Answer:
the antelope is 5280 feet the cougar is 4224 feet the hare is 4136 feet the kangaroo is 3520 feet and the coyote is 3773 feet the ostrich is 3773 feet hope this helps the<em> answer is the </em><em>antelope</em>
Answer:
-(6ap³)
Step-by-step explanation:
We have to find the quotients of
We take the common 2a² p from the numerator
= 
= -(6ap³)
Therefore, the quotient after division will be -(6ap³)
Answer:
Tom was speeding.
Step-by-step explanation:
The driving speed of Tom = 20.3 meters per second.
Speed allowed in the zone = 45 miles per hour zone.
Since the unit of speed allowed is given in miles per hour but the speed of tom is given in meters per second. So, first, we have to convert the meter per second into mile per hour then we can compare and find that Tom is speeding or not.
1 mile = 1609.34 meters
1 hour = 3600 second
Now convert 20.3 into mile per hour.
20.3 meters per second. = (20.3 / 1609.34)*3600 = 45.40981 mile per hour.
Since Tom’s speed is more than the allowed speed so he is speeding.
Assume:
Size of sides = x m
Depth of the pool = y m
Therefore, surface area = x^2+4xy =10 m^2
Then, y = (10-x^2)/(4x)
Now,
Volume (V) = x^2*y = x^2*y =x^2(10-x^2)/4x = (10x-x^3)/4 = 1/4(10x-x^3)
For maximum volume, first derivative of volume function is equal to zero.
That is,
dV/dx =0 = 1/4(10-3x^2)
Then,
1/4(10-3x^2) = 0
10-3x^2 = 0
3x^2=10
x= sqrt (10/3) = 1.826 m
And
y= (10-1.826^2)/(4*1.826) = 0.913 m
Therefore,
V= 1.826^2*0.913 = 3.044 m^3
Answer:
x₂ = 7.9156
Step-by-step explanation:
Given the function ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)
If f(x) = ln(x)+x-10
f'(x) = 1/x + 1
f(9) = ln9+9-10
f(9) = ln9- 1
f(9) = 2.1972 - 1
f(9) = 1.1972
f'(9) = 1/9 + 1
f'(9) = 10/9
f'(9) = 1.1111
x₁ = x₀ - f(x₀)/f'(x₀)
x₁ = 9 - 1.1972/1.1111
x₁ = 9 - 1.0775
x₁ = 7.9225
x₂ = x₁ - f(x₁)/f'(x₁)
x₂ = 7.9225 - f(7.9225)/f'(7.9225)
f(7.9225) = ln7.9225 + 7.9225 -10
f(7.9225) = 2.0697 + 7.9225 -10
f(7.9225) = 0.0078
f'(7.9225) = 1/7.9225 + 1
f'(7.9225) = 0.1262+1
f'(7.9225) = 1.1262
x₂ = 7.9225 - 0.0078/1.1262
x₂ = 7.9225 - 0.006926
x₂ = 7.9156
<em>Hence the approximate value of x₂ is 7.9156</em>
