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lara [203]
3 years ago
13

The table shows the highest daily temperature in degrees Fahrenheit averaged over the month for Cosine City, where m is the numb

er of months since January 2001. (m = 0 represents January 2001.)
A sine function is written to represent the data.

What is the amplitude, period, and vertical shift of this equation?

Drag a value into each box to correctly complete the statements.

Mathematics
1 answer:
Bezzdna [24]3 years ago
8 0

Answer:

Part A) The amplitude = 24

Part B) The period = 24

Part C) Vertical shift = 36

Step-by-step explanation:

The general equation of the sine function:

y = A sin (Bx) + C

Where A is the amplitude and B = 360°/Period  and C is the vertical shift

See the attached figure which represents the graph of m and f(m)

So,

<u>Part A:</u>

The function has minimum at 12 and maximum at 60

The difference is = 60 - 12 = 48

So, The amplitude = 48/2 = 24

<u>Part B:</u>

Period: The period of a periodic function is the interval on which the cycle of the graph that's repeated in both directions lies.

We can deduce that the function completes one cycle within 24 months

So, the period = 24

<u>Part C:</u>

Vertical shift is obtained at m = 0

So, f(m) = 36

36 = A sin (0) + C

C = 36 ⇒ Vertical shift

So, The amplitude = 24

The period = 24

Vertical shift = 36

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Answer:

C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

Step-by-step explanation:

You have the quadratic function 2x^2-x+1=0 to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions ax^2+bx+c=0 with a\neq 0 the Bhaskara's Formula is:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}

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It usually has two solutions.

Then we have  2x^2-x+1=0  where a=2, b=-1 and c=1. Applying the formula:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}

Observation: \sqrt{-1}=i

x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i

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x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i

Then the correct answer is option C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

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