uvacowdo

ah, ok. that's understandable

Thadeus’s Thanksgiving feast was just too much! He started cooking the 70-pound turkey at 11:00 A.M.
At 11:30 A.M., he purchased 35 pounds of popcorn. He baked 700 pumpkin pies and half as many giblet-chip cookies. At 8:30 P.M., the turkey was finished cooking. His guests arrived at 6:00 P.M. They immediately ate 2/3 of the cookies! How long did it take Thadeus’s turkey to cook?

if the start time was 11:00 am and the end time was 8:30 PM, it took 1+8.5 = 9.5 -> 9 hrs and 30 minutes for the turkey to cook

hope this helps :)

hey,

this question has been answered before. check out this link: https://web2.0calc.com/questions/decimals_12

hope this helps :)

if this question's already been answered, you can learn from the method they used and plug in the new numbers you have.

if you're still confused, i can give this problem a shot :)

thanks!

here, somebody has answered this question before: https://web2.0calc.com/questions/help-math_22

hope this helps!

thank you for the hint! i originally tried solving the problem like this, but i got stuck, haha. i'm probably just not good enough 

Thank you!

B. Living things respond to stimuli

The alcohol is the stimuli.

Yes, that is correct. Thank you so much!!

(you should create an account so that when you answer questions you can get points) 

oh, i'm sorry.

for example:

sum of digits of 23: 5

sum of digits of 5287: 5+2+8+7 = 22

do you understand or is it unclear? sorry if it was not clear before.

no... sorry because it is not asking for the number itself to be divisible by two, it is asking for the sum of the digits to be div. by 2...thanks for trying!

that is wrong. read the question please, but thanks for trying

yeah...i find that when i use the latex box majority of the times it comes out as the raw form...help! 

because there are 26 options (since there are 26 letters in the alphabet) for the first and second plates, 5 for the third since there are five odd digits, and five for the fourth since there are also 5 even digits. then multiply by two because the odd and even positions could be switched

Let's call one pack of gum X. 

The ratio of # of cherry to # of grape is .\(\\(\dfrac{20}{30} = \dfrac{2}{3})\) Using this, we have the following equality.  

\((\dfrac{20-X}{30} = \dfrac{20}{30 + 5X})\). The ratio to the left is the ratio of gum if chewman loses 1 pack of cherry, and the ratio to the right is the ratio of gum if chewman finds 5 packs of grape. 

Cross multiplying, we have \((20-X)(30+5X) = (20)(30) \Rightarrow (20-X)(30+5X) = 600\). Factoring, we have \(-5x^2 + 70x + 600 = 600 \Rightarrow -5x^2 + 70x = 0 \Rightarrow 5x^2 = 70x \Rightarrow \boxed{x = 14}\).

There are 14 pieces of gum in one pack of gum.