You can do this by drawing one line through parallel to PQS to meet RQ at T
Now calculate length of RT:-
cos 70 = RT / 70 giving RT = 23.94m
sin 70 = ST/70 giving ST = 65.78 m
draw a line from S perpendicular to PQ to meet PQ at U.
PU = 110 - 65.78 = 44.22 m
tan 50 = SU / 44.22 giving SU = 52.70 m
TQ = SU = 52.70 m
So x = TQ + RT = 52.70 + 23.94 = 76.6 m to 1 dec place.
It would be (B) because 450+200x4=2620 fence posts
Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
Answer:
ok i will do that
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
1+3(x-1)-2x
Substitute the value of x into the equation:
1 + 3(4-1) -2 (4)
1 + 3(3)- 8
1 + 9 - 8
2
