GUYS HELPPPP PLEASE I RLY NEED IT RN, I NEED ANSWER #7 IM DESPERATE RN PLSSS

Mathematics

1 answer:

drek231 [11]2 years ago

8 0

Answer:

6 units

Step-by-step explanation:

The questions states that the area = perimeter.

From this we can set up an equation:

$x*y=2x+2y$

Since we know one of the sides is 3 we can substitue one of the variables for 3. In this case I choose $y$

$3*x=2x+2*3\\3x=2x+6$

Solving for x we get $x=6$ .

That's our answer!

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

alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

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

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