Answer:
32
Step-by-step explanation:
Answer:
lol
Step-by-step explanation:
Answer:
Based on the conditions we see that the function never pases through x=2 so then we know that x=2 is not on the domain of f. And we can create the following function who satisfy the conditions:
And we can see that we have a vertical asymptote in x =2 and is negative always on the interval (-∞, 2) and always positive on the interval (2, ∞).
And we can see the function in the figure attached.
Step-by-step explanation:
For this case we need a function that satisfy that is negative on the interval (-∞, 2) and positive on the interval (2, ∞).
Based on the conditions we see that the function never pases through x=2 so then we know that x=2 is not on the domain of f. And we can create the following function who satisfy the conditions:
And we can see that we have a vertical asymptote in x =2 and is negative always on the interval (-∞, 2) and always positive on the interval (2, ∞).
And we can see the function in the figure attached.
Triangle JKL has vertices J(−2, 2) , K(−3, −4) , and L(1, −2) .
Rule: (x, y)→(x + 8, y + 1 )
J’ (-2, 2) → (-2 + 8, 2 + 1 ) → (6, 3 )
K’ (-3, -4) → (-3 + 8, -4 + 1 ) → (5, -3 )
L’ (1, -2) → (1 + 8, -2 + 1 ) → (9, -1)
J’ (6,3)
K’ (5,-3)
L’ (9,-1)
Hope this helps!
Answer:
x=5
Step-by-step explanation:
so we know the final number is 6 so lets look at the rest of the question:
"twice the difference" so we are going to double the difference of two numbers, one we don't know (x) and the other is 2, so lets put this together and solve:
2(x-2)=6, x-2=3, x=5