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3 years ago

In the figure below, segment CD is parallel to segment EF and point H bisects segment DE :

Mathematics

1 answer:

Grace [21]3 years ago

4 0

Answer:

See explanation

Step-by-step explanation:

In the figure below, segment CD is parallel to segment EF, DE is a transversal, then angles DIH and HGI are congruent as alternate interior angles when two parallel lines are cut by a transversal.

Consider triangles DIH and EGH. In these triangles,

$\angle DIH\cong \angle EGH$ as alternate interior angles;
$\angle DHI\cong \angle GHE$ as vertical angles;
$DH\cong HE$ because point H bisects segment DE (given).

Thus,

$\triangle DIH\cong \triangle EGH$ by AAS postulate

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Answer:

Step-by-step explanation:

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2 years ago

You are planning to email to two lists. the first list has 1,200 names. the second list has 900 names. there are 150 names that

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3 years ago

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find x if the distance between points L and M is 15 and point M is located in the first quadrant. L=(-6,2) M=(x,2)

aivan3 [116]

Answer:

Therefore,

$x = 9$

Step-by-step explanation:

Given:

Let,

point L( x₁ , y₁) ≡ ( -6 , 2)

point M( x₂ , y₂ )≡ (x , 2)

l(AB) = 15 units (distance between points L and M)

To Find:

x = ?

Solution:

Distance formula between Two points is given as

$l(LM)^{2} = (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Substituting the values we get

$15^{2}=(x--6)^{2}+(2-2)^{2}\\\\225=(x+6)^{2}$

Square Rooting we get

$(x+6)=\pm \sqrt{225}=\pm 15\\\\x+6 = 15\ or\ x+6 = -15\\\\x= 9\ or\ x = -21$

As point M is located in the first quadrant

x coordinate and y coordinate are positive

So x = -21 DISCARDED

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$x = 9$

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