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Tom [10]
2 years ago
11

What should be added to (-5 a2b - 9 ab) to get a2b + 3 ab?​

Mathematics
1 answer:
Doss [256]2 years ago
5 0

Based on given conditions,

a²b + 3 ab - (-5 a²b - 9 ab)

→a²b + 3 ab + 5 a²b + 9 ab

→(a²b + 5 a²b) + (3ab + 9 ab

Now collecting the coefficients and adding them,

(1 + 5)a²b + (3 + 9)ab

→6 {a}^{2}b + 12ab

Hence, the <u>6</u><u>a</u><u>²</u><u>b</u><u> </u><u>+</u><u> </u><u>1</u><u>2</u><u>a</u><u>b</u> should be added.

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Binomial Theorem, proof
IrinaVladis [17]

It follows from the definition of the binomial coefficient:

\dbinom nr=\dfrac{n!}{r!(n-r)!}

So we have

(n+1)\dbinom nr=(n+1)\dfrac{n!}{r!(n-r)!}=(r+1)\dfrac{(n+1)!}{(r+1)!(n-r)!}

That is, (n+1) gets absorbed into the numerator's factorial, and we introduct (r+1) into the denominator. Now, (n+1)-(r+1)=n-r, so we get

(n+1)\dbinom nr=(r+1)\dfrac{(n+1)!}{(r+1)!((n+1)-(r+1))!}=(r+1)\dbinom{n+1}{r+1}

as required.

4 0
3 years ago
Ik the answer but can you explain how <br> Giving brainliest
natka813 [3]

Answer:

4/5 * 3/7

To multiply the fractions, multiply the numerators and denominators separately

4*3/5*7

multiply the numbers

12/5*7

multiply the numbers

12/35

hope that helps

Step-by-step explanation:

8 0
3 years ago
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amm1812
A^2 + b^2 = c^2.....a and b are the legs and c is the hypotenuse

so u would use : 20^2 + 21^2 = 29^2
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C. 0.0389 I hope this helps
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