Four million five hundred thirty seven thousand five hundred forty five
Explanation:
First you had to square from both sides of equation form.
Then you expanded form.
Solve by equation form.
Add by four from both sides of equation form.
Simplify.
Verified solutions.
Final answer: 
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We will get the number of possible selections, and then subtract the number less than 25 cents.
We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.
That's 5*4*3 = 60 selections
Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels.
To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2 on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).
If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.
And there is 1 other selection less than 25 cents, 2 dimes and no nickels.
So that's 4+3+1 = 8 selections which we must subtract from the 60.
Answer 60-8 = 52 selections of coins worth 25 cents or more.
2(10-3) distribute = 2*10 is 20 and 2*3 is 6
20-6+(5-14/2)
now 20-6 is 14
14+(5-14/2)
now 14/2 is 7
14+(5-7)
now 5-7 is -2
14+-2
is 12
the answer is 12
but if you want to you can do 12/3 to get the greatest common factor
which is 4
hope i help you :P
Hello,
function minmax(int p1,int p2,int p3, int adr_big, int adr_small)
{ int mini=p1,maxi=p1;
if (p1>p2) {mini=p2;}
else {maxi=p2;};
if (p3>maxi) maxi=p3;
if (p3<mini) mini=p3;
*adr_big=maxi;
*adr_small=mini;
};
// main
int a=31,b=5,c=19,big,small;
minmax(a,b,c,&big,&small);
