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xeze [42]
3 years ago
13

Is 0.26 is closer to 0, 1/2 or, 1

Mathematics
2 answers:
professor190 [17]3 years ago
8 0
Here, we know 1/2 = 0.50
So, the difference between 0.50 & 0.26 is 0.24
And difference between 0 and 0.26 is 0.26
1 is very apart from 0.26 as compared to other two

So, It is closer to 1/2

Hope this helps!
stepan [7]3 years ago
3 0
1/2 is the correct answer.....
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Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?
garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right).

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be z (x) = \cos x the base formula, where x is measured in sexagesimal degrees. This expression must be transformed by using the following data:

T = 180^{\circ} (Period)

z_{min} = -4 (Minimum)

z_{max} = 5 (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of 2\pi radians. In addition, the following considerations must be taken into account for transformations:

1) x must be replaced by \frac{2\pi\cdot x}{180^{\circ}}. (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

\Delta z = \frac{z_{max}-z_{min}}{2}

\Delta z = \frac{5+4}{2}

\Delta z = \frac{9}{2}

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

z_{m} = \frac{z_{min}+z_{max}}{2}

z_{m} = \frac{1}{2}

The new function is:

z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)

Given that z_{m} = \frac{1}{2}, \Delta z = \frac{9}{2} and T = 180^{\circ}, the outcome is:

z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

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3 years ago
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konstantin123 [22]

Answer:

Step-by-step explanation:

f(x) = \frac{2x}{3x^2 - 3} \\\\ Domain: \\ 3x^2 - 3 ≠ 0 \\ 3x^2 ≠ 3 \\ x^2 ≠ 1 \\ x ≠ \pm 1

D = {x E R/ x ≠ 1 and x ≠ - 1}

I Hope I've helped you.

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