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Q: If the sum of the squares of two consecutive odd positive integers is 202 what are the integers?

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The square root of 202 equals 14.213 or the square root of 202 (only factors of 202 are 202, 101, 2 and 1)

There are infinite pairs of numbers whose product is 202. To find a pair, choose a number and divide 202 by that number to get its mate. Below are some examples of this method: First number------------------Solve for second number---------------Pair 2--------------------------------202/2=101---------------------------------2 and 101 8--------------------------------202/8=25.25------------------------------8 and 25.25 53-------------------------------202/53=3.811----------------------------53 and 3.811 4005----------------------------202/4005=.050--------------------------4005 and .050 .077 ----------------------------202/.077=2623.377---------------------.077 and 2623.377 *All numbers were rounded to the nearest thousanth

f = 202 because 7*5*202 = 7070

6 feet 7.5 inches

4 rectangles. (I'm assuming you mean 1414 congruent squares) Suppose we made a rectangle out of 1414 squares. Then its area is 1414x2, if x is the length of a square's side, and all its sides are multiples of x. Say the length and width are ax and bx, then we have two ways of finding the area, which must be equal: ax*bx=1414x2, or ab=1414. Now we just need to find the number of ways to get 1414 as a product of two whole numbers. 1414 factors as 2*7*101 (which are all prime), so there are 4 ways to split it up: 1414 = 1*1414 = 2*707 = 7 * 202 = 14 * 101. Each of these ways gives a different rectangle made up of 1414 squares.

Related questions

9 and 11 92 = 81 112 = 121 81+121 = 202

Let n be an integer, then two consecutive odd integers can be expressed as n and n+2. The equation can the be wrote as n2 + (n+2)2 = 202 = n2 +n2 + 4n +4 = 202 which implies 2n2 + 4n - 198 = 0 which implies n2 + 2n - 99 = 0 the quadratic formula can then be used to solve for n as n = (-2 +- (4 - 4*-99)^(1/2))/(2) = -1 +- (1/2)*(400)^(1/2) = - 1 +- 10 take the positive solution and you get n = 9 which implies your two consecutive integers are 9 and 11.

202 + 222 = 400 + 484 = 884

101 and 102 Explanation: I'll use x to represent the first integer and x + 1 to represent the second integer (consecutive integers are one right after another). x + x + 1 = 203 2 x = 202 x = 101 The two numbers are 101 and 102.

x + x+1 = 203 2x+1 = 203 2x = 202 x = 101 x+1 = 102

The Hollywood Squares - 1965 1-202 is rated/received certificates of: USA:TV-G

The Hollywood Squares - 1965 1-202 was released on: USA: 27 July 1967

The Hollywood Squares - 1965 2-202 was released on: USA: 11 June 1968

The answer is 58. 56+57+58 = 171 171 is the largest palindrome less than 220 that can be divided by three. Further palindromes are: 181, 191, 202 and 212.

use this formula: S=(N/2)(F+L) S= the sum of the first 100 positive even integers N= the number of terms F= first term L=last term Since we are using the first 100 positive integers we will replace "N" with 100. so it will look like this : S=(100/2)(F+L) Next we must substitute the "F" with the first positive term which is the number 2. So now it looks like this: S=(100/2)(2+L) We use the number 2 because our positive integers are 2,4,6,8,10,12...... you get the picture. Now we must substitute the "L" with the last positive term which is 200. As you probably guessed it will look like this: S=(100/2)(2+200) We use the number 200 because it is the last positive even integer. If we used 202 then that would have meant we used the 101th positive even integer and we don't want to do that. == == We should solve the numbers inside the parenthesis first. S=(100/2)(2+200) 100 divided by 2 = 50 So: S=50(2+200) 2+200=202 That means: S=50(202) 50 x 202 = 10,100 Finally: S=10,100 There you go! Wasn't that easy? Note: This formula only works if you are looking for the sum of the first 100 positive or negative, even or odd integers. It sucks cause this formula is the Bomb right? Oh well

202 x 202 = 40,804

The positive integer factors of 404 are: 1, 2, 4, 101, 202, 404

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