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The sphere is not written in vector notation in fact there is no equation of sphere. "a" is the radius in this context which is some constant like 5. |r| is the distance from the origin. The definition of |r| is as follows, (r dot_product r)^1/2.

so, |r| = a do you understand that? Substitution is trivial.

Another potential approach: https://www.wikihow.com/Find-the-Area-of-Regular-Polygons

It seems to come from formula for the area of a pentagon. After simplification you end up with the squares. Google "area of pentagon" on google and you will understand.

Why is it urgent? What have you attempted. Give it a go.

You should only round after all calculations have been completed. Otherwise there there is an issue with intermediate rounding error. I am not to sure what would be the protocol for question in multiple stages in highschool e.g. b i) correct to two decimal places b ii)

Before asking a question show your attempt. I would check if your calculator is in degrees mode.

I thought 8i cannot be included because you wouldn't have a vector?

Have you tried just walking in. Assuming there are no class being held you should be able to enter. btw: I am not in Adelaide.

It should be possible to drop out (i.e. withdraw) of the subjects you are failing. So they are not included in WAM or GPA. You will occur fines for doing the subject given the census date has passed but on your testamur it will be shown as withdrawn.

So just find the domain/range of e^-x first. With that you determine domain/range of f(x).

Just consider each case. case 1: all the letters are different case 2: repetition involving S case 3 repetition involving R case 4: repetition involving S and R

Definition: A function is continuous at x=a if lim x->a f(x) = f(a) So we want lim(x->1) a-x^2 = lim(x->1) (x-a)^2 => lim (x->1) a-x^2 - (x-a)^2 = 0 => a-1 - (1-a)^2 = 0 => a-1 -1 -a^2 +2a = 0 => a^2 -3a +2 = 0 => (a-2)(a-1) = 0 => a = 2

Unfortunately I don't think there is an easy way to answer the question. If you calculate the deriative you will find it is cos(x)/|cos(x)|, as an exercise try doing this yourself, it is bit challenging. As a result the answer is d). As a side note you could eliminate some options by...

Almost correct. Drogonski as already mention but it will make it clear. By definition the range of tan^-1 is "-pi/2 < theta < pi/2" so 4pi/3 is not valid answer.

Remember year 9 math. ASTC d) -cos (7pi/8) = cos(pi/8) . Now use the the theorem f^-1(f(x)) = x to get pi/8. f) I presume can figure out the value -2sin(4pi/3). Now if you know the exact value of tan^-1(sqrt(3)) you will get the answer. Remember the special triangles i.e. {1, 1, sqrt(2)} and...

Just a minor note. You will have two triangle with area (1/2)*(10^2)*sin(180-theta). In this case it won't matter because sin(180-theta) = sin(theta) for all value theta that are part of the real numbers.

Good question You may find the following post interesting. https://stats.stackexchange.com/questions/430636/is-a-rating-in-a-set-range-a-categorical-or-numerical-variable

Hi take a look at tyrone97 post

To give an analogy. Let suppose you throw a coin 20 times, and gettng 13 heads. Then proportion of heads is 13/20. 1/20 is use because of 20 weeks. So fr = f * 1/20.