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3 years ago

Help guys!!!! pls & tyyyyy

1 answer:

TEA [102]3 years ago

8 0

Answer:

I think

A:(2,1)

B:(4,-2)

C:(4,-6)

D:(1,-3)

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Alex787 [66]

Answer:

Step-by-step explanation:

C, the answer is exponential function

3 0

2 years ago

lisov135 [29]

The last one (0,0) the middle of the graph is always (0,0)

8 0

3 years ago

Kipish [7]

The simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\frac{x - y}{4x^2 - 8xy + 3y^2} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{x^2 - y^2} -1$

Express x^2 - y^2 as (x + y)(x - y) and factorize other expressions

$\frac{x - y}{(2x - y)(2x - 3y)} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Rewrite the expression as products

$\frac{x - y}{(2x - y)(2x - 3y)} \times \frac{2x - 3y}{2x + y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Cancel out the common factors

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{4x^2 - y^2}{(x + y)} -1$

Express 4x^2 - y^2 as (2x - y)(2x + y)

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{(2x - y)(2x + y)}{(x + y)} -1$

Cancel out the common factors

$\frac{1}{x + y} -1$

Take the LCM

$\frac{1 - x - y}{x + y}$

Hence, the simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$