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2 years ago

PLEASE ANSWER ALL QUESTIONS 50 POINTS BRAINLIEST IT'S EASY

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

6 0

Answer:

Here they are Hope this helps :3

Step-by-step explanation:

1. 52

2. 4

7. (im not sure i think its H or A)

7.(not sure think its F)

8.x=6

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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

2 years ago

Answers? I have no clue what to do

zysi [14]

#2 is answer 4, #3 is answer 3, #4 is answer 4. The answer to #5 is 6xy if it is a possibility.

3 0

2 years ago

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If trapezoid T'U'V'W' represents trapezoid TUVW reflected over the y-axis, the ordered pair of W' is ___________. (Note: You mus

docker41 [41]

the answer is negative 7,2

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2 years ago

If you deposit Tsh 4000 into an account paying 9% annual interest compound monthly ,low long until there is Tsh 10000 in the acc

pav-90 [236]

Answer:

It will take 10.63 years.

Step-by-step explanation:

Giving the following information:

Present value (PV)= $4,000

Future Value (FV)= $10,000

Interest rate (i)= 9% annual compounded

<u>To calculate the time required to reach $10,000, we need to use the following formula:</u>

n= ln(FV/PV) / ln(1+i)

n= number of years

n= ln(10,000 / 4,000) / ln(1.09)

n= 10.63

It will take 10.63 years.

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2 years ago

Amber is on a family cell phone plan with 2 other family members. The family gets 720 shared cell phone minutes a month, each fa

chubhunter [2.5K]

Amber plus 2 family members = 3 total people on the plan.

They share the minutes equally so divide the minutes by 3 to find how many she can use:

720 / 3 = 240 minutes per month.

Now divide the monthly minutes by 30 days:

240 / 30 = 8 minutes per day.

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2 years ago