Hello,
p=>q is equivalent to ~q → ~p
p-------q-------p=>q--~q ----- ~p----~q → ~p
0-------0-------1-------1 ------- 1------- 1
0-------1-------1-------0------- 1------- 1
1-------0-------0-------1------- 0-------0
1-------1-------1-------0------- 0-------1
Column 3= column 6 ==>equivalent
Answer B
Answer:
x=16
y=8.00
8/16=0.5 (50 cents per apple)
y=0.5x+0
Answer:
The correct answer would be the first option
Step-by-step explanation:
Answer:
Edge2020
Step-by-step explanation:
<u>RST≅BAC</u>
<u>YOUS WELCOME</u>
Answer:
option D
Step-by-step explanation:
So we know that 
Applying this to our function, we have that:

We know that the argument of a square root should always BE POSITIVE.
So we need to evaluate in which points the expression (x+2)(x-5) is positive.
So we know that (x+2) is possitive when x>-2 and negative when x<-2.
Also we know that (x-5) is possitive when x>5, if x<5 then x is negative.
Then we have:
if x<-2, then:
(x-5) is negative
(x+2) is negative
Then (x+2)(x-5) is positive.
If x>5 then:
(x-5) is positive
(x+2) is positive.
Then (x+2)(x-5) is positive.
If -2<x<5 then:
(x-5) is negative
(x+2) is positive
Then (x+2)(x-5) is negative, so it's undefined.
So the function is defined for x>5 and x<-2
So the correct option is option D.