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3 years ago

Which equations represent the line that is perpendicular to the line 5x - 2y = -6 and passes through the point

Mathematics

1 answer:

Vinvika [58]3 years ago

6 0

Answer

See below

Step-by-step explanation:

The given line has equation

$5x - 2y = - 6$

The slope intercept form is

$y = \frac{5}{2} x + 3$

The slope of this line is

$m = \frac{5}{2}$

The line perpendicular to this line has slope

$- \frac{2}{5}$

If this line passes through (5,-4), the point-slope form is

$y + 4 = - \frac{2}{5} (x - 5)$

The slope-intercept form is

$y = - \frac{2}{5} x- 2$

The standard form is

2x+5y=-10

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Answer: A

Step-by-step explanation:

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2 years ago

A. 3.4

seropon [69]

A is your answer 3.4

Use Pythagorean Theroum

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3 years ago

WILL GIVE BRAINLIEST!!

VMariaS [17]

Answer:

$A(-1,4)\to A''(-1,-1)$

$B(0,2)\to B''(1,0)$

$C(1,2)\to C''(1,1)$

$D(-2,4)\to D''(-1,2)$

Step-by-step explanation:

The given trapezoid has vertices at A(−1,4), B(0,2), C(1,2) and D(2,4).

The transformation rule for 90° counterclockwise rotation is

$(x,y)\to(-y,x)$

This implies that:

$A(-1,4)\to A'(-4,-1)$

$B(0,2)\to B'(-2,0)$

$C(1,2)\to C'(-2,1)$

$D(2,4)\to D'(-4,2)$

This is followed by a translation 3 units to the right.

This also has the rule: $(x,y)\to (x+3,y)$

$A'(-4,-1)\to A''(-1,-1)$

$B'(-2,0)\to B''(1,0)$

$C'(-2,1)\to C''(1,1)$

$D'(-4,2)\to D''(-1,2)$

Therefore:

$A(-1,4)\to A''(-1,-1)$

$B(0,2)\to B''(1,0)$

$C(1,2)\to C''(1,1)$

$D(-2,4)\to D''(-1,2)$

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3 years ago

What's the difference between partial variation and direct variation

shepuryov [24]

Answer:

Direct Variation is a relation that is of the form y = mx.

and

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Explanation:

This is a simple way to see the difference between them.

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3 years ago

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Fofino [41]

Answer:

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Step-by-step explanation:

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