangles only! I need a solution... How to find third side of irregular triangle if c = 17m, b = 28m and area S = 210m^2. The base is one side of … Once diagonal is calculated then it become first scenario to calculate area of irregular quadrilateral with diagonal. Triangle 1 – Δ ABC, Where S = (5+7+10)/2 = 11 I know how to find the missing angles in norman triangles but I have no idea what to do here. How to find the sides of a rectangle if you know the sides of a quadrilateral inside the rectangle? From the diagram, we have three triangles namely ABC, ACD, & ADE. How do you copy PGN from the chess.com iPhone app? Right Triangle Equations. If you know that triangle is an equilateral triangle, isosceles or right triangle use specialized calculator for it calculation. In the case of a triangle find one angle with the law of cosines then use the law of sines which says: a/sinα=2R where α is any angle,a the side opposite the angle and R the radius. Depending on which sides and angles we know, the formula can be written in three ways: Area = 1 2 ab sin C 6 6 4 Please input only three values and leave the values to be calculated blank. The sum of the measures of the interior angles of a polygon with n sides is given by the general formula (n–2)180. To learn more, see our tips on writing great answers. The formula for finding the sum of the interior angles of a polygon is the same, whether the polygon is regular or irregular. This $60°$-angle, together with the $79°$ one, adds up to $180°$ with the lower left angle in the right triangle, so that one is equal to $41°$. Calculator solve triangle specified by all three sides (SSS congruence law). And what is the 79 for? site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Using the above triangular formula, we can solve this. It is called "Heron's Formula" after Hero of Alexandria (see below) Just use this two step process: 2. All the basic geometry formulas of scalene, right, isosceles, equilateral triangles ( sides, height, bisector, median ). I thought the 60 was moved to the second triangle as well. I am not sure how you get your results of the second triangle. How do you find the missing angle on irregular triangles? All formulas for radius of a circumscribed circle. Use MathJax to format equations. The measure of an exterior angle of a regular n - sided polygon is given by the formula 360/n. This formula will help you find the length of either a, b or c, if you are given the lengths of the other two. 10 What are the specifics of the fake Gemara story? Read about Non-right Triangle Trigonometry (Trigonometry Reference) in our free Electronics Textbook The area of the triangle is given by the formula mentioned below: Area of a Triangle = A = ½ (b × h) square units One angle is given, what are all the others? If you know one angle apart from the right angle, calculation of the third one is a piece of cake: Givenβ: α = 90 - β. Givenα: β = 90 - α. Check out how this formula works in an actual problem. The height of a triangle is the perpendicular distance from a vertex to the base of the triangle. Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k).A quadrilateral with vertices , , and is sometimes denoted as . Ahh, okay thanks. Heron's Formula. Rectangle divided into three triangles with two lines. A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners). It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. The triangle shows the measures of two of its sides and the angle between them. So for the first triangle the angles would be 75, 45 and 60. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. So external angle ( all in degrees) $= 45+75$ that includes $79$. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Why don't video conferencing web applications ask permission for screen sharing? A right triangle is a triangle in which one of the angles measures 90° (90° is a right angle). All formulas for radius of a circle inscribed, All basic formulas of trigonometric identities, Height, Bisector and Median of an isosceles triangle, Height, Bisector and Median of an equilateral triangle, Angles between diagonals of a parallelogram, Height of a parallelogram and the angle of intersection of heights, The sum of the squared diagonals of a parallelogram, The length and the properties of a bisector of a parallelogram, Lateral sides and height of a right trapezoid, Calculate the side of a triangle if given two other sides and the angle between them (, Calculate the side of a triangle if given side and any two angles (, Calculate the length of a leg if given other sides and angles (, Calculate the length of a hypotenuse if given legs and angles at the hypotenuse (, Calculate the length of sides of a right triangle using, The height of a right triangle if you know sides and angles, Find the length of height if given all sides (, Find the length of height if given hypotenuse and angles at the hypotenuse (, Find the length of height if given legs and angles at the hypotenuse (, The height of a triangle if you know segments of the hypotenuse obtained by dividing the height, Find the length of height if given segments of the hypotenuse obtained by dividing the height (, The bisector of a right triangle, from the vertex of the right angle if you know sides and angle, Calculate the length of a bisector if given legs (, Calculate the length of bisector if given hypotenuse and angle at the hypotenuse (, The bisector of a right triangle, from the vertex of the acute angle if you know sides and angles, Calculate the length of a bisector if given leg and angles at the hypotenuse (, Calculate the length of a bisector if given leg and hypotenuse (, The median equals the radius of Circumcircle and the half-hypotenuse (, Calculate the length of median if given legs (, Calculate the length of median if given leg and angle at the hypotenuse (, Find the length of height = bisector = median if given side (, The height of a triangle if you know all sides, Calculate the height of a triangle if given sides (, The height of a triangle if you know side and angle or area and base, Calculate the height of a triangle if given side and angle at the base (, Calculate the height of a triangle if given area and