How many 2-member committees can be formed from a group of 7 people?

Mathematics

1 answer:

Marizza181 [45]3 years ago

6 0

3 groups, unless I'm misunderstanding the question.

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Find the product of the complex numbers. Express your answer in trigonometric form.

Marianna [84]

In order to find the product of two complex numbers their magnitudes are multiplied and their arguments(measure of angles) are added up.

For z₁
Magnitude = 4
Argument = π/3

For z₂
Magnitude = 3
Argument = 2π/5

So, the product of z₁ and z₂ will be:

$z_{1} z_{2} =4*3(cos( \frac{ \pi }{3}+\frac{2 \pi }{5})+isin( \frac{ \pi }{3}+\frac{2 \pi }{5})) \\ \\ 
z_{1} z_{2} =12(cos( \frac{11 \pi }{5}+isin( \frac{11 \pi }{5}))$

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3 years ago

What is the simplified form of24 y to the fifth power over 15 x to the eighth powerdivided by8 y squared over 4 x to the fourth

lions [1.4K]

(24y⁵/15x⁸)÷(8y²/4x⁴).
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4 years ago

Help me due tomorrow at 4

Korolek [52]

4 for the first one and 1.5 for the second one

5 0

3 years ago

For the matrices below. Find all (real) eigenvalues. <br><br> A= [7 9]<br> [0 9]

Jet001 [13]

Answer:

The answer is "9 and 7".

Step-by-step explanation:

Given:

$A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]$

Using formula:

$|A-\lambda \cdot I|= 0\\\\$

$\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\$

$\to \lambda=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda=9\\\\$

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3 years ago

Kinsey adds -1 7/8 and it's opposite

djyliett [7]

Answer:

Step-by-step explanation:

the opposite is 1 -7/8 when you add them together they cancel out to zero

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3 years ago