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malfutka [58]
2 years ago
15

The number of mosquitoes in anchorage, alaska (in millions of mosquitoes) as a function of rainfall (in centimeters) is modeled

by: m(x)=-x^2+14xm(x)=−x 2 +14xm, left parenthesis, x, right parenthesis, equals, minus, x, start superscript, 2, end superscript, plus, 14, x how many centimeters of rain will produce the maximum number of mosquitoes?
Mathematics
1 answer:
bekas [8.4K]2 years ago
3 0
M (x) = - x ^ 2 + 14x
 To find the maximum function, the first thing to do is to obtain the first derivative:
 m '(x) = - 2x + 14
 We match zero:
 0 = -2x + 14
 We clear x:
 x = 14/2 = 7
 x = 7
<span>
 We verify that this value is a maximum through the second derivative:
 m '' (x) = - 2
 m '' (7) = - 2 (<0 therefore it is a maximum)

 Answer:
<span> 7 centimeters of rain will produce the maximum number of mosquitoes.</span></span>
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Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

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d = 9.22

Hence, the distance to the x axis is 9.22 units

<em>(e) The distance from (4, -7, 6) to the y axis</em>

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

<em>(f) The distance from (4, -7, 6) to the z axis</em>

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

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