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Kisachek [45]
3 years ago
5

The sum of a number and 2 times its reciprocal is -3

Mathematics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

(-2,-1)

Step-by-step explanation:

let the number=x

its reciprocal=1/x

x+2(1/x)=-3

x+2/x=-3

x²+2=-3x

x²+3x+2=0

(x+2)(x+1)=0

x=-2,-1

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allsm [11]

(n+2)^4 =

(n²+4n+4)(n²+4n+4) =

n^4 + 4n³ + 4n² + 4n³ + 16n² + 16n + 4n² + 16n + 16 =

n^4 + 4n³+24n²+ 4n + 16


Hope this helps

8 0
3 years ago
Total employment for sheet metal workers in 2016 is projected to be 201,000. If this is a 6.3% increase from 2006, approximately
natka813 [3]
201,000 = (2006 employment) +6.3%*(2006 employment)

201,000 = (2006 employment)*(1.063)

201,000/1.063 = (2006 employment) ≈ 189,087 . . . . . matching selection D
7 0
3 years ago
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3.8 repeating as a fraction
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Ans=3.77777777...
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4 0
3 years ago
3/5a = 5 1/2 rational number equation​
photoshop1234 [79]

Answer:

a = 55/6

Step-by-step explanation:

<u>Solving in steps:</u>

  • 3/5a = 5 1/2
  • 3/5a = 11/2
  • a = 11/2 : 3/5
  • a = 11/2*5/3
  • a = 55/6  or  9 1/6
7 0
3 years ago
Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

7 0
3 years ago
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