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3 years ago

Who can tell me what all the prime numbers are,please!!!!!!!!!!!!!

Mathematics

2 answers:

Oxana [17]3 years ago

8 0

There are an infinite amount of prime numbers plz be more clear on the range

Volgvan3 years ago

6 0

Prime numbers are numbers that only have one pair of multiples; 1 and the number itself. I can't give you all the possible prime numbers, but I'll at least give you the prime numbers between 1 and 100.

1
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97

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What is the common factor of the numerator and denominator in the expression <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%282x%2

mixer [17]

Answer:

Step-by-step explanation:

(x - 4)

What you are being asked to do is find the exact same binomial in the top as is in the bottom.

But there's a small catch. You must stipulate that x cannot equal 4. If it does, then you will get 0/0 which is undefined. You can't have that happening -- not at this level.

Any other value for x is fine.

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2 years ago

Aldon and Jamal raised the same amount of money for the school fundraiser. Aldon

Naddika [18.5K]

Answer: $5

Step-by-step explanation: This can be represented by the equation

40 + 12T = 25 + 15T

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2 years ago

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kupik [55]

I believe it would be 12/18 so that would be 2/3 and 0.6 in decimal form

4 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=13%20%5Ctimes%2011%20%3D%3F" id="TexFormula1" title="13 \times 11 =?" alt="13 \times 11 =?" al

gizmo_the_mogwai [7]

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7 0

2 years ago

Need step-by-step. Please answer correctly. This determines what grade I get. I will mark brainliest to the first person who ans

tatyana61 [14]

Answer:

3√3

Step-by-step explanation:

For the problem shown here, your answer 3√3 is correct.

When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:

$\dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}$

If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.

<u>Example</u>:

$\dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}$

It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.

9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.

Please discuss this question with your teacher.

3 0

2 years ago