Whats the problem for this?
Answer:
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem,
a
2
+
b
2
=
c
2
, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. The relationship of sides
|
x
2
−
x
1
|
and
|
y
2
−
y
1
|
to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example,
|
−
3
|
=
3
. ) The symbols
|
x
2
−
x
1
|
and
|
y
2
−
y
1
|
indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.
c
2
=
a
2
+
b
2
→
c
=
√
a
2
+
b
2
It follows that the distance formula is given as
d
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
→
d
=
√
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
We do not have to use the absolute value symbols in this definition because any number squared is positive.
A GENERAL NOTE: THE DISTANCE FORMULA
Given endpoints
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
, the distance between two points is given by
d
=
√
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
Step-by-step explanation:
Answer:
x < -3
Step-by-step explanation:
Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.
−4x−3> 36/4
Divide 36 by 4 to get 9.
−4x−3>9
Add 3 to both sides.
−4x>9+3
Add 9 and 3 to get 12.
−4x>12
Divide both sides by −4. Since −4 is negative, the inequality direction is changed.
x< 12/-4
Divide 12 by −4 to get −3.
x < -3
Answer:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
Step-by-step explanation:
Let the dimensions of the box be x, y and z
The rectangular box has a square base.
Therefore, Volume of the box
Volume of the box
The material for the base costs 
, the material for the sides costs 
, and the material for the top costs 
.
Area of the base 
Cost of the Base 
Area of the sides 
Cost of the sides=
Area of the Top
Cost of the Base 
Total Cost, 
Substituting 
To minimize C(x), we solve for the derivative and obtain its critical point
![C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2](https://tex.z-dn.net/?f=C%27%28x%29%3D%5Cdfrac%7B0.6x%5E3-4.8%7D%7Bx%5E2%7D%5C%5CSetting%20%5C%3AC%27%28x%29%3D0%5C%5C0.6x%5E3-4.8%3D0%5C%5C0.6x%5E3%3D4.8%5C%5Cx%5E3%3D4.8%5Cdiv%200.6%5C%5Cx%5E3%3D8%5C%5Cx%3D%5Csqrt%5B3%5D%7B8%7D%3D2)
Recall: 
Therefore, the dimensions that minimizes the cost of the box are:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
