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Arturiano [62]
4 years ago
9

Consider the universal set R,R, define the interval A=[−7,1],A=[−7,1], interval B=(−1,5),B=(−1,5), and CC be the negative real n

umbers. Complete the following exercises in interval notation.
1. A∪B=A∪B=

2. A∩B∩C=A∩B∩C=

3. A∩(B∪C)=A∩(B∪C)=

4. A^c ∪ B^c∪ C^c=

5. (C−A)∪(B−C)=(C−A)∪(B−C)=

Mathematics
1 answer:
AURORKA [14]4 years ago
8 0
Draw a diagram representing the real number line, and the sets A  and B.

According to the diagram:

<span>1. A∪B= [-7, 5)

2. </span><span>2. A∩B∩C= (-1, 0)

</span>3. A∩(B∪C)= [-7, 1] <span>∩ (-INFINITY, 5) = [-7, 1]

4. </span>4. A^c ∪ B^c∪ C^c= (not A) ∪ (not B) ∪ (not C) = 

(-infinity, -7) ∪ (1, infinity)   ∪  (-infinity, -1] ∪ [5, infinity)   ∪ [0, infinity)

= (-infinity, -1] ∪ (1, infinity ) ∪ [0, infinity) = (-infinity, -1] ∪ [0, infinity) 

5. (C−A)∪(B−C)= "in C but not in A" union "in B but not in C"

=(-infinity, -7) ∪ [0, 5) 

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the corresponding point on the graph of  y = f(x) where g(x) = \frac{1}{3(f(x))} is (-12,\frac{1}{36}) .

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

We have , The point (-12,12) is on the graph of y=f(x).  We need to Find the corresponding point on the graph of y=g(x) where g(x)= 1/3f(x) . Let's find out:

On function y = f(x) , a point p(-12,12) lies . Another function g(x) = \frac{1}{3(f(x))} .

According to question ,  as On function y = f(x) , a point p(-12,12) lies :

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Now, Putting x = -12 in function g(x) = \frac{1}{3(f(x))} we get:

⇒ g(x) = \frac{1}{3(f(x))}

⇒ y=g(-12) = \frac{1}{3(f(-12))}     { 12= f(-12) }

⇒ y= \frac{1}{3(12)}    

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4 years ago
On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th
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Answer:

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Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, a = \sqrt{(3-0)^{2}+(4-0)^{2}}

Or, a =  \sqrt{9+16}

Or, a =   \sqrt{25}

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Similarly

The measure of length BC = b = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, b = \sqrt{(3-3)^{2}+(2-4)^{2}}

Or, a =  \sqrt{0+4}

Or, b =   \sqrt{4}

∴ b = 2 unit

And

So, The measure of length CA = c = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, c = \sqrt{(3-0)^{2}+(2-0)^{2}}

Or, c =  \sqrt{9+4}

Or, c =   \sqrt{13}

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Now, area of Triangle written as , from Heron's formula

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Or, A = \frac{3}{2} × \sqrt{1+\sqrt{13} }

∴  Area of triangle = 1.5\sqrt{4.6}

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