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cricket20 [7]
3 years ago
7

This is probably so easy but im dumb lol plz help.

Mathematics
2 answers:
Luba_88 [7]3 years ago
6 0

Answer:

(5x)^2=5x^2

Step-by-step explanation:

you just remove the parenthsis.

Airida [17]3 years ago
5 0
Don’t say you’re dumb ! we always need help sometimes.

(5x)^2 = ____x^2

distribute the exponent ‘2’ to both terms inside the parentheses.

so it will be (5)^2(x^2)

simplify. 5^2=25, (x^2)=x^2

so the answer is 25x^2
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Which of the numbers 19,21,23, and 25 has the most factors
guapka [62]
First let's find the factors of each number:
19: 1,19
21: 1,3,7,21
23: 1,23
25: 1,5,25
21 has 4 factors, 25 has 3 and 23 and 19 are both prime numbers (with two factors).  21 has the most factors.
5 0
3 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
Factor completely 3x4 − 30x3 + 75x2
k0ka [10]
3x^2(x^2-10x+25) hope this helps
5 0
4 years ago
Read 2 more answers
Help me on this question pleaseeeee1
miskamm [114]

Answer:

75 degrees?

Step-by-step explanation:

It's an acute angle that is almost an right angle (90 degrees) so my best guess is 75 degrees.

3 0
3 years ago
Divide x3-3x2-10x+24 by x-2 fill in the missing number
Ierofanga [76]

Answer:

For step 1, the missing number will be

k = 1

Step-by-step explanation:

Hope this helps! :)

5 0
3 years ago
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