There are 12 sixth graders, 11 seventh

Kevin is three years older than Daniel. Two years ago, Kevin was 4 times as old as Daniel.

GarryVolchara [31]

Answer:

I believe that Kevin is 6 years old now.

Step-by-step explanation:

I used the guess and check method to figure this out.

First, I began by making Daniel 1 years old, two years ago. Kevin was 4 times as old as Daniel two years ago, so he would be 4 at the time. Now, Daniel would be 3 and Kevin would be 6 years old. The statement that Kevin is three years older than Daniel would confirm this, because 6-3=3. Hope this helps, this is my first time answering:)

3 0

2 years ago

Find the slope of line a and line b from two points

Degger [83]

Answer: Line A has a slope of 3/4 and Line B has a slope of 5/6

Step-by-step explanation: Look at the attachment

5 0

2 years ago

Read 2 more answers

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

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

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

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

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

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

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

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