Answer:
B.25$
Step-by-step explanation:
to make the money less one of the freinds will carry 9kg and the other 12 kg the last one carries the left luggage weighing 7kg and 6kg which add up to 13kg.
the first freind won't pay anything as he is passing 9kg luggage
the second will pay : 12-10 =2 so 2kg*5$=10$
the last freind will pay 13-10=3 so 3kg*5$=15$
to find the total money the pay we will add 10$ and 15$ which became 25$.
hope this helped.
its q and c c c c c c c c c c c a a a a a a a a a a. a
Answer:
Mikey weighs 110 pounds.
Step-by-step explanation:
Because Mikey weighs 15 pounds more, just add 95 pounds and 15 pounds to get Mikey's weight:
95 + 15 = 110
I hope this helps you!
Please rate my answer and give me Brainliest!
Have a nice day! :)
Answer:
17
Step-by-step explanation:
Here in this question for finding the numbers that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively we have to first subtract the remainder of the following. By this step we find the highest common factor of the numbers.
And then the required number is the HCF of the following numbers that are formed when the remainder are subtracted from them.
Clearly, the required number is the HCF of the numbers 398−7=391,436−11=425, and, 542−15=527
We will find the HCF of 391, 425 and 527 by prime factorization method.
391=17×23425=52×17527=17×31
Hence, HCF of 391, 4250 and 527 is 17 because the greatest common factor from all the numbers is 17 only.
So we can say that the largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively is 17.
Note: - whenever we face such a type of question the key concept for solving this question is whenever in the question it is asking about the largest number it divides. You should always think about the highest common factor i.e. HCF. we have to subtract remainder because you have to find a factor that means it should be perfectly divisible so to make divisible we subtract remainder. because remainder is the extra number so on subtracting remainder it becomes divisible.