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2 years ago

Answer quick for brainliest

Mathematics

1 answer:

AVprozaik [17]2 years ago

6 0

Answer:

Step-by-step explanation:

y = mx + c

Here, m is the slope & c is the y-intercept

y = -3x - 1

Slope = -3

y intercept = -1

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What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

#learnwithBrainly

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How do I solve this?

vodomira [7]

I’m assuming what you’re asking here is how to *factor* this expression. For that, let’s rearrange the expression into a more familiar form:

-c^2-4c+21

From here, we’ll factor out a -1 so that we have:

-(c^2+4c-21)

Let’s focus on the quadratic expression inside the parentheses. To find our factors (c + x)(c + y), we’ll need to find two terms x and y that multiply together to make -21 and add together to make 4. It turns out that the numbers -3 and 7 work out perfectly for that purpose (-3 x 7 = -21 and 7 + (-3) = 4), so substituting them in for x and y, we have:

(c + (-3))(c + 7)
(c - 3)(c + 7)

And adding back on the negative from a few steps earlier:

-(c - 3)(c + 7)

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