Answer:
11628
Step-by-step explanation:
Thanks!
Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
________________________________
now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P
Answer:
Yes, you can conclude that ΔJKL ≅ XYZ.
Step-by-step explanation:
I forget how to describe it in geometric terms but, if you read the similarity statement again, and follow each angle, you'll notice that the shapes match, and they correspond to each other, it would be different if it said something like ΔLKJ ≅ XYZ because it does not match.
If you're looking for a specific reason, we'd say that they're congruent because of SAS (Side-Angle-Side)
If this still confuses you, hit me up.
Answer:
Step-by-step explanation:
We use formula of BODMAS we starting by removing blackets by multiply (2)(3)and w get 6 and then by 2 we get 12 and add by 5 we get 17 and cross 17 to -36 and 17 change to -17 and then take -17-36=-53..
-36=(2)(3)2+5
-36=(6)2+5
-36=12+5
-36=17
-17-36
=-53
Answer:
C) P=56cm^2
Step-by-step explanation:
P=(a*h)/2 a=14, h=8
P=(14*8)/2
P=112/2
P=56