Please help I have 3 more questions

Mathematics

2 answers:

USPshnik [31]3 years ago

8 0

Answer: There is no remander

↓↓↓Using Long division ↓↓↓

0 0 6

1 9 1 1 4

− 0

1 1

− 0

1 1 4

− 1 1 4

______________________________________

Hope this helps!

chubhunter [2.5K]3 years ago

3 0

Answer:

Step-by-step explanation:

no remainder 6 with the remainder of 0

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If the graph of the function g defined by g(x)=-2x^2-8 is translated vertically upward by 6 units it becomes the graph of a func

Aleks [24]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
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}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, let's see yours $\bf \begin{array}{lclcclll}
g(x)=&-2(&1x^2&+0)&-8\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D 
\end{array}$

if it shifts up vertically by 6 units, well, then just add 6 to D

5 0

4 years ago

What is the antiderivative of 1-cos^2(x)

Lyrx [107]

Answer:

$\frac{1}{4}$ sin2x + $\frac{1}{2}$ x + c