Hi
5/7 × 11/7 × 5/6 = 55/49 × 5/6 = 275/294
5/7 × 11/7 = (5×11)/(7×7) = 55/49
55/49 × 5/6 = (55×5)/(49×6) = 275/294 (irreducible)
Answer: 275/294
Sa= bw+hw+lw+
(2)
SA=(12*4)+(16*4)+(20*4)+(
)
SA= 28+64+80+(6*16*2)
SA=28+64+80+(96*2)
SA=28+64+80+192
SA=364
Answer:
7 buses
Step-by-step explanation:
51 divided by 8 = 6 and you still have 3 students letf so you have to use one more bus
Answer:
11
Step-by-step explanation:
The distance that a is to b is b-a=17-2=15.
The line segment from a=2 to b=17 has length 15.
We need to know what is 3/5 of 15.
3/5 of 15 means what is 3/5 times 15?
(3/5)(15)=3(3)=9
So this means we are looking to make a line segment that is 9 units from 2 which is 2 to 11.
So 11 is 3/5 the way from a=2 to b=17.
Let's check from 2 to 11 that is a length of 9 and from 2 to 17 that is a length of 15.
Is 9/15 equal to 3/5?
Yes, 9/15 can be reduced to 3/5.
