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slava [35]
2 years ago
7

**i would appreciate it if someone helped! ** Here is a number pyramid puzzle.

Mathematics
1 answer:
solniwko [45]2 years ago
3 0

Answer:

*

Step-by-step explanation:

                   - 96

                -8         12

          - 4         2          6

     4         -1          -2       -3

16       .25     -4        .5           -6

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Factorize the following expressions
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Step-by-step explanation:

1. a² - ( b ² - 2bc + c² ) = a ² - b ² + 2bc - c²

2. 8p² - 18 q²

3. 3ab² - c²d + 3ab - b²c²

4. x² - 2x + 1

8 0
3 years ago
Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

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3 years ago
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Rudiy27
Yes, because the probability of choosing an odd number is equal...
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Answer:

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6 dimes are 60 cents

5 nickels are 25 cents

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60 + 25 + 4= 89

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3 years ago
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