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If you ate only two each day, how many would you have eaten over the course of the twelve days? How many sweets would be left over? So how many days must you have eaten a third sweet?

Has your class not covered exponents and radicals yet, so you need an explanation of the topic? If so, there are loads of lessons available online, such as in this listing. Please study at least two lessons from the listing, and then attempt the exercise. If you get stuck, you can then reply...

Yes, after an application of relevant rules, there can be only one answer. But, because the posted expression is ambiguous, we can't be sure which rules are being used. Therefore, people can arrive at two different answers, depending upon which rule(s) they have assumed were to be applied. This...

What are your thoughts? What have you tried? How far have you gotten? What recent results/rules/theorems have been covered in class? ("Read Before Posting") Please be complete. Thank you!

Think about the vertical multiplications you've done by hand. For each digit in the lower number, you'd get a row below that you would then add. Each row's start was shifted one place to the left. Thinking of those rows, you would have (as many digits as the top number, or maybe one more because...

It's not that it's a totally different question. It's that we're used to thinking in terms of positive infinity, and we forget that, just because there is no "minus" sign on the exponent, doesn't mean that it isn't negative. It's a human failing. But yes, you *will* need to be careful because...

What are "waits"?

You're thinking in terms of x trending toward *positive* infinity. But the limit is for *negative* infinity. When you're on the negative side of the x-axis, e^{2x} is going to get very small, because the 2 is positive and the x is getting strongly negative, so the base is being raised to a...

Does this mean that they overlap for at least 25% of the time? I don't know. What algorithm are you supposed to apply? What are the rules?

I get the same result. Now try simplifying the second half of the expression by breaking the one fraction into two: \qquad \dfrac{1}{2}\left(\dfrac{e^x\, sqrt{e^x - 1\,}}{e^x - 1} - \dfrac{e^x \cdot e^x}{2\sqrt{e^x - 1\,}(e^x - 1)}\right) \qquad \dfrac{1}{2}\left(\dfrac{2e^x\, \sqrt{e^x -...

What you have posted means the following: \qquad y = \dfrac{x^2 + 3x}{x} + a Was this what you meant? Because, if so, I can understand your confusion; the value of a has nothing to do with any discontinuity of the function. If not, what *did* you mean? When you reply, please clarify at what...

You've posted a picture containing four named shapes, various parts of which are shaded, and given us a cryptic reference to whatever were the original instructions. However, we need further information, starting with what you mean by "operations" that you are supposed to "write" (as opposed to...

What are your thoughts? What have you tried? How far have you gotten? ("Read Before Posting") Please be complete, so the helpers can see where things are going sideways. Thank you!

What are your thoughts? What have you tried? Where are you getting stuck? ("Read Before Posting") Please be complete, so that we can see where things are going sideways. Thank you!

Thread moved. Would it be correct to assume that the first four lines are the "givens", and that the starred line indicates the beginnings of computations based on those givens? So the problem statement was along the lines of the following? \qquad \textrm{Given that } \alpha + \beta =...

Is there a typo in the first row? Shouldn't it be x_1 + 2x_2 - x_3 = 6? You can either operate *with* a row or else *on* a row, but you can't do both. And it appears that what you did was operate *on* the row, by multiplying by -\frac{1}{5}, which is fine. I'm not sure what you did here, but...

If you're not sure of your answer, check it. If each child gets \frac{5}{12} of a package, then the twelve children will, in total, get 12\left(\frac{5}{12}\right) packages, which is all 5 [packages. As for the "how" for the "proper" (that is, the expected) way to arrive at the answer, I cannot...

You have three equations in six unknowns. Kinda by definition, you cannot solve for unique numerical values.

What are your thoughts? What have you tried? How far have you gotten? ("Read Before Posting") When you reply, please include a clarification of the expression for (f \circ g)(x). In particular, it looks as though you are squaring the sine twice: \left(\sin^2(x)\right)^2. Is this correct? Thank...

Please reply with a clear listing of your thoughts and efforts so far, so the helpers can see where you are getting stuck. Thank you!