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3 years ago

Can someone please help me with this

Mathematics

2 answers:

galben [10]3 years ago

4 0

Yes thats a function, it passes the vertical line test!

svet-max [94.6K]3 years ago

4 0

Answer:

Yes it is a function

Step-by-step explanation:

If it passes the vertical & horizontal line test, then it is a one to one. if it passes the vertical line test but fails the horizontal, then it is a function, but not one to one.

Hoped this helped you!

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Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

Here’s the riddle! <br> I’m aware it sayas “step 3” but i found it on a riddles website

Mazyrski [523]

Answer:

l have no clue

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the area.<br> help pls

Neko [114]

Answer:

30 square feet

Step-by-step explanation:

The area of a trapezoid is equal to (a+b)/2 * h
In this case, the value of a is 5, b is 10, and h is 4. You can substitute those numbers into the formula and get the answer of 30.

4 0

2 years ago

Read 2 more answers

Let ABCDEFGH be a cube and M the midpoint of GH. It is known that the distance between the lines BM and AD is a<img src="https:/

vovikov84 [41]

Answer:

√5

Step-by-step explanation:

We suppose the vertices are named clockwise around the top of the cube, then clockwise around the bottom (looking down from above the cube), with vertex E below vertex D. Then line AD is in plane ADEF, and line BM is in plane BCHG.

The distance between the named parallel planes is the distance between the lines. That distance is AB, which is given as √5.

_____

A diagram helps.

8 0

3 years ago

The Taylors Have recorded their weekly grocery expenses for the past 12 weeks and determined that the mean weekly expense was 60

kherson [118]

We are told that the Taylors Have recorded their weekly grocery expenses for the past 12 weeks and determined that the mean weekly expense was 60.26. Later , Mrs.Taylor discovered that 1 weeks expense of $74 was incorrectly recorded as $47.

We can find the correct mean weekly expenses by calculating the mean of difference between 74 and 47, then adding it to the given mean weekly expenses.

$\frac{74-47}{12}$

$\frac{27}{12}=2.25$

Now let us add 2.25 in 60.26 to find the correct mean weekly expenses.

$\text{Correct mean weekly expenses}=60.26+2.25$

$\text{Correct mean weekly expenses}=62.51$

Therefore, the correct mean weekly expenses for the Taylors will be 62.51.

8 0

3 years ago