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Let's first multiply the two functions to make our lives really easy. We have

\(( -x^2+ 8x - 5)(x^3 - 11x^2 + 2x)\\ -\left(x^{5}-11x^{4}+2x^{3}\right)+8x\left(x^{3}-11x^{2}+2x\right)-5\left(x^{3}-11x^{2}+2x\right)\\-x^{5}+19x^{4}-95x^{3}+71x^{2}-10x\)

The leading term is \(-x^5\)

The leading coefficient is \(-1\)

The degree of the polynomial is the highest power appearing in the polynomial, which in this case, is 5. 

The constant term is the one with no x value. Since there is no constant, the constant is just 0. 

The coefficient of x^2 is \(71\)

Thanks! :)

Let's note something really important before we begin. 

To solve this problem, we must first do something quite important.

It's crucial that we split this inequality into two seperate ones before combining them back again. Thus, we have

\(x^4+4x^2>-4\\ x^4+4x^2<21\)

Now, note something about the first inequality. 

We have the terms x^4 and x^2 which is guarenteed to be positive. Since we are adding them, all terms of x work, so we don't worry about anything. 

Next, we have

\(x^4+4x^2<21\)

Subtracting both sides by 21 and rewriting in standard form, we have the inequality

\(x^4+4x^2-21<0\)

Factoring out +-\sqrt3, we have

\(\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\left(x^2+7\right)<0\)

Now, let's note the intervals. If x is greater than square root 3, then we have a positive number. If x is less than square root 3, we also have a positive number. 

If x is either, then we have 0. Thus, the interval is

\(-\sqrt{3} <\sqrt{3}\)

In interval notation, we have

\((-\sqrt{3},\sqrt{3})\)

So our answer is \((-\sqrt{3},\sqrt{3})\)

Thanks! :)

Modulo is essentially the remainder we get when we divide two numbers.

For example, we have

\(10 \equiv 1 (\mod 3)\)

In this case, 1 is the remainder, 3 is the number we divide into 10. 

Thus, we say that 10 is congruent to 1 modulo 3. 

To solve the problem, we have the equation

\(9n+1 \equiv 1 (\mod 9)\)

We simply can write the equation

\(9n+1 \leq 1000\)

Now, we simplfy solve for n. Setting the two equations togther, we have

\(9n = 999\\ n=111\)

Thus, our answer is 111. 

Thanks! :)

Thanks, CPHill. 

I just realized you could have just done \(100-25 = 75\%\) to find the total amount of orange and dark. 

Then, we multiply the reciprocal of that by \(78\)

It gets the same answer of \(104\) since it's essentially the same steps. We still have

\(x = 78 * 4/3\\ x = 26*4\\ x = 104\)

It does save a few steps and is more efficient, but hey, same answer...

Thanks! :)

Slope =  [ b^2 - a^2 ] / [ b - a]  = [ (b + a) (b -a) ] / (b - a)  = 2

So

a + b  =  2

Very nice, NTS !!!!

I like your solution....

https://web2.0calc.com/questions/geometry_35384

Let's set up some variables to solve this problem. 

Let's first set that \(P = (x,y)\)

Now, from the problem, we can write the complicated equation of

\(PA^2 + PB^2 + PC^2 = (x-7)^2 + (y + 11)^2 + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2 \)

Now, we have to simplify this gigantic equation. By expanding and combining lots of like terms, we get that

\(3 x^2 - 70 x + 3 y^2 + 40 y + 1247 \\ 3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247\)

Now, we simplfy complete the square for x and y to achieve our x and y values for point Q, which will be crucial. 

Completing the square, we have

\(3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12 \\ 3 [ (x - 70/6]^2 + (y + 40/6)^2 ] + 2116 / 3\)

Now, from this, we can tell what Q has for their x and y value. 

We have that Q = (70/6, -20/3)

Thus, we know that

\(k = 2116 / 3 \)

So 2116/3 is our final answer. 

Thanks! :)

Let remainder to be r

And k to be a non-negative integer.

