How do you expand brackets

5-6n=2n+5 <br>Solve for n

olganol [36]

<span>5-6n=2n+5 , 5 cancels out
-6n -2n = 0
-8n = 0
n =0</span>

7 0

3 years ago

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Which is the intersection of the sets {0, 1, 5, 7, 8} and {0, 5, 7, 9}?

MrRissso [65]

The intersection of the steps is just a fancy way of asking wi\hich numbers are in both equations so lets look and see :)

The 1st = 0...1...5...7...8
The 2nd =0........5...7........9
Answer = D {0, 5, 7}
Hope this helps have any questions let me know :)

6 0

3 years ago

18=b/3+3 what's the solution

madam [21]

b= 63 and that is your answer

5 0

2 years ago

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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

17.5 how to make this into a fraction

BARSIC [14]

Take the .5 of the 17.5 and convert it to a fractin.
we know .5 is equal to 50% or 1/2

now we have a mixed number $17 \frac{1}{2}$

in order to convert it into an improper fraction you need to multiply the whole number by the denominator and then add the numerator all while keeping the denominator constant.

$W \frac{N}{D} = \frac{(W*D)+N}{D}$

so $17 \frac{1}{2}= \frac{(17*2)+1}{2}= \frac{35}{2}$

5 0

3 years ago

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