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elena-14-01-66 [18.8K]
3 years ago
14

Which polynomial is equal to (2x^2-5x-4)-(4x^2-3x-2) , and how is it equal?

Mathematics
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

Option A

Step-by-step explanation:

Hi.

Let's simplify the given expression:-

(2x^2-5x-4)-(4x^2-3x-2)

When we remove the second parentheses we must distribute the negative over the contents:-

= 2x^2 - 5x - 4 - 4x^2 + 3x + 2

Now adding like terms, we get:-

- 2x^2 - 2x - 2    

So it's option A










Mademuasel [1]3 years ago
3 0

Answer:

A. -2x^2-2x-2

Step-by-step explanation:

(2x^2-5x-4)-(4x^2-3x-2)

distribute the minus sign

(2x^2-5x-4)-4x^2+3x+2

2x^2-5x-4

-4x^2+3x+2

-------------------------

-2x^2 -2x-2

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\large\begin{array}{l} \mathsf{\displaystyle\int\!3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{1}{2}\cdot 2\cdot 3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^{2x-1}\cdot 2\,dx\qquad(i)} \end{array}


\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{2x-1=u\quad\Rightarrow\quad 2\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^u\,du}\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{1}{\ell n\,3}\,3^u+C}\\\\ =\mathsf{\dfrac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}


\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!3^{2x-1}\,dx=\frac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}}\qquad\checkmark \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2154165


\large\textsf{I hope it helps. :-)}
</span>

Tags: <em>integrate indefinite integral substitution exponential base logarithm log ln composite integral calculus

</em>
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