2n+2(n-4)=56
2n+2n-8=56
4n-8=56
4n=64
n=16
1st side is 16
2nd side is 12
Answer:
(9,5)
Step-by-step explanation:
10+8/2=18/2=9
6+4/2=10/2=5
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
Answer:
the connection is at 4 on x axis and 5 on y axis, so the solutions are x=4; y=5
Answer:
yeah
Step-by-step explanation: