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3 years ago

8

Can someone pleaseeee help me with this?? I sort of understand, but every time I think I have it, I something confuses me. So ca

n someone help? Thank you.
Answers:
HL
LL
LA
HA

1 answer:

KiRa [710]3 years ago

4 0

HL is the correct answer the others are not possible

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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.

8 0

3 years ago

A rectangle has an area of 20 ft.² in a similar rectangle has an area of 180 ft.² what is the ratio of areas of the similar

vovikov84 [41]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{cccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array}\\\\
-----------------------------\\\\$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s^2}{s^2}=\cfrac{20}{180}\implies \left( \cfrac{s}{s} \right)^2=\cfrac{20}{180}\implies \cfrac{s}{s}=\sqrt{\cfrac{20}{180}}
\\\\\\
\cfrac{s}{s}=\cfrac{\sqrt{20}}{\sqrt{180}}\implies \cfrac{s}{s}=\cfrac{2\sqrt{5}}{6\sqrt{5}}\implies \cfrac{s}{s}=\cfrac{2}{6}\implies \cfrac{s}{s}=\cfrac{1}{3}$

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Maria bought a swimming pool with a circumference of 24 feet. She wants to buy a cover for her pool. What is the approximate siz

Angelina_Jolie [31]

24 feet I think but it might be like 25 so u can attach it

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The area of a circle is less than 1 dm². Find the radius of the circle. The area of a circle is less than 1 dm². Find the rad

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$\bf \textit{area of a circle}\\\\
A=\pi r^2\qquad 
\begin{cases}
\textit{let's say the Area is 1 exactly}\\
A=1
\end{cases}\implies 1=\pi r^2
\\\\\\
\cfrac{1}{\sqrt{\pi }}=r$

if you drop the radius to less than that, then the Area will also drop to less than 1 unit, in this case it happens to be a decimeter

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A Air Freshener Used 3 3/4 milliliters perfume wanted to make 3 air freshners how many milliliters of perfume woud she use

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