NotThatSmart

When we have f(x+3), the graph shifts to the left by 3 units. 

When we have -8, the graph shifts down 8 units. 

The vertex will not be affect by the 3 and the 2. 

Thus, the vertex of the new parabola is (-2, -10) when we shift it 3 units to the left and 8 units down!

Thanks! :)

First, let's combine some like terms!

\(x^2 - 22c + c\). 

In order for this to be the square of a binomial, we need c to be the square of half the value of -22. 

Solving this, we get c = 121. 

Indeed, if we have \(x^2-22x+121=(x-11)^2\). 

So we have c = 121. 

Thanks! :)

First off, let's combine like terms to get \(x^2 + sx + 100\). 

For this to be a square of a binomial, we must recognize that s is equal to double the square root of 100. 

Solving this we get s=20. 

Indeed, \(x^2+20x+100 = (x+10)^2\). 

However, we also get s=-20. 

We have \(x^2-20x+100 = (x-10)^2\). 

This means s=-20, 20

Thanks! :)

(Haha, this is the post that got me to 200 points!)

Alright! I'll tackle this problem!

First off, we notice there is an f(x) function inside the second equation. 

We know f(3)=4, we let's let x/3 in the second equation be 3 so that we have an integer. 

Setting x=9 would get us this desired equation. 

We have \(y=2f(3) + 17\), which is essentially \(2(4)+17\). 

This simplifies to 25, meaning when x equals 9, we have y equals 25. 

This means that (9, 25) HAS to be on the graph for y=2f(x/3)+17!

Thanks! :)

We can use constructive counting or casework to solve this problem. 

CASE 1

First, let's see all the two digit numbers that satisfy these conditions. 

We only have two, with 23 and 32. 

CASE 2

Now, let's check all the 3 digit numbers. 

We have 311, 131, 113, giving us 3. 

We also have 221, 212, and 122, giving us another 3. 

Case 3

4 digit number time! This is where it gets a bit complicated, but it's still not that bad. 

We have numbers with three 1s and one 2, which has 4 different possibilities. 

Case 4

Last case ! 5 digit numbers only have 1, with 11111. 

Now, we just have to add up all our cases, and we get 2+3+3+4+1 = 13. 

We have 13 numbers that satisfy the conditions. 

NOTICE: This problem didn't have a lot of cases, so it would be easy to count. Bigger and wider ranges would require permutations, which I really really dislike! :)

I hope I answered your question!

Thanks! :)

This might seem like a weird way to do it, but like others, I would reccommend the Cauchy-Schwarz inequality.

Essentially, what this inequality states is that where a and b are real numbers, we can write the inequality \((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\)

We could apply this to this problem!

We have \((x^2+3y^2)(1+\frac{1}{3}) \geq (x+\sqrt{3}(\frac{1}{\sqrt{3}})y)^2\). 

\(18*\frac{4}{3}\ge(x+y)^2\)

\(x+y\le\sqrt{24}\)

Clearly, from this equation, we find that the maximum value is \(\sqrt{24}\)

We could probably brute force this problem, but this trick is defintely easier!

Thanks! :)

First, let's simplify the expression by combining like terms. 

\(x^2 -14x + y^2 + 16y + 26\). 

Ok! Now all we have to do is to take the derivative with respect to x and y and set that to 0!

We have 

\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)

and 

\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)

Now, all we have to do is to plug in these values for x and y into the orginial function. 

We get \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87\). 

The lowest temperature on the plane is -87 degrees, which is quite cold!

Thanks! :)

First, let's combine some like terms to simplify this expression!

\(x^2 -14x + y^2 + 16y + 26 \)

Now, all we have to do is to take the derivative with respect to x and y and set that to 0 before plugging the values back into the function. 

We have 

\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)

and 

\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)

Now, plugging the value back into the function, we have \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87 \). 

So the lowest temperature is just -87!

Thanks! :)

First, let's isolate y in the first equation. 

\(xy = 2/5 \\ y = 2 / (5x) \)

Plugging this value into the second expression, we get 

\(6x + 3 / (2 / (5x) ) = \\ 6x + (15/2)x = \\ (27/2) x\)

However, notice that there is no mimumum value or maximum value because the smaller x is, the smaller the number is to no limit. 

There are absolutely no restrictions for this expression. 

Thanks! :)