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Papessa [141]
3 years ago
8

What is a simpler form of the expression? (2n 2 + 5n + 3)(4n – 5) A. 8n 3 + 10n 2 – 13n – 15 B. 8n 3 + 30n 2 – 37n – 15 C. 8n 3

– 10n 2 + 37n – 15 D. 8n 3 + 13n 2 – 10n – 15
Mathematics
1 answer:
Sonja [21]3 years ago
3 0
<span>(2n^2 + 5n + 3)(4n – 5)
= 8n^3 -10n^2 + 20n^2 - 25n + 12n - 15
= 8n^3 + 10n^2 - 13n - 15 

answer is </span><span>A. 8n^3 + 10n^2 – 13n – 15

hope that helps</span>
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What is the value of "c" in the following quadratic? (Make sure the equation is in
garri49 [273]

               \rule{50}{1}\large\blue\textsf{\textbf{\underline{Given question:-}}}\rule{50}{1}

           <em>What is the value of c in the quadratic </em>\large\text{$x^2+28=-11x$}?

          \rule{50}{1}\large\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

           Before starting to solve, you should notice something - the

quadratic is not in its standard form!

We can easily fix it by adding \large\textit{11x} on both sides:-

\large\text{$x^2+28-11x=0$}

We can switch the order of 28 and -11x:-

\large\text{$x^2-11x+28=0$}

  Now, the quadratic is in its standard form, so we can get down to

finding the value of "c".

Remember, the standard form of a quadratic looks like so:-

  •  \large\text{$ax^2+bx+c=0$}

Now we can just write our <u>quadratic</u> here:-

  • \large\text{$x^2-11x+28=0$}

Now, can you see what the value of "c" is?

An easy way to <u>remember</u> "c" in quadratics is:-

The "c" in quadratics is the constant.

   

Henceforth, we conclude that the value of "c" in the given quadratic is:-

 \Large\textbf{28}\Large\checkmark

<h3>         Good luck with your studies.</h3>

        \rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}    

5 0
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