Connect with a knowledgeable community and get your questions answered on IDNLearn.com. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.

A football player weighed 170 2/3 pounds in May. During the summer, he gained 25 5/12 pounds. During the first week of fall practice, he lost 10 1/4 pounds, and during the second week he lost another 3 1/2 pounds. How much did he weigh at that point?

A: 182 1/3 lb
B: 181 1/2 lb
C: 182 5/12 lb
D: 183 2/3 lb

Help please?


A Football Player Weighed 170 23 Pounds In May During The Summer He Gained 25 512 Pounds During The First Week Of Fall Practice He Lost 10 14 Pounds And During class=

Sagot :

Answer:

182 1/3 lbs

MAY:  weight = 170 2/3

Summer: gained 25 5/12

New weight : 170 2/3 + 25 5/12  

Get a common denominator of 12

170 2/3 *4/4  + 25 5/12

170 8/12 + 25 5/12 = 195 13/12

The he lost 10 1/4 during the first week of fall practice

195 13/12 - 10 1/4

We need to get a common denominator of 12

195 13/12 - 10 1/4*3/3

195 13/12 - 10 3/12

185 10/12

The he lost another 3 1/2 pound during the second week

185 10/12 - 3 1/2

We need to get a common denominator of 12

185 10/12 - 3 1/2 * 6/6

185 10/12 - 3 6/12

182 4/12

Simplify the fraction

182 1/3 lbs

Hope this helps!

have a great day! :))))))

We need to find the weight of a person at a certain period of time.

The weight of the person at the required time is A. [tex]182\dfrac{1}{3}\ \text{lb}[/tex]

Initial weight of person = [tex]170\dfrac{2}{3}[/tex] pounds.

Weight gained = [tex]25\dfrac{5}{12}[/tex] pounds

Weight lost = [tex]10\dfrac{1}{4}+3\dfrac{1}{2}[/tex] pounds

So, weight of the person is

[tex]170\dfrac{2}{3}+25\dfrac{5}{12}-(10\dfrac{1}{4}+3\dfrac{1}{2})\\ =170+\dfrac{2}{3}+25+\dfrac{5}{12}-(10+\dfrac{1}{4}+3+\dfrac{1}{2})\\ =195+\dfrac{8+5}{12}-13-\dfrac{1+2}{4}\\ =182+\dfrac{13}{12}-\dfrac{3}{4}\\ =182+\dfrac{13-3}{12}\\ =182+\dfrac{10}{12}\\ =182+\dfrac{1}{3}=182\dfrac{1}{3}[/tex]

The weight of the person at the required time is A. [tex]182\dfrac{1}{3}\ \text{lb}[/tex]

Learn more:

https://brainly.com/question/17767863

https://brainly.com/question/24901550