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telo118 [61]
2 years ago
8

Which inequalities would have an open circle when graphed? Check all that apply. T Less-than-or-equal-to 25 -2. 5 Less-than-or-e

qual-to m x > 5. 4 One-half less-than x x > 0.
Mathematics
1 answer:
Alexandra [31]2 years ago
4 0

The inequalities would have an open circle when graphed are as follows;

The inequality shows that it does not contain boundary point 5.

        \rm x>5.4

The inequality shows that it does not contain the boundary point 1/2.

        \rm \dfrac{1}{2}>x

The inequality shows that it does not contain the boundary point 0.

        \rm x>0

<h3>What is a circle?</h3>

A round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the center):

The graph will have an open circle only if it has the symbol < or > because it indicates that the boundary is not included.

  • The inequality shows that it does not contain boundary point 5.

        \rm x>5.4

  • The inequality shows that it does not contain the boundary point 1/2.

        \rm \dfrac{1}{2}>x

  • The inequality shows that it does not contain the boundary point 0.

        \rm x>0

To know more about circles click the link given below.

brainly.com/question/11833983

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For this case we have the following equation:

(2x + 3) ^ 2 + 8 (2x + 3) + 11 = 0

Let u = 2x + 3

We have:

u ^ 2 + 8u + 11 = 0

By definition, given an equation of the form ax ^ 2 + bx + c = 0

The quadratic formula, to find the solution can be written as:

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In this case we have:

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Substituting in the quadratic formula we have:

See attached image

Answer:

Option B

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