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2 years ago

Can You Please Help I Don't Get This

Mathematics

2 answers:

Mila [183]2 years ago

8 0

I think c is your correct answer!!

saw5 [17]2 years ago

7 0

Answer:

A. Incorrect

B. Incorrect

C. Correct

D. Incorrect

Step-by-step explanation:

y = 5x + 6 means that Player A had 6 points and got 5 points per question.

The equation for the linear function is y = 2x + 6 because Player B had 6 points and gains 2 points per question.

A is incorrect because Player B gains 2 points and Player A gains 5 points

B. is incorrect because both of them have 6 points

C. is correct because Player A gains 5 points while Player B gains 2 points

D. is incorrect because there is already a correct answer and the evidence we needed was not given

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$\bf (g\circ f)(x)=x\implies g(~~f(x)~~)=x \\\\\\ \begin{cases} f(x) = 3x\\ g(x)=\cfrac{1}{3}x \end{cases}\qquad \qquad g(~~f(x)~~)=\cfrac{1}{3}[f(x)]\implies g(~~f(x)~~)=\cfrac{1}{3}(3x)$

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2 years ago

I need help I'm so confused

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2 years ago

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2 years ago

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2 years ago

Explain how to find each root, with its multiplicity, for the polynomial y=x^5-27x^2.

Nuetrik [128]

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

The characteristics of the polynomial can be derived from algebraic techniques. Roots and multiplicity can be found by <em>factoring</em> the polynomial. The multiplicity is represented by a power binomial of the form:

$(x-r_{i})^{m},\,m \le n$ (1)

Where $n$ is the grade of the polynomial.

Now we proceed to factor the formula:

$y = x^{5}-27\cdot x^{2}$

$y = x^{2}\cdot (x^{3}-27)$

$y = x^{2}\cdot (x^{2}+3\cdot x +9)\cdot (x-3)$

Please notice that $x^{2}+3\cdot x + 9$ have two complex roots.

$y = x^{2}\cdot \left(x+\frac{3}{2}-i\,\frac{3\sqrt{3}}{2} \right)\cdot \left(x+\frac{3}{2}+i\,\frac{3\sqrt{3}}{3} \right)\cdot (x-3)$

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

We kindly invite to see this question on polynomials: brainly.com/question/1218505

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2 years ago