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3 years ago

Solve inequality ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍

Mathematics

2 answers:

Marta_Voda [28]3 years ago

7 0

X=64

BIDMAS
B= BRACKETS
I=INDICES
D=DIVISION
M=MULTIPLICATION
A=ADDITION
S=SUBTRACTION

artcher [175]3 years ago

4 0

Answer:

x < 64

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

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

$\displaystyle \frac{x}{8} < 8$

Step 2: Solve for x

[Multiplication Property of Equality] Multiply 8 on both sides: $\displaystyle x < 64$

Here we see that any number x less than 64 would work as a solution to the inequality.

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jasenka [17]

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Step-by-step explanation:

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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

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