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2 years ago

Write equations for the horizontal and vertical lines passing through the point (-6,1)

Mathematics

1 answer:

Oksanka [162]2 years ago

4 0

Answer:

y = 1 and x = - 6

Step-by-step explanation:

The equation of a horizontal line parallel to the x- axis is

y = c ( where c is the value of the y- coordinates the line passes through )

The line passes through (- 6, 1 ) with y- coordinate 1 , then

y = 1 ← equation of horizontal line

The equation of a vertical line parallel to the y- axis is

x = c ( where c is the value of the x- coordinates the line passes through )

The line passes through (- 6, 1 ) with x- coordinate - 6 , then

x = - 6 ← is equation of vertical line

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$Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2$

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$We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\$

$As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2$

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