sum of squares: 32 + 52 = 9 + 25 = 34
square of sum (3 + 5)2 = 82 = 64
This is a version of the Cauchy-Schwarz inequality.
The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. 1. squares, not square roots 2. right triangle, not isosceles 3. sides opposite the hypotenuse, not any two 4. What are the mistakes, not what is
No, because the biggest length (hypotenuse) has to be equal to the square root of the sum of the squares of the other two sides, which it is not
Top row: 3-17-7 Middle row: 13-9-5 Bottom row: 11-1-15
First you would square the length of each of the sides. You'd add the squares of the smaller two numbers. If they are smaller than the square of the biggest side, then the triangle is obtuse. If their sum is bigger, then the triangle is acute. If they are equal, then the triangle is a right triangle.Example:length of sides-23, 34, 49232+342__492529+1156__24011685
3 and 5
3
The question is ambiguous.Does it want the sum of the squares, or the square of the sum ? They're different.Here are both:1). Sum of the squares: . (1)2 + (2)2 + (3)2 + (4)2 + (5)2 = 1 + 4 + 9 + 16 + 25 = 552). Square of the sum: . (1 + 2 + 3 + 4 + 5 )2 = (15)2 = 225
132,344
The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. 1. squares, not square roots 2. right triangle, not isosceles 3. sides opposite the hypotenuse, not any two 4. What are the mistakes, not what is
3 is the smallest integer that cannot be written as the sum of two squares. This is easy to see, since the only squares less than or equal to 3 are 0 and 1.
Most of the time it is False...You can prove this buy example. Using 2 & 3 the sum of the squares is 13 but adding 2 & 3 then squarring it gives a result of 25. Zero and one will work. I'm not sure if there are any others.
The number of squares in an n-by-n square is 1^2 + 2^2 + 3^2 + ... + n^2 This sum is given by the formula n(n + 1)(2n + 1)/6 Jai
No, because the biggest length (hypotenuse) has to be equal to the square root of the sum of the squares of the other two sides, which it is not
Infinitely many, but only 30 squares within a 1 unit grid. 4*4 square: 1 3*3 squares: 4 2*2 squares: 9 1*1 squares: 16
If they are width and length of a rectangle or triangle, where the angle between them is 90°, then it's time to call on our mate Pythagoras. He said that the square of the diagonal = the sum of the squares of the other two sides. 1. Square the width 2. Square the length 3. Add these squares together 4. Find the square root
Top row: 3-17-7 Middle row: 13-9-5 Bottom row: 11-1-15
If a question says solve the sum of the squares of 3 and 10, you would multiply 3 by 3 to get 9 and 10 by 10 to get 100, and add the two numbers to get 109. 32 + 102 = 9 + 100 = 109