Answer: (0, -1 ), (1, -1), (2, 1)
<u>Step-by-step explanation:</u>
Set the exponent equal to zero - <em>that is your anchor point</em>.
Then choose an x-value less than and greater than the anchor point.
I guess you mean the derivative of:
√7x + √2x, if so then:
Remember that the derivative of √u = 1/(2√u) . u'
if 7x = u, then u' = 7 , hence the derivative of √7x is 7/(2√7x)
Use same logic for √2x and you will find:
y' = 7/(2√7x) + x/√2x
Answer:
Sqrt (11/8)
Step-by-step explanation:
Sec^2 - Tan^2 = 1
- Tan^2 = 1 - Sec^2
Tan^2 = -1 + Sec^2
Tan^2 = 3/8
3/8 = -1 + Sec^2 Theta
1 + 3/8 = Sec ^2 Theta
11/8 = Sec^2 Theta
Sqrt ( 11/8) = Sec of theta
-- Gage Millar, Algebra 1/2 Tutor (Pending Pre-Calc Tutor)
Answer:
1060 feet
Step-by-step explanation:
A boat looks up at a light house at an angle of elevation of 23.If the top of the lighthouse is 450 feet higher than sea level, what is the horizontal distance from the boat to shore?
We solve the above question using the Trigonometric function of Tangent
tan x = Opposite/Adjacent
x = Angle of Elevation = 23°
Opposite = 450 feet
Adjacent = The horizontal distance from the boat to shore = y
Hence,
tan 23 = 450/y
Cross Multiply
tan 23 × y = 450
y = 450/tan 23
y = 1060.1335646 feet
Approximately = 1060 feet
Therefore, the horizontal distance from the boat to shore is 1060 feet
Completing the function process square solving the quadratic circle 363(y)