Im having trouble with this :(

Mathematics

1 answer:

MrRa [10]3 years ago

5 0

The answer is plus 4 because a positive plus a negative is a negative

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Can you show work for 8.01/0.9

Gnoma [55]

Answer:8.01/0.9=8.9

Step-by-step explanation:

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3 years ago

What kind of shift does the graph have? y = -sin (x-1) from y = sin(x)

Mademuasel [1]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

now, with that template in mind, let's see

$\bf \begin{array}{lllll}
y=&-1sin(&1x&-1)&+0\\
&A&B&C&D
\end{array}
\\\\\\
\textit{horizontal shift of }\cfrac{C}{B}\implies \cfrac{-1}{1}\implies -1
\\\\\\
\textit{A is negative, so is flipped upside-down}$

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3 years ago

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iris [78.8K]

Answer:

9,600

Step-by-step explanation:

100-4 is 96

96 times 100 is 9600

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