jfan17

This is true, as the domain of arctan can accept any real number. the range however is limited to [-pi/2, pi/2] so it is continuous and 1-1 as a function.

Hey guest!

Since we know that \(g(x) = x^3+5x^2+9x-2\), we already have part of the desired expression of f(g(x)).

The question then becomes reduced to: 

\(f(x^3+5x^2+9x-2)\)

since \(f(x) = x^3-6x^2+3x-4\), this may seem really complicated at first glance. However, remember what they ask us for; the constant term. That means we can ignore everything except for the constant term of the expression \(x^3+5x^2+9x-2\), which is -2.

Plugging -2 into the function \(f(x)\), we get:

\(f(-2) = (-2)^3-6(-2)^2+3(-2)-4 = -8-24-6-4 =-42\)

Thus, the constant term of the function f(g(x)) is -42

True, as the inverse sine allows any x value between -1 and 1(that's the max and minimum of sine). The range is also all real numbers, as sine can go negative or positive.

No problem!

Hey guest!

Let's first start off with an observation:

\(9x^{2}+24xy + 16y^2 = (3x+4y)^2\)

As such,

Part A) Because it says he arranges the marble as a square array, it makes sense that the total is a square number(of which it is). Since we know that 

\(9x^{2}+24xy + 16y^2 = (3x+4y)^2 \), that's how many total marbles there are. To get the number of marbles on each side of the square array, we just take the square root of given number, which gives us:

\(\sqrt{(3x+4y)^2} = 3x+4y\) marbles on each side of the array. The key takeaway here is that since the total number of marbles is a square number, that means since in a square array each side has an equal amount of marbles, the number squared must equal the total. To get the number, we just square root the total to get how many are on each side.

Part B) Since we already have the equation \(3x+4y\) as the number of marbles on each side of the square array, this problem becomes trivialized: it then becomes a simple question of substitution, where we get: 

\(3*(2) + 4*(5) = 6 + 20 = 26\)marbles on each side of the array

Hey guest!

I'm not really sure what the question means when the line has a distance of 1 unit from the origin. Does that mean the y intercept or x intercept, or what exactly? If it means that the line stops at 1 unit from the origin, that would just be a vector, so I'm not really sure what the question is asking here. If you have any ideas, feel free to lmk!

Hey there guest! First, let's start the problem with the knowledge of a (pretty common) theorem in geometry: The power of a point theorem. There are three variations to this theorem, but in this problem, we only need to look at 1 : the variation which involves the tangent of a circle and a secant(a line intersecting two points on the circle. 

Firstly, notice we can extend the line QS to intersect the other side of the circle at Z. This is where power of a point comes in handy:

By power of a point:

\(QP^{2} = QR*QZ\)

Substituting known values in:

\(6^{2} = 3 * QZ\)

Dividing by 3, we get:

\(12 = QZ\)

Then RZ = QZ - 3(since that's the length of QR),

RZ = 12 - 3 = 9

Now that we have the length of a chord of Circle O(RZ is a chord), we can use power of a point once again, except another of its variations: with two intersecting chords. 

A handy trick to realize is that the circle's diameter is also a chord, meaning we could find the length with power of a point theorem. 

If we draw a diameter intersecting S, we could make power of a point easier to work with. By power of a point:\((R+2)*(R-2) = RS*SZ = 3*6 = 18\)

where R is the radius of the circle

This simplifies to: 
\(R^{2}-4 = 18\)

\(R^{2} = 22 \)

\(R= \sqrt{22}\)

Hey there guest! I'm doing this at 12:30 rn so I didn't get to finish the whole question. I will share what I've gotten thru so far:

Let's first start off with some basic logarithm properties: first,

\(c\log_{a}b = \log_{a}b^c\)

Next:

\(\log_{a}b + \log_{a}c = \log_{a}b+c\)

With these 2 properties, we have enough to finish our question!

For the first term, a * log 10 of 2, we can rewrite that as:

\(\log_{10}2^a\)

Repeating this for all the terms in the given equation, we eventually reach:

\(\log_{10}2^a + \log_{10}3^b + \log_{10}5^c + \log_{10}7^d\)= 2005

This simplifies to:

\(\log_{10}(2^a*3^b*5^c*7^d) = 2005\)

Using the basic definition of a log, we can rewrite this as:

\(10^{2005} = (2^a*3^b*5^c*7^d)\)

That's about as far as I got! :/ sorry for the inconvenience!

Now adding on to what I did yesterday:

As far as we're concerned, this problem is now just asking us for which of the prime factor "four-tuples"(in 2, 3, 5, and 7) multiply to equal \(10^{2005}\). We clearly see that there is only 1 distinct tuple that gives us \(10^{2005}\), because 10 factors out into \((2*5)^{2005}\), meaning that a must be 2005, c must be 2005, and b and d are 0.

Hey nice to see you again Magnus! This is another case of stars and bars like I was explaining to you earlier. Because we have 8 cherry candies, let's first look at how to distribute that to 4 children. That would be:

\({8+4-1 \choose 8}\)= \({11 \choose 8}\)= 165

Now that we have the number of ways to distribute the cherry candies, we can multiply that with the number of ways to distribute the melon candies(correct me if I'm wrong with this reasoning).

Distributing 7 things to 4 people, we get:

\({7+4-1 \choose 7}\)= \({10 \choose 7}\)= 120

165 * 120 = 19800 ways. Now that I think about it, I feel like my number is a bit high, but feel free to double check this reasoning!

yep!