csc(2x) = csc(x)/(2cos(x))
1/(sin(2x)) = csc(x)/(2cos(x))
1/(2*sin(x)*cos(x)) = csc(x)/(2cos(x))
(1/sin(x))*1/(2*cos(x)) = csc(x)/(2cos(x))
csc(x)*1/(2*cos(x)) = csc(x)/(2cos(x))
csc(x)/(2*cos(x)) = csc(x)/(2cos(x))
The identity is confirmed. Notice how I only altered the left hand side (LHS) keeping the right hand side (RHS) the same each time.
x = -2
function value at x = -2 is intersection of the x value and graph
which is, 3
Hope it is clear:)
Answer:
80p
if you are adding 4p every time you add a drink you will be adding 7 drinks therefore you have to do (7x4p) this will equal 28p then you have to add 28p to 52p
Hope this helps
Answer:
Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Step-by-step explanation:
Given the figure with dimensions. we have to find the area of given figure.
Area of figure=ar(1)+ar(2)+ar(3)
Area of region 1 = ar(ANGI)+ar(AIB)
![=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha](https://tex.z-dn.net/?f=%3DL%5Ctimes%20B%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height%5C%5C%5C%5C%3D%5B1500%5Ctimes%20%285000-2000-1500%29%5D%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%283000-1500%29%5Ctimes%20%285000-2000-1500%29%5C%5C%5C%5C%3D3375000m%5E2%3D337.5ha)
Area of region 2 = ar(DHBC)

Area of region 3 = ar(GFEH)

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha
=987.5 ha
Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.
Let the fencing be done through x m downward from B which divides the two into equal area.
⇒ Area of upper part above fencing=Area of lower part below fencing
⇒
Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Answer:
{4, 4.5, 5, 5.5, 6, 6.5, 7} {0, 1, 2, 3, 4, 5, 6}. 4 ≤ h ≤ 7 0 ≤ d ≤ 6. 4 ≤ d ≤ 7. {4, 5, 6, 7}. 0 ≤ h ≤ 6 0 ≤ h ≤ 7