The edge bisector divides the segment either next of it proportionally. So, you have ...

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x/4 = 2.25/3

x = 4·2.25/3 . . . . Main point by 4

x = 3

Since DB = 3 and DA = 2.25, point D is clearly not the midpoint that AB.







Q.1 x-2/5=7

divide -2/5 i beg your pardon is -0.40

x-0.40=7. Move -0.40 to the right side of the equal sign and also that i do not care +0.40

x=7+0.40 . 7+0.40 is 7.4

x=7.4

Q.2 2.1r=6.174

divide 2.1 by both political parties which 2.1r divided by 2.1 is 1r and 6.174 separated by 2.1 is 2.94

r=2.94

Q.3

equation: 2/3 = 32 2/3 d

Q.4 x9=3/5

3/5 is 0.60

x9=0.60

divide 9 by both sides. 9x separated by 9 is 1x and 0.60/9 is

0.06666 (the 6 is continuing)

x=0.06666

Q.5

x=-3

Q.6

7.25 =0.5x

divide 0.5 by both sides. 0.5x split by 0.5 is 1x and 7.25 separated by

0.5 is 14.5

x=14.5

Q.7

the expression: 7-9m

Q.8

answer is C

Q.9

(x2) + (10(2))


Answer from: myhomeacc32

question 1 7 2/5

question 3. 2/3 * d = 32 2/3

question 9 2(x+10)

srry ns dont know all of the answers due to the fact that i didnt have these on my test but the ones ns answered space all exactly tm on this have actually a great day.


Answer from: IsabellaGracie
Answers:

These difficulties can be addressed by the Thales’s Theorem, i beg your pardon states:

Two triangles are comparable when they have equal angles and proportional sides

In addition, if two triangles space similar, their angle are comparable as well. This means that the relation between two sides of the big triangle is same to the relation between two political parties of the little triangle, as follows:

*

Knowing this, lets’s begin with the answers:

1. Because that this very first problem, see the very first figure attached. Over there are displayed the two triangles, and we room told both room similar, this method (according the front explanation above) that we have the right to use the Thale’s organize to find .

Therefore, we can create a relation between two political parties of the big triangle i m sorry is equal to the relation between two sides of the tiny triangle; in this instance let’s use

*
and
*
for the an initial triangle and
*
and
*
for the second:

*

*

Now we uncover :

*

*

*

*

*

Finally, the worth of is 12 units.

2. For this problem, view the 2nd figure attached. In stimulate to uncover the values of the segment BE and also EC, we will certainly use 2 sides of each triangle (ABE and also DCE) follow to the Thale’s Theorem:

For the segment BE:

*

*

Simplifying:

*

For the segment EC:

*

*

Simplifying:

*

3. For this problem, view the third figure attached:

*

Solving for :

*

Finally:

*

4. For this problem, check out the fourth figure attached.

See more: Who Sings Keeper Of The Stars " Was A Near Miss, The Keeper Of The Stars

*

Solving for :

*

*
*
*
*
*

Answers:

These problems can be resolved by the Thales’s Theorem, i beg your pardon states:

Two triangle are similar when they have equal angles and also proportional political parties

In addition, if two triangles space similar, their angles are comparable as well. This method that the relation between two political parties of the huge triangle is same to the relation in between two sides of the tiny triangle, as follows:

*

Knowing this, lets’s begin with the answers:

1. Because that this first problem, check out the an initial figure attached. There are presented the two triangles, and also we room told both room similar, this means (according the prior explanation above) that we deserve to use the Thale’s organize to uncover .

Therefore, us can develop a relation in between two political parties of the huge triangle i beg your pardon is same to the relation between two sides of the tiny triangle; in this situation let’s use

*
and also
*
because that the first triangle and also
*
and
*
because that the second:

*

*

Now we find :

*

*

*

*

*

Finally, the value of is 12 units.

2. Because that this problem, see the second figure attached. In stimulate to discover the worths of the segments BE and also EC, we will use 2 sides of every triangle (ABE and DCE) according to the Thale’s Theorem: