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mel-nik [20]
3 years ago
15

X^2 + 32x + 256 = 1please show steps​

Mathematics
1 answer:
Elena L [17]3 years ago
3 0

\boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}

<h2>Explanation:</h2>

In this case, we have the following equation:

x^2+32x+256=1

But we can write this equation as:

x^2+32x+256=1 \\ \\ Subtract \ -1 \ from\ both \ sides: \\ \\ x^2+32x+256-1=1-1 \\ \\ x^2+32x+255=0

So this final result is a quadratic equation written in Standard Form (ax^2+bx+c=0). We need to find the solutions to this equations, so let's use quadratic formula:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ b=32 \\ c=255 \\ \\ \\ x=\frac{-32 \pm \sqrt{(32)^2-4(1)(255)}}{2(1)} \\ \\ x=\frac{-32 \pm \sqrt{1024-1020}}{2} \\ \\ x=\frac{-32 \pm \sqrt{4}}{2} \\ \\ x=\frac{-32 \pm 2}{2} \\ \\ Finally, \ two \ solutions: \\ \\ \boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}

<h2>Learn more:</h2>

Quadratic Equations: brainly.com/question/10278062

#LearnWithBrainly

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5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ... 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ... 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...

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3 years ago
Write answer as an inequality solved for c ​
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Given the function :

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So, substitute with each value of x to find the corresponding value of f(x)

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So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)

\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\  \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\  \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\  \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}

the graph of the function and the points will be as shown in the following image :

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1 year ago
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Len [333]
I believe the answer is 11/15 .
5 0
3 years ago
Read 2 more answers
Greatest to least<br> 2/7, 1/4, 2/9
Stella [2.4K]

<u>1/3; 2/7; 2/9</u>

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2/9 --> 14/63

As you can see I am correct, but I found the LCM by multiplying 1/3 by 21 to get to 63 and the numerator as well

7 0
3 years ago
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