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goldenfox [79]
3 years ago
6

How do you solve literal equations like 4y+x=40 and y=-3+2/3

Mathematics
1 answer:
baherus [9]3 years ago
5 0
4y+x=40
4y=40-x
y=10-1/4x (divide all by 4 to get y by itself)



y=-3+2/3
y=(-9+2)3 (distribution)
y=-7/3
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Answer:

I believe the answer is C :)

Step-by-step explanation:

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Suppose f (x) = x^2. Find the graph of f(x+2)
NISA [10]
\bf ~~~~~~~~~~~~\textit{function transformations}
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% templates
f(x)={{  A}}({{  B}}x+{{  C}})+{{  D}}
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~~~~y={{  A}}({{  B}}x+{{  C}})+{{  D}}
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\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
~~~~~~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}}\\
~~~~~~if\ {{  D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{  D}}\textit{ is positive, upwards}\\\\
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with that template in mind,

\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}

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8 0
3 years ago
Do you know how to use the Pythagorean identity?
goblinko [34]

The Pythagorean theorem or identity is given by

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Where

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8 0
2 years ago
3(x-4)= -5 slove for x
suter [353]
<span>3(x-4)= -5
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3 years ago
Solve for the indicated variable.
marta [7]

Answer:

Option (c) "r=\dfrac{q}{4c}-h"

Step-by-step explanation:

The given equation is :

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We need to solve it for r.

Dividing both sides of the given equation by 4c.

\dfrac{q}{4c}=\dfrac{4c}{4c}(h+r)\\\\(h+r)=\dfrac{q}{4c}

Now, subtract h both sides of the equation.

h+r-h=\dfrac{q}{4c}-h\\\\r=\dfrac{q}{4c}-h

Hence, this is the required solution.

3 0
3 years ago
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