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katen-ka-za [31]
3 years ago
9

If z1 = 12(cos55° + isin55°) and z2 = 4(cos95° + isin95°), then |z1z2|= ???

Mathematics
1 answer:
zaharov [31]3 years ago
6 0
\bf \begin{cases}
z_1=12[cos(55^o)+i~sin(55^o)]\\\\
z_2=4[cos(95^o)+i~sin(95^o)]
\end{cases}\\\\
-------------------------------

\bf ~~~~~~\textit{ product of two complex numbers}\\\\
{{ r_1}}[cos({{ \alpha}})+isin({{ \alpha}})]\quad \cdot \quad {{ r_2}}[cos({{ \beta}})+isin({{ \beta}})]
\\\\\\
\implies {{ r_1\cdot r_2}}[cos({{ \alpha}} + {{ \beta}})+isin({{ \alpha}} + {{ \beta}})]\\\\
-------------------------------\\\\
z_1\cdot z_2\implies 12\cdot 4[cos(55^o+95^o)+i~sin(55^o+95^o)]
\\\\\\
48[cos(150^o)+i~sin(150^o)]
\\\\\\
|~48[cos(150^o)+i~sin(150^o)]~|\implies 48[cos(150^o)+i~sin(150^o)]
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Find the 1st , 2nd (median) and 3rd quartile of the set of numbers 6,8,2,6,7,9 ?
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A soccer ball is kicked toward the goal. The height of the ball is modeled by the function h(t) = −16t2 + 48t where t equals the
zlopas [31]
1. The function h(t)=-16t^{2}+48t is a parabola of the form ax^{2}+bx. The the formula for the axis of symmetry of a parabola is x= \frac{-b}{2a}. We can infer from our function that a=-16 and b=48, so lets replace those values in our formula:
x= \frac{-b}{2a}
x= \frac{-48}{-2(16)}
x= \frac{-48}{-32}
x= \frac{3}{2}
x=1.5
We can conclude that to the left of the line of symmetry the ball is reaching its maximum height, and to the right of the line of symmetry the ball is falling.

2. Lets check how much time the ball takes to reach its maximum height and return to the ground. To do that we are going to set the height equal to zero:
0=-16t^{2}+48t
-16t(t-3)=0
t=0 or t-3=0
t=0 or t=3
From our previous point we know that the ball reaches its maximum time at t=1.5, which means that <span>it takes 1.5 seconds to reach the maximum height and 1.5 seconds to fall back to the ground.</span>

7 0
3 years ago
Read 2 more answers
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Hatshy [7]

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\(◎o◎)/

8 0
3 years ago
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In each of the following equations, what is
timama [110]

The graph will cross at the coordinates (-2, 9)

<h3>How to solve equations?</h3>

y = 3x + 15

y = 3 - 3x

y = 3x + 15

Hence,

when x = 2

y = 3(2) + 15 = 21

when x = 0

y = 3(0) + 15 = 15

y = 3 - 3x

when x = 2

y = 3 - 3(2)

y = 3 - 6

y = -3

when x = 0

y = 3 - 3(0)

y = 3

Therefore, let's check if the equation will cross.

y = 3x + 15

y = 3 - 3x

using substitution,

3 - 3x = 3x + 15

3  - 15 = 3x + 3x

- 12 = 6x

x = -12 / 6

x = -2

y = 3 - 3(-2)

y = 3 + 6

y = 9

Therefore, the graph will cross at the coordinates (-2, 9)

learn more on equations here: brainly.com/question/19297665

#SPJ1

8 0
2 years ago
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