13.) (t⁻⁴)⁻⁹t² Now if the exponents outside of the parenthesis, that means you multiply them. t⁻⁴⁽⁻⁹⁾=t⁴⁵ Next, When it look like this: t⁴⁵ · t², don't multiply! When it comes to this, you had to add the exponents because it has the same base number: t⁴⁵⁺²=t⁴⁷.
Here's your answer to your question, t⁴⁷.
y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
Answer:
Slope is 3/2
Step-by-step explanation:
3x-2y=6
-2y=-3x+6
y=3/2-3
M∠E ≅ m∠J
m∠F ≅ m∠K
m∠G ≅ m∠L
So, m∠ L is 58.
