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Since gcd(1-99) = 1, lcm(1-99)=99!

Heyyy homework cheater, if you spent five seconds searching you would find that your question has been asked before on different sites, proving that it is homework. Please do it yourself! Thank you! :))

He in fact is! Guest!

Praise from Caesar.  Thank you.  Say goodnight, Ginger.  :-)   hehe a little joke 

_.

I believe Cphill is working at "Brainly".

LCM(1 to 99) = 6,972,037,522,971,247,716,453,380,893,531,230,3556,800

GCD(1 to 99) = 1

The line of symmetry passes through the minimum point on the curve and that will be at x = -b/2a, so -b/2a = 2, 

4a + b = 0 .....................(eq 1).

The curve passes through (1, 1), so

a + b + c = 1 .........................(eq 2)

and through (4 - 7) so

16a + 4b + c = -7 .....................(eq 3).

Solve the three equations simultaneously, begin by subtracting 4 times eq 1 from eq 3, to arrive at 

a = -8/3, b = 32/3, c = -7.

Putting y = 0 and solving the resulting quadratic,

\(\displaystyle x = \frac{-32/3 \pm \sqrt{(32/3)^{2}-4(-8/3)(-7)}}{2(-8/3)} \\ \displaystyle =2 \mp(3/16)\sqrt{352/9} = 2 \mp(3/16)(4/3)\sqrt{22} \\ \displaystyle 2 \mp\sqrt{22/16}= 2 \mp \sqrt{11/8},\)

so n = 11/8.

Ron, this is an excellent presentation!

 GA

--. .- 

This place already has a lot of homework cheaters. You do not need to care for these homework cheaters, as they are hurting themselves asking others for answers. This is not a list of questions that come from me, my teacher, or others. This is my way of helping others improve on their skills. Thank you. 

This question has already been asked on https://web2.0calc.com/questions/algebra_20884#:~:text=For%20the%20heck%20of%20it......let%27s%20let%20the%20total,pct.%20games%20left%20that%20we%20need%20to%20win

You mean me or Reallystupidguy? 

I am mandarian 🤣😂 我怎么没看到这个问题。🤣😂

There are 3 ways to get a \(y^4\) term: \((2y)^4\), \((3y^2) ^2\), and \((2y)^2 \times (3y^2)\)

For the first way, we need 4 \(2y\)'s  and 2 \(-5\)'s, so we have\((2y)^4 \times (-5)^2 = 400y^4\). But we also need to multiply by \({6 \choose 2} = 15\) because there are 15 ways to order them. This makes for \(400y^4 \times 15 = 6000y^4\)

For the second way, we need 2  \(3y^2\)'s and 4 \(-5\)'s, so we have \((3y^2)^2 \times (-5)^4 = 5625y^4\). But like last time, we need to multiply by 15 for the same reason. Thus we have \(5625y^4 \times 15 = 84375y^4\). 

For the final way, we need 2 \(2y\)'s, 1 \(3y^2\)'s, and 3 \(-5\)'s, so we have \((2y)^2 \times (3y^2) \times (-5)^3 = -1500y^3\). But this time, we need to multiply by \({6! \over 3!2!} = 60\). Thus we have \(-1500y^4 \times 60 = -90000y^4\)

So in total, the coefficient of \(y^4\) is \(6,000y^4 + 84,375y^4 - 90,000y^4 = \color{brown}\boxed{375y^4}\)

Nice Job! I did not want to do this question anyway, but the fact that you did it is cool. 

Their are many sayings, such that he is on vacation, while other people think that it is because web2.0calc.com is failing. Check out this link:     https://web2.0calc.com/questions/for-those-of-you-wondering-about-cphill-and-others

Their are many sayings, such that he is on vacation, while other people think that it is because web2.0calc.com is failing. Check out this link:     https://web2.0calc.com/questions/for-those-of-you-wondering-about-cphill-and-others

In triangle PQR, M is the midpoint of PQ. Let X be the point on QR such that PX bisects angle QPR, and let the perpendicular bisector of PQ intersect AX at Y. If PQ = 36, PR = 36, QR = 16, and MY = 50, then find the area of triangle PQR   

We don't need all those midpoints and angle bisectors and whatever MY is.   

We have the three sides of a triangle, and that's all we need to find its area.  

We could still get the area, but it's an isoceles triangle, and that makes it easier.  

The sides:     PQ = 36, PR = 36, QR = 16     

This tells us that triangle PQR is an isoceles triangle, with angle P at the apex.  

So drop a perpendicular line from P down to base QR, intersect at point J.   

This does two things.  It bisects QR and it creates two right triangles.    

So use Pythagoras' formula to determine the length of this perpendicular,  

which, conveniently for us, is also the height of PQR.  

PJ² = 36² – 8²    ----->   PJ² = 1296 – 64   ----->    PJ = 35.1   

Area of PQR  =  (1 / 2) • base • height  =  (1 / 2) • 16 • 35.1  =  280.8    

check answer    

per Heron's Formula, Area  =  sqrt of (44)(8)(8)(28)  = sqrt(78,848)  =  280.8   

Confirmed.  I love it when a plan comes together.    

_.

As if I need the answers. 😂🤣😂🤣😂🤣😂🤣 I do not care about the answer. I just want to let people solve problems. 

All members of our painting team paint at the same rate. If 20 members can paint a 2000 square foot wall in 30 minutes, then how long would it take the  members to paint a 2000 square foot wall, in minutes?      

If the painting team can paint a 2000 square foot wall in 30 minutes, then  

long would it take them to paint a 2000 square foot wall, in minutes?   

There must be something missing from this problem.     30 minutes.  

_.

Nvm I forgot lines can't be diagonal sorry :(

I didn't know, but someone else said it on a previous question here (plus people started randomly insulting each other) :https://web2.0calc.com/questions/what-is-the-number-of-units-in-the-length-of-overline-ab

you seem to know an awful lot about what a cheated question looks like