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Ivan
3 years ago
11

Choose the function whose graph is given by:

Mathematics
2 answers:
xeze [42]3 years ago
6 0

Answer:

Option B. y = cosx + 1

Step-by-step explanation:

To understand the given graph we will note down it's features as below.

1). Since it's a periodic function and y value of graph is maximum at x=0 so its a cosine function.

2). Vertical shift of the graph is 1 means y=0 is the mid line of the graph.

3). Amplitude of the graph is = (2-0)/2 = 1

4). Phase shift of the given graph = 0

The graph satisfying all properties is option B. y = cosx + 1

Daniel [21]3 years ago
4 0

Answer:

The correct option is B.

Step-by-step explanation:

From the given graph it is clear that the maximum value of the function is 2 at x=0, so it a cosine function.

The general form of a cosine function is

y=a\cos(bx+c)+d       .... (1)

Where, a is amplitude, 2π/b is period, c is phase shift and d is midline.

Since maximum value is 2 and minimum value is 0, so

Amplitude=a=\frac{2-0}{2}=1

Midline=b=\frac{2+0}{2}=1

Period=2\pi\Rightarrow b=1

Since maximum value is at x=0, therefore the phase shift is c=0.

Put these values in equation 1.

y=1\cos(1x+0)+1

y=\cos(x)+1

Therefore the correct option is B.

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Explanation:

The given rule for the arithmetic sequence is A(n)=-6+(n-1)(\frac{1}{5} )

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute n=1,4,10 in the general rule for the arithmetic sequence.

To find the first term, substitute n=1 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(1)=-6+(1-1)(\frac{1}{5} )

A(1)=-6+(0)(\frac{1}{5} )

A(1)=-6

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute n=4 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(2)=-6+(4-1)(\frac{1}{5} )

A(2)=-6+(3)(\frac{1}{5} )

A(2)=\frac{-30+3}{5}

A(2)=\frac{-27}{5}

Thus, the fourth term of the arithmetic sequence is -\frac{27}{5}

To find the tenth term, substitute n=10 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(10)=-6+(10-1)(\frac{1}{5} )

A(10)=-6+(9)(\frac{1}{5} )

A(10)=-6+\frac{9}{5}

A(10)=-\frac{21}{5}

Thus, the tenth term of the arithmetic sequence is -\frac{21}{5}

Hence, the first, fourth and tenth terms of the arithmetic sequence is -6, -\frac{27}{5} and -\frac{21}{5}

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