Answer:
40 minutes.
Step-by-step explanation:
- time spent on the thread mill as a function of d(day) = t(d)
= 30 +2(d-1) minutes
[where d represents day number]
(when d is 1 the time spent should be 30
so substitute d=1 and check it will be 30 only.
on first day he will spend 30 minutes so, I added 30 .
and on every additional day( these additional days are excluding first day), he will increase the time by 2 minutes.so, I added 2(d-1) to initial 30 minutes)
- so, T(6), the minutes he will spend on the treadmill on day 6=
=30+2(6-1)
=30+2(5)
=30+10
=40 minutes
<span>To solve these GCF and LCM problems, factor the numbers you're working with into primes:
3780 = 2*2*3*3*3*5*7
180 = 2*2*3*3*5
</span><span>We know that the LCM of the two numbers, call them A and B, = 3780 and that A = 180. The greatest common factor of 180 and B = 18 so B has factors 2*3*3 in common with 180 but no other factors in common with 180. So, B has no more 2's and no 5's
</span><span>Now, LCM(180,B) = 3780. So, A or B must have each of the factors of 3780. B needs another factor of 3 and a factor of 7 since LCM(A,B) needs for either A or B to have a factor of 2*2, which A has, and a factor of 3*3*3, which B will have with another factor of 3, and a factor of 7, which B will have.
So, B = 2*3*3*3*7 = 378.</span>
Answer:
Step-by-step explanation:20
20 130 3 8.12
- 1)
2:2 12-1 21=-11 # 1 - 2
1212
II + 3)
472
I + 3 2(1 - 1)
-1 2:2
3(12 - 1)
32 – 23– 3+.
