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AnnZ [28]
2 years ago
13

X-2y=18 2x+y=6 parallel perpendicular or neither

Mathematics
1 answer:
sergey [27]2 years ago
4 0
<h3>Therefore they are perpendicular.</h3>

Step-by-step explanation:

A equation of line is

y =mx +c

Here the slope of the line is m.

Given equations are

x - 2y = 18

⇔-2y = -x +18

\Leftrightarrow y =\frac{1}{2} x -9............(1)

and 2x + y = 6

⇔y = -2x +6 ............(2)

Therefore the slope of equation (1) is(m_1)= \frac{1}{2}

Therefore the slope of equation (2) is(m_2)= -2

If two lines are perpendicular, when we multiply their slope we get -1.

therefore,

m_1. m_2 =\frac{1}{2}. (-2) = -1

Therefore they are perpendicular.

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mrs_skeptik [129]

Answer:

D

Step-by-step explanation:

i think its D :)

8 0
3 years ago
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Hey I’ve got the answer for the first question but I’m stuck on this one 5^-3 X 2^-1
Phantasy [73]

Answer:

  • See below

Step-by-step explanation:

a)

  • 6ˣ = 1/216
  • 6ˣ = 1/6³
  • 6ˣ = 6⁻³
  • x = -3

b)

  • 5⁻³ × 2⁻¹ =
  • 1/5³ × 1/2 =
  • 1/(125×2) =
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8 0
2 years ago
Determine the range of the following graph
Ierofanga [76]

Given:

The graph of a function is given.

To find:

The range of the graph.

Solution:

We know that, the domain is the set of input values and range is the set of output values.

In a graph, domain is represented by the x-axis and range is represented by the y-axis.

From the given graph it is clear that there is an open circle at (-8,-8) and a closed circle at (3,4). It means the function is not defined at (-8,-8) but defined for (3,4).

The graph of the function is defined over the interval -8. So, the domain is (-8,3].

The values of the function lie in the interval -8. So, the range is (-8,4].

Therefore, the range of the function are all real values over the interval (-8,4].

7 0
3 years ago
When you subtract two positive intergers, is the result always a positive integer.
GaryK [48]

thanks for the fact, that was really helpful !!!

5 0
3 years ago
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I have a series of 2 questions.
Margarita [4]
1)
I:y=3x-4
II:9x-3y=14


substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14

this is obviously not equal so there is no solution, the lines are parallel

2)
I:y=4x+6
II:5x-y=6

substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54

the solution is (12,54)
3 0
3 years ago
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