Do the multiplication, then take the square root. After the multiplication, you get 28x4y2. The square root of 28 is 2 times the root of 7, the square root of x4 is x2, and the root of y2 is y. So, the final result is 2 Times Square root of 7, times x2y.
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To find the product of 2xy and x, you multiply the coefficients (2 and 1) to get 2, then multiply the variables x and x to get x^2, and finally multiply the variables y and 1 to get y. Therefore, the result of 2xy times x is 2x^2y.
yes * * * * * No, it isn't. 2xy = 2 times x times y. Depending on which you mean: 2x2y = 2 times x times 2 times y = 4xy; OR 2x2y = 2 times x times x times y.
it means two times x times y.
8-x2y 8-2xy 2xy-8
2xy substitute 3 for y and 5 for x 2(5)(3) = 30
√18xy = √(9*2xy) = √9*√2xy = 3*√(2xy)
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
4xy - 2xy
To find the product of 2xy and x, you multiply the coefficients (2 and 1) to get 2, then multiply the variables x and x to get x^2, and finally multiply the variables y and 1 to get y. Therefore, the result of 2xy times x is 2x^2y.
how can we convert algebraic expression into QBASIC a square + b square i = pTR/100 2xy mx+c a=r square a+b
(x + 4y)(x - 6y)
Can't be solved without knowing what x and y are. You can factor out 6y though. The expression then becomes: 6y * squareroot(2xy)
(x - y)2 >= 0 since the left hand side is a square. ie x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy
(x + y)2 = x2 + 2xy + y2
x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.
Since 2xy is a factor of 2x2y, it is automatically the GCF.
it would equal 2xy