Answer:
The solutions of the equation are 0 , π
Step-by-step explanation:
* Lets revise some trigonometric identities
- sin² Ф + cos² Ф = 1
- tan² Ф + 1 = sec² Ф
* Lets solve the equation
∵ tan² x sec² x + 2 sec² x - tan² x = 2
- Replace sec² x by tan² x + 1 in the equation
∴ tan² x (tan² x + 1) + 2(tan² x + 1) - tan² x = 2
∴ tan^4 x + tan² x + 2 tan² x + 2 - tan² x = 2 ⇒ add the like terms
∴ tan^4 x + 2 tan² x + 2 = 2 ⇒ subtract 2 from both sides
∴ tan^4 x + 2 tan² x = 0
- Factorize the binomial by taking tan² x as a common factor
∴ tan² x (tan² x + 2) = 0
∴ tan² x = 0
<em>OR</em>
∴ tan² x + 2 = 0
∵ 0 ≤ x < 2π
∵ tan² x = 0 ⇒ take √ for both sides
∴ tan x = 0
∵ tan 0 = 0 , tan π = 0
∴ x = 0
∴ x = π
<em>OR</em>
∵ tan² x + 2 = 0 ⇒ subtract 2 from both sides
∴ tan² x = -2 ⇒ no square root for negative value
∴ tan² x = -2 is refused
∴ The solutions of the equation are 0 , π
Answer:
The perimeter of a circle can be found by using the followinfg expression
P = 2*π*r
where
π = 3.14
r = radius of the circle = half the diameter of the circle
In this case, if we are given the radius, we use
P = 2*π*r
If we are given the diameter, we use
P = 2*π*(D/2) = π*D
1) 27in
radius = 27in
P = 2*(3.14)*(27 in) = 169.56 in
diameter = 27 in
P = (3.14)*(27 in) = 84.78 in
2) 79 in
radius = 79 in
P = 2*(3.14)*(79 in) = 496.12 in
diameter = 79 in
P = (3.14)*(79 in) = 248.06 in
3) 1809 in
radius = 1809 in
P = 2*(3.14)*(1809 in) = 11360.52 in
diameter = 1809 in
P = (3.14)*(1809 in) = 5680.26 in
4) 152 in
radius = 152 in
P = 2*(3.14)*(152 in) = 954.56 in
diameter = 152 in
P = (3.14)*(152 in) = 477.28 in
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Yes, if all 8 of the students shared the cost of $100 dollars equally, that would mean each student paid $12.50.
