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lozanna [386]
3 years ago
11

What is the amplitude, period, and phase shift of f(x) = −4 sin(2x + π) − 5?. .

Mathematics
1 answer:
slamgirl [31]3 years ago
4 0

Answer:

Amplitude = 4; period = π; phase shift: x = negative pi over two

Step-by-step explanation:

In this problem you have to use the form <em>a sin (bx-c)+d </em>to find the amplitude, period, and phase shift.

Amplitude:4

Period : π

Phase shift: -\frac{\pi }{2}

I hope this helped! Have an amazing day :)

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Perform the operation. Enter your answer in scientific notation.   9 × 102 − 5.6 × 102 =
Arada [10]

Answer:

1489.2

Step-by-step explanation:

9*102-5.6*102=

102(9+5.6) =

102(9.0+5.6) =

102(14.6) =

102*14.6=

14.6*(100+2) =

(14.6*100+14.6*2) =

(1460+(14+.6) *2) =

1460+(28+1.2) =

1460+(28.0+1.2) =

1460+29.2=

1460.0+

29.2

=1489.2

3 0
3 years ago
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Identify all rays and lines in the picture below.<br> (Full question above)
Arte-miy333 [17]

Answer:

\bold{Rays:}\ \overrightarrow{EA},\ \overrightarrow{EB},\ \overrightarrow{EC},\ \overrightarrow{ED},\ \overrightarrow{DE},\ \overrightarrow{BE}\\\\\bold{Lines:}\ \overleftrightarrow{ED},\ \overleftrightarrow{BE}

Step-by-step explanation:

Look at the picture.

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2 years ago
HELP PLEASE, I will mark brainless
ahrayia [7]
To find the slope take two ordered pairs and put them in form of x2-x1/y2-y1 and then you can find the slope.

(-2,17) (-3,20)

(-2)-(-3) = 1
(17)-(20)=-3

slope is 1/-3
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Solve for x: 4 − (x + 2) −7<br> x −9
olga2289 [7]

Answer:

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2 years ago
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solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. x-2y+
Alekssandra [29.7K]

Answer:

x = -5, y = -6, z = -3

Step-by-step explanation:

Given the system of three equations:

\left\{\begin{array}{l}x-2y+3z=-2\\6x+2y+2z=-48\\x+4y+3z=-38\end{array}\right.

Write the augmented matrix for the system of equations

\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)

Find the reduced row-echelon form of the augmented matrix for the system of equations:

\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)\sim \left(\begin{array}{ccccc}1&-2&3&|&-2\\0&-14&16&|&36\\0&-6&0&|&36\end{array}\right)\sim \left(\begin{array}{ccccc}1&3&-2&|&-2\\0&16&-14&|&36\\0&0&-6&|&36\end{array}\right)

Thus, the system of three equations is

\left\{\begin{array}{r}x+3z-2y=-2\\16z-14y=36\\-6y=36\end{array}\right.

From the last equation:

y=-6

Substitute it into the second equation:

16z-14\cdot (-6)=36\\ \\16z=36-84\\ \\16z=-48\\ \\z=-3

Substitute y = -6 and z = -3 into the first equation:

x+3\cdot (-3)-2\cdot (-6)=-2\\ \\x=-2+9-12\\ \\x=-5

7 0
3 years ago
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