When two parallel lines are intersected by a transversal, the same-side exterior angles are supplementary. That means that their sum is 180.
Using that logic, if the two roads were parallel, then the sum of their same-side exterior angles will add up to 180. Yet their same-side exterior angles add up to 170 (130 + 40 = 170), hence they can't be parallel.
See the drawing attached below.
Using supplmenatry angles (two angles whose sum of measures add up to 180 or a straight line), we can say that:
m<DIE + m<HID = 18
40 + m<HID = 180
m<HID = 140
Similarly:
m<BHC + m<CHI = 180
130 + m<CHI = 180
m<CHI = 50
Using verticle angles therome, (when two lines intersect, the angles opposite to eachother are congruent, or have the same measure), we can say that:
m<DIE = m<GIH = 40
m<GIE = m<HID = 140
m<CHI = m<AHB = 50
m<BHC = m<AHI = 130
Answer:
0.75
Step-by-step explanation:
Question 1 Answer: both of them equal 20
Step-by-step explanation: when you add 9+3 it equals 12 then you add 8 and that makes 20. 9+8 also equals 17 then you add 3 which equals 20
<span>a) We know that the correct answer will be the square root of 256 since the competition area is a square with an area of 256 meters. And since 10^2 = 100 which is less than 256, the answer has to be greater than 10. And since 20^2 = 400 which is greater than 256, the answer also has to be less than 20. Therefore the answer has to be between 10 and 20.
b) The last digit has to be either a 4 or a 6. The units digit is the only digit that will contribute to the units digit of the square. And 0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49, 8^2 = 64, 9^2 = 81. Of the 10 possible digits, only the values 4 and 6 have a square that has an units digit of 6.
c) The square root of 256 based up (a) and (b) above has to be either 14, or 16. So the dimensions are either 14x14 meters or 16x16 meters.</span>
