squares  
29 results



1, 9, 25, 49, .......... What is next
1, 9, 25, 49, .......... What is next 1^2 = 1 3^2 = 9 5^2 = 25 7^2 = 49 So this pattern takes odd numbers and squares them. Our next odd number is 9: 9^2 = [B]81[/B]

A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter
[SIZE=4]A rectangle is cut in half to create two squares that each has an area of 25. What is the perimeter of the original rectangle? A. 20 B. 25 C. 30 D. 50[/SIZE] [SIZE=4]Area of a square: A = s^2 We're given A = 25: s^2= 25 s = 5 This means the rectangle width is 5. The rectangle length is 2(5) = 10. Perimeter of a rectangle: P = 2l + 2w P = 2(10) + 2(5) P = 20 + 10 P = [B]30 (choice C)[/B] [B][/B] [B][MEDIA=youtube]tKpS1gQY68o[/MEDIA][/B][/SIZE]

All squares are rectangles and all rectangles are parallelograms, therefore all squares are parallel
All squares are rectangles and all rectangles are parallelograms, therefore all squares are parallelograms. Is this true? [B]Yes.[/B] This is similar to A implies B and B implies C so A implies C also known as transitive property

Covariance and Correlation coefficient (r) and Least Squares Method and Exponential Fit
Free Covariance and Correlation coefficient (r) and Least Squares Method and Exponential Fit Calculator - Given two distributions X and Y, this calculates the following:
* Covariance of X and Y denoted Cov(X,Y)
* The correlation coefficient r.
* Using the least squares method, this shows the least squares regression line (Linear Fit) and Confidence Intervals of α and Β (90% - 99%)
Exponential Fit
* Coefficient of Determination r squared r2
* Spearmans rank correlation coefficient
* Wilcoxon Signed Rank test

difference between 2 positive numbers is 3 and the sum of their squares is 117
difference between 2 positive numbers is 3 and the sum of their squares is 117 Declare variables for each of the two numbers: [LIST] [*]Let the first variable be x [*]Let the second variable be y [/LIST] We're given 2 equations: [LIST=1] [*]x - y = 3 [*]x^2 + y^2 = 117 [/LIST] Rewrite equation (1) in terms of x by adding y to each side: [LIST=1] [*]x = y + 3 [*]x^2 + y^2 = 117 [/LIST] Substitute equation (1) into equation (2) for x: (y + 3)^2 + y^2 = 117 Evaluate and simplify: y^2 + 3y + 3y + 9 + y^2 = 117 Combine like terms: 2y^2 + 6y + 9 = 117 Subtract 117 from each side: 2y^2 + 6y + 9 - 117 = 117 - 117 2y^2 + 6y - 108 = 0 This is a quadratic equation: Solve the quadratic equation 2y2+6y-108 = 0 With the standard form of ax2 + bx + c, we have our a, b, and c values: a = 2, b = 6, c = -108 Solve the quadratic equation 2y^2 + 6y - 108 = 0 The quadratic formula is denoted below: y = -b ± sqrt(b^2 - 4ac)/2a [U]Step 1 - calculate negative b:[/U] -b = -(6) -b = -6 [U]Step 2 - calculate the discriminant ?:[/U] ? = b2 - 4ac: ? = 62 - 4 x 2 x -108 ? = 36 - -864 ? = 900 <--- Discriminant Since ? is greater than zero, we can expect two real and unequal roots. [U]Step 3 - take the square root of the discriminant ?:[/U] ?? = ?(900) ?? = 30 [U]Step 4 - find numerator 1 which is -b + the square root of the Discriminant:[/U] Numerator 1 = -b + ?? Numerator 1 = -6 + 30 Numerator 1 = 24 [U]Step 5 - find numerator 2 which is -b - the square root of the Discriminant:[/U] Numerator 2 = -b - ?? Numerator 2 = -6 - 30 Numerator 2 = -36 [U]Step 6 - calculate your denominator which is 2a:[/U] Denominator = 2 * a Denominator = 2 * 2 Denominator = 4 [U]Step 7 - you have everything you need to solve. Find solutions:[/U] Solution 1 = Numerator 1/Denominator Solution 1 = 24/4 Solution 1 = 6 Solution 2 = Numerator 2/Denominator Solution 2 = -36/4 Solution 2 = -9 [U]As a solution set, our answers would be:[/U] (Solution 1, Solution 2) = (6, -9) Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution

