<span>y(1) = 0
y'(1) = e
y" = c1ex + c2e-x
y' = c1ex - c2e-x
for solving c1, 0 = c1e1 + c2e-1
this implies that c1 = - (c2/e2) and
to solve c2
e1 = (-c2e-2)e1 - c2e-1
e1 = (-2c2e-1)
c2= - (e1/2e-1) = - (e2/2)
c1 = - (c2/e2) = (e2/2e2)
Therefore y =(e2/2e2)ex - (e2/2)e-x</span>
Correct answer for the above question is - option B. 86°
<u>Step-by-step explanation:</u>
Given:
∠NOP = 24°
∠NOQ = 110°
∠NOP and ∠POQ are adjacent angles
To Find:
∠POQ = ?
Solution:
Hereby, we can say that ∠POQ lies between line OQ and ON as given (∠NOP and ∠POQ are adjacent angles )
∠NOQ - ∠NOP = ∠POQ
∠POQ = 110° - 24°
<u>∠POQ = 86°</u>
Angle POQ is 86°
Thus we can conclude option B as correct answer.
Answer:
B
Step-by-step explanation:
-(x^2+4x-21)
If you were to factor this, it would be,
-(x+7)(x-3)
Please mark me Brainliest
A) -3
B) 5
C) -2
D)-6
i have to write 20 characters so sisiasskklsa
Answer:
6
Step-by-step explanation:
70/84 = 5/N
70N = 420
N = 420/70 = 6
