Work out the formula for the nth term of thequadratic sequence:5, 11, 19, 29
1 answer:
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Answer:
a) not proportional
b) proportional; k =
Step-by-step explanation:
a) for any proportional equation, the line must pass through the origin. The equation in a) is y = 4x + 1, and the '+1' is the y-intercept. This means that the line does not pass through the origin, so x and y cannot increase by the same amount (i.e. they are not proportional).
Another way to determine this is is to use the y = kx base. If you have an equation that fits that it's likely proportional.
Here, if the equation was only y = 4x then it'd be proportional because u can see that k = 4. This is not the equation though, and the 4x + 1 doesn't fit to the y = kx formula so it can't be proportional.
b) straight away you can see that there's no 'c' term (y = mx + c) which means the y-intercept is 0, so the line passes through the origin. While this does not immediately mean the line is proportional, you can make sure that it is by checking it fits with the y = kx equation.
y = -(3/5)x fits with y = kx, with k being -3/5
Answer:
Step-by-step explanation:
We have been given that vector v has a direction of (5,3). We are asked to find direction angle for v to nearest degree.
We know that direction of vector with components can be determined by
.
Upon substituting the components of given vector in above formula, we will get:
Upon rounding to nearest degree, we will get:
Therefore, the direction angle for vector v is 31 degrees.
First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
= Solution
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.