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3 years ago

What is the value of 3|2x + y|when x = 6 and y = -4? -48 -24 24 48

Mathematics

1 answer:

Gelneren [198K]3 years ago

8 0

Answer:

Step-by-step explanation:

i answered it and it was correct

your welcome! :)

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Hat term can you add to StartFraction 5 over 6 EndFraction x minus 4 to make it equivalent to One-half x minus 4?

kolezko [41]

Answer:

$\dfrac{-x}{3}$ should be added to the given equation.

Step-by-step explanation:

Let a is added to the given equation.

According to the given condition,

$\dfrac{5x}{6}-4+a=\dfrac{x}{2}-4\\\\a=\dfrac{x}{2}-4+4-\dfrac{5x}{6}\\\\a=\dfrac{x}{2}-\dfrac{5x}{6}\\\\a=\dfrac{3x-5x}{6}\\\\a=\dfrac{-2x}{6}\\\\a=\dfrac{-x}{3}$

So, $\dfrac{-x}{3}$ should be added to the given equation.

8 0

3 years ago

Simplify, state all restrictions.

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 $y \ne -x$

Read more about expressions at:

brainly.com/question/723406

#SPJ1

7 0

2 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$