There are 8 values of z.
The product we are interested in is 12/7.
The smallest possible value is 18
a + d = -4
S/T = 2/5
x^2-6x+k=0
\(x = {-(-6) \pm \sqrt{(-6)^2-4(1)(k)} \over 2(1)}=3\pm1\sqrt{9-k}\)
If k gets any bigger than 9, there both roots will be zero. Hence the largest value k can be such that that equation has at least one real solution is 9.
Q(x)=(lx^2+mx+n)
Q(x)^2=(l^2x^2 + 2mlx^3 + (2ln+m^2)x^2+2mnx+n^2)
Therefore
l^2=1
2ml=2
ml=1
2ln+m^2=-4
a=2mn
b=n^2
If l is one, so is m.
If l is -1, so is m.
l has to be negative, because of 2ln+m^2=-4, which means 2ln is negative.
2ln=-5
n=2.5
b=6.25
a=-5
a+b=1.25
(I think)
There is none!
n = 0
See the answer here: https://web2.0calc.com/questions/algebra_70308
Let the sales tax rate = R
R x 380 = 11.4
R = 11.4 /380
R =0.03 x 100 = 3% - sales tax rate.
]12! + 15! + 18! + 21! + 24!]= 2^10 * 3^5 * 5^2 * 7 * 11 * 53 * 129169 * 189221423
