Tan x /(1 +sec x) + (1+sec x) /tan x
Tan x=sin x / cos x
1+ sec x=1 +1/cos x=(cos x+1)/cos x
Therefore:
tan x /(1 +sec x) =(sin x/cos x)/(cos x+1)/cos x=
=(sin x * cos x) / [cos x* (cos x+1)]=sin x /(Cos x+1)
(1+sec x) /tan x=[(cos x+1)/cos x] / (sin x/cos x)=
=[cos x(cos x+1)]/(sin x *cos x)=(cos x+1)/sin x
tan x /(1 +sec x) + (1+sec x) /tan x=
=sin x /(Cos x+1) + (cos x+1)/sin x=
=(sin²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
Remember: sin²x+cos²x=1⇒ sin²x=1-cos²x
=(1-cos²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
=2 cos x+2 / [sin x(cos x+1)]=
=2(cos x+1) / [sin x(cos x+1)]=
=2 /sin x
Answer : tan x /(1 +sec x) + (1+sec x) /tan x= 2/sin x
Step-by-step explanation:
1/2 × (p+9) is the algebraic expression.
Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)
Answer: 3
You can find slope by using the 'rise over run' method. Find 2 clear points and count the number of units between those points, across the x axis and up the y axis. You should have 3/1 (y/x) as your slope.