For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
Answer:
-15
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Step-by-step explanation:
<u>Step 1: Define</u>
(4x + 2)(x - 5) / 2
x = 2
<u>Step 2: Evaluate</u>
- Substitute: (4 · 2 + 2)(2 - 5) / 2
- Parenthesis - Multiply: (8 + 2)(2 - 5) / 2
- Parenthesis - Add: 10(2 - 5) / 2
- Parenthesis - Subtract: 10(-3) / 2
- Multiply: -30 / 2
- Divide: -15
False
A function can’t have 2 y-intercepts so y is not a function of x
Answer:
Step-by-step explanation:
From the given information,
Suppose
X represents the Desktop computer
Y represents the DVD Player
Z represents the Two Cars
Given that:
n(X)=275
n(Y)=455
n(Z)=405
n(XUY)=145
n(YUZ)=195
n(XUZ)=110
n((XUYUZ))=265
n(X ∩ Y ∩ Z) = 1000-265
n(X ∩ Y ∩ Z) = 735
n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)
145 = 275+455 - n(X ∩ Y)
n(X ∩ Y) = 585
n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)
195 = 455+405-n(Y ∩ Z)
n(Y ∩ Z) = 665
n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)
110 = 275+405-n(X ∩ Z)
n(X ∩ Z) = 570
a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)
n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735
n(X ∪ Y ∪ Z) = 50
c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)
n(X ∪ Y ∪ C') = 145-50
n(X ∪ Y ∪ C') = 95
