Answer:
400 m^2.
Step-by-step explanation:
The largest area is obtained where the enclosure is a square.
I think that's the right answer because a square is a special form of a rectangle.
So the square would be 20 * 20 = 400 m^2.
Proof:
Let the sides of the rectangle be x and y m long
The area A = xy.
Also the perimeter 2x + 2y = 80
x + y = 40
y = 40 - x.
So substituting for y in A = xy:-
A = x(40 - x)
A = 40x - x^2
For maximum value of A we find the derivative and equate it to 0:
derivative A' = 40 - 2x = 0
2x = 40
x = 20.
So y = 40 - x
= 40 - 20
=20
x and y are the same value so x = y.
Therefore for maximum area the rectangle is a square.
This is the concept of trigonometry, we are required to calculate the number of floors the building has given the information above;
# floors=[height of the building]/[height of each floor]
height of each floor=17 ft
let the height of the building be,h.
h is given by;
tan theta=opposite/adjacent
theta=80
opposite=h
adjacent=75 ft
thus
tan 80=h/75
h=75 tan 80
h=425.35 ft
thus the number of floors will be:
425.35/17
=25.020≈25 floors
Answer:
See explanation and hopefully it answers your question.
Basically because the expression has a hole at x=3.
Step-by-step explanation:
Let h(x)=( x^2-k ) / ( hx-15 )
This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.
Solving for x in that equation:
Adding 15 on both sides:
hx=15
Dividing both sides by h:
x=15/h
For it be a hole, you also must have the numerator is zero at x=15/h.
x^2-k=0 at x=15/h gives:
(15/h)^2-k=0
225/h^2-k=0
k=225/h^2
So if we wanted to evaluate the following limit:
Lim x->15/h ( x^2-k ) / ( hx-15 )
Or
Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.
We were ask to evaluate
Lim x->3 ( x^2-k ) / ( hx-15 )
Comparing the two limits h=5 and k=225/h^2=225/25=9.
Please see the pic, I'd solved your question in it.
Answer:
Answer:
Step-by-step explanation:
The manufacturer requires that the weight differs from 20 oz in at most 0.4 oz, therefore we can write this difference (which can be either above or below 20 oz, to be smaller than or equal to 0.4 oz.
The weight can be:
1) smaller than or equal to 20 oz + 0.4 oz: then
2) larger than or equal to 20 oz - 0.4 oz: then
and which combined, can be written as a double inequality;
This double condition can also be written using the absolute value symbol as:
Step-by-step explanation:
