Mathematics, 16.07.2019 00:30 miller5452

What is the general equation of a sine function with an amplitude of 6, a period of π/4 and a horizontal shift of π/2?

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The general equation of a sine function with an amplitude of 6, a period of π/4, and a horizontal shift of π/2 is $\rm y = 6\;sin(8x-4\pi)$ .

Given :

Sine function with an amplitude of 6, a period of π/4, and a horizontal shift of π/2.

The following steps can be used in order to determine the general equation of a sine function:

Step 1 - The generalized equation of sine function is given below:

$\rm y = Asin(cx+d)$

where A is the amplitude, c is the time period and d is the horizontal shift.

Step 2 - According to the given data, the amplitude of 6, a period of π/4, and a horizontal shift of π/2.

Step 3 - The value of b is calculated as:

$\rm \dfrac{\pi}{4}=\dfrac{2\pi}{b}$

b = 8

Step 4 - The value of 'c' is calculated as:

$\rm 8\pi = -2c$

$c = -4\pi$

Step 5 - Substitute the value of the known terms in the above expression.

$\rm y = 6\;sin(8x-4\pi)$

Therefore, the correct option is C).

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Answer from: keeks11

The formula for the general sin function is y=Asin(bx+c)

where b is the period and c is the phase shift. We are given a period of $\frac{ \pi }{4}$ and the formula to solve for b is $period= \frac{2 \pi }{b}$ . We set up to solve for b: $\frac{ \pi }{4} = \frac{2 \pi }{b}$ . Cross multiply to get $b \pi =8 \pi$ and b = 8. Now for the phase shift. The formula for that is -c/b, and since we already know b, and we have a phase shift value of pi/2, we solve for c: $\frac{ \pi }{2}= \frac{-c}{8}$ and $8 \pi =-2c$ . Therefore, $c=-4 \pi$ . Putting all that together into the equation with the A value of 6, we have $y=6sin(8x-4\pi)$ . Factoring out the 8 then, the final equation is $y=6sin(8(x- \frac{\pi}{2} ))$ , third choice down.

Answer from: Quest

answer: the first graph

step-by-step explanation:

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.