Answer:
tobin should have divided by three
Step-by-step explanation:
Answer:
Step-by-step explanation:
Step 1: Write the equation in the correct form. In this case, we need to set the equation equal to zero with the terms written in descending order.
Step 2: Use a factoring strategies to factor the problem.
Step 3: Use the Zero Product Property and set each factor containing a variable equal to zero.
Answer:
A) Not mutually exclusive
Step-by-step explanation:
Given



Required
Determine if they are mutually exclusive or not
Mutually exclusive are defined by:

So, we have:

Take LCM


By comparing:
---- Calculated
---- Given
We can conclude that A and B are not mutually exclusive because:

I'm thinking this is what the problem looks like:

. The first thing to do is to move the

over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this:

. The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.