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yarga [219]
2 years ago
10

Show if √5×√5×√8 is rational or irrational numbers​

Mathematics
2 answers:
kobusy [5.1K]2 years ago
8 0
√5 x√5 x√8= √(5x5x8) = √200 = 10 √2
It’s an irrational number
Mekhanik [1.2K]2 years ago
8 0

Answer:

Irrational number

Step-by-step explanation:

<h3>Rational and irrational number:</h3>

  Prime factorize the number and then simplify.

\sf \sqrt{5}*\sqrt{5}*\sqrt{8}=\sqrt{5*5*8}

                     \sf = \sqrt{5*5*2*2*2}\\\ = 5* 2* \sqrt{2}\\\\= 10\sqrt{2}is an irrational number.

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×1 = 3 : 8.

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Step-by-step explanation:

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Step-by-step explanation:

they have different slopes and are linear

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atroni [7]

Answer:

<h2>3x² - 9x + 8</h2>

Step-by-step explanation:

(2x^2-5x+3)-(-x^2+4x-5)\\\\=2x^2-5x+3-(-x^2)-4x-(-5)\\\\=2x^2-5x+3+x^2-4x+5\qquad\text{combine like terms}\\\\=(2x^2+x^2)+(-5x-4x)+(3+5)\\\\=3x^2-9x+8

4 0
3 years ago
A survey of 279 SPC students was taken at registration. Of those surveyed: 51 students had signed up for a Language Arts course
stepladder [879]

Treating the amounts as Venn sets, we have that 161 students signed up for only Math, considering that 40 did not sign for any.

<h3>What are the Venn sets?</h3>

For this problem, we consider the following sets:

  • Set A: Students that have signed up for Arts.
  • Set B: Students that have signed up for Humanities.
  • Set C: Students that have signed up for Math.

4 students had signed up for all three courses students, hence:

(A ∩ B ∩ C) = 4.

8 students had signed up for both a Math and Humanities, hence:

(B ∩ C) + (A ∩ B ∩ C) = 8

(B ∩ C) = 8.

18 students had signed up for both a Math and Language Arts, hence:

(A ∩ C) + (A ∩ B ∩ C) = 18

(A ∩ C) = 14.

9 students had signed up for both a Language Arts and Humanities, hence:

(A ∩ B) + (A ∩ B ∩ C) = 9

(A ∩ B) = 5.

36 students had signed up for a Humanities course, hence:

B + (A ∩ B) + (B ∩ C) + (A ∩ B ∩ C) = 36

B + 5 + 8 + 4 = 36

B = 19.

51 students had signed up for a Language Arts course, hence:

A + (A ∩ B) + (A ∩ C) + (A ∩ B ∩ C) = 36

A + 5 + 14 + 4 = 51

A = 28.

Considering that there are 279 students, and supposing 40 did not sign for any course, we have that:

A + B + C + (A ∩ B) + (B ∩ C) + (A ∩ C) + (A ∩ B ∩ C) + 40 = 279.

28 + 19 + C + 5 + 8 + 14 + 4 + 40 = 279

118 + C = 279

C = 161

161 students signed up for only Math, considering that 40 did not sign for any.

More can be learned about Venn sets at brainly.com/question/24388608

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3 0
2 years ago
Consider the two triangles shown. Which statements are correct?
shutvik [7]

Answer:

A

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3 years ago
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