You just multiply the numbers and if it’s less than that number you it it in the bud hood this helps!
Answer:
2
Step-by-step explanation:
For any positive numbers a,b we always have the following identity:
(gcd(a,b) denotes the greatest common divisor between a and b, and lcm(a,b) denotes the least common multiple between a and b)
In our case, we are given that 
and that 
. Plugging that in into our identity, we get:
And so solving for 
:
Answer:
12 inches
Step-by-step explanation:
Let b represent the base
h represents the height
area of the parallelogram = base * height = 216 square inches
From the question'
b = 18 + 3h
Slot in the value of b
216 = (18 + 3h) * h
expand
216 = 18h + 3h^2
subtract 216 from both sides
0 = 18h + 3h^2 - 216
rearrange
3h^2 + 18h - 216 = 0
divide through by 3
h^2 + 6h - 72 = 0
Now, lets solve!
h^2 + 6h - 12h - 72 = 0
h( h + 6 ) - 12(h + 6) = 0
(h - 12) (h + 6) = 0
h - 12 = 0
h = 12
and
h + 6 = 0
h = - 6
Taking the positive value of h
Hence, the height is 12 inches
Lets check
when h = 12 inches
Area of the parallelogram = 18* 12 = 216 square inches .... correct
when h = -6inches
A = 18 * -6 ≠ 216 square inches
So height is 12 inches
Answer:
positive charge...............
Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.
Answer:
Dale live 315 miles from the mountains
Step-by-step explanation:
Let y be the speed of Dale to the mountains
Time taken by Dale to the mountains=7 hrs
Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1
Time taken by Dale back home = 5hours
Since it speed increased by 18 miles per hour back home it speed = y+18
So distance traveled home =speed × time = (y+18)5 ...... eqn 2
Since distance cover is same in both the eqn 1 and eqn 2.
Eqn 1 = eqn 2
7y = (y+18)5
7y = 5y + 90
7y - 5y = 90 (collection like terms)
2y = 90
Y = 45
Substitute for y in eqn 1 to get distance away from mountain
= 7y eqn 1
= 7×45
= 315 miles.
∴ Dale leave 315 miles from the mountains
