Answer:
a.) April has 30 days
If you for option No.2, you would have:
[day, amount($)]
1, 0.01
2, 0.02
3, 0.04
4, 0.08
5, 0.16
6, 0.32
7, 0.64
8, 1.28
9, 2.56
10, 5.12
11, 10.24
12, 20.48
13, 40.96
14, 81.82
15. 163.48
16, 327.68
17, 655.36
18, 1310.72
19, 2621.44
20, 5242.88
21, 10,485.76
22, 20,971.52
23, 41,943.04
24, 83,886.08
25, 167,772.16
26, 335,544.32
27, 671,088.64
28, 1,342,177.28
29, 2,684,354.56
30, 5,368,709.12
Option 2 grants you more than 5 times as much as Option A and is therefore obviously better.
b.) A diagram would show first a slow rise, than a steeper and steeper rise, then would almost growvertically. exponential growth
#coronatime
The most elegant form to describe the given numbers would simply be
$=1*2^x
You start with one, eich doubles after a day (x=1). x is the number of days and how often you multiply by 2
c.) Wasn't sure without calculating, but I guessed Opt.2, because it seemed that one should be tricked into choosing Opt.1
Have a nice day
Brainliest would be appreciated
If there are questions left, feel free to ask them
Answer:
see picture with work shown
Write the equation of the line passing through the points (-7,5) and (-5,9):

.
You also have another two points (-3,13) and (-1,17). Look whether coordinates of these points satisfy the line equation:
1. For (-3,13) you have

;
2. For (-1,17) you have

.
Conclusion: All four points lie on the line y-5=2(x+7), so <span>the relationship shown by the data is linear.</span>
Answer: Correct choice is D.<span />
14 - 2n = -32
32 + 14 = 2n
46 = 2n
46/2 = n
23 = n
Checking:
14 - 2(23) = -32
14 - 46 = -32
-32 = -32 (Matched, answer is correct n = 23)
1 step (B): raise both sides of the equation to the power of 2.
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2 step (A): simplify to obtain the final radical term on one side of the equation.
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3 step (F): raise both sides of the equation to the power of 2 again.
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4 step (E): simplify to get a quadratic equation.
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5 step (D): use the quadratic formula to find the values of x.
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6 step (C): apply the zero product rule.
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Additional 7 step: check these solutions, substituting into the initial equation.