Set equation: p = 17k + r

The answer would be: \(\boxed{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}\)

To check: 

r = 0, k = 1

r = 1, k = 6

r = 2, k = 0

r = 3, k = 0

r = 4, k = 5

r = 5, k = 0

r = 6, k = 1

r = 7, k = 0 

r = 8, k = 3

r = 9, k = 2

r = 10, k = 3

r = 11, k = 0

r = 12, k = 1

r = 13, k = 0

r = 14, k = 1

r = 15, k = 4

r = 16, k = 3

The answer is 0.

We can use Apollonius's Theorem*

*Apollonius's theorem states that in any triangle, the square of the length of a median is equal to the sum of the squares of the lengths of the two sides containing the median, minus one-quarter of the square of the third side.

So: PM^2 = (2PQ^2 + 2PR^2 − QR^2)/4​

Since PQ = 5, PR = 8, QR = 11

PM^2 = (2*25 + 2*64 - 121)/4

PM^2 = 57/4

PM = \(\frac{\sqrt{57}}{2}\)

So the answer is \(\boxed{\frac{\sqrt{57}}{2}}\)

Very smart way to do it!

I totally forgot about factoring a^3 + b^3 as \((a+b)(a^2-ab+b^2)\)

It makes a lot of sense and is probably the most efficient way to complete this problem. 

Subsituting in \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\) work, I geuss, but good work!

Both solutions do eventually reach the desired goal of 24, so that's awesome. :)

Good job to you too, @aboslutelydestroying!

For this question, use formula: (a+b)^2 = a^2+b^2+2ab

since a+b=4

a^2+b^2+2ab = 16

a^2+b^2 = 6 + ab

substitute:

6 + ab + 2ab = 16

3ab + 6 =16

3ab = 10

ab= 10/3

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a+b = 4, a^2+b^2 = 6 + 10/3 = 28/3 , and -ab = -10/3

28/3 - 10/3 = 6

=> 4*6 = 24

here's another way to understand :)

(good job, @NotThatSmart)

We don't actually have to find what a and b is. 

First, let's note that we have \((a+b)^2 = a^2+2ab+b^2\)

Squaring both sides of the first equation, we find that

\(a^2+2ab+b^2 = 16\)

Subtracting the second equation from this equation we just computed, we have

\(3ab = 10\\ ab = 10/3\)

Now, let's note something else real quick. 

We have that \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\)

Since we also know that \((a+b)^3 = 4^3=64\), we have the equation \( a^3 + 3ab(a+b)+b^3=64\)

Wait! we already have all the numbers needed to isolate a^3 and b^3. We just found what ab was and a+b is given, so we have

\(a^3+b^3=64-3(10/3)(4) \\a^3+b^3 = 64-40\\ a^3+b^3=24\)

So 24 is our final answer. 

Thanks! :)

I'm not the best at these questions, but heres my understanding. We have

\(f_Q(x)=\sqrt{5^2-x^2}\\ f_R(x)=\sqrt{8^2-(x-11)^2}\\ 25-x_P^2=64-x_P^2+22x_P-121\\ x_P=\dfrac{121+25-64}{22}\\ x_P=3.\overline{72}\\ y_P=3.333\)

We are using coordinate geometry to solve this problem. Now, since we know these two values, we find that

\(P(3.\overline{72},\ 3.333)\\ M(5.5,\ 0)\\\)

Now, we can use the distance formula to find PM. We have

\( \overline{PM}=\sqrt{(x_M-x_P)^2+y_P^2}=\sqrt{(5.5-3.727)^2+3.333^2}\\\overline{PM}=3.775\)

I don't think this was the clearest explenation, but that's all I could understand. 

So 3.775 is our answer. 

Thanks! :)

We can use a handy equation and the fact that it's a right triangle to solve this problem. 

First, let's note some equations. Since AM and BN is the median, let's set a and b to be the two legs of the triangle. we have the equation

\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

Now, these two equations are quite important, since we want to know what a^2+b^2 is. Plugging in 1 and 1, and isolating all terms except for a and b, we find that

\(2^2 = 4a^2+b^2\\ 2^2 = 4b^2 + a^2\)

Now, adding the two equations, we have the formula

\(5a^2+5b^2 = 8\\ a^2+b^2 = 8/5\)

Thus, the hypotenuse, which is equal to the square root of a^2 + b^2 is just \(\sqrt{8/5}\)

So our answer is just \(\sqrt{8/5}\)

Thanks! :)

We can first note somethings really important to solve this problem. 

