Asked by OvO?
6.
(01.03 HC)
Tom simplified an expression in three steps, as shown:
x to the power of negative 2 multiplied by y to the power of 4, over y multiplied by x to the power of 4 multiplied by x to the power of 4 multiplied by y to the power of negative 2, the whole to the power of 4 equal to x to the power of 2 multiplied by y to the power of 8, over y to the power of 4 multiplied by x to the power of 8 multiplied by x to the power of 8 multiplied by y to the power of 2 equal to x to the power of 2 multiplied by y to the power of 8, over y to the power of 6 multiplied by x to the power of 16.
Which is the first incorrect step and why? (5 points)
Step 1, all the exponents are increased by 4
Step 2, the exponents of the same base are added during multiplication
Step 3, the exponents of the same base are subtracted during division
Step 3, all the exponents are added during division
(01.03 HC)
Tom simplified an expression in three steps, as shown:
x to the power of negative 2 multiplied by y to the power of 4, over y multiplied by x to the power of 4 multiplied by x to the power of 4 multiplied by y to the power of negative 2, the whole to the power of 4 equal to x to the power of 2 multiplied by y to the power of 8, over y to the power of 4 multiplied by x to the power of 8 multiplied by x to the power of 8 multiplied by y to the power of 2 equal to x to the power of 2 multiplied by y to the power of 8, over y to the power of 6 multiplied by x to the power of 16.
Which is the first incorrect step and why? (5 points)
Step 1, all the exponents are increased by 4
Step 2, the exponents of the same base are added during multiplication
Step 3, the exponents of the same base are subtracted during division
Step 3, all the exponents are added during division
Answers
Answered by
OvO?
I think it's the last one but I'm not sure could someone check if I'm right?
Answered by
Anonymous
type it again, using math notation, and maybe my eyes will stop bleeding.
y^8, x^2/y^3, etc. Lose all the words cluttering things up.
recall that
x^a * x^b = x^(a+b)
and so on
y^8, x^2/y^3, etc. Lose all the words cluttering things up.
recall that
x^a * x^b = x^(a+b)
and so on
Answered by
OvO?
The problem is: (x^-2y4 / yx^4 *x^4y^-2) ^4 = step 1: ( x^2y^8 / y^4x^8 * x^8y^2) = step 2: (x^2y^8 / y^6x^16) = step 3: x^-14 * y^2
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