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prohojiy [21]
3 years ago
14

t%7B2%7D%20%2B%20%20%5Csqrt%7B3%7D%20%20%7D%20%20%2B%20...%20%2B%20%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7Bn%20%2B%201%7D%20%20%2B%20%20%5Csqrt%7Bn%7D%20%7D%20%20%20%5Cleqslant%2020" id="TexFormula1" title=" \frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{n + 1} + \sqrt{n} } \leqslant 20" alt=" \frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{n + 1} + \sqrt{n} } \leqslant 20" align="absmiddle" class="latex-formula">
help....................​
Mathematics
1 answer:
statuscvo [17]3 years ago
8 0

Answer:

all i see is ..........

Step-by-step explanation:

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A triangle with sides of 14 15 and 29 does it form a right triangle 
Morgarella [4.7K]

Answer: No.

Step-by-step explanation: This triangle does not form a right triangle because it does not have a 90 degree angle.

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5x-2x-1x=<br> A.4x<br> B.2x<br> C.8x <br> What is it
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Answer:

2x

Step-by-step explanation:

We can first pull the x out of every term:

5x-2x-1x=x(5-2-1)

Then, simplify:

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A salad dressing recipe includes 4/5 cup of lime jucie for every two cups of greek yogurgt
amid [387]
And where is the rest of question?
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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
It's in the attachment <br> URGENT
maxonik [38]

Step-by-step explanation:

\frac{2a}{a + b - c}  =  \frac{2b}{b + c - a}  =  \frac{2c}{a + c - b}  = k \\  \\ by \: theorem \: on \: equal \: ratios \\   \\ \frac{2a + 2b + 2c}{a + b - c + b + c - a + a + c - b}  = k \\  \\  \frac{2(a + b + c)}{a + b + c}  = k \\ \\   \therefore \: k = 2

5 0
3 years ago
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