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jok3333 [9.3K]
3 years ago
9

What is the range of this piecewise function

Mathematics
1 answer:
White raven [17]3 years ago
6 0

Answer as a compound inequality: -4 \le y < 2

Answer in interval notation:  [-4, 2)

=============================================

Explanation:

The range is the set of all possible y outputs of a function. When dealing with a graph like this, we just look at the highest and lowest points to determine which y values are possible.

The lowest point occurs when y = -4. We include this value. So far we have y \ge -4 which is the same as -4 \le y

The upper ceiling for the y value is y = 2. We can't actually reach this value because of the open hole at (-3,2). So we say that y < 2

Combine -4 \le y and y < 2 to get the compound inequality -4 \le y < 2

This says y is between -4 and 2, including -4 but excluding 2.

To convert this to interval notation, we write [-4, 2) where the square bracket says to include the endpoint and the curved parenthesis says to exclude the endpoint.

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XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3
nataly862011 [7]
<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

A = \frac{1}{2} x b x h      -------------(i)

Where;

A = 18\sqrt{3}

b = XZ = base of the triangle

h = YZ = height of the triangle = 6

<em>Substitute these values into equation(i) and solve as follows:</em>

18\sqrt{3} =  \frac{1}{2} x b x 6

18\sqrt{3} =  3b

<em>Divide through by 3</em>

6\sqrt{3} =  b

Therefore, b = XZ = 6\sqrt{3}

<em>Now, assume that the circle is centered at O;</em>

Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180°   [sum of angles in a triangle]

=>  XOZ + 30° + 30° = 180°

=>  XOZ + 60° = 180°

=>  XOZ = 180° - 60°

=>  XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}

\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}

\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}

\frac{1}{12}  = \frac{1}{2|OZ|}

\frac{1}{6} = \frac{1}{|OZ|}

|OZ| = 6

The radius of the circle is therefore 6.

<em>Now, let's calculate the length of the arc XZ</em>

The length(L) of an arc is given by;

L = θ / 360 x 2 π r          ------------------(ii)

Where;

θ = angle subtended by the arc at the center.

r = radius of the circle.

In our case,

θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

5 0
3 years ago
Determine the intercepts of the line <br> The x and y intercepts please help
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Answer:

See below in bold.

Step-by-step explanation:

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Isolate the h in c=25h+350
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Subtract 350 from both sides then divide both sides by 25.

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The answer would be iii: 10-15%
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