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2 years ago

HELP

Mathematics

1 answer:

Semmy [17]2 years ago

8 0

The second and third functions both decrease with a slope of -4, but the second fiction has a y intercept of -3 and the third has a y intercept of +3. The first and fourth functions are both increasing with a slope of positive 4. The first one has a y intercept of -3, but the fourth has a y- intercept of -3. all of the functions are linear.

it helps to write all as equations

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Is it is 250.1?

Step-by-step explanation:

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2 years ago

Simplify the radical

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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

Please help i’m so stuck i can’t do calc for the life of me

KATRIN_1 [288]

Answer:

$-3x^2+29x-150+\frac{946x^2-341x-756}{x^3+6x^2-3x-5}$ is the answer. See steps below.

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}\\\\-3x^5+11x^4+33x^3-26x^2-36x-6\\\\\mathrm{and\:the\:divisor\:}x^3+6x^2-3x-5:\\\\\mathrm{Quotient}=\frac{-3x^5}{x^3}=-3x^2\\\\\mathrm{Multiply\:}x^3+6x^2-3x-5\mathrm{\:by\:}-3x^2\:\:\rightarrow\:\:-3x^5-18x^4+9x^3+15x^2\\\\\mathrm{Subtract\:}-3x^5-18x^4+9x^3+15x^2\mathrm{\:from\:}-3x^5+11x^4+33x^3-26x^2-36x-6\mathrm{\:to\:get\:new\:remainder}.\\\\\mathrm{Remainder}=29x^4+24x^3-41x^2-36x-6$

$=-3x^2+\frac{29x^4+24x^3-41x^2-36x-6}{x^3+6x^2-3x-5}$

Repeat the steps and you will reach a point where no further division is possible.

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2 years ago