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DanielleElmas [232]
2 years ago
7

Please solve this question fast.

Mathematics
1 answer:
Alexxandr [17]2 years ago
3 0

Answer:

See Explanation

Step-by-step explanation:

{a} =  {6}^{x},  \:  \: b = {6}^{y}  ...(given) \\  {a}^{y}. {b}^{x}  = 36..(1) \\ pluggig \: the \: values \: of \: a \: and \: b \: in \: (1) \\  ({6}^{x}) ^{y} . ({6}^{y}) ^{x} =  {6}^{2}  \\ {6}^{xy}.{6}^{xy} =  {6}^{2}  \\  {6}^{xy + xy}  =  {6}^{2}  \\  {6}^{2xy}  =  {6}^{2}  \\ bases \: are \: equal \: so \: exponentes \: will \: also \: be\\ \: equal \\  \therefore \: 2xy = 2 \\ xy =  \frac{2}{2}  \\ \huge \red { \boxed{ xy = 1}} \\ hence \: proved \\

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Read 2 more answers
WILL MARK BRAINLIEST!!!!20 POINTS!!!!URGENT!!!
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the firsts third and last one

Step-by-step explanation:

hope this helped

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3 years ago
The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k
stepan [7]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;% left side templates&#10;\begin{array}{llll}&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}\\\\&#10;--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.
8 0
2 years ago
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