Answer:
The given functions are
The sine function has a standard period of 
by definition. However, this might change if we use a factor as coefficient of the x-varible, but in this case we don't have that.
Therefore, the period of both trigonometric functions is .
Now, the images of each function is the y-variable set values that defines each function.
So, the function has an image defined by the set ![[-3,5]](https://tex.z-dn.net/?f=%5B-3%2C5%5D)
. It's impotant to notice that the range of a standard function is [-1,1], however, in this case, the function was shifted 1 unit up and it was streched by a factor of 4, that's why the standard image changes to .
About the second function , the image set is ![[-2,2]](https://tex.z-dn.net/?f=%5B-2%2C2%5D)
, because the function was streched by a factor of 2.
Additionally, the image attached shows the graph of the given functions.
Answer:
The correct answer is C. That problem best represent x - 15 =60.
Answer:
The point-slope equation of the line is y - 2 = 3(x + 9)
Step-by-step explanation:
The form of the point-slope equation is y - y1 = m(x - x1), where
- m is the slope of the line
- (x1, y1) is a point on the line
∵ The slope of a line is 3
∴ m = 3
∵ The line passes through point (-9, 2)
∵ x1 = -9
∴ y1 = 2
→ Substitute the values of m, x1, and y1 in the point-slope form
∵ y - y1 = m(x - x1)
∴ y - 2 = 3(x - (-9))
→ Remember (-)(-) = (+)
∴ y - 2 = 3(x + 9)
∴ The point-slope equation of the line is y - 2 = 3(x + 9)
The answer is =, because in addition, it doesn't matter what order you add things in, it comes out the same.
