Functions cannot have the same X value (the first number), but they can have the same Y value (the second number).
<span>A. {(1,2),(2,3),(3,4),(2,1),(1,0)}
B. {(2,−8),(6,4),(−3,9),(2,0),(−5,3)}
C. {(1,−3),(1,−1),(1,1),(1,3),(1,5)}
D. {(−2,5),(7,5),(−4,0),(3,1),(0,−6)}
Choice A. has two repeating X values [(1,2) and (1,0), (2,3) and (2,1)]
Choice B. has one repeating X value [(2, -8) and (2,0)]
Choice C. all has a repeating X value (1)
Choice D doesn't have any repeating X values.
In short, your answer would be choice D [</span><span>{(−2,5),(7,5),(−4,0),(3,1),(0,−6)}] because it does not have any repeating X values.</span>
Hey there!
Let's break this expression into two parts:
6(3x-1) and -10x
To solve the first part, we need to use the distributive property which states:
a(b+c) = ab+ac
Applying that to this problem, we have:
6(3x) + 6(-1) =
18x - 6
Now, we can take that -10x and put it right back in:
18x - 6 - 10x
Combine like terms and subtract the 10x from the 18x to get:
8x - 6
Hope this helps!
Answer:
3, 15, 14
Step-by-step explanation:
x + 5x + ( 5x -1) = 32
11x - 1 = 32
11x = 32 + 1
11x = 33
11x/11 = 33/11
x = 3
therefore
the first number is 3
the second is 5 × 3 = 15
the third number is 15 - 1 = 14
F(t)= 7.00 x t + 55
A reasonable domain is (0,1,2,3).
The range is ($55,$62,$69,$76)
