Solve this inequality

Mathematics

2 answers:

Evgen [1.6K]3 years ago

5 0

Answer is x< -14.28
The < has a line under

nadya68 [22]3 years ago

5 0

$\dfrac{x}{-28} \geq 0.51$

This is the inequality. Let's multiply both sides by -28.
When multiplying/dividing both sides of an inequality by a negative number, we reverse the symbol.

$x \leq 0.51 \times (-28)$
$x \leq -14.28$

That's your solution to the inequality. Hope this helps! :)

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The corner of a room where two walls meet the floor should be a right triangle. Jeff makes a mark along each wall. One mark is 9

Aleksandr [31]

ab^2 = 15^2 + 8^2
ab^2 = 289
ab = 17

4 0

2 years ago

Read 2 more answers

What is the range of the function y = x 2?

Leno4ka [110]

Answer:

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

Step-by-step explanation:

Given the function

$y=x^2$

We know that the range of a function is the set of values of the dependent variable for which a function is defined.

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

$a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)$

$f\left(x\right)\ge \:0$

Thus,

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.