$Simplifying
x + -8y = 40

Solving
x + -8y = 40

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '8y' to each side of the equation.
x + -8y + 8y = 40 + 8y

Combine like terms: -8y + 8y = 0
x + 0 = 40 + 8y
x = 40 + 8y

Simplifying
x = 40 + 8y$

8 0

2 years ago

Read 2 more answers

A line plot shows the shape of the data, but not individual values.<br><br> true or false

kondor19780726 [428]

Safari has proven that this is false

8 0

2 years ago

Maria takes 25% of $26.50 and subtracts that amount from the normal price. She takes 10% of the discounted selling price and add

Irina18 [472]

I got 1.9875. Just do exactly what the problem says. 26.50*25%=6.625 26.50-6.625=19.875 19.875*10%=1.9875

6 0

2 years ago

I don't understand exponents

Firlakuza [10]

You must apply the law of exponets.

You must apply only at the same letter. X or Y

$\frac{x^{m}}{x^{n}} = x^{m-n}
$

$\frac{x^{3}}{x^{6}} = x^{3-6} =x^{-3}$

$\frac{y^{3}}{y^{-6}} = y^{(3-(-6))} =y^{3+6}=y^{9}$

$\frac{20x^{3}y^{3}}{32x^{6}y^{-6}} = \frac{20x^{(3-6)}y^{(3-(-6))}}{32}=\frac{20x^{(-3)}y^{(3+6)}}{32}=\frac{20x^{(-3)}y^{(9)}}{32}
$

But
$x^{-n} = \frac{1}{x^{n}}$
Then

$\frac{20x^{3}y^{3}}{32x^{6}y^{-6}} = \frac{20x^{(-3)}y^{(9)}}{32}=\frac{20y^{9}}{32x^{3}}=\frac{10y^{9}}{16x^{3}}=\frac{5y^{9}}{8x^{3}}$

5 0

3 years ago