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Whitepunk [10]
3 years ago
8

A Trapezoid has an area of 5 1/2 ft sq. If the trapezoid has a base that measures 2 3/5 ft and a height of 1 1/2 ft, what is the

length of the other base?
Mathematics
1 answer:
GalinKa [24]3 years ago
7 0
First of all, let us change all the mixed number into improper fraction:
5 1/2=11/2
2 3/5=13/5
1 1/2=3/2
Area of the trapezoid=
1/2(b1+b2)×h

11/2=1/2(13/5+b2)×3/2
Divided 3/2 or multiply 2/3 for both side
11/2×2/3=1/2(13/5+b2)×3/2×2/3
11/3=1/2(13/5+b2)
Multiply 2 for both side
11/3×2=1/2(13/5+b2)×2
22/3=13/5+b2
Subtract 13/5 for both side
b2= 4 11/15. As a result, the length of other base is 4 11/15 feet. Hope it help!
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deff fn [24]

Answer:

the minimum is located in x = -5/3 , y= -5/3

Step-by-step explanation:

for the function

f(x,y)=2x + 2y

we define the function g(x)=9x² - 9xy + 9y² - 25  ( for g(x)=0 we get the constrain)

then using Lagrange multipliers f(x) is maximum when

fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →

fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0

g(x) =0 → 9x² - 9xy + 9y² - 25 = 0

subtracting the second equation to the first we get:

2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0

- 18*y + 9*x + 18*x - 9*y = 0

27*y = 27 x  → x=y

thus

9x² - 9xy + 9y² - 25 = 0

9x² - 9x² + 9x² - 25 = 0

9x² = 25

x = ±5/3

thus

y = ±5/3

for x=5/3 and y=5/3 →  f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 →  f(x)= -20/3  (minimum)

finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y

thus the minimum is located in x = -5/3 , y= -5/3

4 0
3 years ago
Someone help? 25 points
kirill115 [55]

Answer:

A: scalene: non have the same angel

B: isosceles: two sides have the same angel

C: isosceles:two sides have the same angel

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4 0
4 years ago
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Bad White [126]

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Step-by-step explanation:

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