We will take the common terms in the expressions-
Here in the expression, 3y is a common term.
So,
So, the answer is
Answer:
Step-by-step explanation:
For the first 30 hours, Andrew will earn 4(30) = $120. For each additional hour beyond 30 that Andrew works, he will earn 5(h - 30). Therefore, the function that gives Andrew's total wages when he works more than 30 hours is:
w = 120 + 5(h - 30).
So, if he works 35 hours, he will earn:
120 + 5(35 - 30) = 120 + 5(5) = 120 + 25 = $145.
If he works 42 hours, he will earn:
120 + 5(42 - 30) = 120 + 5(12) = 120 + 60 = $180.
I hope this helps!
Step-by-step explanation:
You want to set one of the equations equal to either x or y. For this, I put x+y=-2 equal to y which would be y=-x-2. Then you plug this in to the other equation for y. This would give you 5x+2(-x-2)=2. When you factor the two into the parenthesis, combine like terms, and keep the x values on ones side and the other numbers on the other side, you get 3x=6
Answer: m ⇒ e ∨ p
Step-by-step explanation:
First look at the statement and locate the connectives ( these are words which connect sentences such as: and, or, implies, is equivalent and not.) Each of these connectives have a symbol/ variable.
- In this case the connectives are "only if" and "or" these are important because we will not attach propositional variables to them, where:
⇒ = " only if"
∨ = "or"
- Convert the phrases into variables where:
m = " You can see the movie"
e = "You are over 18 years old"
p = " you have the permission of a parent"
- Construct the statement sequentially.
You can see the movie (m) only if (⇒) you are over 18 years old (e) or (∨) you have the permission of a parent (p)
- Remove the phrases and you'll have your answer:
m ⇒ e ∨ p
Answer:
Step-by-step explanation:
Option A. All the real values of x where x < -1
Procedure
Solve the inequality:
(x -3)(x+1)>0
That happens in two cases.
1) When both factors >0
x-3>0 and x+1>0
x>3 and x >-1
The intersection is x >3
2) When both factors <0
x-3<0 and x+1<0
x<3 and x<-1
the intersection is x<-1.
We have obtained that the function is positive for the intervals x < -1 and x > 3. But in one of those intervals the function is decresing and in the other is increasing.
You can recognize that the function given is a parabola and, because the coefficient of the quadratic term is positive, the parabola opens upward. Then the function is decreasing in the first interval and increasing in the second interval.
