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Ksju [112]
3 years ago
5

Please help me answer all parts of the question in the photo.

Mathematics
1 answer:
goldenfox [79]3 years ago
3 0
z_1=4+2i\\z_2=-1+2i\\\overline(z_2)=\sqrt{(-1)^2+(2i)^2}=\sqrt{1-4}=i\sqrt{3}\\ z_1+\overline(z_1)=4+2i+\sqrt{4^2+(2i)^2}=4+2i+\sqrt{16-4}=4+\sqrt{12}+2i\\z_2+\overline(z_2)=-1+2i+\sqrt{(-1)^2+(2i)^2}=-1+2i+\sqrt{1-4}=-1+(2+\sqrt{3})i
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Answer:

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Step-by-step explanation:

We have been given that a particle moves according to the velocity equation v(t)= 6t^2-18t+12. We are asked to find the distance that the particle will travel in its first 2 seconds.

s(t)=\int |v(t)|dt

s(t)=\int\limits^2_0 |6t^2-18t+12|dt

Now, we will eliminate the absolute value sign as:

s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt

s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1

s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1

s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))

s(2)=2-9+12-(0)-16+36-24-(-2+9-12)

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s(2)=6

Therefore, the particle will travel 6 feet in first 2 seconds.

   

 

7 0
3 years ago
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