Note that (a^b)^c=a^(bc) so in this case we have:
(27^(1/3))^2 the cube root of 27=±3
(±3)^2 and ±3^2=9
So the equivalent expression is 9 or if you need an exponent 9^1 :)
I am assuming that you meant 27^(2/3)
Answer:
p^2 + q^2 = 22/9.
Step-by-step explanation:
For a quadratic function ax^2 + bx + c if the zeroes are A and B then
A + B = -b/a and AB = c/a.
2x^2 - 7x + 3
Now p and q are the zeroes of the above function so
p + q = 7/3 and pq = 3/2.
Now p^2 + q^2 = (p + q)^2 - 2pq
= (7/3)^2 - 2* 3/2
= 49/9 - 3
= 49/4 - 27/9
= 22/9.
It has to be at lest 9u2-4?
6y-5=11
Move -5 to the other side. Sign changes from -5 to +5.
6y-5+5=11+5
6y=11+5
6y=16
Divide by 6 for both sides to get y by itself.
6y/6=16/6
Cross out 6 and 6, divide by 6 and then becomes 1*1*y=y
y=16/6
Reduce 16/6 by dividing by 2
16/2=8
6/2=3
Answer: y=8/3 or y=2 2/3
= 0.02 is the answer