Question 3 of 20

Mathematics

1 answer:

olchik [2.2K]2 years ago

3 0

I think it’s C but it could be D.. i would go with C tho

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(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

valentinak56 [21]

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1 ==> z = -4x + 2y - 1

3x - 2y + 2z = 1 ==> z = (1 - 3x + 2y)/2

==> -4x + 2y - 1 = (1 - 3x + 2y)/2

==> -8x + 4y - 2 = 1 - 3x + 2y

==> -5x + 2y = 3

==> y = (3 + 5x)/2

==> z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

3 0

2 years ago

Given the piecewise function shown below, select all of the statments that are true.

timurjin [86]

The given the piecewise function is :

$f(x) = \[ \begin{cases} 
 2x & x \ \textless \ 1 \\
 5 & x=1 \\
 x^2 & x\ \textgreater \ 1 
 \end{cases}
\]$

To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25

To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4

To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4

To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:
$\framebox {B. f(2) = 4} \ \framebox {D. f(1) = 5}$

8 0

3 years ago

Logariths, please help

Valentin [98]

Answer:

{ - 2 , 8 }

Step-by-step explanation:

$log_{4}$ ( x² - 6x ) = 2