6-2.......Algebra 2 Help, Please?

2/8=1/4

X^2/X^6=1/X^4

Y^5/Y=Y^4

Z^-4/Z^3=1/Z^7

THEREFORE COMBINING THESE

ANSWER=Y^4/(4 X^4 Z^7)

choose letter c

those believe those who said d

but tell it to your classmates if they want to copy your homework

Everyone using PEMDAS and getting 9 seem to be making the same assumption that 6÷2(1+2) == (6÷2) * (1+2), whereas in fact you would get (6÷2)*1÷(1+2). If you treat it as a fraction PEMDAS becomes irrelevant and you can solve it in any order you like. 6÷2(1+2) ==6/ 2 * (1+2) == 6 * 1 / 2*(1+2) == 6 * 1/2 * 1/ (1+2) You can solve this is any order you like If you have complaints that there's fractions causing problems, take them out of the brackets 1st 6 / 2*1 + 2*2 = 6 / 2 + 4 = 6 / 6 = 1 or solve the brackets 1st then solve the denominator 6 / 2* (1+2) = 6/ 2* (3) = 6 / 6 = 1 If you want to treat the 6 ÷ 2 separately fine you just have to remember that a / b * c = a * (b * c)^-1 = a * b^-1 * c^-1 hence (6 / 2) * (1 + 2)^-1 = 3 * (1 + 2)^-1 = 3 ÷ (1 + 2) = 3 ÷ 3 = 1 You just have to remember the question is asking what is a÷b where a = 6 and b = 2(1+2), and that you cannot in any circumstances just move the (1+2) from the denominator to the numerator as this is solving the question, 6(1+2)÷2, which is completely different!

z^-4 in the numerator becomes z^4 in the denominator. So, we have: 2x^2 y^5/8x^6 y z^7.

Now divide both numerator and denominator by 2 x^2 y to get c) y^4/4x^4 z

Answer is c)

y^4/ 4x^4 z^7

2x^2 y^5 z^-4 / 8x^6 y z^3

2x^2 y^5 / 8x^6 y z^7

y^4 / 4x^4 z^7

which is C

when you divide exponents, they just subtract

2x^2 y^5 z^-4 / 8x^6 y z^3

C. (y^4) / (4x^4)(z^7) <----Final answer

(c) is the correct option. i.e. y^4 / 4x^4 z^7

2 * x^2 * y^5 * z^-4 y^4

--------------------------- = ------------------

8 * x^6 * y * z^3 4 * x^4 * z^7

So the answer is (c)

d

these are all fully simplified

the denominators and numerators do not have any gcfs