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Kaylis [27]
3 years ago
13

Describe how you regroup when you add the sum of 64+ 43

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
8 0
    6    4
   +
     4    3
     _____ 
   1  0    7
   ______


3 + 4 = 7      Put down the 7.

6 + 4  = 10    Put down the 0 first, and then 1 is left.  Put down the 1 also to the extreme left.

So the answer is    107      
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Your answer should be 86.68.
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3 years ago
What is the fourth term in the binomial expansion (a+b)^6)
Dafna11 [192]

Answer:

20a^3b^3

Step-by-step explanation:

<u>Binomial Series</u>

(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example:  4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3

\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3

\implies \dfrac{6!}{3!3!}a^{3}b^3

\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3

\implies \left(\dfrac{120}{6}\right)a^{3}b^3

\implies 20a^3b^3

7 0
2 years ago
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AleksAgata [21]

Step-by-step explanation:

1 \frac{13}{20}

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3 0
3 years ago
This maximization linear programming problem is not in "standard" form. It has mixed constraints, some involving ≤ inequalities
stellarik [79]

Answer: See Annex

z (max)  = 190

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Step-by-step explanation:

z = 5*x  +  7*y     to maximize

Subject to:

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