The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ. What is the minimum value of the sum of the lengths of AP, BP, and C
Answers
The question is missing parts. Here is the complete question.
The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ.
What is the minimum value of the sum of the lengths of AP, BP and CP?
Answer: The sum is 4.73.
Step-by-step explanation: Height of a triangle is a perpendicualr line linking a vertex and its opposite side.
Because triangle ABC is isosceles, point Q divides the base in 2 equal parts:
BQ = CQ = 1
Suppose QP = x
To calculate minimum value of the sum:
AP = AQ - QP
AP = 3 - x
Since triangles BQP and CQP are congruent and right triangles, use Pythagorean Theorem to figure out the value of BP and CP:
BP = CP = [tex]\sqrt{BQ^{2}+PQ^{2}}[/tex]
BP = [tex]\sqrt{1+x^{2}}[/tex]
Then, sum of AP, BP and CP is
[tex]f(x)=3-x+2\sqrt{1+x^{2}}[/tex]
The minimum value is calculated using first derivative:
[tex]f'=-1+\frac{2x}{\sqrt{x^{2}+1} }[/tex]
The value of x is limited: it can assume value of 0, when P=A and x=3, when P=Q. So, interval is [0,3].
x has value:
[tex]-1+\frac{2x}{\sqrt{x^{2}+1} }=0[/tex]
[tex]2x=\sqrt{x^{2}+1}[/tex]
[tex]4x^{2}-x^{2}-1=0[/tex]
[tex]3x^{2}=1[/tex]
x = ± [tex]\frac{1}{\sqrt{3} }[/tex]
x can't assume negative value because is not in the interval:
x = [tex]\frac{1}{\sqrt{3} }[/tex]
To find the minimum value of the sum, substitute x in the function above:
f([tex]\frac{1}{\sqrt{3} }[/tex])=[tex]3-\frac{1}{\sqrt{3} } +2\sqrt{1+(\frac{1}{\sqrt{3} })^{2} }[/tex]
[tex]f(\frac{1}{\sqrt{3}} )=3-\frac{1}{\sqrt{3}}+2(\sqrt{\frac{4}{3} } )[/tex]
[tex]f(\frac{1}{\sqrt{3}} )=3-\frac{1}{\sqrt{3}} +\frac{4}{\sqrt{3}}[/tex]
[tex]f(\frac{1}{\sqrt{3}} )=3+\frac{3}{\sqrt{3} }[/tex]
[tex]f(\frac{1}{\sqrt{3}} )=4.73[/tex]
The minimum value of the sum of AP, BP and CP is 4.73.
The minimum value of the sum of the lengths of AP, BP and CP is [tex]3 + \frac{\sqrt{3}}{3}[/tex] units.
According to this statement we must determine the sum of the line segment lengths so that a minimum is found. By Pythagorean theorem and given data we have the following expression:
[tex]y = AP + BP + CP[/tex]
[tex]y = 3-x +2\sqrt{x^{2}+1}[/tex] (1)
Now we proceed to find the critical values by performing first and second derivative tests.
FDT
[tex]-1 +\frac{2\cdot x}{\sqrt{x^{2}+1}} = 0[/tex]
[tex]2\cdot x = \sqrt{x^{2}+1}[/tex]
[tex]4\cdot x^{2}-x^{2}-1=0[/tex]
[tex]3\cdot x^{2} = 1[/tex]
[tex]x^{2} = \frac{1}{3}[/tex]
[tex]x = \frac{\sqrt{3}}{3}[/tex]
SDT
[tex]y'' = \frac{2\cdot \sqrt{x^{2}+1}-2\cdot x \cdot \left(\frac{2\cdot x}{\sqrt{x^{2}+1}} \right)}{x^{2}+1}[/tex]
[tex]y'' = \frac{2\cdot x^{2}+2-4\cdot x^{2}}{(x^{2}+1)^{3/2}}[/tex]
[tex]y'' = 2\cdot \frac{1-x^{2}}{(x^{2}+1)^{3/2}}[/tex]
[tex]y'' \approx 0.866[/tex]
A minimum exists when [tex]y'' > 0[/tex], then we conclude that [tex]x = \frac{\sqrt{3}}{3}[/tex] lead to a relative minimum. And by (1) we have the minimum sum:
[tex]y = 3-\frac{\sqrt{3}}{3}+2\sqrt{\frac{4}{3} }[/tex]
[tex]y = 3 - \frac{\sqrt{3}}{3} +\frac{4\sqrt{3}}{3}[/tex]
[tex]y = 3+\frac{\sqrt{3}}{3}[/tex]
The minimum value of the sum of the lengths of AP, BP and CP is [tex]3 + \frac{\sqrt{3}}{3}[/tex] units.
