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Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
an = a1r^(n-1)
a5 = a1 r^(5-1)
-6 =a1 r^4
a2 = a1 r^(2-1)
-48 = a1 r
divide
-6 =a1 r^4
---------------- yields 1/8 = r^3 take the cube root or each side
-48 = a1 r 1/2 = r
an = a1r^(n-1)
an = a1 (1/2)^ (n-1)
-48 = a1 (1/2) ^1
divide by 1/2
-96 = a1
an = -96 (1/2)^ (n-1)
the sum
Sn = a1[(r^n - 1/(r - 1)]
S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]
=-96 [ (1/2) ^ 17 -1 /-1/2]
= 192 * [-131071/131072]
approximately -192
