If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.
So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.
(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.
(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)
= x^4 + 29x^2 + 100
So the equation would be x^4 + 29x^2 + 100 = 0.
Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.
287/8.40=34.1666667
34 hours would be too short so you would need to round it up so he would need to work 35 hours but yoh would earn extra money
Answer:
1 and -0.6
Step-by-step explanation:
As you are selling each candy in a packet of 5, you should First divide 360 by 5. This would give you how many let’s say packets you would sell. This would give you 72. Each once you sells at one dollar. Therefore if you sell a total of 72 packets you will do 72 multiplied by 1, which would give you a total of $72z
Let x represent the shoes and let y represent the clothes.
So x+y=132
x+54=132 by substituting the Y value, so we don't the x value and that is what we have to solve.
So x+54=132, subtract 54 from both sides to get x value
so X=78
The cost of shoes is $78. and that is the answer. I hope it helps you.
