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2 years ago

Find the equation of plane that contains P(1,-1,2) and has the normal n=2i+3j+5k (please answer quickly )

Mathematics

1 answer:

mixas84 [53]2 years ago

8 0

Answer:

2x + 3y + 5z = 9

Step-by-step explanation:

The normal defines all the planes perpendicular to that vector. To isolate to a single plane, just apply the known point.

ax + by + cz = d

2x + 3y + 5z = d

2(1) + 3(-1) + 5(2) = 9

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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

$x^2-48=0$

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

$x^2=48$

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

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Solving

$x=\sqrt{48}$

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

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

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

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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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2 years ago

You are running a fuel economy study. One of the cars you find is red. It can travel 24 4/5 miles on 4/5 gallon of gasoline. Wha

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Answer:

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