Distance between A and B
= Distance between C and D
= sqrt((4 - 1)^2 + (5 - 2)^2)
= sqrt(3^2 + 3^2)
= sqrt(2 * 3^2)
= sqrt(3^2) * sqrt(2)
= 3sqrt(2)
Distance between B and C
= Distance between A and D
= sqrt((4 - 3)^2 + (5 - 0)^2)
= sqrt(1^2 + 5^2)
= sqrt(26)
Since sqrt(26) is more than 3sqrt(2), the length must be sqrt(26).
Hope this helps you.
Answer:
Start from (0,0) and do rise over run
Step-by-step explanation:
Rise 4
Run 3
For the III (3rd) quadrant use
Rise -4
Run -3
Answer:
C. ∠SRT≅∠VTR and ∠STR≅∠VRT
Step-by-step explanation:
Given:
Quadrilateral is a parallelogram.
RS║VT; RT is an transversal line;
Hence By alternate interior angle property;
∠SRT≅∠VTR
∠STR≅∠VRT
Now in Δ VRT and Δ STR
∠SRT≅∠VTR (from above)
segment RT= Segment RT (common Segment for both triangles)
∠STR≅∠VRT (from above)
Now by ASA theorem;
Δ VRT ≅ Δ STR
Hence the answer is C. ∠SRT≅∠VTR and ∠STR≅∠VRT
Answer:
1/3
Step-by-step explanation:
Range of data set: 12 - 6 = 6
Given that there is a uniform distribution:
P(6<x<7) = 1/6
P(7<x<8) = 1/6
P(8<x<9) = 1/6
P(9<x<10) = 1/6
P(10<x<11) = 1/6
P(11<x<12) = 1/6
where x is weight loss
Probability that weight loss is more than 10 pounds:
P(10<x<11) + P(11<x<12) = 1/6 + 1/6 = 1/3
