Find like terms and group them
3n and 5n are same and 4n^2 and 4n^2 are same
but 3mn and 5n are not and 3n and 2n^3 are not
group like terms
(-2m^2)+(-2mn+1mn)+(8)+(-10m)+(10n-5n)+(2n^2)
add like terms
(-2m^2)+(-mn)+(8)+(-10m)+(5n)+(2n^2)
-2m^2+2n^2-10m+5n-mn+8 is simplest form
Answer:
where is simplify question
9514 1404 393
Answer:
Step-by-step explanation:
The difference in ages remains the same over time. Enoch, at twice Martin's age, is still 15 years older. That means Martin is 15 and Enoch is 30.
Answer:
f(x) = 4.35 +3.95·sin(πx/12)
Step-by-step explanation:
For problems of this sort, a sine function is used that is of the form ...
f(x) = A + Bsin(2πx/P)
where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.
It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...
f(x) = 4.35 + 3.95·sin(2πx/24)
The argument of the sine function can be simplified to πx/12, as in the Answer, above.
Answer:
Step-by-step explanation:
<h2><u>Part A</u></h2>
in interval ( 0 ; 2)
<h2><u>Part B</u></h2>
in interval (2; 4)
<h2><u>Part C</u></h2>
in interval (4 ; 6 )
<h2><u>Part D</u></h2>
The graph shows that at first the ball rises up ; and then it is seen that it goes down and loses height to zero , from which it can be concluded that the height after 10 seconds remains unchanged and therefore the height of the ball after 16 seconds will be zero
