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AnnZ [28]
3 years ago
13

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

Mathematics
1 answer:
sergey [27]3 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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CD has endpoints C(-3, 4) and D(1, -2). Find the coordinates<br> of its midpoint.
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Step-by-step explanation:

Formula: (midpoint)

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x=\frac{x_c+x_D}{2} =\frac{-3+1}{2} =\frac{-2}{2} =-1

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3 years ago
Consider the expression. (StartFraction 2 m Superscript negative 1 baseline n Superscript 5 Baseline Over 3 m Superscript 0 Base
suter [353]

Answer:

The correct option is option (3) 4 ÷ 25.

Step-by-step explanation:

The expression in terms of <em>m</em> and <em>n</em> is:

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Exponent rule of division:

a^{x}\div a^{y}=a^{x-y}

Compute the value of the expression for <em>m</em> = -5 and <em>n</em> = 3 as follows:

F(m,n)=[\frac{2m^{-1}n^{5}}{3m^{0}n^{4}}]^{2}

F(-5,3)=[\frac{2\csdot (-5)^{-1}\cdot (3)^{5}}{3\cdot (-5)^{0}\cdot (3)^{4}}]^{2}

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4 years ago
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$480
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