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2 years ago

Find the third term in the expansion of (a+b)^9 in the simplest form

Mathematics

1 answer:

forsale [732]2 years ago

6 0

Answer:

$The\ third\ term\ is\ \boxed{36a^7b^2}\\$

Step-by-step explanation:

$\left(\begin{array}{c}9\\ 1\end{array}\right)=\dfrac{9}{1} =9\\\\\left(\begin{array}{c}9\\ 2\end{array}\right)=\dfrac{9*8}{2*1} =36\\\\\\(a+b)^9=a^9+\left(\begin{array}{c}9\\ 1\end{array}\right)a^8b+\left(\begin{array}{c}9\\ 2\end{array}\right)a^7b^2+....+b^9\\\\\\The\ third\ term\ is\ \boxed{36a^7b^2}\\$

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otez555 [7]

Answer:

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Step-by-step explanation:

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2 years ago

There are 10 girls and 8 boys in Mr Augellis class. If 5 more girls and 5 more boys join the class, will the ratio of girls to b

zimovet [89]

Yes they will stay the same

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3 years ago

HELP ME PLEASE, it’s confusing

kakasveta [241]

The number at the end of the equation shifts the graph up or down.

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2 years ago

Read 2 more answers

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g100num [7]

Answer:

Option B: 3x + 3y - 3

Step-by-step explanation:

In an isosceles triangle, the two congruent sides are equal.

We are told the base is; x - y - 2 units

Now let each of the congruent sides be represented as A.

Thus the perimeter equation will be;

P = 2A + x - y - 2

Now, we are told that the perimeter is; 7x + 5y - 8 units

Thus;

7x + 5y - 8 = 2A + x - y - 2

Rearranging gives;

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2 years ago

In this exercise, we estimate the rate at which the total personal income is rising in the Richmond-Petersburg, Virginia, metrop

babymother [125]

Answer:

The rate at which the total income is increasing is$1627415600

Step-by-step explanation:

From the given data

Population in the area=P=961400

The rate of increase in population=dP/dt=9200

The Average Income=I=$30593

The rate of increase in Average Income=dI/dt=$1400

Now the total income is given as

Total Income=TI=PI

So the rate of increase of total personal income is given as $\frac{d}{dt}(PI)$

From the product rule of derivatives

$\frac{d}{dt}(P*I)=I*\frac{dP}{dt}+P*\frac{dI}{dt}$

Here

I is the Average Income which is given as $30593

dI/dt is the rate of increase of income which is given as $1400.

P is the population of the area which is given as 961400

dP/dt=9200 is the rate of increase of population which is given as 9200

By substituting the values,

$\frac{d}{dt}(P*I)=I*\frac{dP}{dt}+P*\frac{dI}{dt}\\\frac{d}{dt}(P*I)=30593*9200+961400*1400\\\frac{d}{dt}(P*I)=\$1627415600$

So the rate at which the total income is increasing is$1627415600

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3 years ago