y is 2 less than the product of 3 and x
When multiplying powers, you add them! y4 times y6 = y10. Try it with y = 2: 2 to the fourth = 16, 2 to the sixth = 64 16 x 64 = 1024 = 2 to the tenth.
2y2
To write this algebraically: 7(y^3)|y = 2 Substitute 2 for y: 7(2^3) 2^3 = 8, so substitute 8 for (2^3): 7*8 = 56
(4y)-2
y is 2 less than the product of 3 and x
It depends what the special product is. Common special products are: - perfect square trinomials ... x^2 + 2ax + a^2 = (x + a)^2 - difference of squares ... x^2 - y^2 = (x - y)(x + y)
When multiplying powers, you add them! y4 times y6 = y10. Try it with y = 2: 2 to the fourth = 16, 2 to the sixth = 64 16 x 64 = 1024 = 2 to the tenth.
(x+y)/2
2y2
2(xy)
It is 2y.
To write this algebraically: 7(y^3)|y = 2 Substitute 2 for y: 7(2^3) 2^3 = 8, so substitute 8 for (2^3): 7*8 = 56
It is: 7y+2
(4y)-2
(y^2)x8
let this integer be (xy) (xy) = 10x + y 10x+y = 2.x.y 10x + y - 2xy = 0 2x.(5-y) +y= 0 x= y / 2(y-5) when the integer provides this condition, it is equal to twice the product of its digits. And there is such only one integer. 36