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Lorico [155]
3 years ago
5

Determine whether the graphs of the given equations are parallel, perpendicular, or neither.

Mathematics
2 answers:
Zigmanuir [339]3 years ago
5 0

Answer: c perpendicular


Step-by-step explanation:

y=x+12 y=mx+b m=1 slope 1

Y=-x+1 m=-1 slope -1

slopes are perpendicular obviously, or

(1,y(1))-(0,y(0)) = (1,13)-(0,12) = (1,1)

(1,Y(1))-(0,Y(0)) = (1,0)-(0,1) = (1,-1)

(1,-1) dot (1,1) = 1×1+1×-1 = 0 therefore perpendicular.

Two vectors are perpendicular if and only if their dot (inner) product is zero, or the projection of one onto the other is zero, or if the cosine of the angle between them is zero.


Kay [80]3 years ago
4 0

Answer:

They are perpendicular

Step-by-step explanation:

These lines are in fact perpendicular when plotted, and actually have a intersection point of (-5 1/2, 6 1/2)

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<h3>hello!</h3>

We're given a point that the line intersects and its slope.

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\boxed{y-y1=m(x-x1)}

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<h3>note:-</h3>

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Answer:  The given logical equivalence is proved below.

Step-by-step explanation:  We are given to use truth tables to show the following logical equivalence :

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We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q      ∼P     ∼Q     P⇔ Q    ∼P ∨ Q     ∼Q ∨ P        (∼P ∨ Q)∧(∼Q ∨ P)

T     T         F        F             T            T                   T                       T

T     F         F        T             F             F                   T                       F

F     T         T        F             F            T                   F                       F

F     F         T        T             T            T                   T                       T

Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.

Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).

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