4
You'll need to be a little more specific in the way you phrase your question. It could be read any number of ways, including:
√x + 9√x - 5
√(x + 9)√(x - 5)
√(x + 9)√x - 5
(√x + 9)(√x - 5)
√(x + 9)(√x - 5)
√[(x + 9)√(x - 5)]
And several more. The ambiguity of your question makes it impossible to answer with certainty.
Now, I'll take a guess here, and assume that you mean:
(√x + 9)(√x - 5)
That then could be expanded like so:
(√x + 9)(√x - 5)
= √x√x - 5√x + 9√x - 45
= x + 4√x - 45
If however that's not the actual format you meant, you will need to clarify your question.
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.
Since the root is in the denominator of the exponent, just divide the 5 by the square root value (2), so the solution is x5/2.
x3/x1/2 = x5/2.
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)
300 = 2 x 2 x 3 x 5 x5 and sqrt(2) is irrational then sqrt(300) is irrational