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AysviL [449]
3 years ago
6

(2x2y4)3 Simplify the exponential expression

Mathematics
1 answer:
Gwar [14]3 years ago
4 0

Answer:

6x^2y^4

Step-by-step explanation:

I hope you find this helpfull, btw I really need a brainliests, i just need one more to rank up, nd i would really appriciate it if you let me <em>borror</em> it ;D

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Solve the linear system ALGEBRAICALLY. Show All Steps for solving! <br> y=x2-4x-1<br> x+y=-3
Galina-37 [17]

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3 years ago
HELLPP ASAP!!!What are the exact solutions of x2 − 5x − 1 = 0 using x equals negative b plus or minus the square root of the qua
Mnenie [13.5K]

Answer: x = the quantity of 5 plus or minus the square root of 29 all over 2

Step-by-step explanation:

The given quadratic equation is expressed as

x² - 5x - 1 = 0

The equation is already in the standard form of ax² + bx + c

The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the given quadratic equation,

a = 1

b = - 5

c = - 1

Therefore,

x = [- - 5 ± √(- 5² - 4 × 1 × - 1)]/2 × 1

x = [5 ± √(25- - 4)]/2

x = [5 ± √29]/ 2

x = ( 5 + √29)/- 2 or x = (5 - √29)/2

6 0
3 years ago
The question is on the picture
Radda [10]

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Step-by-step explanation:

5 0
3 years ago
This graph shows the solutions to the inequalities y &gt; -
Goryan [66]
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7 0
3 years ago
In this task, you will perform multiplication on rational and irrational numbers and determine whether the sets are closed under
velikii [3]

Answer:

  • <em>The </em><em>set of rational numbers</em><em> is </em><em>closed under multiplication</em><em>.</em>
  • <em>The </em><em>set of irrational numbers</em><em> is </em><em>not closed under multiplication</em><em>. </em>
  • Although are only few instances where the set of irrational numbers may show closure, but it must be noted that closure property would not extend to the entire set of irrational numbers.
  • <em>The Product of two rational numbers will be rational.</em>

Step-by-step explanation:

The set of rational numbers contains the set of numbers that can be written as a fraction or ratio between two integers - integer values in the nominator and denominator (denominator ≠ 0).

The set of integers - denoted by Z - can be termed as rational numbers as integer, for instance, 3 can be written as the ration 3/1.

Therefore, when we multiply two rationals, it basically means we are just multiplying two such fractions which will yield another fraction of the same form as integers are closed under multiplication.

<em>Part A) Determining multiplication on rational numbers</em>

Let a/b and c/d be rational numbers where a, b, c and d are integers and (b, d ≠ 0).

For example, a = 2, b = 3, c = 4, and d = 5    a, b, c and d are integers

Lets multiple 2/3 × 4/5 to check what the multiplication of two rational number can bring.

So,

2/3 × 4/5 = (2)(4) / (3)(5) = 8/15

As 8 and 15 are the integers, and can be written in a ratio 8/15 in the nominator and denominator which make it a rational number. Hence, <em>the </em><em>set of rational numbers</em><em> is </em><em>closed under multiplication</em><em>.</em>

<em>Determining multiplication on irrational numbers</em>

The set of irrational numbers contains the set of numbers that can not be written as a fraction or ratio between two integers.

The product of two irrational numbers will be sometimes irrational.

There will be certain cases sometimes where the product of two irrational numbers will yield irrational number. But, it must be noted that few irrational numbers may multiply to yield a rational number.

For example,

                     √5 × √2 = √10 which is termed as irrational

                      √8 × √2 = √16 =  4 which is termed as rational

So, there are only few instances that may show closure, but it must be noted that closure property would not extend to the entire set of irrational numbers.

So, we can safely say that <em>the </em><em>set of irrational numbers</em><em> is </em><em>not closed under multiplication</em><em>.</em>

<em>Part B) Is the Product of two rational numbers irrational or rational?</em>

Let a/b and c/d be rational numbers where a, b, c and d are integers and (b, d ≠ 0).

Lets multiple a/b × c/d to check if the product of two rational numbers can be rational or irrational.

So,

a/b × c/d = ac / bd      where b, d  ≠ 0

As ac and bd are also integers, so the sets are closed under multiplication.

<em>So, ac/bd is a ratio having integers in the nominator and denominator which make it a </em><em>rational number</em>. Hence, <em>the Product of two rational numbers will be rational.</em>

<em>Proving closure property</em>

Let x and y are rational numbers such that

x = a/b and y = c/d  where b, d  ≠ 0 and a, b, c and d are integers

x × y = a/b × c/d

xy = ac / bd

As ac and bd are also integers, so the sets are closed under multiplication.

<em>So, ac/bd is a ratio having integers in the nominator and denominator which make it a </em><em>rational number</em>. Hence, <em>the product of two rational numbers will be rational.</em>

<em>For example,</em>

x = a/b = 6/7

y = c/d = 8/9

xy = a/b × c/d

xy = ac / bd = (6)(8) / (9)(7)

xy = 48/63

As 48 and 63 are the integers, and can be written in a ratio 48/63 in the nominator and denominator which make it a rational number. Hence, <em>the </em><em>set of rational numbers</em><em> is </em><em>closed under multiplication</em><em>.</em>

<em>Keywords: closure property, rational number, irrational number</em>

<em>Learn more about closure property and rational number from brainly.com/question/7667707</em>

<em>#learnwithBrainly</em>

4 0
3 years ago
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