Question

@Honda

Answer 1

any idea of what it is?

Answer 2

I thought it might be A

Answer 3

close but your missing one thing. the dog is 25% more than the other so the .25p would be added to something

Answer 4

ok so e ?

Answer 5

yep that is correct

Answer 6

ok for the other one tho i think its b

Answer 7

well b would give you the incorrect answer because you are adding the .25p because the dog is the original weight plus 25% more.

Answer 8

ik but it is more than 25% when u look at it and it does include the variable b,

Answer 9

what other can i choose< the one we crossed off (A) don't include a variable next to it

Answer 10

I mean it does,but no other answer option does tho

Answer 11

Hood still need an explanation, or you're good?

Answer 12

I still need an explination

Answer 13

Okay, so lets say this dog is 20 pounds. right?

Answer 14

yes

Answer 15

So, in that case it's the same thing as doing
\[20+(1/4)20=?\]

Answer 16

(1/4)=25%

Answer 17

25%(20)=5

Answer 18

So, the second dog is 25 pounds.

Answer 19

yea

Answer 20

C?

Answer 21

no, you should add the variable to 0.25

Answer 22

so 0.25 plus p =0.25p so A?

Answer 23

I can see why you said A, but your missing the other dogs weight.

Answer 24

so p for the other dogs weight plus 0.25p = p + 0.25p = C?

Answer 25

p+0.25p should be your final answer.

Answer 26

thanks

Answer 27

(Weight of dog #1)+ 25%(Weight of dog #1)= (Weight of dog #2)
Anytime