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Murljashka [212]
3 years ago
11

The diagram represents a difference of squares.

Mathematics
2 answers:
harkovskaia [24]3 years ago
8 0
The factors are
(m-6)(m+6)

Explanation:
Group the first two terms and factor out the common factor:
i.e m(m-6)
Repeat the procedure for terms 3 and 4.
6(m-6)

Regrouping:
m(m-6)+6(m-6)

On factoring out (m-6), we get:
(m-6)(m+6)
Naily [24]3 years ago
8 0

Answer: The answer is D. (m – 6)(m + 6).

Step-by-step explanation:

This is 100% correct for the online course edge.nuity.

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I need the steps to this question.
yanalaym [24]

im not sure but maybe look up how to calculate volume of a cone.. then go from there :D

3 0
3 years ago
Approximate: log 1/2 5
zubka84 [21]

Answer:

-2.32193...

Step-by-step explanation:

log.5 (5)

5 0
4 years ago
Read 2 more answers
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
____ [38]

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.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

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

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

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

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
To add a positive number move right on the number line
hammer [34]

Answer:

Yes, to add a positive number you ALWAYS move right on the number line.


6 0
4 years ago
Read 2 more answers
ASAP
denis-greek [22]
<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.

4 0
3 years ago
Read 2 more answers
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