If you intend for the problem to be equivalent to \ \ \dfrac{x^{\frac{8}{2}}}{3x^{\frac{1}{2}}}, \ \ then you must put grouping symbols around the denominator,
such as in x^(8/2)/[3x^(1/2)].
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Type: Posts; User: lookagain
If you intend for the problem to be equivalent to \ \ \dfrac{x^{\frac{8}{2}}}{3x^{\frac{1}{2}}}, \ \ then you must put grouping symbols around the denominator,
such as in x^(8/2)/[3x^(1/2)].
...
Nope. You are talking about your question which doesn't apply. You are off track and ignoring my points.
Why would I ask that, when it's not an appropriate analogy?
I might write "x = 13 OR 14" for the solutions to a quadratic equation, where the plus sign of \pm corresponds
to the solution of...
I don't know what correspondence issue you're referring to. "Plus or minus" is the same as "minus or plus," given that there is the
connector word "OR."
Example:
Show \pm\dfrac{f}{g} \ \ ...
The plus-minus sign is a mathematical symbol with multiple meanings across mathematics and non-mathematics areas. And, it is used in different ways within mathematics itself,
sometimes, depending...
x-8 = \pm\sqrt{67}
No, you supposedly solved for one variable in an equation and substituted in back into the same equation.
That would have given you an identity. You would have needed at least one other...
y = \dfrac{2x}{\sqrt{x^2 - 9}}
Regarding the range of the original function, y cannot belong to the interval [-2, 2].
No, it does not necessarily refer to a plurality. It refers to the number of multiples of it, whether that be 0, 1, or a
whole number more than 1. A good definition is given in the beginning of...
You have a right angle in each of those two triangles.
What do you conclude about the remaining two corresponding angles?
(x - 3)^4(x - 4)(x - 5)(x - 8)^2 \ \ is a polynomial that has four distinct zeros.
The zero 3 has multiplicity 4, because it occurs four times.
The zero 4 has multiplicity 1, because it...
You must have meant when x = 2.001, because ln(5*2.001 - 10) < 0.
cired2002, the numerical values for the domain of the original function are those
for the range of its inverse function. And...
No, the proof does not conclude that.
The proof concludes that either u is prime, or u has prime factors larger than the supposed largest prime in the list.
5x^(-3/6) + x^(-11/6)
5x^(-1/2) + x^(-11/6) . . . . I recommend you leaving it in this form. However, check with the instructions/instructor
to see if negative exponents are allowed in the...
Type grouping symbols around your fractional exponents. Use other grouping symbols for clarity is a recommendation. Space out your characters for more readability.
(I don't need the asterisk...
No, show it as the following for your instructor/course work and work it through. You should show your work here if you get stuck.
The area portion from x = -2 to x = 0:
...
Not necessarlly. What the OP gave to you is flawed.
A right triangle is inscribed in a circle. The problem needs that.
It's 29/(2pi). You need grouping symbols.
Here, the domain of n is necessarily being taken as that of the integers, because the context is mathematical induction.
f(x) = -10x2+1 has no inverse function, because it is not a one-to-one function.
f(x) = -10x2+1, \ x \ge 0 is a one-to-one function, and it has an inverse function.
Likewise, f(x) = -10x2+1, ...
If you wanted to use conjugates of the numerator and the denominator, you could begin with these steps:
\displaystyle\lim_{x\to \ 2} \ \dfrac{\sqrt{6 - x} \ - \ 2}{\sqrt{3 - x} \ - \ 1} \ = ...
Do not be name-calling on here. You must not be clear that I was responding to a post that existed where yours is now.
I was civil toward you. I immediately stated the truth about it, and then...
No, it was done there, and it should not have been done. And clearly I stated that already. "Was" is past tense as concerns that person's post.
You're making up false arguments. You're making...
No, you don't understand. That will never be done that way, because it needlessly and instantly throws the problem into more complexity
instead of going with the straightforward simplification. ...
No, ksdhart2, that is not how the first one is to be done. It's on the same level of complexity as the work for f^{-1}(f(x)) that you showed below it.
f(f^{-1}(x)) = (\left( \sqrt[3]{x} + 3...
You are missing required grouping symbols. Before you work on stapel's strategy, this is what you really intended:
1/(x + 1) - 2/(x + 1)^2 - (x^2 - 1)/(x+1)^3
That is, \ \ \dfrac{1}{x +...