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nikitadnepr [17]
2 years ago
11

Help please I will give brainlist

Mathematics
1 answer:
pogonyaev2 years ago
7 0

Answer:

B y+8 = 2(x-4)

Step-by-step explanation:

perpendicular slope is a negative reciprocal of the x

which is 2x

then use the formula

y-y1 = m(x-x1)

y + 8 = 2(x-4)

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Find unit rate round to nearest hundredth if necessary.
S_A_V [24]

Answer:

Step-by-step explanation:

0.57 . im guessing

3 0
2 years ago
PLEASE HELP, please write it out in the system of equations (step by step)
gizmo_the_mogwai [7]

Answer:

{x,y}={-2,3}

Step-by-step explanation:

[1]    2x + y = -1

  [2]    x - 2y = -8  <------- given linear equations

Graphic Representation of the Equations : ----> given in attatchment

y + 2x = -1        -2y + x = -8  < ----- point where they connect is shown in graph

Solve by Substitution :

// Solve equation [2] for the variable  x

 [2]    x = 2y - 8

// Plug this in for variable  x  in equation [1]

  [1]    2•(2y-8) + y = -1

  [1]    5y = 15

// Solve equation [1] for the variable  y

  [1]    5y = 15

  [1]    y = 3

// By now we know this much :

   x = 2y-8

   y = 3

// Use the  y  value to solve for  x

   x = 2(3)-8 = -2

Solution :

{x,y} = {-2,3}

7 0
2 years ago
Read 2 more answers
For the vectors Bold uequalsleft angle negative 8 comma 0 comma 1 right angleand Bold vequalsleft angle 1 comma 3 comma negative
trapecia [35]

Answer:

Step-by-step explanation:

Given that there are two vectors U = (-8,0,1)

V = (1,3,-3)

To find projection of u on V and also projection of V on U

First let us find the dot product U and V

U.V = -8(1)+0(3)+1(-3)\\= -11

Projection of U on V =\frac{U.V}{|V|} \\=\frac{-11}{\sqrt{1+9+9} } \\=\frac{-11}{\sqrt{19} }

Similarly projection of V on U

=\frac{U.V}{|U|} \\=\frac{-11}{\sqrt{64+0+1} } \\=\frac{-11}{\sqrt{65} }

4 0
3 years ago
What is the domain of h?
sergeinik [125]

Answer:

C

Step-by-step explanation:

  • The domain is all the x numbers for each point
  • So -5,-4,0,2,4
3 0
2 years ago
What is the sum of the matrices shown below?
damaskus [11]

Answer:

\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]

Step-by-step explanation:

To add matrices, we add the corresponding components.

The given matrices is

\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]

We add the corresponding components to get;

\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}3+6&9+0\\5+-8&-2+4\end{array}\right]

We simplify to get:

\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]

5 0
2 years ago
Read 2 more answers
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