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2 years ago

SOMEONE HELP what is 975687x56475+5475-567

Mathematics

2 answers:

Elena-2011 [213]2 years ago

8 0

Answer:

$the \: answer \: for \: your \: question \: is \: = \: 55101928 233$

<h3>Hope it helps you.. </h3>

scoray [572]2 years ago

7 0

Answer:

55101928233

Step-by-step explanation:

i simplified it

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chubhunter [2.5K]

Answer:

2x + .8 = 58.9

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Step-by-step explanation:

2x + .8 = 58.9

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We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$

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xxMikexx [17]

Answer:

Step-by-step explanation:

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