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3 years ago

Expand the binomial (y+1)^8

Mathematics

1 answer:

gogolik [260]3 years ago

5 0

Using Pascal's triangle, we can determine the coefficients and degree of each term.

The coefficients for the formula will be: 1, 8, 28, 56, 70, 56, 28, 8, and 1.

We can set up our formula this way:

(a + b)^8 = (a^8 + 8a^7b + 28a^6b^2 + 56a^5b^3 + 70a^4b^4 + 56a^3b^5 + 28a^2b^6 + 8ab^7 + b^8)

We can plug our values for a and b into the equation to expand the binomial.

After plugging in our values, we're left with:

The reason the expanded value looks nearly identical to the formula is because our 2nd term is 1, and 1 raised to any power is equal to 1.

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4 years ago

Which represents the inverse of the function f(x) = 4x?

Nimfa-mama [501]

<h3>Answer: D) h(x) = (1/4)x</h3>

This is the same as h(x) = x/4

In decimal form, it would be h(x) = 0.25x

=============================================

Explanation:

Inverses undo the original function. The original function tells us to take any input (x) and multiply by 4 to get the output y = f(x) = 4x

To get the inverse, we reverse the operation applied here. The opposite of multiplication is division. So we'll divide by 4 instead of multiply by 4. This is why the answer is x/4 or (1/4)x

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Another approach is to do the following: Replace f(x) with y. Swap x and y. Solve for y

f(x) = 4x

y = 4x

x = 4y

4y = x

y = x/4

h(x) = x/4

We end up with the same result as before.

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Extra info:

For any real number x, the following two equations are true

h( f(x) ) = x

f( h(x) ) = x

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Find the given derivative by finding the first few derivatives and observing the pattern that occurs. d103 dx103 (sin(x))

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To find $\frac{d^{103}}{dx^{103}} \left(\sin{(x)}\right)$ , we find the first few derivatives and observe the pattern that occurs.

$\frac{d}{dx} (\sin{(x)})=\cos{(x)} \\ \\ \frac{d^2}{dx^2} (\sin{(x)})= \frac{d}{dx} (\cos{(x)})=-\sin{(x)} \\ \\ \frac{d^3}{dx^3} (\sin{(x)})= -\frac{d}{dx} (\sin{(x)})=-\cos{(x)} \\ \\ \frac{d^4}{dx^4} (\sin{(x)})= -\frac{d}{dx} (\cos{(x)})=-(-\sin{(x)})=\sin{(x)} \\ \\ \frac{d^5}{dx^5} (\sin{(x)})= \frac{d}{dx} (\sin{(x)})=\cos{(x)}$

As can be seen above, it can be seen that the continuos derivative of sin (x) is a sequence which repeats after every four terms.

Thus,

$\frac{d^{103}}{dx^{103}} \left(\sin{(x)}\right)= \frac{d^{4(25)+3}}{dx^{4(25)+3}} \left(\sin{(x)}\right) \\ \\ = \frac{d^3}{dx^3} \left(\sin{(x)}\right)=-\cos{(x)}$

Therefore,

$\frac{d^{103}}{dx^{103}} \left(\sin{(x)}\right)=-\cos{(x)}$ .

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