2 quavers=1crochet and 2crochets=1minim
2q ÷ q = 2
2q
Indeed. 74 cents can be made with the following set of 7 coins: 50c 10c 10c 1c 1c 1c 1c
7p + 2q = 46 . . . . (A) 5p + 3q = 36 . . . . (B) 3*(A): 21p + 6q = 138 2*(B): 10p + 6q = 72 Subtracting gives 11p = 66 so that p = 6 Substitute for p in (A): 7*6 + 2q = 46 or 42 + 2q = 46 which gives 2q = 4 so that q = 2 Solution: (p, q) = (6,2)
2 quavers=1crochet and 2crochets=1minim
18 1i 2q 1i 2q -1i,-18 1i -2q 1i -2q -1i,0 1i -2q 18,-a 1i -2q u,-k 1i -2q k,-u 1i -2q a,-18 1i -2q 0,-1i 1i -2q -a,-1s 1i -2q -k,-26 1i -2q -u,-2g 1i -2q -18,k 1i 2q k,u 1i 2q a,18 1i 2q 0,1i 1i 2q -a,1s 1i 2q -k,26 1i 2q -u,2g 1i 2q -18,2q -42 2q -1i,2q -42 -2q -42 -2q -18,0 -42 -2q -3o,-a -42 -2q -3e,-k -42 -2q -34,-u -42 -2q -2q,-18 -42 -2q -2g,-1i -42 -2q -26,-1s -42 -2q -1s,-26 -42 -2q -1i,-2g -42 -2q -18,0 -42 2q -3o,a -42 2q -3e,k -42 2q -34,u -42 2q -2q,18 -42 2q -2g,1i -42 2q -26,1s -42 2q -1s,26 -42 2q -1i,2g -42 2q -18,0 1i 0 i4 1s qs,0 i4 -1s qs,0 5k 34 h6,0 5k -2q h6,-1i 1i 2q 1i,2q u a 1i,2q 18 0 1i#-18 -34 -u -34 -k -2q -k -2g -u -26 -18 -26 -1i -2g -1i -2q -18 -34,u -34 18 -34 1i -2q 1i -2g 18 -26 u -26 k -2g k -2q u -34,-1i -a -u 0 18 0 1s -a,0 -1i -a -u 0 -u,-a 1i 0 1s#B 0 -3e 8e,B 0 u 2q,G -1s -34 9r,G -1s k 71,G 1s k 47,G 1s -34 1d,G 0 -42 b8,G -2q -18 8e,G 0 1i 5k,G 2q -18 2q
simply add the like terms: 4p -3p +2q -2q = p
2q ÷ q = 2
Let f(X)=2X2+6X+3 So f(-p)=f(2q) or 2p2-6p+3=8q2+12q+3 or p2-3p=4q2+6q or p2-4q2=3p+6q or (p+2q)(p-2q)=3(p+2q) so p-2q=3
If you mean: 10q-155(2q+4) then it is -300q-620
8(2q) = (8*2)q = 16*q = 16q.
I'm not sure about the answer but what is 4p - 2q = 18? I need to know what p is.
3
2q
2q
Answer is 30pqr