If Triangle J K L is congruent to Triangle B C A, which statement must be true?

What is the solution to the system of equations? 3x-6y=-12 " " x-2y=10 Use the substitution method to justify that the given sys

Thepotemich [5.8K]

The two lines in this system of equations are parallel

Step-by-step explanation:

Let us revise the relation between 2 lines

If the system of linear equations has one solution, then the two line are intersected
If the system of linear equations has no solution, then the two line are parallel
If the system of linear equations has many solutions, then the two line are coincide (over each other)

∵ The system of equation is

3x - 6y = -12 ⇒ (1)

x - 2y = 10 ⇒ (2)

To solve the system using the substitution method, find x in terms of y in equation (2)

∵ x - 2y = 10

- Add 2y to both sides

∴ x = 2y + 10 ⇒ (3)

Substitute x in equation (1) by equation (3)

∵ 3(2y + 10) - 6y = -12

- Simplify the left hand side

∴ 6y + 30 - 6y = -12

- Add like terms in the left hand side

∴ 30 = -12

∴ The left hand side ≠ the right hand side

∴ There is no solution for the system of equations

∴ The system of equations represents two parallel lines

The two lines in this system of equations are parallel

Learn more:

You can learn more about the equations of parallel lines in brainly.com/question/8628615

#LearnwithBrainly

6 0

3 years ago

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

2 years ago

PLZ LOOK AT THE IMAGE AND PLZ TELL ME THE ANSWER I WILL GIVE BRAINLESS

raketka [301]

Answer:

a) Unit price :

12 cupcakes for $29 = $2.42

50 cupcakes for $129 = $2.58

Reasoning:

12 cupcakes for $29 = $29 ÷ 12 cupcakes which gives 2.416666667 and if rounded off gives you $2.42.
50 cupcakes for $129 = $2.58 accurately.

b) 12 cupcakes for $29 gives the lowest unit price as $2.42 is less than $2.58.

I hope this answer will help you !!!!

7 0

3 years ago

Last year there were 520 students enrolled at Olsen Middle School . This year there are 572 students enrolled. What is the perce

RSB [31]

Answer:

10%

Step-by-step explanation:

We have procent:

520 —-> 100%

572 ——> x%

So, it is:

520*x=572*100

X=57200/520=110%

110%-100%=10%

6 0

3 years ago

Solve x + 3y = 9 3x – 3y = -13 (1 point)

zavuch27 [327]

Answer:

(-1, 10/3)

x = -1

y = 10/3

Step-by-step explanation:

To solve a system, one of the methods you can use is elimination. To use elimination, you need a variable to have the same coefficient in BOTH equations. Since both equations have "3y", with the same coefficient (3), we can use this method.

We want to eliminate a variable by cancelling it out. Since positive 3y PLUS negative 3y is 0, the variable in eliminated. ADD each of the terms in the equations together.

. x + 3y = 9

+ 3x – 3y = -13

. 4x – 0y = -4 3y - 3y = 0

. 4x = -4 Divide both sides by 4 to isolate 'x'

. x = -1 Solved for the x-coordinate

Since we know one variable, 'x', we can easily find the other, 'y'. Substitute 'x' with -1 in any of the equations. Then, isolate 'x'.

x + 3y = 9

(-1) + 3y = 9

-1 + 1 + 3y = 9 + 1 Add 1 to both sides to cancel out left side.

3y = 9 + 1 Add on the right side (9 + 1 = 10)

3y = 10 Divide both sides by 3 to isolate 'y'

y = 10/3 Solved for the y-coordinate

Put the 'x' and 'y' coordinates together in an ordered pair, which you write as (x, y).

The solution to the system is (-1, 10/3).

5 0

3 years ago