Question

What is the fourth term in this multiplication pattern?



multiplication pattern using the formula 2 • 6n – 1



A.
12


B.
72


C.
432


D.
2592

Answer 1

@crazysingh

Answer 2

well substiute n = 1, then n= 2 then n = 3 etc

and evalaute

Answer 3

huh

Answer 4

im in 6th grade

Answer 5

and is the question actually

\[2 \times 6^{n - 1}\]

Answer 6

yes

Answer 7

@crazysingh

Answer 8

on so here is

n = 1

\[2 \times 6^{ 1 - 1} = 2 \times 6^{0}\]

just find the value of the equation above

then multiply it by 6 to get a new answer

repeat the process twice more for your answer...

Answer 9

72

Answer 10

wait

Answer 11

216

Answer 12

so you need

2 x 6 x 6 x 6 =

Answer 13

432

Answer 14

so its C. thx so much!!!

Answer 15

Glenn invested $18,000 at 5% interest compounded annually.



How much interest will Glenn earn in 3 years?



A.
$20,837.25


B.
$19,826.24


C.
$2837.25


D.


$2340.23

Answer 16

@campbell_st

Answer 17

wow... I didn't think compound interest was a 6th grade problem...

Answer 18

it is

Answer 19

lol...ok... then do you know the formula...

Answer 20

no

Answer 21

ok... well here is a link

https://qrc.depaul.edu/StudyGuide2009/Notes/Savings%20Accounts/Compound%20Interest.htm

I think it explains it well...

good luck

Answer 22

\(\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\\ \quad \\
r=rate\qquad t=years\qquad n=periods\qquad P=principal\)

Answer 23

k thx

Answer 24

this is hard!

Answer 25

Glenn invested $18,000 at 5% interest compounded annually.



How much interest will Glenn earn in 3 years?



A.
$20,837.25


B.
$19,826.24


C.
$2837.25


D.


$2340.23

Answer 26

is the question

Answer 27

*sigh*

Answer 28

oh yes i knew the formula i just didnt get what u wre saying

Answer 29

the only different your n value = 1

Answer 30

P= principal or original amount = 18,000
r = rate in decimal format 3% = 0.03
t = years
n = periods per year, well is annually, thus once a year, thus period = 1

thus \(\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\\ \quad \\
r=0.03\qquad t=3\qquad n=1\qquad P=18,000
\\ \quad \\
A=18,000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\)

Answer 31

umm hard so hard!!!

Answer 32

thus campbell_st 's comment -> wow... I didn't think compound interest was a 6th grade problem...

Answer 33

here is a link to an online calculator that will help you

http://web2.0calc.com/

Answer 34

huh???!!!

Answer 35

ok

Answer 36

what do i do though on the calculator??!!

Answer 37

b

Answer 38

@jdoe0001

Answer 39

plug in your values I gather

Answer 40

whatever im leaving cya ugh im gonna get a F!

Answer 41

im dum!

Answer 42

stupid

Answer 43

i hate myself!!!

Answer 44

BYE!

Answer 45

keep in mind that -> \(\bf A=18,000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=18,000\left(1+0.03\right)^3
\\ \quad \\
\implies A=18,000\left(1.03\right)^3\)

you know maybe as campbell_st said, it may just be too premature of a chapter, which would not be your fault really

assuming the section on compound interests hasn't been covered yet

Answer 46

give me the answer plz!!!

Answer 47

b im just putting b

Answer 48

angel u r in which degree or say class

Answer 49

if i get under an 80% its fault

Answer 50

im in 6th grade!

Answer 51

usa education

Answer 52

i got a 40% WHAT THE HECK THAT feather HE GOT ME WRONG!

Answer 53

i said that b it ch not feather!

Answer 54

cya!

Answer 55

woo cool!
good keep going