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3 years ago

What is the value of the coefficient of the fourth term of the expansion of (a+b)^5 ?

Mathematics

1 answer:

melomori [17]3 years ago

6 0

Answer:

I don't know if you wanted it expanded using "Pascal's Triangle" or "Binomial Theorem" but either way is the same. Your expanded equation is:

a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

I hope this is any help to you, have a nice day.

Step-by-step explanation:

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Which statement describes the graph?

Leya [2.2K]

Answer:

Step-by-step explanation:

The answer is A, the graph raises, crosses at (0, 5) and then remains constant.

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3 years ago

Find a potential function for F or determine that F is not conservative. (If F is not conservative, enter NOT CONSERVATIVE.) F =

xxMikexx [17]

For $F$ to be conservative, we need to have

$\dfrac{\partial f}{\partial x}=\cos z$

$\dfrac{\partial f}{\partial y}=10y$

$\dfrac{\partial f}{\partial z}=-x\sin z$

Integrate the first PDE with respect to $x$ :

$\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx+\int\cos z\,\mathrm dx\implies f(x,y,z)=x\cos z+g(y,z)$

Differentiate with respect to $y$ :

$\dfrac{\partial f}{\partial y}=10y=\dfrac{\partial g}{\partial y}\implies g(y,z)=5y^2+h(z)$

Now differentiate $f$ with respect to $z$ :

$\dfrac{\partial f}{\partial z}=-x\sin z=-x\sin z+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

So we have

$f(x,y,z)=x\cos z+5y^2+C$

so $F$ is indeed conservative.

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3 years ago

Michelle buys a motorcycle for $10,000. The motorcycle loses 20% of its value each year. The depreciation of the motorcycle is s

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After two years, the motorcycle is worth $6,400.

I hope this answer helped you! If you have any further questions or concerns, feel free to ask! :)

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3 years ago

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Anastasy [175]

Step-by-step explanation:

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3 years ago

The zero of F -1(x) in F(x) = x + 3 is<br><br> -3<br> 0<br> 3

Effectus [21]

Answer:

The only zero of $F^{-1}(x)$ is $3$ .

Step-by-step explanation:

The function $F^{-1}$ is the inverse of function $F$ .

For a given $x$ and a given $y$ , $F^{-1}(y) = x$ if and only if $F(x) = y$ .

Let $y$ be a zero of $F^{-1}(y)$ . That is, $F^{-1}(y) = 0$ .

By the reasoning above, since $x = 0$ and this particular $y$ satisfy $F^{-1}(y) = x$ , it must be true that $F(x) = y$ for the same $x = 0\!$ and $y\!$ .

Since $x = 0$ and $F(x) = x + 3$ , $F(0) = 3$ . Therefore, $y = F(0) = 3$ since $y = F(x)$ .

Thus, the zero of $F^{-1}(y)$ would be $y = 3$ .

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2 years ago