The Angle Angle Side postulate (AAS) states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.
The Angle Side Angle (ASA) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
Consider <u>two right triangles</u> MNL and QNL. Note that
1. In triangle MNL, m∠NLM=90°-m∠LMN=90°-58°=32°.
2. In triangle QNL, m∠NQL=90°-m∠QLN=90°-32°=58°.
In these triangles:
- m∠MNL=m∠QNL=90° (given);
- m∠NLM=m∠NQL=58° (proved);
- m∠MLN=m∠QLN=32° (proved);
- Side LN is common (given).
Then triangles MNL and QNL are congruent by ASA (1, 3 and 4 conditions) or by AAS (2, 3 and 4 conditions).
Answer: correct choice is A
<span>Let n, n+2, and n+4 represent the three consecutive even integers. Or 2 in your case </span>
Answer:
y = -2x
Step-by-step explanation:
If two lines are parallel to each other, they have the same slope.
The first line is y = -2x + 3. Its slope is -2. A line parallel to this one will also have a slope of -2.
Plug this value (-2) into your standard point-slope equation of y = mx + b.
y = -2x + b
This question only asks for a line parallel to y = -2x + 3.
This means we can plug in any number (except 3) for b. We cannot plug in 3 because then it will be the same equation!
To make things simple, we can use 0 and therefore leave it out.
y = -2x
This line is parallel to y = -2x + 3.
Hope this helps!
Answer: 27! - [22! * 6!]
Step-by-step explanation:
Decorations available = 12 blue ballons, 9 Green Lanterns and 6 red ribbons
To determine the number if ways of arrangement for if the ribbons must not be together, we must first determine the number of possible ways to arrange these decorations items if there are no restrictions.
Total number of ways to arrange them = [12+9+6]! = 27! = 1.089 * 10^28
If this ribbons are to be arranged distinctively by making sure all six of them are together, then we arrange the 6 of them in different ways and consider the 6ribbons as one entity.
Number of ways to arrange 6 ribbons = 6!.
To arrange the total number of entity now that we have arranged this 6ribbons together and taking them as one entity become: = (12 + 9 + 1)!. = 22!
Number of ways to arrange if all of the ribbons are taken as 1 and we have 22 entities in total becomes: 22! * 6!.
Hence, to arrange these decoration items making sure no two ribbons are together becomes:
= 27! - (22! * 6!)
= 1.08880602 * 10^28
Answer: infinitely many solutions
Step-by-step explanation:
Simplify left side using distributive property to 6x+16. Combine like terms (2x and 4x) on the right side to 6x+16. Both sides simplify to the same expression (left side = right side) so there are infinitely many solutions. You can plug in any real number for x and the left side will always equal the right side.