We have,
(2y)^3 × y^-1
Simplify the term (2y)^3
= 8y^3 × y^-1
Now, multiply the terms with the same base y by adding their exponents.
Note: Using exponent product rule x^y × x^z = x^(y+z)
= 8y^{3+(-1)}
= 8y^(3-1)
= 8y^(3-1)
= 8y^2
1/7x
Factorize the numbers (4 and 28) as they have a common factor of 4 so you can simplify it to 1/7. Cancel out the 3 in the x (numerator) from the denominator and you will have x^1 remaining in the denominator. That is also the same as x, therefore making it 1/(7x)
Answer:
Step-by-step explanation:
m = 1/2, (18,10)
y - y = m(x-x)
y - 10 = 1/2 (x - 18)
y - 10 = 1/2x -9
y = 1/2x - 9 + 10
y = 1/2x + 1
Fourth line in 3rd section
