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Otrada [13]
3 years ago
7

A life guard in a tower 20 ft above sea level spots a struggling surfer at an angle of depression of 15 . How far is the surfer

from the base of the tower?
Mathematics
1 answer:
GalinKa [24]3 years ago
7 0

Answer: 19.31\ ft

Step-by-step explanation:

Given

The tower is at a height of h=20\ ft

the angle of depression is 15^{\circ}

Suppose the surfer is x ft away from the base of the tower

\therefore \text{From the figure, we can write}\\\\\Rightarrow \cos 15^{\circ}=\dfrac{x}{h}\\\\\Rightarrow x=h\cos 15^{\circ}\\\Rightarrow x=20\cos 15^{\circ}\\\Rightarrow x=19.31\ ft

Therefore, the surfer is at a distance of 19.31 ft from the base of the tower.

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Consider the polynomial operation 3x(x − 2)(5x + 2). Is the expression equivalent? Select Yes or No for A‐D.
Sav [38]

Answer:

see attached

Step-by-step explanation:

3x(x − 2)(5x + 2)

  • (x − 2)(5x + 2) expand
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  • 3x ( 5x² - 8x - 4)  expand
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Translate the sentence into an equation.
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Answer:

Eight times the sum of a number and 2 is equal to 7.

times = multiply

sum = ()

and = +

SO..

8(w + 2 ) = 7

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2 years ago
3^x grows faster than x^3. Explain why.
Allisa [31]
Because 3 is a number and the x is unknow so its just an exponent the x ^ 3 grows three each time

3 0
3 years ago
F(x) = x4 - x3 - 2x- 4 factor the polynomial function over the complex numbers
olchik [2.2K]

Answer:

<u>f(x) = = (x + √2 i) (x - √2 i) (x - 2 ) (x + 1)</u>

Step-by-step explanation:

The given function is f(x) = x⁴ - x³ -2x -4

factor the polynomial function

f(x) = x⁴ - x³ -2x -4 = (x⁴ - 4) - (x³ + 2x )  ⇒ take (-) as a common from (- x³ -2x)

     = (x² + 2 ) (x² - 2) - x (x² + 2)   ⇒ take (x² + 2)  as a common

     = (x² + 2 ) ( x² - x - 2)

     = (x + √2 i) (x - √2 i) (x -2 ) ( x+1)

Notes:  (x⁴ - 4) = (x² + 2 ) (x² - 2)

            (x² + 2)= (x + √2 i) (x - √2 i)

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8 0
3 years ago
A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per
HACTEHA [7]

Answer:

A) Revenue function = R(x) = (580x - 10x²)

Marginal Revenue function = (580 - 20x)

B) Fixed Cost = 900

Marginal Cost function = (300 + 50x)

C) Profit function = P(x) = (-35x² + 280x - 900)

D) The quantity that maximizes profit = 4

Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

8 0
3 years ago
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