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sergejj [24]
2 years ago
6

A={ 0,1,3,5,6}; B={1,2,3,7} draw the venn diagram of AUB Writ​

Mathematics
1 answer:
Sati [7]2 years ago
8 0

Answer:

Here,

n(A)={0,1,3,5,6}

n(B)={1,2,3,7}

n(AUB)=?

we know that,

AUB={0,1,2,3,5,6,7}

Step-by-step explanation:

is it okay ?

hopefully it works <3 ❣️

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Urgent help needed...............
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Answer:

Option b is correct

\{x | x \neq \pm 7, x\neq 0\}.

Step-by-step explanation:

Domain is the set of all possible values of x where function is defined.

Given the function:

h(x) = \frac{9x}{x(x^2-49)}

To find the domain of the given function:

Exclude the values of x, for which function is not defined

Set denominator = 0

x(x^2-49) = 0

By zero product property;

x = 0 and x^2-49= 0

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\{x | x \neq \pm 7, x\neq 0\}

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3 years ago
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To get from his house to the lecture hall at school, Lin walked west 651 feet. After class, he walked northeast 910 feet to the
Ahat [919]

The general direction that Lin walked from the gym to his house is; B: Lin walked southwest, creating an obtuse triangle.

<h3>How to interpret distance bearing?</h3>

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Distance between the lecture hall and gym = 910 feet.

Distance between the gym and Lin apostrophe's house = 615 feet.

Distance between the lecture hall and Lin apostrophe's house = 651 feet.

Now, since this 3 distances form a triangle and the 3 distances are unequal, then we can call it an obtuse triangle since he walked west and then walked northwest.

Now, since he walked back to his house from the gym, we can say that he walked southwest if we picture the bearing of his first two directions.

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2 years ago
Describe the steps on how to graph the ordered pair ( - 3, 4)​
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Step-by-step explanation:

6 0
2 years ago
A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s
andre [41]
Answer:

x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation: 

Let 

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio, 

Length of the pool (with the patio) 
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio) 
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
=  (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0    (1)

Let 

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

x = \frac{-b \pm  \sqrt{b^2 - 4ac}}{2a}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l + w)^2) + 16lw}}{8} &#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}&#10;\\ = \frac{-2(l + w) \pm  \sqrt{(4(l^2 + 6lw + w^2)}}{8}
= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}&#10;\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}


Since (l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \  0, -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right) is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by 

\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}




7 0
3 years ago
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