How do you solve 5 5/8 - 2 6/7 - 2/7 - 5/8

1 answer:

7 0

first make 5 5/8 a improper fraction so 45/8 and then do the same thing to 2 6/7 so now 45/8- 20/7 and then subtract that to 2/7 and that subtract that by 5/8 to finally get the answer

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The graph of g(x) = -x^2 is shifted 3 units left and 1 unit up. If this new graph is f(x), then what is the value of f(-1.5)

kondor19780726 [428]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

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

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

with that template in mind, let's see.

to the left by 3 units, C = +3
up by 1 unit, D = 1

$\bf g(x)=-x^2\\\\\\ g(x)=-1(1x+\stackrel{C}{0})^2+\stackrel{D}{0}\implies g(x)=-1(1x+3)^2+1
\\\\\\
g(x)=-(x+3)^2+1\impliedby f(x)\qquad \qquad f(-1.5)=-[(-1.5)+3]^2+1$

and surely you know how much that is.

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Step-by-step explanation:

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