1)which is not a major factor resulting in pollution from increased economic activity in china?
D) more bycicles are being used for traveling purposes.
2) Economic trends in chinas Gross Domestic Product over a 50-Year period can be best described as (1point)
B)relatively stable with low levels of economic activity from the 1950's to the late 1990s.
Answer : The value of
is
.
Explanation :
As we are given 6 right angled triangle in the given figure.
First we have to calculate the value of
.
Using Pythagoras theorem in triangle 1 :




Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 2 :





Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 3 :





Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 4 :





Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 5 :





Now we have to calculate the value of
.
Using Pythagoras theorem in triangle 6 :





Therefore, the value of
is
.
Sorry, I won't understand the question what you asked now.
I believe it’s b race riots