Pls help and look at the screenshot will give brainlist

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

At Candyland Daycare there are 7 children to every 2 caretakers. If there are 35 children at the daycare, then how many caretake

OLga [1]

10
because you would do 35 divided by 7 which is 5.
5 x 2 is 10

8 0

3 years ago

Read 2 more answers

There are 33 questions on Allison's

Brums [2.3K]

Answer:

x=15

y=18

Step-by-step explanation:

then x+y=33

and 2x+5y=120

then x=33–y

since 2x+5y=120

then 2(33–y)+5y=120

then 66–2y+5y=120

then 66+3y=120

then 3y=120–66

then 3y=54

then y=54/3

then y=18

then x=33–18=15

4 0

3 years ago

PLEASE PLEASE PLEASE HELP ME. IM STUCK IN A SUMMER SCHOOL AND IF I DONT GET GOOD GRADES ILL GET IN BIG TROUBLE AND IM TOTALLY LO

Mrrafil [7]

Answer:

We know that in the box there are:

4 twix

3 kit-kat

Then the total number of candy in the box is:

4 +3 = 7

a)

Here we want to find the probability that we draw two twix.

All the candy has the same probability of being drawn from the box.

So, the probability of getting a twix in the first drawn, is equal to the quotient between the number of twix and the total number of candy in the box, this is:

p = 4/7

Now for the second draw, we do the same, but because we have already drawn one twix before, now the number of twix in the box is 3, and the total number of candy in the box is 6.

this time the probability is:

q = 3/6 = 1/2

The joint probability is the product of the individual probabilities, so here we have

P = p*q = (4/7)*(1/2) = 2/7

b) same reasoning than in the previous case:

For the first bar, the probability is:

p = 3/7

for the second bar, the probability is:

q = 2/6 = 1/3

The joint probability is:

P = p*q = (3/7)*(1/3) = 1/7

c) Suppose that first we draw a twix.

The probability we already know that is:

p = 4/7

Now we want another type, so we need to draw a kit-kat, the probability will be equal to the quotient between the remaining kit-kat bars (3) and the total number of candy in the box (6)

q = 3/6

The joint probability is:

P = p*q = (4/7)*(3/6) = 2/7

But, we also have the case where we first draw a kit-kat and after a twix, so we have a permutation of two, then the probability in this case is:

Probability = 2*P = 2*2/7 = 4/7

3 0

3 years ago

Who is really good with math? I need serious help i am failing math

Oduvanchick [21]

Answer: I am really good at math so if you need help just say so

3 0

2 years ago