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2 years ago

(10 points)I REALLY NEED HELP NOW!! (A) positive (b)negative (c)zero

Mathematics

2 answers:

tatyana61 [14]2 years ago

8 0

Hello from MrBillDoesMath!

Answer:

Cchoice B is the correct answer.

Discussion:

As stated by an earlier respondent, choice B is the correct answer. But why?

From the diagram, point B lies between 0 and 1 which means B is positive

From the diagram, point C lies between -1 and -2 which means C is negative

The quotient of a negative number by a positive number is negative, giving the answer as choice c

Regards,

MrB

P.S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!

Sindrei [870]2 years ago

4 0

The answer is B because b is positive and C is negative so if you divide it you will get a negative answer

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3 years ago

Read 2 more answers

write a function g whose graph represents a translation 2 units to the right followed by a horizontal stretch by a factor or 2 o

Alenkinab [10]

$\bf f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}\\\\
f(x)=&{{ A}} \left|{{ B }}x+{{ C}} \right|+{{ D}}
\\\\
--------------------\\\\$

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

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

with that template in mind, let's see,

$\bf f(x)=|x| \implies \begin{array}{lllccll}
f(x)=&1|&1x&+0|&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.

the smaller A is, the wider it opens, the larger it is, the more it shrinks.

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3 years ago