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ki77a [65]
3 years ago
8

This table represents a relationship between x and y, where x is the independent variable.

Mathematics
2 answers:
kotykmax [81]3 years ago
6 0
2x because this is the answer
chubhunter [2.5K]3 years ago
4 0
2x, that should be the answer I can assume, hopefully that helps
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happy birthday

Step-by-step explanation:

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The one with the reflection over the line HI and then the 6 unit translation to the right
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A projectile is fired into the air with an initial vertical velocity of 160 ft/sec from ground level. How many seconds later doe
djverab [1.8K]

The maximum height of the projectile is the maximum point that can be gotten from the projectile equation

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\mathbf{t = \frac{- 160}{-32}}

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Hence, the projectile reaches the maximum after 5 seconds

Read more about maximum values at:

brainly.com/question/6636648

8 0
3 years ago
Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;% left side templates&#10;\begin{array}{llll}&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}\\\\&#10;--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see    \bf f(x)=\sqrt{x}\qquad &#10;\begin{array}{llll}&#10;g(x)=&\sqrt{0.5x}\\&#10;&\quad \uparrow \\&#10;&\quad  B&#10;\end{array}

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
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Step-by-step explanation:

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5 0
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