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notsponge [240]
2 years ago
7

Let U={x: x is an integer and 2≤x≤10}. In each of the following cases, find A,B and determine whether A⊆B,B⊆A, both or neither:

A={x: 2x+1>7},B={x: x^2>20}. A={x:x^2-3x+2=0},B={x:x+7 is a perfect square}.
Mathematics
1 answer:
12345 [234]2 years ago
8 0

A is not a subset of B but B is a subset of A (that is can be found in A) that is B⊆A is correct

<h3>Set theory</h3>

Set is defined as the arrangement of elements. They can be represented using the venn diagram.

Given the following sets

U = {x: x is an integer and 2≤x≤10} = {3, 4, 5, 6, 7, 8, 9}
A = {x: 2x+1>7} = {x > 3}

B={x: x^2>20} = {x >± 20}

From the set, can see that A is not a subset of B but B is a subset of A (that is can be found in A) that is B⊆A is correct

Learn more on sets here: brainly.com/question/13458417

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3 years ago
The volume of a cylinder of radius 3r and height 6r is given by TT(3r)2(6r). Simplify the expression.
OLga [1]

Answer: \pi 36r^{2}

Step-by-step explanation:

ALRIGHT ARE YOU READY FOR EPIC MATH TIME?

because i know i am. so lets get this bread.

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interesting... that first character is called pi. maybe i should have said, "lets get this pi".

so now lets do some algebra. i am going to work from right to left. lets multiply 6r by 2. that gives us 12r. now we have... \pi (3r)(12r)

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6 0
2 years ago
Which does not show a direct variation between x and y?
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A direct variation is a mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. In other words, a direct variation is where y = x * (constant).

In answer a, y = x*.5, so it is a direct variation.
In answer c, y = x* \frac{1}{9}, so it is a direct variation.
In answer d, y = x*2, so it is a direct variation,

Only answer b is left, which means the answer must be 'b'. 

We also know 'b' is the answer because it cannot be expressed as <span>y = x * (constant). Instead, it is expressed as </span>y =  \frac{constant}{x}, which is not the same thing and is therefore not a direct variation.

Hope I helped, and let me know if you have any questions :)
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