529 square feet = 49.15 square meters.
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Yes because 232 = 529
529
There is more than one perfect square between 100 and 1000! All the perfect squares between those two numbers: 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 and 961.
Suppose the height is 15x cm. Then the width is 8x cm. Then the surface area is 2*(15x*8x + 8x*22 + 22*15x) cm2 Therefore 2*(15x*8x + 8x*22 + 22*15x) = 2506 or 15x*8x + 8x*22 + 22*15x = 1253 120x2 + 506x - 1253 = 0 so that x = -529/30 or 13/4 But x cannot be negative so x = 1.75. Therefore, volume = (15*1.75)*(8*1.75)*22 = 8085 cm3
Find its square root and raise it in the second power.64 = 8281 = 92841 = ?2 you need to work a little bit here1. base would be in two digits (odd number with 3 digits, (3+1)/2)2. the first digit would be 2 (the square of 2 is less than 8), so the base would be between 20 and 30.3. the second digit would be 9 (the square of 9 ends in 1)4. the answer is 29, 841 = 292529 = ?21. base would be in two digits (odd number with 3 digits, (3+1)/2)2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 20 and 30.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 23 or 27?4. 529 - 2232 = 529 - 409 = 120; it is 3, because 2(2*3)=12. So 529 = 2322809 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 5 (the square of 5 is less than 28,), so the base would be between 50 and 60.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 53 or 57?4. 2809 - 5232 = 2809 - 2509 = 300; it is 3, because 2(5*3)=30. So 2809 = 5327056 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 8 (the square of 8 is less than 70,), so the base would be between 80 and 90.3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?4. 7056 - 8242 = 7056 - 6416 = 640; it is 4, because 2(8*4)=64. So 7056 = 8427396 = ?21. base would be in two digits (even number with 4 digits, 4/2)2. the first digit would be 8 (the square of 8 is less than 73), so the base would be between 80 and 90.3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?4. 7396 - 8262 = 7396 - 6436 = 960; it is 6, because 2(8*6)=96. So 7396(if you test 4 first, you'll see that it doesn't work, so it will be 6)55696 = ?21. base would be in three digits (odd number with 5 digits, (5+1)/2)2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 200 and 300.3. the last digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 2_4 or 2_6?4. let's work a little bit here:2_4; 22_42 = 4001640016+ ... = 4569600080 (101*2*4*_, we need to find the middle digit)01600 (102*2*2*4)04000 (103*2*2*_)55696 - 45696 = 10000400160008001600040001000000080 doesn't change, this tell us the middle digit can be 1 or 6. But one is too small and 6 is too big (2*2*1= 4, 2*2*6=32). So the last digit is not 4, it is 6. Let's find the middle digit.2_6; 22_62 = 4003640036+ ... = 4655600120 (101*2*6*_, we need to find the middle digit)02400 (102*2*2*6)04000 (103*2*2*_)55696 - 46556 = 9140400360012002400040000914000120 becomes 00160, and tell us the middle digit can be 3 or 8, but 8 is too big (2*2*8=32), so it is 3Thus, 55696 = 2362.263169 = ?21. base would be in three digits (even number with 6 digits, 6/2)2. the first digit would be 5 (the square of 5 is less than 26), so the base would be between 500 and 600.3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 5_3 or 5_7?4. let's work a little bit here:5_3; 52_32 = 250009250009+ ... = 263069000060 (101*2*3*_, we need to find the middle digit)003000 (102*2*5*3)010000 (103*2*5*_)263169 - 263069 = 10025000900006000300001000000010000060 doesn't change so the only possibility is when the middle digit is 1.Thus, 2631169 = 5132