Difference of Two Squares
Free Difference of Two Squares Calculator - Factors a difference of squares binomial in the form a2 - b2 or multiplies 2 binomials through in the form (ax + by)(ax - by).

divide the sum of the square of a and b by thrice c
divide the sum of the square of a and b by thrice c Sum of the squares of a and b is found as follows: [LIST] [*]a squared means we raise a to the power of 2: a^2 [*]b squared means we raise b to the power of 2: b^2 [*]Sum of the squares means we add both terms: a^2 + b^2 [*]Thrice c means we multiply c by 3: 3c [/LIST] Divide means we have a quotient: [B](a^2 + b^2)/3c[/B]

Divide the sum of the squares of a and b by the square of c
Divide the sum of the squares of a and b by the square of c square of a: a^2 square of b: b^2 Sum of the squares of a and b: a^2 + b^2 square of c: c^2 Divide the Sum of the squares of a and b by the square of c: [B](a^2 + b^2)/c^2[/B]

Factoring and Root Finding
Free Factoring and Root Finding Calculator - This calculator factors a binomial including all 26 variables (a-z) using the following factoring principles:
* Difference of Squares
* Sum of Cubes
* Difference of Cubes
* Binomial Expansions
* Quadratics
* Factor by Grouping
* Common Term
This calculator also uses the Rational Root Theorem (Rational Zero Theorem) to determine potential roots
* Factors and simplifies Rational Expressions of one fraction
* Determines the number of potential positive and negative roots using Descarte’s Rule of Signs

Find two consecutive integers if the sum of their squares is 1513
Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29

Find two consecutive odd integers such that the sum of their squares is 290
Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Find two consecutive positive integers such that the difference of their square is 25
Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Find two consecutive positive integers such that the sum of their squares is 25
Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25

Football Squares
Free Football Squares Calculator - Generates a Football Squares grid

If all A's are B's, then all B's are A's. Is this true?
If all A's are B's, then all B's are A's. Is this true? [B]No.[/B] Example: All dogs are mammals, but not all mammals are dogs. All squares are rectangles, but not all rectangles are squares.

Jennifer is playing cards with her bestie when she draws a card from a pack of 25 cards numbered fro
Jennifer is playing cards with her bestie when she draws a card from a pack of 25 cards numbered from 1 to 25. What is the probability of drawing a number that is square? The squares from 1 - 25 less than or equal to 25 are as follows: [LIST=1] [*]1^2 = 1 [*]2^2 = 4 [*]3^2 = 9 [*]4^2 = 16 [*]5^2 = 25 [/LIST] So the following 5 cards are squares: {1, 4, 9, 16, 25} Therefore, our probability of drawing a square is: P(square) = Number of Squares / Number of Cards P(square) = 5/25 This fraction can be simplified. So [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F25&frac2=3%2F8&pl=Simplify']we type in 5/25 into our search engine, choose simplify[/URL], and we get: P(square) = [B]1/5[/B]

Lagrange Four Square Theorem (Bachet Conjecture)
Free Lagrange Four Square Theorem (Bachet Conjecture) Calculator - Builds the Lagrange Theorem Notation (Bachet Conjecture) for any natural number using the Sum of four squares.

Prove that the difference between alternate consecutive squares as always even
Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is always odd Since both cases are even, we've proven our statement. [MEDIA=youtube]J_E9lR5qFY0[/MEDIA]

Prove the difference between two consecutive square numbers is always odd
Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have even * odd = Even Add 1 we have Odd Since both cases are odd, we've proven our statement. [MEDIA=youtube]RAi0HbH5bqc[/MEDIA]

ratio of the squares of t and u
ratio of the squares of t and u Ratio is also known as quotient in algebraic expression problems. The square of t means we raise t to the power of 2: t^2 The square of u means we raise u to the power of 2: u^2 ratio of the squares of t and u means we divide t^2 by u^2: [B]t^2/u^2[/B]

Squares
Free Squares Calculator - Solve for Area of a square, Perimeter of a square, side of a square, diagonal of a square.

sum of the squares of u and v
sum of the squares of u and v The square of u means we raise u to the power of 2 u^2 The square of v means we raise v to the power of 2 v^2 The sum means we add v^2 to u^2: [B]u^2 + v^2[/B]