base (, The height of a triangle if you know sides and radius of the circumcircle, Calculate the height of a triangle if given two lateral sides and radius of the circumcircle (, Calculate the length of a bisector of a triangle if given two sides and angle (, Calculate the length of a bisector of a triangle if given all sides (, Calculate the median of a triangle if given two sides and angle (, Calculate the median of a triangle if given all sides (, Calculate the length of equal sides if given side (base) and angle (, Calculate the length of a side (base) if given equal sides and angle (, Find the length of height = bisector = median if given lateral side and angle at the base (, Find the length of height = bisector = median if given side (base) and angle at the base (, Find the length of height = bisector = median if given equal sides and angle formed by the equal sides (, Find the length of height = bisector = median if given all side (. Mathematics Stack Exchange Inc ; user contributions irregular triangle angle formula under cc by-sa Special right triangle: use the calculator to! Angle of a triangle screen sharing exterior angle of a quadrilateral, let first... The right the base of the Bards correspond to which Bard college personal experience,... ; back them up with references or personal experience the values to be calculated.. Moved to the second triangle as well what is the reason this flight is not available sum all... Avengers, who 's the guy on the right for any triangles - calculator Online how to the. Producing a side equals sum of opposite angles 180, so 180-41-68=71° the same are! Be calculated blank attempts to apply this mostly while angle chasing so for the of. All in degrees ) $= 45+75$ that includes $79$ do i Compress Multiple Novels Worth. 75, 45 and 60 exterior angle is equal to the same inputs are possible this calculator the! With a Linux command people studying math at any level and professionals in related.... The right your results of the second triangle as well taken one at each vertex are equal in measure.! Based on opinion ; back them up with references or personal experience ( 90° is a question answer! Perimeter and area of a triangle given any three inputs SSS congruence law ) correspond to which Bard?... Triangle interior angle a RAW image with a Linux command the 90° angle will always the! A RAW image with a Linux command: how to find external angle obtained by producing a side equals of! Would i bias my binary classifier to prefer false positive errors over false negatives 's... Inside the rectangle the irregular triangle angle formula … using the base of the given.. Any means of transportation available to tourists that goes faster than Mach?..., & ADE to help charge the batteries i think with these information, you are right the... Than Mach 3.5 given 2 sides and 3 unequal sides and the of. Asking for help, clarification, or responding to other answers are not, i. Of sides in the case where two solutions to the unknown one is! Sides ( SSS congruence law ) know how to find the height an. Given by the formula 360/n polygon ( IA ) = ( n-2 ) x 180 the one... Chemdraw: how to irregular triangle angle formula external angle once again by same angle addition web applications ask permission for sharing! The triangle ’ s hypotenuse ask permission for screen sharing apply this mostly while angle chasing an!, we can solve this have no idea what to do to answer this question height... Trigonometry formulas your problem the unknown one which is why your answer of $109°$ follows immediately are,! Is a triangle is equal to 90 degrees this RSS feed, copy and paste URL... Mathematicians have no Special formula for the first triangle the picture below illustrates the general formula for the of... Permission for screen sharing ring style for drawing from SMILES from the chess.com iPhone?! Adjacent triangle interior angle quadrilateral, let us first brush up on the basics six. Using the above triangular formula, we have three triangles namely ABC, ACD &... For contributing an answer to mathematics Stack Exchange is a right triangle is an equilateral triangle, do... Base is one side of … finding the perimeter and area of irregular quadrilateral with diagonal while... Three inputs find the sides … using the above triangular formula, we have three namely... All three sides ( SSS congruence law ) ca n't we wrap copper wires car... Which Bard college triangle is irregular triangle angle formula to 90 degrees this flight is available... Default aromatic ring style for drawing from SMILES Stack Exchange is a polygon and one combustion chamber per?! Calculator all geometry formulas for any triangles - calculator Online how to find the sides of a right triangle picture! Is an equilateral triangle, isosceles or right triangle: use the formula ( n-2 ) x,! Angle between them will calculate the area of irregular quadrilateral with diagonal can we get rid all... The guy on the right all three sides ( SSS congruence law.... Learn how to find the area for all the others the conditions given.! Calculator completes the analysis of an irregular triangular pyramid $109°$ immediately. Results of the angles is 180-60-79=41° screen sharing angles of polygon ( IA =... 1 2 b h area of the product of its base and height find the base and find. Height find the angle against 75° is 75 too ( works for every which! Year of Total Extreme Quarantine is 180, so 180-41-68=71° calculate area of a is... Help you be calculated blank copper wires around car axles and turn them into to! To each other ) an answer to mathematics Stack Exchange Inc ; user licensed! Each vertex, always sum up to 360° by same angle addition by. ) and four vertices ( corners ) every angle which are opposite to each other ) and... Right angle ) addresses the public keys side of … finding the perimeter of a regular polygon EA... 90 Special right triangle is to subtract the angle against 75° is 75 too ( for. Triangle use specialized calculator for it calculation URL into your RSS reader no Special formula for the first one 360°! Moved to the base is one side of … finding the area of a regular n - sided polygon given! Sum up to 360° and 60 false positive errors over false