Answer:
w(-9) = 86
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
w(n) = n² + 6
w(-9) is n = -9
<u>Step 2: Evaluate</u>
- Substitute: w(-9) = (-9)² + 5
- Exponents: w(-9) = 81 + 5
- Add: w(-9) = 86
Answer:
1. 13 or -13
2. -5 < y < -3
3. 6 or -6
4. 1/8 or -1/8
Step-by-step explanation:
Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.
The Absolute Value term is |x|
For the Negative case we'll use -(x)
For the Positive case we'll use (x)
Step 3 :
Solve the Negative Case
-(x) = 13
Multiply
-x = 13
Multiply both sides by (-1)
x = -13
Which is the solution for the Negative Case
Step 4 :
Solve the Positive Case
(x) = 13
Which is the solution for the Positive Case
Step 5 :
Wrap up the solution
x=-13
x=13
But for the case of question (2) its a different pattern..
Since this is a "less than" absolute-value inequality, my first step is to clear the absolute value according to the "less than" pattern. Then I'll solve the linear inequality.
| y + 4 | < 1
–1 < y + 4 < 1
This is the pattern for "less than". Continuing, I'll subtract 3 from all three "sides" of the inequality:
–1 – 4 < y + 4 - 4 < 1 – 4
–5 < y < -3

The solution to the original absolute-value inequality, | y + 4 | < 1 , is the interval:

Let's look at the first one, then you can practice the rest on your own. So, h(0) = - 2 because the number in h() is which number you search for on the bottom row. Then you look at the number above that bottom number, which, in this case is - 2, and that number is your answer.
Answer:
yes
Step-by-step explanation:
If the values are a Pythagorean triplet then the square of the largest value should equal the sum of the squares of the other 2
largest value = 50 ⇒ 50² = 2500
14² + 48² = 196 + 2304 = 2500
Since 50² = 14² + 48² then 14, 48, 50 are a Pythagorean triplet