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Kisachek [45]
3 years ago
6

What is the value of the modulus of 2−6i ? Answer as a simplified radical.

Mathematics
2 answers:
lana66690 [7]3 years ago
8 0

Answer:

2sqrt(10)

Step-by-step explanation:

| 2 - 6i |

sqrt[2² + (-6)²]

sqrt(4 + 36)

sqrt(40)

2sqrt(10)

sqrt: square root

Alja [10]3 years ago
6 0

Answer:

6.3244

Step-by-step explanation:

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When buying snacks for her classroom, Ms. Davis bought 4 bags of mixed nuts and 8 bags of pretzels. A bag of pretzels costs $1.5
vfiekz [6]

Answer:

C

Step-by-step explanation:

The snacks altogether (Pretzels and Nuts) cost 57 dollars.

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8.<br> Find the arc length of a central angle of 7pi/3 in a circle whose radius is<br> 3inches
pashok25 [27]

\textit{arc's length}\\\\ s = r\theta ~~ \begin{cases} r=radius\\ \theta =\stackrel{radians}{angle}\\[-0.5em] \hrulefill\\ \theta =\frac{7\pi }{3}\\ r=3 \end{cases}\implies s=(3)(\frac{7\pi }{3})\implies s=7\pi \implies s\approx 21.99~in

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2 years ago
The diagram shows squares arranged around the edge of a right triangle. What is the area of square B
Lina20 [59]

Step-by-step explanation:

take the right triangle

15^2 = 9^2 + x^2

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3 years ago
I really don't understand all of this
PolarNik [594]

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Distribute each one like a robot

Or factor it like a human.

5(x+2y - 3)

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Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.
Gekata [30.6K]

Take the augmented matrix,

\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]

Swap the row 1 and row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]

Add -2(row 1) to row 2, and -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]

Multiply through row 3 by 1/9:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]

Multiply through row 2 by -1/3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

Add -2(row 2) and -1(row 3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

So we have \boxed{x=-3,y=-2,z=4}.

3 0
3 years ago
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