Nota - The statement reports typographical issues, the correct form is presented below:
The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ. What is the minimum value of the sum of the lengths of AP, BP and CP.
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Answer:
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Answers
Answer:
Step-by-step explanation:
first we need to get the side length of a square.
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Answers
Answer:
a) The number is:
0.73652913
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0.73652913 = 0.73652913*(100,000,000)/(100,000,000) = 73,652,913/100,000,000
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b) The number is:
0.746746...
Where the periodic part is 746.
the periodic part has 3 digits, then we can do:
0.746746...*1000 = 746.746746....
Now we can subtract the number itself to get:
746.746746... - 0.746746... = 746
And what we did is:
N*1000 - N = N*999
Then to transform the number into a quotient of integers, we can multiply and divide by 999.
0.746746... = 0.746746...*(999)/(999) = 746/999
That is a quotient of two integers, so 0.746746... is a rational number.
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Answer:
1
Step-by-step explanation:
The point is located in the first quadrant because x and y are both positive. The quadrants are labeled in counter-clockwise order, starting in the upper-right.
what is the value of x in 1.5=x+0.25
Answers
Answer:
x = 1.25
Step-by-step explanation:
1.5 = X + 0.25
1.25 = x you subtract 0.25 from both sides
Solve the system:
x - y = 5
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Answer:
8,3
Step-by-step explanation:
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Answer:
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Answers
Answer:
x = 62
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Step-by-step explanation:
x+(0)+(0-2)=60
x-2=60
x=62
(0)+y+(y-2)=60
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A parabolic arch sculpture is on top of a city bank. A model of the arch is y = −0.005x2 + 0.3x where x and y are in feet.
The image is of a rectangle which represents the building of a Bank and its height is 30 feet. On top of it a semi circle is placed whose diameter is equal to the width of the rectangle.
a. What is the distance from the highest point of the arch to the ground?
b. What is the width of the bank?
A. a. 34.5 feet
b. 60 feet
B. a. 4.5 feet
b. 60 feet
C. a. 4.5 feet
b. 30 feet
D. a. 34.5 feet
b. 30 feet
Answers
Answer:
A. a. 34.5 feet
b. 60 feet
Step-by-step explanation:
Parabola equation
[tex]y=-0.005x^2+0.3x[/tex]
Differentiating with respect to x we get
[tex]\dfrac{dy}{dx}=-0.01x+0.3[/tex]
Equating with zero
[tex]-0.01x+0.3=0\\\Rightarrow -0.01x=-0.3\\\Rightarrow x=\dfrac{0.3}{0.01}\\\Rightarrow x=30[/tex]
Double derivative of the parabolic equation
[tex]\dfrac{d^2y}{dx^2}=-0.01<0[/tex]
So, [tex]x=30[/tex] is maximum.
[tex]y=-0.005\times 30^2+0.3\times 30\\\Rightarrow y=4.5[/tex]
So, the maximum height of the arch will be 4.5 feet.
From the ground the highest point of the arch will be [tex]30+4.5=34.5\ \text{ft}[/tex]
We are taking the x axis as the width of the bank.
[tex]0=-0.005x^2+0.3x\\\Rightarrow 0.005x^2=0.3x\\\Rightarrow x=\dfrac{0.3}{0.005}\\\Rightarrow x=60[/tex]
So, the width of the bank will be 60 feet.
Helppp plzzzz N+5n=
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Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When carrying gallons of fuel, the airplane weighs pounds. When carrying gallons of fuel, it weighs pounds. How much does the airplane weigh if it is carrying gallons of fuel?
Answers
Step-by-step explanation:
Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When carrying 15 gallons of fuel, the airplane weighs 2187 pounds. When carrying 35 gallons of fuel, it weighs 2303 pounds. How much does the airplane weigh if it is carrying 50 gallons of fuel?
----
You have two points relating fuel and weight.
(15,2187) and (35,2303)
------------
slope = (2303-2187)/(35-15) = 5.8
----
intercept = ?
2187 = 5.8*15 + b
b = 2100
-------
Equation:
f(x) = 5.8x + 2100
---
f(50) = 2390 lbs.
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