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3 years ago

HELP ASAP!!! what similarities and differences did you notice between the three types of compound inequalities?

Mathematics

2 answers:

rodikova [14]3 years ago

8 0

Similarities

Adding or subtracting a number from both sides.

Multiplying or dividing both sides by a positive number.

Applying an increasing function to both sides.

Simplifying one or both sides.

Differences

Multiplying or dividing both sides by a negative number changes the direction of an inequality. So we cannot multiply or divide both sides by a variable unless we know that the variable is either always positive or always negative. This is not a concern when solving equalities.

When applying a decreasing function to both sides of an inequality, the direction of the inequality changes.

Swapping both sides of an inequality also changes the direction of an inequality. Again we don't have to worry about this with equalities.

Setler79 [48]3 years ago

4 0

Answer:Double inequalities are similar to and compound inequalities, in that their solutions need to satisfy two inequalities. However, solutions to or compound inequalities can include values that are included in either one of the two solution regions.

Step-by-step explanation:

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3 years ago

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2 years ago

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2 years ago

PLEASE HURRY! this is financial algebra

Zina [86]

Answer:

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Step-by-step explanation:

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The expressions which represents the slope of a tangent to the curve $y=\frac{1}{x+1}$ at any point (x, y) are:

$f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}$

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Mathematically, the slope of a tangent line to the curve is given by this equation:

$f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}$

Given the function:

$f(x)=y=\frac{1}{x+1}$

When (x + h), we have:

$f(x+h)=y=\frac{1}{x+h+1}$

Next, we would find the derivative of f(x):

$f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}$

Read more on slope of a tangent here: brainly.com/question/26015157

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2 years ago