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Lera25 [3.4K]
3 years ago
13

Find the equation of the line specified. The slope is 6, and it passes through ( -4, 4). A. Y = 6x 4 c. Y = 12x 28 b. Y = 6x - 2

0 d. Y = 6x 28.
Mathematics
1 answer:
Tresset [83]3 years ago
7 0

Answer: A

Step-by-step explanation:

You have your slope (mx) which is 6, so it would look like 6x. We have our points and can see that y is 4. Put it all together using the slope formula to get y = 6x + 4

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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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