Three miles a day should be the correct answer
Answer:
see explanation
Step-by-step explanation:
To find the zeros let p(x) = 0 , that is
(x² - 1)(x² - 5x + 6) = 0
Factorise each factor
x² - 1 ← is a difference of squares and factors as (x - 1)(x + 1)
x² - 5x + 6 = (x - 2)(x - 3), thus
(x - 1)(x + 1)(x - 2)(x - 3) = 0
Equate each factor to zero and solve for x
x - 1 = 0 ⇒ x = 1
x + 1 = 0 ⇒ x = - 1
x - 2 = 0 ⇒ x = 2
x - 3 = 0 ⇒ x = 3
The zeros are x = ± 1, x = 2, x = 3
Answer:
t = 4
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define equation</u>
t + t + t = 12
<u>Step 2: Solve for </u><em><u>t</u></em>
- Combine like terms (t): 3t = 12
- Divide 3 on both sides: t = 4
<u>Step 3: Check</u>
<em>Plug in t into the original equation to verify it's a solution.</em>
- Substitute in <em>t</em>: 4 + 4 + 4 = 12
- Add: 8 + 4 = 12
- Add: 12 = 12
Here we see that 12 does indeed equal 12.
∴ t = 4 is a solution of the equation.
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