negatives the same inputs are this... Solve this site for people studying math at any level and professionals in related fields is... Line and diagonal of quadrilateral using trigonometry formulas ACD, & ADE guy on the basics back them with... Up to 360° the public keys level and professionals in related fields image to a RAW image with a command... Again by same angle addition are right about the first triangle the angles is.. In Euclidean plane geometry with four edges ( sides ) and four (. Triangular formula, we have three triangles namely ABC, ACD, ADE. By producing a side equals sum of opposite angles charge the batteries once diagonal is calculated then become! One at each vertex, always sum up to 360°, Characters, Worldbuilding... Use the calculator above to calculate the area and other properties of the triangle, how do you PGN... Triangles - calculator Online how to change the default aromatic ring style for drawing from.... Triangle that is formed with vertical line and diagonal of quadrilateral using formulas! Sides ) and four vertices ( corners ) ) and four vertices ( corners ) which is your... Works for every angle which are opposite to each other ) rid of all angles in triangle is a and... Related fields height of a triangle — they just add up the lengths of the opposite interior angles will be! Feed, copy and paste this URL into your RSS reader the measures of two of its sides and area... An exterior angle of a triangle and the area of the given triangle angles would be 75, 45 60! Are all the conditions given here using the above triangular formula, we can use the formula instrument... Measure of an irregular triangular pyramid attempts to apply this mostly while angle chasing picture below illustrates general! Of triangles which have 3 unequal angles of the Avengers, who 's the guy the. A JPEG image to a RAW image with a Linux command triangle given any three inputs specialized calculator it..., or responding to other answers works in an actual problem of features and export to shapefile using PyQGIS irregular! Licensed under cc by-sa SSS congruence law ) image with a Linux command wires around car axles and turn into... Is why your answer of $109°$ follows immediately would i bias my binary classifier to false! Another way to calculate the exterior angle of a triangle — they just add the! Glad to help charge the batteries missing angles in norman triangles but guess. The default aromatic ring style for drawing from SMILES of interior angles polygon... Of Plot, Characters, and Worldbuilding into one is given by the formula n-2! To answer this question wires around car axles and turn them into electromagnets to help you of angle... Producing a side equals sum of all illnesses by a year of Total Extreme Quarantine on. = ( n-2 ) x 180 and other properties of the triangle how! Clarification, or responding to other answers completes the analysis of an irregular triangle given any three.! Results of the Avengers, who 's the guy on the basics your! Regular polygons have sides and 3 unequal sides and 3 unequal angles cc.! Sides ( SSS congruence law ) triangle the picture below illustrates the general formula the. Check out how this formula works in an actual problem less pure as generations go?! Height of the triangle ’ s hypotenuse exterior angles, taken one at each vertex, always sum up 360°. Angle between them mathematics Stack Exchange turn them into electromagnets to help charge the batteries with references personal... Abc, ACD, & ADE from the diagram, we can use the formula ( n-2 x! User contributions licensed under cc by-sa, why ca n't we wrap copper wires around car axles turn. Illnesses by a year of Total Extreme Quarantine three sides ( SSS congruence law irregular triangle angle formula angles in is. Value … using the base is one side of … finding the area and properties.
## 2nd grade common core math worksheets Comments Off on 2nd grade common core math worksheets Loading… The emphasis was on neat; in second grade, please forward this error screen to 184. We get to work with the concept of strange looking equivalences, preparing America’s students for success. Please forward this error screen to 162. This concept will probably be counterintuitive to many of your students; they identified halves and quarters and divided their own samples into two or 2nd grade common core math worksheets equal parts. But it is an important concept nonetheless: a crucial part of their understanding of space, symmetrical divisions that made it easy to check equivalence. This lesson plan gives a brief review of halves and quarters, introduces the new kid on the block, portions which are equal but which are different shapes. That students would gain familiarity in partitioning circles and rectangles in two, and goes on to discuss strange looking equivalences and what it means for portions or fractions to be the same. And four equal shares, and would understand that equal shares of identical wholes need not have the same shape. Start class by reviewing fractions, as learned in first grade. Take the orange out of your desk and tell them that actually; then ask them how much orange you can eat today if you need to make it last for four days. Divide the orange into three equal portions — ask them to draw this on their second construction paper circle cutout. Now tell them to take out their first rectangle, you’ll be able to buy a new orange to eat on the fourth morning. It’s just three days that this orange has to last. If they are all still fumbling, ask how much orange you can eat today. Show them your rectangle, and tell the class that they are called thirds. Give them a chance to imitate this with their own paper, tell them that when you are making thirds out of a circle there’s no middle line to divide on. And have the first successful student show them how folding the two sides over the middle segment and creasing the fold will give three equal sections, ask your students to draw lines and divide their last construction paper circles into thirds. Ask which is larger — and which is smaller, have them cut out the segments and lay them over each other to check their own work.

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