Answer:
Step-by-step explanation:
Problem is HCF:
Of;
a²b⁵ and a³b³
The highest common factor, HCF is used to factor expressions in order to simplify them.
H.C.F is a factor that is common to both expressions.
here we have
a²b⁵ and a³b³
a²b⁵ = a x a x b x b x b x b x b
a³b³ = a x a x a x b x b x b
In both expressions, for a:
a x a and b x b x b x b x b
a x a x a and b x b x b
common factors are a and b
both entities contains a and b
Highest common factor H.CF;
a² and b³
this makes <em><u>a²b³ </u></em>
<em><u>a²b³ </u></em> is common to both of them and it is the highest term that can be factored from the the two.
To determine the total distance traveled by an object in constant speed, just multiply the speed by the time. So, for the car, the total distance traveled is,
distance by car = (20 m/s)(5s) = 100 m
For decelerating object, the distance traveled is,
distance = Vt - (at^2)/2
For the truck,
distance by truck = (20 m/s)(10s) - (2m/s^2)((10s)^2)/2 = 100 m
Relatively, both have traveled the same distance.
Answer:
I think the answer should be B
Since he can run 1/4 of a mile in 2.5 minutes, he will run a full mile in 4*2.5 =10 minutes, assuming he runs at a steady speed.
If you know that the car travels 0.75 per minute and that there are 60 minutes are in a hour, you could multiply 0.75 with 60, which gives you an answer of 45.
