Which set of side lengths is a pythagorean triple?
a.1, 3, 10
b.4, 5, 9
c.9, 40, 41
d.16, 30, 44

Ok so just do
a^2+b^2=c^2
c=longest side
a and b are the other sides

input for a and b and c and see which ones are true

A. 1^2+3^2=10^2
1+9=100
10=100
false

B, 4^2+5^2=9^2
16+25=81
41=81
false

C. 9^2+40^2=41^2
81+1600=1681
1681=1681
true

D. 16^2+30^2=44^2
256+900=1936
1156=1936
false

answer is C

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to

(m+11n)^2=(m+11n)(m+11n)=m^2+11mn+11mn+121n^2=m^2+22mn×121n^2 for example; use foil: (a+b)(a+b)first. a^2outside. abinside. ablast. b^2