Question

Help with equation please

Answer 1

The amount of water, W, that collects in an outside birdbath after t hours is given by the following function

Answer 2

Determine the percent decrease in the amount of water in the birdbath between 3 and 6 hours.

Answer 3

so we first need to find W(3) and W(6)

Answer 4

hokay

Answer 5

any ideas on how to do what freckles said?

Answer 6

sorry im back @freckles had to help my grandparents move their luggage inside

Answer 7

would you just fill in 3 and 6 for x?

Answer 8

yea

Answer 9

would it be a decimal?

Answer 10

now your problem mentioned a decrease from t=3 to t=6
so percentage decrease would be
\[\frac{W(3)-W(6)}{W(3)} \cdot 100 \%\]

Answer 11

well w3 would cancel out

Answer 12

\[(\frac{W(3)-W(6)}{W(3)})\cdot 100 \% \\ ( \frac{W(3)}{W(3)}-\frac{W(3)}{W(6)}) \cdot 100 \% \\ (1-\frac{W(3)}{W(6)}) \cdot 100 \%\]

Answer 13

could you simplify that to (1-\[\frac{ W(1) }{ W(3) }\]

Answer 14

W(3) does not equal W(1)
W(6) does not equal W(3)
And you can also not say W(3)/W(6)=W(3/6)=W(1/2)=W(1)/W(2)
And I'm saying you cannot do that

Answer 15

That is why at the beginning I asked you to find W(3) and W(6) because we will definitely need those values

Answer 16

how would you work out w(3)/w(6)

Answer 17

Well earlier you said to find w(3) you would use the function given
just replace x with 3
so why not do that?

Answer 18

W(3)=4(3^2)+3+2/2(3^2)+1

Answer 19

w(3)=36+5/18+1
w(3)=41/19

Answer 20

2.16 in decimal

Answer 21

now find w(6)

Answer 22

w(6)=4(6^2)+6+2/2(6^2)+1

Answer 23

\[(\frac{W(3)-W(6)}{W(3)})\cdot 100 \% \\ ( \frac{W(3)}{W(3)}-\frac{W(3)}{W(6)}) \cdot 100 \% \\ (1-\frac{W(3)}{W(6)}) \cdot 100 \% \\ (1-\frac{\frac{41}{19}}{W(6)})100 \%\]

Answer 24

w(6)=152/73

Answer 25

\[(\frac{W(3)-W(6)}{W(3)})\cdot 100 \% \\ ( \frac{W(3)}{W(3)}-\frac{W(3)}{W(6)}) \cdot 100 \% \\ (1-\frac{W(3)}{W(6)}) \cdot 100 \% \\ (1-\frac{\frac{41}{19}}{W(6)})100 \% \\ (1-\frac{\frac{41}{19}}{\frac{152}{73}})100 \%\]

Answer 26

oops I didn't notice I flipped that second fraction in the second line

Answer 27

\[(\frac{W(3)-W(6)}{W(3)})\cdot 100 \% \\ ( \frac{W(3)}{W(3)}-\frac{W(6)}{W(3)}) \cdot 100 \% \\ (1-\frac{W(6)}{W(3)}) \cdot 100 \% \\ (1-\frac{W(6)}{\frac{41}{19}})100 \% \\ (1-\frac{\frac{152}{73}}{\frac{41}{19}})100 \%\]

Answer 28

\[(1-\frac{152}{73} \cdot \frac{19}{41})100 \%\]
ok see you can finish simplifying this

Answer 29

wait why did you switch 41/19 around

Answer 30

well I did say
"oops I didn't notice I flipped that second fraction in the second line"

Answer 31

but it was originally 41/19 i thought you were talking about switching it to the numerator

Answer 32

I'm not sure are you talking (a-b)/c=a/c-b/c
or you talking about why is a/b divided by c/d equal a/b*d/c

Answer 33

i got 2888/2993

Answer 34

and i meant you put 41/19 in the equation but then switched it to 19/41

Answer 35

\[(\frac{W(3)-W(6)}{W(3)})\cdot 100 \% \\ ( \frac{W(3)}{W(3)}-\frac{W(6)}{W(3)}) \cdot 100 \% \\ (1-\frac{W(6)}{W(3)}) \cdot 100 \% \\ (1-\frac{W(6)}{\frac{41}{19}})100 \% \\ (1-\frac{\frac{152}{73}}{\frac{41}{19}})100 \% \\ (1-\frac{\frac{152}{73}}{\frac{41}{19}} \cdot \frac{\frac{19}{41}}{\frac{19}{41}}) 100 \% \\ (1-\frac{\frac{152}{73}}{\frac{\cancel{41}}{\cancel{19}}} \cdot \frac{\frac{19}{41}}{\frac{\cancel{19}}{\cancel{41}}}) 100 \% \\ (1-\frac{2888}{2993}) 100 \%\]

Answer 36

ohhhh ok

Answer 37

would the final answer be 105/2993

Answer 38

multiply that by 100 to get the percentage

Answer 39

though that is the equivalent in un-percentage form

Answer 40

.035

Answer 41

well that is what 105/2993 is approximately
and multiplying .035 by 100 gives?

Answer 42

3.5%

Answer 43

Way to hang in there
Those were pretty big numbers

Answer 44

yus ><