Since we know that he ate 1/4 pound one time, and 1/6 another time, we have to add up the fractions.
1/4+1/6
LCM of 4 and 6 = 12
3/12 + 2/12
Add the numerators:
5/12
So, he ate 5/12 of a pound of bars in both breaks.
Hope this helps!

We have, Discriminant formula for finding roots:

Here,
- x is the root of the equation.
- a is the coefficient of x^2
- b is the coefficient of x
- c is the constant term
1) Given,
3x^2 - 2x - 1
Finding the discriminant,
➝ D = b^2 - 4ac
➝ D = (-2)^2 - 4 × 3 × (-1)
➝ D = 4 - (-12)
➝ D = 4 + 12
➝ D = 16
2) Solving by using Bhaskar formula,
❒ p(x) = x^2 + 5x + 6 = 0



So here,

❒ p(x) = x^2 + 2x + 1 = 0



So here,

❒ p(x) = x^2 - x - 20 = 0



So here,

❒ p(x) = x^2 - 3x - 4 = 0



So here,

<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:

Step-by-step explanation:

Step 1: Divide the numbers

Step 2: Simplify

Step 3: Simplify

Therefore, the simplified answer is 
Answer:
no enough info
Step-by-step explanation:
The last one because the other ones only go to the positive side and not the negative too since the cube root