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Stolb23 [73]
2 years ago
7

What is the value of "c" in the following quadratic? (Make sure the equation is in

Mathematics
1 answer:
garri49 [273]2 years ago
5 0

               \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}    

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6 0
3 years ago
Please solve and show working for 13-((4/5)+(6/8))=
UNO [17]
<span>13-((4/5)+(6/8))
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6 0
3 years ago
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Read 2 more answers
Check all apply 3x+4=10
garri49 [273]

Answer:

Solution is x=2. sorry if this is not what you are looking for!

Step-by-step explanation:

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