I believe the 3rd answer down if the information you provided is correct
Squaring the sum on the top, rules out 1 and 2 (not squaring the 7 only)
then 4 times the difference. so I believe 4 (x-1) on the bottom.
Explanation:
Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that
a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.
Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.
Answer:
instincts
Step-by-step explanation:
i luv u!
I think you want the least common multipule (exg LCM of 2,3,4 is 12)
so we find the factors and include all of them so like this
15=3 times 5
12=2 times 2 times 3
8=2 times 2 times 2
so to include all of them we include 3 2's, 1 3 and 1 5 or 2 times 2 times 2 times 3 times 5 or 8 times 15 or 4 times 30 or 120
the LCM is 120
so 4/15=32/120
1/12=10/120
3/8=45/120
add 32/120+10/120+45/120=(32+10+45)/120=87/120
