Answer:
1) By SAS theorem, ΔADE≅ΔCDF
2) By SSS theorem, ΔBDE≅ΔBDF
Step-by-step explanation:
Consider isosceles triangle ABC (see diagram).
1. In triangles ADE and CDF:
- AD≅DC (since BD is median, then it divides side AC in two congruent parts);
- AE≅CF (given);
- ∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).
By SAS theorem, ΔADE≅ΔCDF.
2. In triangles BDE and BDF:
- side BD is common;
- DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);
- BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).
Be SSS theorem, ΔBDE≅ΔBDF.
Step 1: -12x - 2y ≥ -42
Step 2: -2y ≥ -12 - 42
Step 3: y ≤ -6x + 21
So the answer is
D) <span>y ≤ -6x + 21</span>
Answer:
46 .....................then 7
Answer:
C I think
Step-by-step explanation:
Total number of ways to make a pair:
The first player can be any one of 7 . For each of those . . .
The opponent can be any one of the remaining 6 .
Total ways to make a pair = 7 x 6 = 42 ways .
BUT ... every pair can be made in two ways ... A vs B or B vs A .
So 42 'ways' make only (42/2) = 21 different pairs.
If every pair plays 2 matches, then (21 x 2) = <em><u>42 total matches</u></em> will be played.
Now, is that an elegant solution or what !
