I’m stuck on this can anyone help me out
1 answer:
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Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
22. Alternate angles
23. 4: (
, substitute b with 10 and a with 20 and you get 4.)
24. 0.28: 0.12 + 0.60 = 0.72. Probability is always out of 1, so 1 - 0.72 should give you the probability of losing, which is 0.28.
23. 4: (
24. 0.28: 0.12 + 0.60 = 0.72. Probability is always out of 1, so 1 - 0.72 should give you the probability of losing, which is 0.28.