Math  /  Algebra

Question13. Given that Z1=(32i),Z2=(2+4i)Z_{1}=(3-2 i), Z_{2}=(-2+4 \mathrm{i}) a. workout i. Z1+Z2Z_{1}+Z_{2} (2 marks) ii. Z1Z2Z_{1}-Z_{2} (2 marks) \qquad \qquad \qquad \qquad b. Represent the following in triangle diagram i. Z1+Z2Z_{1}+Z_{2} (2 marks) \qquad \qquad \qquad ii. Z1Z2Z_{1}-Z_{2} \quad (2marks) \qquad \qquad Page 5 of 10 \qquad \qquad \qquad \qquad c. Put 2i1i\frac{2-i}{1-i} in the form a+ib\mathrm{a}+\mathrm{ib} and hence find the values of a\mathbf{a} and b\mathbf{b}. (4 marks) \qquad \qquad \qquad \qquad
14. a. Differentiate from the first principles y=4x2y=4-x^{2} (3 marks) \qquad \qquad \qquad \qquad b. A point moving in a straight line such that the distance SS covered after tt seconds from the fixed point on the line is given by the equation S=(t32t2+t)\mathrm{S}=\left(t^{3}-2 t^{2}+t\right) meters, find i. Velocity after three seconds. (3 marks) \qquad \qquad \qquad

Studdy Solution
Calculate the velocity after three seconds for the given distance function S=t32t2+t S = t^3 - 2t^2 + t .
Velocity is the derivative of the distance with respect to time:
v(t)=ddt(t32t2+t) v(t) = \frac{d}{dt}(t^3 - 2t^2 + t) =3t24t+1 = 3t^2 - 4t + 1
Substitute t=3 t = 3 :
v(3)=3(3)24(3)+1 v(3) = 3(3)^2 - 4(3) + 1 =3×912+1 = 3 \times 9 - 12 + 1 =2712+1 = 27 - 12 + 1 =16 = 16
The velocity after three seconds is 16 16 meters per second.

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