Use the the graph to determine the domain and range . Then state whether the relation is a function

Use substitution to solve the following system of linear equations and fill in the following blanks:

marissa [1.9K]

Answer:

$x=-4$

$y=3$

$z=-5$

Step-by-step explanation:

Given:

$x+y+z=-6$

$x-6y-7z=-29$

$-7y-5z=4$

Solve for $x$ in the 1st equation:

$x+y+z=-6$

$x+y=-z-6$

$x=-y-z-6$

Substitute the value of $x$ into the 2nd equation and solve for $z$ :

$x-6y-7z=-29$

$(-y-z-6)-6y-7=-29$

$-7y-z-13=-29$

$-7y-z=-16$

$-z=-16+7y$

$z=16-7y$

Substitute the value of $z$ into the 3rd equation and solve for $y$ :

$-7y-5z=4$

$-7y-5(16-7y)=4$

$-7y-80+35y=4$

$28y-80=4$

$28y=84$

$y=3$

Plug $y=3$ into the solved expression for $z$ and evaluate to solve for $z$ :

$z=16-7(3)$

$z=16-21$

$z=-5$

Plug $z=-5$ into the solved expression for $x$ and evaluate to solve for $x$ :

$x=-(3)-(-5)-6$

$x=-3+5-6$

$x=2-6$

$x=-4$

Therefore:

$x=-4$

$y=3$

$z=-5$

6 0

2 years ago

Could someone help me with number 6 ? The directions say what are the variables? which I know the variables are the height and t

kramer

As the time go the grass height go through increasing and decreasing

6 0

2 years ago

How do you solve x^4? Is it (a) x times x times x times x or (b) x plus x plus x plus x

andriy [413]

Step-by-step explanation:

${x}^{4} = x \times x \times x \times x$

Multiply x as much time as its degree.

8 0

2 years ago

Read 2 more answers

Which pair of angles are corresponding?

tatyana61 [14]

It's C because 1 and 5 are the same angles so they are parallel.

8 0

3 years ago

Find two unit vectors orthogonal to both 8, 5, 1 and −1, 1, 0 .

Elina [12.6K]

In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:

‹1, -1, 1› × ‹0, 1, 1›

You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.

So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...

In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0

That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:

a - b + c = 0
b + c = 0

This is two equations, three unknowns, so you can solve it with one free parameter:

b = -c
a = c - b = -2c

The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›

The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:

|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6

Then we divide that vector by its magnitude to yield one solution:

‹ -2/√6 , -1/√6 , 1/√6 ›

And take the negative for the other:

‹ 2/√6 , 1/√6 , -1/√6 ›

7 0

2 years ago