Can someone help me with this?

Mathematics

2 answers:

Step2247 [10]2 years ago

6 0

the square root below is between the integers of 9 and 10

Step-by-step explanation:

9×9 is equal to 81

10*10=100

nadya68 [22]2 years ago

4 0

Answer:

Step-by-step explanation:

It is not E becuase SQRT(97) is less than SQRT(100), SQRT(100) is 10,

SQRT(81) is 9 SO SQRT(97) is between 9 and 10, C.

Hope it helps :)

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If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

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Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$