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3 years ago

Algebra two graph problem!

Mathematics

1 answer:

Paladinen [302]3 years ago

5 0

It has no real solutions. To be precise, no real zeros (or roots).

The zeros (here called "solutions", I guess) are the points where the function intersects the x-axis: at those points the value of the function is zero. So they are solutions to the equation f(x)=0.

This function does not intersect the x-axis. Hence, no solutions.

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Converting 52ft2 into meters sq

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3 years ago

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

Evaluate f (2) if f (x)= 10-x^3

weeeeeb [17]

<span> f (x)= 10-x³.
f(2) means the value of the function when x=2.
So, we need to substitute 2 instead of x.

f(2) = 10 - 2³ = 10-8 =2

f(2) = 2

</span>

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3 years ago

I don't know what is 4309/73

Alisiya [41]

Step-by-step explanation:

59 2/73

or 59.02 as the answer

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