Question

find the product y⁴×(-2/y³)×y​

Answer 1

Answer:

y^4 = y*y*y*y

y^3 = y*y*y

y^2 = y*y

Now, when computing y^3 * y^3 we can just do:

y^3 * y^3 = (y*y*y) * (y*y*y) = y*y*y*y*y*y = y^6

Next, when computing y^2 * y^4 we can just do:

y^2 * y^4 = (y*y) * (y*y*y*y) = y*y*y*y*y*y = y^6

Finally, we have the equation:

y^2 * y^4 =y^6

y^3 * y^3 + y^2 * y^4 = y^6 + y^6 = y^12

y^6/y^6=1

the 2 y’s cancel out as well as the power symbol leaving us with 6/6=1.

when the 1’s cancels out as well as the power symbol you are left with y/y=y which will be equivalent to 6/6 which will equal 1 making y equal 1.

Y^2=y*y

y^6=y*y*y*y*y*y

y^8=y*y*y*y*y*y*y*y

y*y+ y*y*y*y*y*y+y=y*y*y*y*y*y*y*y*y(y^8+y)

this is y^2+y^6+y which equals y^8+y which in our case is y^9 but in other cases it wouldn’t be the same since it would have a higher value at the end when you add y than if you timesed by all the values of y and in other cases when you times by all the values of y it would be bigger than just adding y on, after you times y.

Y^2=y*y

y^6=y*y*y*y*y*y

the Simplified numbers are 1 i.e y^1

the 2nd solution is to add ^1 to all y^3, and add ^2 to y^2 so they all are consistent and can be simplified.

y^3+^1=y^4

y^3+^1=y^4

y^4*y^4+y^4*y^4/4=y^1*y1+y^1*y^1

Answer 2

answer is -2y² y⁴-2y³×y¹=-2y¹×y¹=-2y²