Alright bud the best answer to this question will be that the mode is 12 because it appears the most
General Idea:
Domain of a function means the values of x which will give a DEFINED output for the function.
Applying the concept:
Given that the x represent the time in seconds, f(x) represent the height of food packet.
Time cannot be a negative value, so
The height of the food packet cannot be a negative value, so
We need to replace 
for f(x) in the above inequality to find the domain.
![-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0](https://tex.z-dn.net/?f=%20-15x%5E2%2B6000%5Cgeq%200%20%5C%3B%20%5C%3B%20%20%5BDivide%20%5C%3B%20by%5C%3B%20-15%5C%3B%20on%5C%3B%20both%5C%3B%20sides%5D%5C%5C%20%5C%5C%20%5Cfrac%7B-15x%5E2%7D%7B-15%7D%20%2B%5Cfrac%7B6000%7D%7B-15%7D%20%5Cleq%20%5Cfrac%7B0%7D%7B-15%7D%20%5C%5C%20%5C%5C%20x%5E2-400%5Cleq%200%5C%3B%5BFactoring%5C%3Bon%5C%3Bleft%5C%3Bside%5D%5C%5C%20%5C%5C%20%28x%2B200%29%28x-200%29%5Cleq%200%20)
The possible solutions of the above inequality are given by the intervals ![(-\infty , -2], [-2,2], [2,\infty )](https://tex.z-dn.net/?f=%20%28-%5Cinfty%20%2C%20-2%5D%2C%20%5B-2%2C2%5D%2C%20%5B2%2C%5Cinfty%20%29%20)
. We need to pick test point from each possible solution interval and check whether that test point make the inequality 
true. Only the test point from the solution interval [-200, 200] make the inequality true.
The values of x which will make the above inequality TRUE is 
But we already know x should be positive, because time cannot be negative.
Conclusion:
Domain of the given function is 
Answer:
The numbers are 10 and 5
Step-by-step explanation:
x+y=15
(1/x)+(1/y)=(3/10)
(1/(15-y))+(1/y)=(3/10)
x=10, y=5
Answer:
add, subtract, multiply and divide complex numbers much as we would expect. We add and subtract
complex numbers by adding their real and imaginary parts:-
(a + bi)+(c + di)=(a + c)+(b + d)i,
(a + bi) − (c + di)=(a − c)+(b − d)i.
We can multiply complex numbers by expanding the brackets in the usual fashion and using i
2 = −1,
(a + bi) (c + di) = ac + bci + adi + bdi2 = (ac − bd)+(ad + bc)i,
and to divide complex numbers we note firstly that (c + di) (c − di) = c2 + d2 is real. So
a + bi
c + di = a + bi
c + di ×
c − di
c − di =
µac + bd
c2 + d2
¶
+
µbc − ad
c2 + d2
¶
i.
The number c−di which we just used, as relating to c+di, has a spec
