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Maslowich
2 years ago
6

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics
1 answer:
luda_lava [24]2 years ago
7 0

Answer:

Step-by-step explanation:

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In triangle abc shown below side ab is 6 and side ac is 4
kaheart [24]

Question is Incomplete, Complete question is given below:

In Triangle ABC shown below, side AB is 6 and side AC is 4.

Which statement is needed to prove that segment DE is parallel to segment BC and half its length?

Answer

Segment AD is 3 and segment AE is 2.

Segment AD is 3 and segment AE is 4.

Segment AD is 12 and segment AE is 4.

Segment AD is 12 and segment AE is 8.

Answer:

Segment AD is 3 and segment AE is 2.

Step-by-step explanation:

Given:

side AB = 6

side AC = 4

Now we need to prove  that segment DE is parallel to segment BC and half its length.

Solution:

Now AD + DB = AB also AE + EC = AC

DB = AB - AD also EC = AC - AE

Now we take first option Segment AD is 3 and segment AE is 2.

Substituting we get;

DB = 6-3 = 3 also EC = 4-2 =2

From above we can say that;

AD = DB and EC = AE

So we can say that segment DE bisects Segment AB and AC equally.

Hence From Midpoint theorem which states that;

"The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side."

Hence Proved.

7 0
3 years ago
Three ways to write 3 to the 5th power
cestrela7 [59]

First expression

=>  3 x  3 x  3 x 3 x 3 – In this expression, we multiplied 3 5 times and the product is 243

Second expression

=> 3 ^ 5 , in where ^ read as raised to the power. , the product is also 243

Third expression

=> 3^2 x 3 ^3

=> (3 x 3) x (3 x 3 x 3)

=> 9 x 27, the product is also equals to 243.

7 0
3 years ago
Read 2 more answers
What is the common factor of each term in the expression 2b + 5b?”
stellarik [79]
The common factor is b since they both share it.
Hope this helps
7 0
3 years ago
Find an equation of the ellipse that has center (-2,0), a minor axis of length 2, and a vertex at (8,0)
Viefleur [7K]

Answer:

Step-by-step explanation:

(-2,0) and (8,0) are horizontally aligned, so the ellipse is horizontal.

6 0
3 years ago
Rewrite \sqrt((1+cos45)/(2)) using a half-angle identity
aleksklad [387]

\stackrel{\textit{Half-Angle Identities}}{cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}} \\\\[-0.35em] ~\dotfill\\\\ \sqrt{\cfrac{1+cos(45^o)}{2}}~~ = ~~cos\left( \cfrac{45^o}{2} \right)\implies \sqrt{\cfrac{1+cos(45^o)}{2}}~~ = ~~cos(22.5^o)

7 0
2 years ago
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