We can find the percentage of milk chocolates there is from the ratio. We have

\(25\% \cdot 7/5 = 5\% \cdot 7 = 35\%\)

Thus, 35% of the box is milk chocolate. Finally we know the amount of orange chocoaltes there are. We have

\(100\% - 25\% - 35\% = 40\%\)

Thus, 40% of the box is orange chocolates. Now, we can use the third condiditon stated in the problem. 

Since dark and orange chocolates formed 78 pieces, let's find the percentage it is in total. We add the two up and we get

\(35\% + 45\% = 75\%\)

Thus, 75% of the box is equal to 78 pieces. We can now right the equation

\(3/4x = 78\) where x is the total amount of pieces. Thus, we have

\(x = 78 * 4/3\\ x = 26*4\\ x = 104\)

Thus our answer is 104 chocolates. 

So choice 1 is our answer .

Thanks! :)

To solve this problem, let's note something really important first. 

Since in the problem, we have \( AD = AB \), then we can conclude that \(\angle ABD = \angle ADB\)

Thus, we can easily solve the problem from here. We can write the equation

\( ABD = (180 - 17) / 2 = 81.5° \)

Thus, our answer is 81.5

Thanks! :)

CA = 19

BA = 13

BC = 3

This isn't possible

ABC must form a triangle but  BC + BA < CA  which violates the triangle inequality

1/17  is a repeating decimal with a period of 16  =  0.0588235294117647......

The 4000th digit  is  7

1n^5 - 38 n^4 - 188 n^3 + 12006 n^2 - 11781 n + 0

You can use Stirling's formula:

n! = sqrt(2*pi*n)*(n/e)^n

You can use Stirling's formula:

n! = sqrt(2*pi*n)*(n/e)^n

Let's analyze the scenario step-by-step:

Initial Flips: We are given that after 2 flips, Erica has more heads than Alan. This can happen in 3 ways:

Erica gets HH and Alan gets TT (2 heads for Erica, 0 for Alan).

Erica gets HT and Alan gets TT (1 head for Erica, 0 for Alan).

Erica gets TH and Alan gets TT (1 head for Erica, 0 for Alan).

Remaining Flips: There are 3 remaining flips for each person. Since the coin is fair (each flip has a 50% chance of heads or tails), there are 23=8 possible outcomes for each person's remaining flips.

Favorable Outcomes: We only care about the outcomes where Erica maintains more heads than Alan after 5 flips. Consider the 3 initial scenarios:

HH vs TT: In this case, Erica needs any combination of heads and tails in her remaining 3 flips. There are 8 possibilities for her flips. For Alan, he cannot have any heads, so he needs all tails. There is only 1 possibility for his flips. So, there are 8⋅1=8 favorable outcomes.

HT vs TT: Here, Erica needs at least 2 heads in her remaining flips. There are (23​)+(33​)=3+1=4 ways for her to achieve this (2 heads and 1 tail, or 3 heads). Again, Alan needs all tails, so there is only 1 possibility for his flips. This gives us 4⋅1=4 favorable outcomes.

TH vs TT: Similar to HT vs TT, Erica needs at least 2 heads. There are 4 favorable outcomes. Alan still needs all tails, resulting in 1 possibility. So, there are 4⋅1=4 favorable outcomes.

Total Favorable Outcomes: Summing up the favorable outcomes from each initial scenario, we get a total of 8+4+4=16 successful outcomes for Erica.

Total Outcomes: There are 23=8 possibilities for Erica's remaining flips and 23=8 possibilities for Alan's remaining flips, leading to a total of 8⋅8=64 total outcomes.

Probability: The probability that Erica has more heads than Alan after 5 flips is the number of favorable outcomes divided by the total number of outcomes: $ \frac{16}{64} = \frac{1}{4}$.

Converting the fraction to relatively prime integers, we have m=1 and n=4. Therefore, m+n=5​.