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kirill [66]
3 years ago
12

Consider the following argument:

Mathematics
1 answer:
mariarad [96]3 years ago
3 0

Answer:

r \Rightarrow p\\ q \Rightarrow p\\ r \vee p\\ (r \vee q) \Rightarrow \neg r\\ s

Step-by-step explanation:

r \Rightarrow p \equiv (\neg r \vee p), p is false.

q \Rightarrow p \equiv (\neg q \vee p), p is false.

(r \vee q), r and q are falses.

(r \vee q) \Rightarrow \neg r \equiv (\neg (r \vee q) \vee \neg r). this sentence is true.

s

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sergiy2304 [10]

Answer:

1. Karen

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4 0
3 years ago
Find the value of sin T rounded to the nearest hundredth, if necessary.
katovenus [111]
<h3>Answer:  0.6</h3>

========================================

Work Shown:

sin(angle) = opposite/hypotenuse

sin(T) = VU/VT

sin(T) = 3/5

sin(T) = 0.6

6 0
3 years ago
Read 2 more answers
Solve the system of equations and choose the correct answer from the list of
Leya [2.2K]

Answer:

x = -19/2      y = -33/2

Step-by-step explanation:

x - (3x + 12) = 7

x - 3x - 12 = 7

-2x - 12 = 7

-2x = 19

x = -19/2

x - y = 7

-19/2 - y = 7

-y = 33/2

y = -33/2

4 0
2 years ago
Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2) – 6 = 0?
Darya [45]
Quadratic is in the form
ax^2+bx+c=0

so distribute and stuff and simplify
remember
a(b+c)=ab+ac

(x+2)^2+5(x+2)-6=0
remember order of opertaions
(x+2)(x+2)+5(x+2)-6=0
x^2+4x+4+5x+10-6=0
add like terms
x^2+9x+8=0
3 0
3 years ago
Read 2 more answers
What is the graph for {3x+5y&lt;10 {x-y&lt;=1
Nostrana [21]
Sounds to me as tho you are to graph 3x+5y<10, and that after doing so you are to restrict the shaded answer area created by the "constraint" inequality x≤y+1.    OR x-1 ≤ y   OR  y≥x-1.  If this is the correct assumption, then please finish the last part of y our problem statement by typing   {x-y<=1}.

First graph 3x+5y = 10, using a dashed line instead of a solid line.

x-intercept will be 10/3 and y-intercept will be 2.  Now, because of the < symbol, shade the coordinate plane BELOW this dashed line.

Next, graph y=x-1.  y-intercept is -1 and x intercept is 1.  Shade the graph area ABOVE this solid line.

The 2 lines intersect at (1.875, 0.875).  To the LEFT of this point is a wedge-shaped area bounded by the 2 lines mentioned.  That wedge-shaped area is the solution set for this problem.
6 0
3 years ago
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