All categories

2 years ago

Pls help! will give brainlist!

Mathematics

2 answers:

xz_007 [3.2K]2 years ago

6 0

The answer is 564 three digit numbers you can form with 1,2,3,4,5,6

ira [324]2 years ago

3 0

564 is the amount of 3 digit numbers that can be formed

You might be interested in

PLZZ HELPP YOU WILL BE THE BRAINLEIST!!!!!!!

sasho [114]

i need a graph so i can see then i will solve it

4 0

2 years ago

Read 2 more answers

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

Can any math genius help me with this

vovangra [49]

Here's my working for 1) You need to find the exterior angle, then divide by 360 to find the number of sides:

Applying these steps : 

180 (Interior Angles) - 162 = 18 (Exterior angle)

360 ÷ 18 is 20 sides 

For 2)

Its the same method, so apply the steps:

180 - 175 = 5

360 ÷ 5 = 72 sides 

Hope it helps! :) 

8 0

2 years ago

6-2x=3+5x Whats the solution set?

Dimas [21]

Answer:

x = 3/7

Step-by-step explanation:

Step 1:

6 - 2x = 3 + 5x Equation

Step 2:

6 = 3 + 7x Add 2x on both sides

Step 3:

3 = 7x Subtract 3 on both sides

Answer:

x = 3/7 Divide

Hope This Helps :)

8 0

2 years ago

Read 2 more answers

_____ less than a number is 51

Kryger [21]

Answer:

W number - blank is 51.

Step-by-step explanation:

I don't understand what the question is. What is the number unless you want something random then here:

Blank - 52

a number = 1

52 - 1 = 51

3 0

2 years ago