Answer:
∠CBJ ≅ ∠ DHG
∠ FJH ≅ ∠ AJB
m∠JAB=75°
m∠JBC=130°
Step-by-step explanation:
Please refer to the image uploaded with this answer.
Here we are given two parallel line C and D and two transverses to them E and G , there by forming a number of angles. Before answering them we must under stand the characteristics of the angles formed by them and the various properties of them. For that please refer to the image attached,
i) we are asked to find the congruent angle to ∠CBJ has a corresponding angle names as ∠ DHG. And corresponding angles are same so ∠CBJ ≅ ∠ DHG
ii) Also ∠ FJH ≅ ∠ AJB , as they are vertically opposite angles and vertically opposite angles are equal.
iii) m∠EFD =75°
m∠EFD ≅ m∠JAB=75° ( corresponding angles )
iv) m∠GHF=130°
m∠GHF=m∠JBC=130° ( corresponding angles )
Solve for x in the second equation first
Answer:
PT = 17
Step-by-step explanation:
Since T is the midpoint of PQ, then
PT = TQ , substitute values
4x + 5 = 8x - 7 ( subtract 4x from both sides )
5 = 4x - 7 ( add 7 to both sides )
12 = 4x ( divide both sides by 4 )
3 = x
Thus
PT = 4x + 5 = 4(3) + 5 = 12 + 5 = 17
Answer:
d = 167.7 m
Step-by-step explanation:
We have,
Mrs. Tyner drove about 150 km east from Pearland, to Houston, Texas. She drove about another 75 km north to Dallas.
It is required to find the approximate air distance from Pearland to Dallas, Texas.
The resultant vector will give the approximate air distance. So,
So, the approximate air distance from Pearland to Dallas, Texas is 167.7 m.
CD, CE, HG, GF, AL, LB, AB, HF, DE. A line segment is comprised of any two points. The ones I listed are the ones specifically named in the picture.