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julia-pushkina [17]
3 years ago
14

Use the rectangle to the right. Write your answer in simplest form.

Mathematics
1 answer:
DochEvi [55]3 years ago
4 0

Answer:

perimeter: 12x-6

area: 6x-10

Step-by-step explanation:

PERIMETER:

6x-5(2)+2(2)

12x-10+4

12x-6

AREA:

6x-5•2

6x-10

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Find the function h(x) = f(x) ∘ g(x) if f(x) = x(2 - x) and g(x) = 3x.
Harrizon [31]
F(x) = x(2 - x)
g(x) = 3x
Substitute 3x for x in the function f(x)

h(x) = f(x) <span>∘ g(x) = f(g(x)) = 3x(2 - 3x)
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100 POINTS IF U GET THIS RIGHT!
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9.2 m.

Step-by-step explanation:

6.2 m.

7.2 m.

8.2 m.

9.2 m.

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2 years ago
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How do you put q fraction into a decimal
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Death Valley has an altitude that is below sea level. Which situation describes a hiker in Death Valley who hikes until he reach
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3 years ago
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Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?
-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>
  • In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

  1. U = {x | x is a real number}
  2. A = {x | x is an odd integer}
  3. R = {x | x = 3, 7, 11, 27}

So,

  • A = 1, 3, 5, 7, 9, 11... are the elements of set A.
  • R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
  • Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(C) no because each element in set A is not represented in set R

(D) no, because each element in set R is not represented in set A

8 0
1 year ago
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