Resource allocation: each day an organism has 100 j of energy to divide between growth and reproduction. each millimeter of growth costs 3 j and each egg produced costs 5 j. denote the amount of growth per day by x1 and the number of eggs produced per day by x2. (a) suppose that the organism divides its energy so that for every millimeter of growth that occurs each day, it produces (on average) two eggs. what is the amount of growth and number of eggs produced on each day? (b) suppose that the organism divides its energy each day so that the total energy spent on eggs is twice that spent on growth. what is the amount of growth and number of eggs produced on each day?

a. 7.69 mm and 15 eggs

Step-by-step explanation:

a. First, we have to write down the equivalent equations given by the statement. Therefore, the equations can be written as follows:

E1:

E2:

If we replace E2 inside E1 we can get:

Now replacing this result in E2 we get:

The organism produces 15 eggs on average and growths 7.69 mm per day.

b. Now, we need to define E2 in terms of energy as follows:

E3:

If we replace E3 in E1 we get:

Now replacing this result in E3 we get:

The organism produces 13 eggs on average and growths 11.11 mm per day.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: