Question

Help

Answer 1

oh wait

Answer 2

it's recursive, so you would start with a1 = 2

then apply the rule, which tells you to multiply the term by 3 and subtract 1 to get the next term

Answer 3

then take the result from that operation and then find the next term using the same rule

Answer 4

recursive formulas stink but thankfully we don't use these very often

Answer 5

a1=2
a2=5
a3=14
a4=41
a5=122

Answer 6

good

Answer 7

so for 9 you want to write an arithmetic equation to solve for d

Answer 8

an = a1 + (n-1)d

an = 11
a1 = 2
n = 4

solve for d

Answer 9

d=3

Answer 10

awesome, now we use the sum formula

Answer 11

okay

Answer 12

since we want n = 50 we will need to find a50 first

Answer 13

use this: an = a1 + (n-1)d

to find an when n = 50, keeping the same a1 and d value as before

Answer 14

149/50

Answer 15

a50 = 2 + (50-1)3 = ?

Answer 16

297

Answer 17

ah think I see what the misunderstanding is

the 50 in a50 is not treated like a number, it's just a label

Answer 18

149

Answer 19

good, 149

Answer 20

now we go back to our sum formula

Answer 21

then find sn using these values

Answer 22

3775

Answer 23

awesome, that's the final answer

Answer 24

for 10, this is what we are given:

3x + 2y = 14
2x - 4y = 4

start by doubling the first eqution

Answer 25

okay

Answer 26

then, you should be able to eliminate y by adding the equations together

Answer 27

wait by doubling do you mean 6x+4y=28

Answer 28

yes

Answer 29

since there's +4y on the top and -4y on the bottom these cancel when you add the equations

Answer 30

12x=112

Answer 31

6x+4y=28
2x - 4y = 4

start by adding the x terms. 6x + 2x = ?

Answer 32

oh we are adding I thought we were multiplying

Answer 33

x=4

Answer 34

as a general rule you can multiply equations by a constant, but not multiply the equations together

Answer 35

yes, x = 4

Answer 36

now solve for y

2x - 4y = 4

when x = 4

Answer 37

y=1

Answer 38

good so (4,1) = your ans