I can't remember how to solve this problem, but I'm pretty sure the answer is 14.
yeah that's what i initially thought too, but the answer is actually 5 - sqrt(15)·i or 5 + sqrt(15)·i
\(\boxed{z=3+2i} \text{ and } \boxed{z=4-i}\)
yeaaa i'm an army too lol
\(4 \frac{3}{10} + 3 \frac{2}{5}+ \frac{1}{3} + \frac{1}{2}=\boxed{\frac{128}{15}} \text{ or } \boxed{8\frac{8}{15}} \)
\(ab + b^2 - ac -bc\)
\(= b(a+b)b(a+b)\)
\(=\boxed{ (a+b)(b-c)}\)
lmao wish i could help u but im going into 8th grade :(
good luck in high school tho!
umm actually i think the gcd of 36 and 104 is 4.
gwenspooner did the prime factorization right, but what she missed is that 3^2 is not a factor of 104. so, the gcd is 2^2=4.
+738 