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11Alexandr11 [23.1K]
3 years ago
8

Whats the definition for an open circle, a closed circle, a solution set, a dashed boundary line, a solid boundary line, a syste

m of inequalities, and a number line?
Mathematics
1 answer:
likoan [24]3 years ago
3 0
<span>Open and closed circles are used to mark the end points of intervals. If the circle is open, it means the end point is NOT included in the interval, but if the circle is closed, it means the end point is included. Consider the interval . In this interval, the low endpoint, the 3, is not included because you have a 'less than' sign. The high endpoint, the 6, is included because you have a 'less than or equal' sign. It is the presence or absence of the 'or equal' part that determines whether the circle is closed or open. The interval would be marked on the number line with an open circle at 3, a closed circle at 6, and a heavy line connecting the two.

</span>
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nevsk [136]

the 4.89 for 1 gallon is cheaper because it saves you money

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3 years ago
How would get the area?
yuradex [85]

Answer:

x > 5

Step-by-step explanation:

Formula: length x width

First, you must substitute in the numbers.

16-2x*10 < 60

Work:

(16-2x)10 < 60

Then, distribute the 10 to the 16 and to the 2x.

160-20x < 60

Next, substract 160 from both sides.

-20x < -100

Now, divide -20 from both sides.

x < 5

This isn't the answer yet because you have to flip the sign because you divided by a negative.

x > 5

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3 years ago
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The distance AB= [?] Round to the nearest tenth.
Law Incorporation [45]

Answer:

  • 3.6
<h3><u>Step-by-step explanation:</u></h3>

A = ( -2, 1 )

B = ( 1, -1 )

\boxed{\bf x_1  : - 2} \boxed{\bf x_2 : 1}

\boxed{\bf y_1 : 1} \boxed{\bf y_2 : - 1}

______________________________

  • \sf d =  \sqrt{(x_2 -x_1) ^{2} + ( y_2 - y_1) ^{2} }
  • \sf d =  \sqrt{(1 -  ( - 2)) ^{2} + ( - 1 - 1) ^{2}  }
  • \sf d =  \sqrt{ {3}^{2} +  {( - 2)}^{2}  }
  • \sf d =  \sqrt{(9 + 4)}
  • \sf d =  \sqrt{13}
  • \sf d = 3.6

<u>________________</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

7 0
2 years ago
Question 17 of 25
iogann1982 [59]

Answer:

I believe it’s C but not 100% sure

Step-by-step explanation:

8 0
3 years ago
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Find the argument of the complex number -3–9 in the interval 0°&lt; ϴ &lt; 360°,
Bingel [31]

Answer:

\theta=251.6^\circ

Step-by-step explanation:

<u>Complex Numbers</u>

They are expressed as the sum of a real part and an imaginary part:

Z = a+b\mathbf{ i}

Complex numbers can also be expressed in polar form:

Z = r(\cos\theta+\sin\theta \mathbf{ i}) = r. cis(\theta)

Where r is the modulus of the complex number and θ is the argument.

The argument can be calculated by:

\displaystyle \tan\theta=\frac{b}{a}

The angle θ must be calculated in the appropriate quadrant depending on the signs of the real and imaginary parts.

The complex number is given as:

Z = -3 -9\mathbf{ i}

Here: a=-3, b=-9

Since both components are negative, the argument lies in the third quadrant (180° < θ < 270°).

\displaystyle \tan\theta=\frac{-9}{-3}

\displaystyle \tan\theta=3

\theta=\arctan(3)

The calculator gives the answer 71.6°, we need to adjust the angle to the third quadrant by adding 180°, thus

\mathbf{\theta=251.6^\circ}

6 0
3 years ago
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