All Answers 356003 Answers

The distance between the foci is 8*sqrt(5).

For what values of K does the circle (x-2k )^2 + (y-3k)^2=10 pass through the point (1,0) ? 

f(10) = -0.5.

The answer is 14.

(r,s,t,u) = (2,-1/4,1/3,3/2).

The area is 100*pi.

The set of possible values is [7/2,infinity).

p(1) = 10.

The sum of the solutions is 37.

1) The sides of a triangle are $16,$ $17,$ and $17.$ Find the circumradius.

2)The sides of a triangle are 6,8  and  10.If x is the area of the circumcircle, and y is the area of the incircle, then find x-y

3) The sides of a triangle are $16,$ $17,$ and $17.$ Find the inradius.

4) In triangle  ABC, M is the midpoint of  line AB. Let D be the point on line BC such that line AD bisects angle BAC and let the perpendicular bisector of line AB intersect line AD at P. If AB = 36 and MP = 9 then find the distance from P to line AC.

let each integer be z 

z,z+1,z+2,z+3 are 4 consecutive integers 

Their sum is equal to n

z+z+1+z+2+z+3=4z+6=n 

Well largest integer in terms of n is 

n=4z+6 

Well, the question didn't say a and b are two distinct numbers right? 

So a possibility would be a=b 

(2,2) works ,(0,0) works 

2*2=2+2 

Hence: 

\(a=\frac{b}{b-1}\)

\(b=\frac{a}{a-1}\)

Notice a and b can't equal to 1 

In other words, 

a+b= \(-\frac{b-2ab+a}{(a-1)(b-1)}\)

a-1 can't equal to 1 

b-1 can't equal to 1

Leads to the same. 

a=ab-b = b(a-1)

a=b(a-1)

a/b=(a-1) 

ab=a+b

ab-a=b

Notice a and b can equal to 0 and it works as well. 

No other integer solutions other than these. 

https://www.wolframalpha.com/input/?i=ab%3Da%2Bb Graph is a hyberbolic 

9996 + 9997 + 9998 + 9999 = 39,990

Wow that was some BIG latex! cool!

do you HAVE to add 1 at least once?

otherwise you just multiply 2 five times and that is the least because there is no way to get a negative number.

If you do, 

then try to avoid multiplying by 2 AFTER you added 1.

That is only 1 possibility, another possiblity is angle A=angle C 

so angle B=100 

Depends on how you draw the triangle. 

other possibility as the Guest said B=70

Angle B = Angle C = [180 - 40] / 2 =70 degrees.

a=1;b=0;c=0;d=0;e=0;f=0;p=0; cycle:n=a*100000+b*10000+c*1000+d*100+e*10+f; if(a+b+c==d+e+f and n%10007==0, goto loop, goto next);loop:printn," ",;p=p+1; next:f++;if(f<10, goto cycle, 0);f=0;e++;if(e<10, goto cycle, 0);f=0;e=0;d++;if(d<10, goto cycle,0);f=0;e=0;d=0;c++;if(c<10, goto cycle,0);f=0;e=0;d=0;c=0;b++;if(b<10, goto cycle,0);f=0;e=0;d=0;c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

150105  180126  300210  330231  360252  390273  450315  480336  600420  630441  660462  690483  720504  750525  780546  870609  900630  930651  960672  990693  Total =  20 - such integers.

Just factor it:

1728 = 2^6 * 3^3

I count 35 ways as follows:

4 + 0+ 0 + 0 = 4.........................4 permutations
3 + 1 + 0 + 0= 4........................12 permutations
2 + 2 + 0 + 0= 4..........................6 permutations
2 + 1 + 1 + 0 = 4........................12 permutations
1 + 1 + 1 + 1  = 4.........................1  permutation
Total =4 + 12 + 6 + 12 + 1 =35 ways.

a=1;b=0;c=0;d=0;p=0; cycle:if(a==b or a==c or a==d or b==c or b==d or c==d, goto loop, goto next); loop:print a*1000+b*100+c*10+d," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT =4,464 integers - have at least 2 repeated digits out of total of 9,000 integers.

So x =4,464 / 9,000 = 62 / 125 =49.6%

The set of all possible values of f(x,y,z) is (0,3/2].