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2 years ago

3y = x - 12 Y= 1/3X -4 parallel, perpendicular or neither

Mathematics

1 answer:

aliina [53]2 years ago

5 0

Answer:

PARALLEL

Step-by-step explanation:

The second equation, y = (1/3)x - 4, has already been solved for y. To compare the slopes easily, we need now to solve the first equation for y, obtaining y = (1/3)x - 4.

We see that these two lines have the same slope, m = 1/3, and thus conclude that these lines are PARALLEL.

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alexandr402 [8]

Answer:

Surface Area=

$100\pi$

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3 years ago

Solve this system of equations: X=? Y=? ( JUST WRITE THE NUMBER ANSWERS WITH A COMMA BETWEEN THEM)

hjlf

Answer:

Therefore the Final Solution is

x = 2 and y = 4

( 2 , 4 )

Step-by-step explanation:

Given:

$y=2x$ ...............Equation ( 1 )

$x=-y+6$ ................Equation ( 2 )

To Find:

x = ?

y = ?

Solution:

Substitute Equation ( 1 ) in Equation ( 2 ) we get

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4 0

3 years ago

Is 10/9 closets to 0, 1/2 ,or 1

VARVARA [1.3K]

1 because 10/9= 1 1/10
which is closer to 1

3 0

3 years ago

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Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

Step 1

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

Step 2

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

Step 3

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

Step 4

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

Step 5

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

Step 6

Segment EF, segment FG, segment GH, and segment EH are congruent.

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

Step 7

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

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Mariulka [41]

Answer:

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3 years ago

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