Answer: 100*(1.032)^t which can be written as 
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Explanation:
b = value of cup after t years
t = time in years (eg: t = 2 means 2 years have passed by)
The value starts at b = 100. After year 1, the value jumps up by 3.2% so we multiply the value $100 by 1.032 which is the proper multiplier to help increase by 3.2%; to see this, notice how 100% + 3.2% = 1 + 0.032 = 1.032
After 2 years, the value jumps another 3.2% so we have another copy of 1.032 multiplied. Then for 3 years, we'll have 3 copies of 1.032 multiplied. And so on.
For t years, we'll have t copies of 1.032 as the multiplier. So we will multiply the initial value 100 by (1.032)^t
That is why the equation is
b = 100*(1.032)^t
which can be written as
Answer:
y < 2
Step-by-step explanation:
Because the shaded in part is after the two, you know that it is y<2
Answer:
Expressions are not equivalent
Step-by-step explanation:
Given expressions are:
Putting w=1 in both expressions
![4(3w+4)\\=4[3(1)+4]\\=4(3+4)\\=4(7)\\=28\\\\16w+12\\=16(1)+12\\=16+12\\=28](https://tex.z-dn.net/?f=4%283w%2B4%29%5C%5C%3D4%5B3%281%29%2B4%5D%5C%5C%3D4%283%2B4%29%5C%5C%3D4%287%29%5C%5C%3D28%5C%5C%5C%5C16w%2B12%5C%5C%3D16%281%29%2B12%5C%5C%3D16%2B12%5C%5C%3D28)
Both expression have value 28 on w=1
Putting w=3 in both expressions
![4(3w+4)\\=4[3(3)+4]\\=4(9+4)\\=4(13)\\=52\\\\16w+12\\=16(3)+12\\=48+12\\=60](https://tex.z-dn.net/?f=4%283w%2B4%29%5C%5C%3D4%5B3%283%29%2B4%5D%5C%5C%3D4%289%2B4%29%5C%5C%3D4%2813%29%5C%5C%3D52%5C%5C%5C%5C16w%2B12%5C%5C%3D16%283%29%2B12%5C%5C%3D48%2B12%5C%5C%3D60)
Both expression have different values at w=3
Hence, in order for the expressions to be equivalent they both should produce same value on w=3 too.
So, it can be concluded that both expressions are not equivalent ..
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>
