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OlgaM077 [116]
3 years ago
9

The real number square root of 23 belongs to which set of numbers?

Mathematics
1 answer:
BARSIC [14]3 years ago
8 0
It  belongs to the set of Irrational Numbers.
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Explain how to simplify this complex fraction. 330pages/3/4hour.
meriva
Answer:
= 330/1 ÷ 3/4

= (330 × 4) / (3 × 1)

= 1320/3

= 440/1

<span>= 440 

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6/5? Could anyone help me I've been stuck on it for a while. 6 over 5 how do I didvide this and get a ratio?
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2 years ago
Sunaina had three pizzas. She shared the pizzas with her friends. Each of them ate
kupik [55]

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6 friends

Step-by-step explanation:

Take 1/7 multiply by 5/6

8 0
3 years ago
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Pls help with 29..tank you
trapecia [35]

Answer:

See below

Step-by-step explanation:

Let the  2 roots be  A and A^2.

Then A^3 = c/a and A + A^2 = -b/a.

Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-

A^3 + (A^2)^3 =  ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2

= (A + A^2)^3 - 3A*3( A + A^2)

Substituting:-

c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)

Multiply through by a^3:-

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Factoring:-

ac(a + c) = 3abc - b^3

This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.

a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3

c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.

These are equal so we have

ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3

The 3acb^2 cancel out so we have:-

a^3c - ac^3 =  3a^2bc - 3abc^2 + b^3c - ab^3

ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)

ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)

Divide through by (a - c):-

ac(a + c) = 3abc - b^3 , which is the result we got earlier.

This completes the proof.

 

3 0
3 years ago
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Please help me<br> How many solutions does the equation<br> 1-(6x - 4) = 3(1 – x) have?
AnnZ [28]

Answer: one solution

this is all i know

3 0
3 years ago
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