I represents
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then there is the imaginary part, and the real part
the i is the imaginary
the 7 is the real part
distribute
i-(7-8i)
i-1(7-8i)
i-7+8i
-7+9i
it is normally writtnen in a+bi where a is the real part and b is the imaginary part
so the answer is -7+9i
Answer:
i havw bno
Step-by-step explanation:
hdaskdjha
Answer:



Step-by-step explanation:

This means:
- when x is equal to zero or less than zero, g(x) will always be 7.
- when x is more than zero, g(x) is




In order to sole the problem you need to create an equation to model what you are trying to do.
You end up with an equation looking like
25000*0.95^t
25000 is the starting value when t=0
95% is the amount remaining after the 5% decay
0.95 is being raised to the power of t to determine how many times it has decayed
When t=3 the equation should be
25000*0.95^3=21434.375
Answer:
D
Step-by-step explanation:
Scatter plot (Use excel with any bivariate data you can find)