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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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Convert the polar coordinates (-3, -60°) to Cartesian coordinates.
notsponge [240]

Answer:

(1.5,-2.6)

Step-by-step explanation:

Given the polar coordinates (-3,60°).

Let our Cartesian coordinates be (x,y)

#We know that when converting the rectangular coordinates (x,y) to polar (r,θ), then:

r=\sqrt{x^2+y^2}\\\\\therefore r^2=x^2+y^2\\\\\theta=tan^{-1}(y/x)\\\therefore tan \theta=y/x

#Using the illustration above, we can express our polar coordinates as:

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#Solve simultaneously to solve for x and y:

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Hence, the Cartesian coordinates are (1.5,-2.6)

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3 years ago
What is 6 1/4 - 3 1/3
Anettt [7]
Hey There!

Remember, first find the gcf of 4,3 in this problem it is 12.

Steps to solve

1.6\frac{1}{4} - 3\frac{1}{3}

2.6\frac{1}{4} - 3 \frac{1}{3} = 2\frac{11}{12}

So, The answer is <span>2\frac{11}{12}</span>

Hope This Helps :)
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