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3 years ago

6

A choir is performing in front of an audience. A microphone is placed in front of the choir to capture sound. Which polar equati

on represents the microphone the choir is using? r = 4cos(2θ) r = 4cos(4θ) r = 4 + cos(θ) r = 4 + 4cos(θ)

2 answers:

Dimas [21]3 years ago

6 0

Answer: D. r = 4 + 4cos(θ)

Step-by-step explanation:

Sonbull [250]3 years ago

5 0

Answer:

not b or c

Step-by-step explanation:

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Read 2 more answers

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

$\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}$

Cramer's rule says that

$x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}$

where $x_i$ is the solution for i-th variable, and $\mathbf A_i$ is a modified version of $\mathbf A$ with its i-th column replaced by $\mathbf b$ .

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

$\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}$

$\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}$

$\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))$

$\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120$

The modified matrices and their determinants are

$\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240$

$\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360$

$\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480$

Then by Cramer's rule, the solution to the system is

$x = \dfrac{-240}{-120} \implies \boxed{x = 2}$

$y = \dfrac{360}{-120} \implies \boxed{y = -3}$

$z = \dfrac{-480}{-120} \implies \boxed{z = 4}$

5 0

2 years ago

It’s this correct 1 2

oksian1 [2.3K]

Make sure that you put a line over the 1s on #2

I can not see the fraction on #1

3 0

4 years ago