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3 years ago

Which Statement is true?

1 answer:

6 0

A) $|-8| \ \textless \ |8|$ is false. There is no solution.
B) $|-2| \ \textgreater \ 1$ is true. There are infinite solutions.
C) $|-7| = -7$ is false. There is no solution.
D) $|12| \ \textgreater \ |-15|$ is false. There is no solution.

Answer: B

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What's a word problem for 10x=2

ohaa [14]

Answer: Divide each side by '10'. x2 = 9 Simplifying x2 = 9 Take the square root of each side: x = {-3, 3

Step-by-step explanation:

8 0

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.

8 0

3 years ago

Someone please help.

umka21 [38]

Let's go:

$\frac{y}{3} = \frac{27}{y} \\\\ y^2 = 27.3 \\\ y^2 = 81 \\\ y = \sqrt{81} \\\ y = 9$

I hope I helped you.

8 0

3 years ago

Someone help me with my math homework pleaseee. Find the volumes of the pyramids and the height is 7cm for the first one.

gregori [183]

$\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad 
\begin{cases}
B=\textit{area of the base}\\
h=height
\end{cases}$

now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12

now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8

now the last one on the far-right

has a height of 8, the Base is a Hexagon, with sides of 6

$\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad 
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=6
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)$

5 0

3 years ago

How I can answer this question, NO LINKS, if you answer correctly I will give u brainliest!

Ahat [919]

Answer:

w is greater than or equal to 358

7 0

2 years ago