Use the identity
sec^2x = 1 + tan^2 x
- so sec x = sqrt(1 + tan^2 x) then:-
tan x + sqrt( 1 + tan^2 x) = 1
sqrt ( 1 + tan^2 x) = 1 - tan x
1 + tan^2 x = 1 + tan^2x - 2 tan x
0 = -2 tanx
tan x = 0
x = 0, π
π is an extraneous root because sec 180 = -1
So the answer is 0 degrees
Answer:
What’s the question
Step-by-step explanation:
Answer with explanation:
→→→Function 1
f(x)= - x²+ 8 x -15
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= - 2 x + 8
Put,f'(x)=0
-2 x+ 8=0
2 x=8
Dividing both sides by , 2, we get
x=4
Double differentiating the function
f"(x)= -2, which is negative.
Showing that function attains maximum at ,x=4.
Now,f(4)=-4²+ 8× 4-15
= -16 +32 -15
= -31 +32
=1
→→→Function 2:
f(x) = −x² + 2 x − 3
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= -2 x +2
Put,f'(x)=0
-2 x +2=0
2 x=2
Dividing both sides by , 2, we get
x=1
Double differentiating the function,gives
f"(x)= -2 ,which is negative.
Showing that function attains maximum at ,x=1.
f(1)= -1²+2 ×1 -3
= -1 +2 -3
= -4 +2
= -2
⇒⇒⇒Function 1 has the larger maximum.
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Step-by-step explanation: I dont know
I am confident that the answer is C. 53.7