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maybe just do trial and error 

 that's what i usually do when i can't figure it out

Hint: It involves a rotation and an enlargement, but the rotation center isn't the origin.

Consider the first car when the second car passes the ramp. 

In that 10 minutes, the first car would have gone 10 miles further because of the speed.

So, the second car has to catch up to the first car, which is already 10 miles away from the ramp.

We let t hours be the time taken for the second car to catch up, after the second car has just passed the ramp.

In t hours, the second car would have travelled 65t miles, and the first car would have travelled 60t miles. But because the first car has a headstart of 10 miles, we have to add 10 and 60t to get the total distance.

60t + 10 = 65t

5t = 10

t = 2

So the second car would catch up to the first car 2 hours after passing that ramp, which means the time is 5:10.

this link might help:

https://www.khanacademy.org/math/geometry-home/transformations

if it doesn't i will try figuring out the answer

thank you

Let r be the radius of the small circle.

\(\sqrt{(3 + r)^2 - 3^2} + r + 2 = \sqrt{5^2 - 3^2}\\ \sqrt{r(6 + r)} + r + 2 = 4\\ \sqrt{r(6 + r)} = 2 - r\\ r^2 + 6r = r^2 - 4r + 4\\ 10r = 4\\ r = \dfrac{2}{5}\)

This is not possible, because gf(x) is of degree 3, not 2.

But if gf(x) really means g(f(x)), then this problem is solvable.

\(g(f(x)) = 3(f(x) - 1) = 3((x + 2)^2 - 1) = 3(x^2 + 4x + 3)\)

Using one digit: 2, 3, 5

Using 2 digits: 23, 25, 32, 35, 52, 53

Using 3 digits: 235, 253, 325, 352, 523, 532
 

In total, she can form 15 numbers.

We can calculate it by considering permutations, which is \(P^3_1 + P^3_2 + P^3_3 = 3 + 6 + 6 = 15\)

First, notice that 24! is an enormous number, and it is the product of 24 consecutive integers, so it has a lot of factors.

Also, notice that if we omit numbers that aren't divisible by 3, we have \(3\times 6 \times 9 \times \cdots \times 24\) as a factor of 24!.

We will find the exponent of 3 in this expression because other terms are not divisible by 3.

\(3\times 6 \times 9 \times \cdots \times 24 = (3\times 1) \times (3\times 2) \times (3\times 3) \times \cdots \times (3\times 8)\)

You may notice that the product every 3 consecutive terms is divisible by \(3^4\).

We have 

\((3^4)^2 \times 3^2 | 24!\)

Simplify to get 

\(3^{10} | 24!\)

And we cannot find any other multiple of 3. So the exponent of 3 is 10.

Coefficients are rational, which means they are real. This means the conjugate of x is also a root.

\(x = 3 - i\sqrt[4]{2}\text{ and }x = 3 + i \sqrt[4]2\text{ are roots.}\)

Now consider the product of roots. The product of roots is just the constant term, which means it must be rational as well.

By Fundamental Theorem of Algebra, the quartic polynomial must have 4 complex roots.

Let a and b be the other roots.

\(ab(3 - i\sqrt[4]2)(3 + i\sqrt[4]2) = ab(9 + \sqrt 2)\)

This means ab is a rational multiple of the conjugate of \(9 + \sqrt 2\).

Notice that the polynomial is not unique, so we can let ab = \(9 - \sqrt 2\) without loss of generality, where k is a rational constant.

Consider the sum of roots. The sum of roots is -1 * (coefficient of \(x^3\)), therefore the sum of roots must be rational as well.

This means that a and b are the conjugates of each other.

It is not hard to see that \((a, b) = ((3 + \sqrt[4]2), (3 - \sqrt[4]2))\) is a possible combination, where k₁ and k₂ are rational.

One of the possible polynomials are \((x - (3 - i\sqrt[4]2))(x - (3 + i\sqrt[4]2))(x - (3 - \sqrt[4]2))(x - (3 + \sqrt[4]2))\), which is \(f(x) = x^4 - 12x^3 + 54x^2 - 108x + 79\)

thx a lot

it has something to do with number theory because I did it before and I forgot the answer

First of all, let's find the multiples of 3 in 24!.

