The answer is D. x < 5.
9x+27 <72
9x <72-27
9x <45
x <45÷9
x <5
Answer:
D Option is the correct one.
Step-by-step explanation:
As,
Distance covered = 40km = 40000m(1k=1000m)
time = 10:15-9:00 = 1:15hours = 75mins
We need to find the rates of the bicyclist in one hour so, for this first we need to calculate the total distance which covered in one hour.
For calculating distance for one hour,
40000/75 = 533.33m/min (means they covered 533.33m distance in one min)
for one hour:
533.33*60= 32000m = 32km
They covered 32km distance in one hour.
Now from the given options D is the correct one because according to give condition north bicyclist covered 4km more that south one. And other reason is we also need to check the sum of distance covered by the both cyclist which must be equal to 32km.
So for A,
Sum = 19 + 15 =34 (wrong not equal to 32)
For B,
Sum = 18 + 13 = 31 (wrong)
For C,
Sum = 16 + 10 = 26(wrong)
For D,
Sum = 18 + 14 = 32 (Right)
And second distance b/w both distance is 4km.
18-14 = 4(Right)
So, D option is the right one.
I hope it will help you.
To find out which inequality is true, we need to plug in n=4.
And since 8 is really greater than 6, this inequality is
true! :)
<u />
7 is not less than 7. So this inequality is
false.
5 is not greater than or equal to 13, so this inequality is
false
<u />
And 1/3 is not greater than 3, so this inequality is also
false.
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
