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3 years ago

Which graph shows the solution to the inequality -x + 2 > 1

Mathematics

1 answer:

Darya [45]3 years ago

8 0

-x+2 > 1
-x+2+x > 1+x .... add x to both sides
2 > 1+x
x+1 < 2
x+1-1 < 2-1 ... subtract 1 from both sides
x < 1

After solving for x, we get x < 1

To graph this, plot an open circle at 1 on the number line and shade to the left of this value. The open circle indicates that 1 is not part of the solution set.

If your teacher requires you to graph this on an xy grid, then draw a vertical line through 1 on the x axis. Make this vertical line a dashed line. Then shade the entire region to the left of this dashed line. Any point in this shaded region will have an x coordinate that is less than 1. The dashed line acts like the open circle. The dashed line tells the reader "any point on this dashed line is not part of the solution set"

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The coordinates of the orthocenter of △JKL is (-4, 8)

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The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

$JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117$

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$\Rightarrow 117=81+36$

$\Rightarrow JL^2=JK^2+KL^2$

$\Rightarrow \text{JKL is a right angle triangle}$

$\Rightarrow \angle K=90^{\circ}$

Therefore, the vertex at K (-4, 8) is the orthocentre.

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