4
Integral of x3/2dx using power rule = (5/2)x5/2 2.5 times the square root of x to the fith.
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
x5+4x4-6x2+nx+2 when divided by x+2 has a remainder of 6 Using the remainder theorem: n = 2
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10*5 = 50 sq ft
This can either be written as x5.5, or x5 times square root of x. √x11 = √x10x = x5√x
Integral of x3/2dx using power rule = (5/2)x5/2 2.5 times the square root of x to the fith.
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
The same way as you find the square root with an even-numbered exponent. For example, the square root of x10 is x5. That is, divide the exponent by 2. Similarly, the square root of x7 is x3.5. Once again, you simply calculate one-half of the exponent. If you prefer to express this with integer exponents and square roots, in this example you can write x3.5 as x3x0.5. The second part, x0.5, is equivalent to the square root of "x".
∛(x4) √(x5) = x4/3x5/2 = x8/6x15/6 = x23/6
The additive inverse is x5 + 2x - 2.
101.59*x5/3approx.
5x + 5y = 5(x+y)
Oh, dude, the square root of x to the power of 5 is x to the power of 5 divided by 2. I mean, like, you're just splitting that power in half, you know? So, if x is raised to the power of 5, taking the square root is like dividing that power by 2. Easy peasy, lemon squeezy.
300 = 2 x 2 x 3 x 5 x5 and sqrt(2) is irrational then sqrt(300) is irrational
is a quintic expression in x (NOT an equation).