How to simplify 7 1\2

Mathematics

2 answers:

Otrada [13]3 years ago

6 0

7 1/2 is already simplified to its lowest terms

malfutka [58]3 years ago

5 0

Its 15/2 becuase 15/2=7 1/2

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Calculate the following limit:

aleksklad [387]

$\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\$
$\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\$
$=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1
$

8 0

3 years ago

Hola como estan?Amblo en español y en portugués

Alexxx [7]

Answer:

Olá, how are you going? Amblo

4 0

3 years ago

Read 2 more answers

I hope it isn’t hard to read if it is don’t hesitate to ask for a better picture :-)

mihalych1998 [28]

Answer: I would say the third one