Write an inequality for the following situation:

What is the solution set of the equation x + 6 = 2(x + 3) - x?

Svetach [21]

Answer:

x = all real numbers

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Distributive Property

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients

Step-by-step explanation:

Step 1: Define

Identify

x + 6 = 2(x + 3) - x

Step 2: Solve for x

[Addition Property of Equality] Add x on both sides: 2x + 6 = 2(x + 3)
[Distributive Property] Distribute 2: 2x + 6 = 2x + 6

Here we see that both sides are identical.

∴ our solution is all real numbers.

6 0

3 years ago

Simplify 24 ÷ (-2)(3) + 7Simplify 24 ÷ (-2)(3) + 7

Mnenie [13.5K]

Solving the expression $24\div (-2)(3) + 7$ we get -29

Step-by-step explanation:

We need to simplify $24\div (-2)(3) + 7$

Solving:

$24\div (-2)(3) + 7$

Using DMAS rule

$=24\div(-2)*(3)+ 7$

$=-12*3+ 7$

$=-36+7$

$=-29$

So, solving the expression $24\div (-2)(3) + 7$ we get -29

Keywords: Solving the expression

Learn more about Solving the expression at:

#learnwithBrainly

7 0

3 years ago

Help me on this radical expression !

n200080 [17]

$\boxed{pd\sqrt[4]{48p^3d}}$

<h2>Explanation:</h2>

Here we have the following expression:

$\sqrt[4]{48p^7d^5}$

So we need to simplify it:

$\sqrt[4]{48p^7d^5} \\ \\ \\ We \ can \ write: \\ \\ p^7=p^4\cdot p^3 \\ \\ d^5=d^4\cdot d \\ \\ \\ So: \\ \\ \sqrt[4]{48p^4\cdot p^3\cdot d^4\cdot d} \\ \\ \\ By \ property: \\ \\ \sqrt[n]{x^n}=x \\ \\ \\ Finally: \\ \\ \boxed{pd\sqrt[4]{48p^3d}}$

<h2>Learn more:</h2>

Mathematical expressions: brainly.com/question/14200575#

#LearnWithBrainly

3 0

4 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

3 0

3 years ago

The rate for one unit of a given quantity is called a(n)

asambeis [7]

The rate for one unit of a given quantity is called a unit rate.

6 0

3 years ago