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3 years ago

8

Which is equivalent to 3 sqr rt 8^1/4

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

12√8ˣ

Step-by-step explanation:

∛ x = x ∧(1/3)

therfore in the question

it changes into 8∧(1/3)(1/4)x

simplifing these

it becomes (1/12)x

8∧(1/12)x

12√8ˣ

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Step-by-step explanation:

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3 years ago

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(a) Find parametric equations for the line through (1, 4, 4) that is perpendicular to the plane x − y + 2z = 7. (Use the paramet

Alex777 [14]

Answer:

$x=1+t\\y=4-t\\z=4+2t$

Step-by-step explanation:

A vector perpendicular to the plane ax+by+cz+d=0 is of the form $(a,b,c)$ .

So, a vector perpendicular to the plane x − y + 2z = 7 is $(1,-1,2)$ .

The parametric equations of a line through the point $(x_0,y_0,z_0)$ and parallel to the vector $(a,b,c)$ are as follows:

$x=x_0+at\\y=y_0+bt\\z=z_0+ct$

Put $(x_0,y_0,z_0)=(1,4,4)$ and $(a,b,c)=(1,-1,2)$

Therefore,

$x=1+t\\y=4-t\\z=4+2t$

xy-plane:

Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6

So, at point (-1,6,0)

yz-plane:

Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2

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3 years ago

The hypotenuse of a right triangle measures 10 inches and the side opposite to <0 one of the non right angles is 5 inches lon

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3 years ago

2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?

7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

The solutions to the quadratic equation $2x^2\:-\:5x\:+\:67\:=\:0$ are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Step-by-step explanation:

Considering the equation

$2x^2\:-\:5x\:+\:67\:=\:0$

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

so

$\frac{2x^2-5x}{2}=\frac{-67}{2}$

$x^2-\frac{5x}{2}=-\frac{67}{2}$

Lets now solve the equation by completeing the remaining steps

Write equation in the form: $x^2+2ax+a^2=\left(x+a\right)^2$

Solving for $a$ ,

$2ax=-\frac{5}{2}x$

$a=-\frac{5}{4}$

$\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}$

Completing the square

$\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}$

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

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

Solving

$x-\frac{5}{4}=\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}$

$x-\frac{5}{4}=i\sqrt{\frac{511}{16}}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}$

$-\frac{5}{4}=i\frac{\sqrt{511}}{4}$

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4}$

Similarly, solving

$x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Therefore, the solutions to the quadratic equation are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

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2 years ago

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