MaxWong

$\text{I'm gonna type everything in }\LaTeX$

$\text{First Question:}\\ \quad \frac{1}{x}+\frac{1}{1-x}\\=\frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}\\=\frac{1-x+x}{x-x^2}\\=\frac{1}{x-x^2}$

$\boxed{\color{red}\therefore \frac{1}{x}+\frac{1}{1-x}=\frac{1}{x-x^2}}$

$\text{Second Question:}\\ \quad 1+\frac{1}{x+1}\\=\frac{x+1}{x+1}+\frac{1}{x+1}\\=\frac{x+2}{x+1}$

$\boxed{\color{red}\therefore 1+\frac{1}{x+1}\neq \frac{2}{x+1}}$

$\text{Third Question:}\\\quad x+\frac{1}{x}\\=\frac{x^2}{x}+\frac{1}{x}\\=\frac{x^2+1}{x}$

$\boxed{\color{red}\therefore x+\frac{1}{x}=\frac{x^2+1}x}$

$\text{Last Question:}\\\quad\frac{1}{x-1}-\frac{1}{x+1}\\=\frac{x+1}{x^2-1}-\frac{x-1}{x^2-1}\\=\frac{x-x+1+1}{x^2-1}\\=\frac{2}{x^2-1}$

$\boxed{\color{red}\therefore\frac{1}{x-1}-\frac{1}{x+1}\neq \frac{2x}{1-x^2}}$

$\color{red}55mm:8cm\\ \color{red}= 5.5cm : 8 cm\\ \color{red}= 11cm:16cm\\\color{red}=11:16$

length of the round edge = $\color{red}\pi\times 50 = 50\pi cm$

Dont forget the straight edge.

length of the straight edge = $\color{red} 2\times 50 = 100 cm$

Perimeter of the semi-circle = $\color{red} 100 + 50\pi$

$\dfrac{5x-1}{4}=5+\dfrac{x}{2}\\ 5x-1 = 20 + 2x\\ 3x = 21 \\ \color{red}\boxed{\color{aqua}x = 7}$

Purplemath, for maths

https://www.purplemath.com/

C.

For A, y = -5/x, the x is in the denominator of the fraction. so it's not a  linear equation.

For B, y = 2 - 5x², that's obviously a quadratic equation.

For D, y = -5x², that's obviously a quadratic equation.

B.

For A. f(x)= 2x², That's obviously a quadratic.

For C, f(x)= 1 / 2x, the x is in the denominator of the fraction, not the numerator, so it's not a linear function.

For D, f(x)= - 1/ 2(x-2), the x is in the denominator of the fraction, not the numerator, so it's not a linear function neither.

So the answer is B

Why there are 2 of the same post help LOL