Take a look at the following sums: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 +
Take a look at the following sums: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 a. Come up with a conjecture about the sum when you add the first n odd numbers. For example, when you added the first 5 odd numbers (1 + 3 + 5 + 7 + 9), what did you get? What if wanted to add the first 10 odd numbers? Or 100? b. Can you think of a geometric interpretation of this pattern? If you start with one square and add on three more, what can you make? If you now have 4 squares and add on 5 more, what can you make? c. Is there a similar pattern for adding the first n even numbers? 2 = 2 2 + 4 = 6 2 + 4 + 6 = 12 2 + 4 + 6 + 8 = 20 a. The formula is [B]n^2[/B]. The sum of the first 10 odd numbers is [B]100[/B] seen on our s[URL='http://www.mathcelebrity.com/sumofthefirst.php?num=10&pl=Odd+Numbers']um of the first calculator[/URL] The sum of the first 100 odd numbers is [B]10,000[/B] seen on our [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=100&pl=Odd+Numbers']sum of the first calculator[/URL] b. Geometric is 1, 4, 9 which is our [B]n^2[/B] c. The sum of the first n even numbers is denoted as [B]n(n + 1)[/B] seen here for the [URL='http://www.mathcelebrity.com/sumofthefirst.php?num=+10&pl=Even+Numbers']first 10 numbers[/URL]

The difference between the squares of two consecutive numbers is 141. Find the numbers
The difference between the squares of two consecutive numbers is 141. Find the numbers Take two consecutive numbers: n- 1 and n Given a difference (d) between the squares of two consecutive numbers, the shortcut for this is: 2n - 1 = d Proof of this: n^2- (n - 1)^2 = d n^2 - (n^2 - 2n + 1) = d n^2 - n^2 + 2n - 1 = d 2n - 1 = d Given d = 141, we have 2n - 1 = 141 Add 1 to each side: 2n = 142 Divide each side by 2: 2n/2 = 142/2 n = [B]71[/B] Therefore, n - 1 = [B]70 Our two consecutive numbers are (70, 71)[/B] Check your work 70^2 = 4900 71^2 = 5041 Difference = 5041 - 4900 Difference = 141 [MEDIA=youtube]vZJtZyYWIFQ[/MEDIA]

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered t
The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. Digit, Probability 1, 0.301 2, 0.176 3, 0.125 4, 0.097 5, 0.079 6, 0.067 7, 0.058 8, 0.051 9, 0.046 [B][U]Fradulent Checks[/U][/B] Digit, Frequency 1, 36 2, 32 3, 45 4, 20 5, 24 6, 36 7, 15 8, 16 9, 7 Complete parts (a) and (b). (a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?
Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

the sum of the squares of a and b
the sum of the squares of a and b Square of a means we raise a to the 2nd power: a^2 Square of b means we raise b to the 2nd power: b^2 The sum of squares means we add these terms together to get our algebraic expression: [B]a^2 + b^2[/B]

The sum of the squares of c and d is 25
The sum of the squares of c and d is 25 The square of c means we we raise c to the power of 2: c^2 The square of d means we we raise d to the power of 2: d^2 The sum of the squares of c and d means we add d^2 to c^2: c^2 + d^2 The word [I]is[/I] means equal to, so we set c^2 + d^2 equal to 25: [B]c^2 + d^2 = 25[/B]

The sum of the squares of two consecutive positive integers is 61. Find these two numbers.
The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].

Two numbers have a sum of 20. Determine the lowest possible sum of their squares.
Two numbers have a sum of 20. Determine the lowest possible sum of their squares. If sum of two numbers is 20, let one number be x. Then the other number would be 20 - x. The sum of their squares is: x^2+(20 - x)^2 Expand this and we get: x^2 + 400 - 40x + x^2 Combine like terms: 2x^2 - 40x + 400 Rewrite this: 2(x^2 - 20x + 100 - 100) + 400 2(x - 10)^2 - 200 + 400 2(x?10)^2 + 200 The sum of squares of two numbers is sum of two positive numbers, one of which is a constant of 200. The other number, 2(x - 10)^2, can change according to the value of x. The least value could be 0, when x=10 Therefore, the minimum value of sum of squares of two numbers is 0 + 200 = 200 when x = 10. If x = 10, then the other number is 20 - 10 = 10.