For the graph given below , choose the correct inequation taking x as the variable

$x > - 1$

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$x \geq - 1$

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$x < - 1$

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$x \leq - 1$

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Solution

The correct option is B
$x \geq - 1$

The graph on the point -1 is darkened , hence it is inclusive, i.e x can be equal to -1 and its extending towards the right . So x can take any value greater than or equal to -1 ; Hence the equation that can be represented is $x \geq - 1$

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Similar questions

Choose the correct answer.
$\int \frac{1}{x^{2} + 2 x + 2} d x$ equals

$(a) x t a n^{- 1} (x + 1) + C (b) t a n^{- 1} (x + 1) + C (c) (x + 1) t a n^{- 1} x + C (d) t a n^{- 1} x + C$

Q. Assertion :Statement-1:

c o t^{- 1} (x) - t a n^{- 1} (x) > 0

for all

x < 1

Reason: Statement-2: Graph of

c o t^{- 1} (x)

is always above the graph of

t a n^{- 1} (x)

for all

x < 1

Q. Solve the following inequality:

\frac{(x - 1) (x - 2) (x - 3)}{(x + 1) (x + 2) (x + 3)} > 1

For each graph given alongside, write an inequation taking x as the variable :

Choose the correct answer in each of the question.
$\int \frac{x}{(x - 1) (x - 2)} d x$ equals
$(a) l o g ∣ ∣ ∣ \frac{(x - 1)^{2}}{x - 2} ∣ ∣ ∣ + C (b) l o g ∣ ∣ ∣ \frac{(x - 2)^{2}}{x - 1} ∣ ∣ ∣ + C (c) l o g ∣ ∣ (\frac{x - 1}{x - 2})^{2} ∣ ∣ (d) l o g | ((x - 1) (x - 2)) | + C$

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For the graph given below , choose the correct inequation taking x as the variable

The correct option is B x≥−1 The graph on the point -1 is darkened , hence it is inclusive, i.e x can be equal to -1 and its extending towards the right . So x can take any value greater than or equal to -1 ; Hence the equation that can be represented is x≥−1

The correct option is B
$x \geq - 1$

The graph on the point -1 is darkened , hence it is inclusive, i.e x can be equal to -1 and its extending towards the right . So x can take any value greater than or equal to -1 ; Hence the equation that can be represented is $x \geq - 1$