im not sure but maybe look up how to calculate volume of a cone.. then go from there :D
Answer:
-2.32193...
Step-by-step explanation:
log.5 (5)
Answer:
The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i
Step-by-step explanation:
1) This claim is mistaken.
2) The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

For example:
3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

We'll have <em>n </em>different complex roots, not necessarily 2i.
For example:
Taking 3 polynomial equations with real coefficients, with


2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

Answer:
Yes, to add a positive number you ALWAYS move right on the number line.
<h2>
Answer:</h2>
The constraint which represent a thriving population of penguins at the zoo is:
0 < x ≤ 10 and 0 < y ≤ 20.
<h2>
Step-by-step explanation:</h2>
Let x = the number of female penguins and y = the number of male penguins.
Now, it is given that:
The group needs two times more males than females to thrive, and the zoo only has room for 10 female penguins.
This means that:
0 < x ≤ 10
( Because there are no room for more than 10 penguins )
Also,
y= 2x
Since,
x>0
This means that: 2x>0
i.e. y>0
and x ≤ 10
i.e. 2x ≤ 20
i.e. y ≤ 20
Hence, the constraints which represent a thriving population of penguins at the zoo is:
0 < x ≤ 10 and 0 < y ≤ 20.