I think we can use the identity sin x/2 = sqrt [(1 - cos x) /2]
cos x - sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1
cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) = cos x - 1 Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1
cos^2 x - 0.5 cos x - 0.5 = 0
cos x = 1 , -0.5
giving x = 0 , 2pi, 2pi/3, 4pi/3 ( for 0 =< x <= 2pi)
because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.
The last 2 results dont fit so the answer is x = 0 , 2pi Answer
26 people. There will be no ice cream left. 19.5/.75=26 people
4. Compute the derivative.
Find when the gradient is 7.
Evaluate 
at this point.
The point we want is then (2, 5).
5. The curve crosses the 
-axis when 
. We have
Compute the derivative.
At the point we want, the gradient is
6. The curve crosses the -axis when 
. Compute the derivative.
When , the gradient is
7. Set 
and solve for . The curve and line meet when
Compute the derivative (for the curve) and evaluate it at these values.
8. Compute the derivative.
The gradient is 8 when 
, so
and the gradient is -10 when 
, so
Solve for 
and 
. Eliminating , we have
so that
.
Answer:2L times 2W
Step-by-step explanation:
10. GCF is 3
11. GCF is 4
12. GCF is 16
13. That the numbers are prime
:)
