solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
Answer: The numbers are 10 and 23
Step-by-step explanation:
Let the smaller number be x
And the larger number be y
Ist sentence;
1. The sum of two numbers is 33.
X + y= 33 .........eqn 1
2. second sentence
the larger number is 3 more than two times the smaller number.
Y = 3 +2x
Put y into eqn 1
X+ 3 +2x = 33
3x= 33- 3
3x= 30
X = 30/3
X= 10 and y= 3+ 2x
Y= 3+ 2(10)
Y = 3+ 20
Y= 23
Therefore the two numbers are 10 and 23
Answer:
40:1
Step-by-step explanation:
To determine the perimeter of the triangle given the vertices, calculate the distances between pair of points. For the first pair (-5,1) and (1,1), the distance is 6. For the next pair, (1,1) and (1, -7), the distance is 8. Lastly, for the pair of points (-5,1) and (1, -7), the distance is 10. Adding all the distance will give the perimeter of the triangle. Thus, the perimeter is 24.