1*3, 2*3 ---> 8*3 = 8 multiples

But we can't forget about the multiples of 3². These have an extra 3.

9*1, 9*2 = 2 multiples

8 + 2 = 10

The exponent is 10: 3¹⁰.

most questions i ask r from aops. lol

That means ab = 2p.

Two integers whose product is a semiprime. That means (a, b) = (2, p) or (p, 2) or (1, 2p) or (2p, 1).

Therefore the number of ordered pair is 4n, where n is the number of prime number less than 100.

You can finish it by counting the number of primes smaller than 100. 

a) We are going to complete the square for the equation y=0.2t²-6.4t+12.

Factorize 0.2 out of the first two terms, y=0.2(t²-32t)+12

Now, complete the square by adding (32/2)²=16²=256 inside of the parenthesis AND subtracting 256x0.2, which equals 51.2. We have to subtract 51.2 to keep the value the same (to see why we don't just subtract 256, use distribution): y=0.2(t²-32t+256)+12-51.2

This simplifies to y=0.2(t-16)²-39.2. Now, we can see that the minimum value of y is -39.2, by making t equal to 16.

To see why this is the minimum, look at (t-16)². The square of any real number > 0. So the minimum value of (t-16)² is 0. 

Therefore, the lowest depth Rosy dived is 39.2 meters.

**My solution is a bit hard to understand if you've never worked with Completing the Square or minimizing quadratics. Most Algebra 2 books cover these topics.

We use binomial theorem.

\((2x + 1)^n = \displaystyle\sum_{k = 0}^n \binom nk (2x)^k\)

Coefficient of \(x^3\) is \(\displaystyle \binom n3 (2)^3 = \dfrac{4n(n - 1)(n - 2)}{3}\) for positive integer n.

Coefficient of \(x^4\) is \(\displaystyle \binom n4 (2)^4 = \dfrac{2n(n - 1)(n - 2)(n - 3)}{3}\) for positive integer n.

Now, the coefficient of \(x^3\) is equal to that of \(x^4 \).

\(\dfrac{4n(n - 1)(n - 2)}{3} = \dfrac{2n(n - 1)(n - 2)(n - 3)}{3}\\ 2n(n - 1)(n - 2) = n(n - 1)(n - 2)(n - 3)\\ n(n - 1)(n - 2)(n - 5) = 0\\ n = 0 \text{ (rej.) or }n = 1 \text{ (rej.) or } n = 2 \text{ (rej.) or } n = 5\)

Therefore n = 5.

it obviously can't be 1 it's even!

Here is another AoPS question in which you can peer edit my response

https://web2.0calc.com/questions/at-my-office-people-own-cats-people-own-dogs-and

I can assure you that this is the right answer.

We let the positive integer solutions to be \(\alpha\) and \(\beta\). Without loss of generality, assume \(\alpha < \beta\).

Then, 

\(x^2 - ax + 2016 = (x- \alpha)(x - \beta)\\ x^2 - ax + 2016 = x^2 - (\alpha + \beta)x + \alpha \beta\)

So, the question is actually asking for the minimum value of \(\alpha + \beta\) for \(\alpha \beta = 2016\). We can list the factors of 2016.

Factors of 2016:

I have listed them in a unique way such that on each column, the numbers multiply to 2016.

We can see that a is the minimum sum of these pair of numbers, which is 42 + 48 = 90.

Find the value of n if the coefficients of \(x^3\) and \(x^4\)are equal in the expansion of \((2x+1)^n\).

I rewrote bc i have a pet peeve where i don't like it when the LaTeX is wrong

it's nothing personal 

There are 2a students in the cooking club. b are in drama so 2a-b only in cooking

add a to drama so 3a-b students in at least one.

Thus we have 3a-b students in at least one of the clubs.

If you don't understand anything feel free to ask.

@ur first question 

it has already been answered here

https://web2.0calc.com/questions/what-is-the-value-of-the-modulus-of-6-4i_1

Here is a diagram that I hope will suit your needs.

I hope we have no more conflicts after this.

Also is EDT GMT-4, Eastern standard?