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vitfil [10]
3 years ago
6

Jamie's father has asked him to paint the fence around their home. If the fence is 119.5 yards long, and Jamie can paint an aver

age of 4.78 yards of the fence per hour, how many hours will it take for Jamie to finish painting the fence?
21
22
23
24
25
Mathematics
1 answer:
MissTica3 years ago
5 0
25.

The total yards of fence (119.5) / the amount of fence Jamie can paint per hour (4.78) = 25.
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Raul is selling coupon books. He started with 50 coupon books and has already won a prize for selling at least 10 coupon books.
Hoochie [10]
The practical domain would be all integers 10 to 50.<span />
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One-half of Heather's age 2 years from now plus one-third of her age three years ago is twenty years. How old is she now?
viktelen [127]
She is 24
24+2=26
26(1/2)=13
13+((24-3)1/3)
13+(21*1/3)
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Therefore she is 24


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What is the unit cost per bottle if a pack of 12 bottles of juice sells for $6?
zvonat [6]
12/6 is 2 bottles for a dollar 
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4 years ago
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Greatest common factor of three univariate monomials. Find the greatest common factor of these three expressions.
Mice21 [21]

Answer:

9w^{2}

Step-by-step explanation:

An easy way to find the GCF of different numbers is to list out all of their factors. In this case, the coefficients share multiple factors, but the largest is 9.

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7 0
3 years ago
Lagrange Multiplier Million Dollar question. Anybody please Use the method of Lagrange multipliers to find the max and min value
jekas [21]

 

We are given the equation: g(s,t) = t^2 e^s

Which is subject to the constraint: s^2 + t^2=3

 

The points (x,y) that will maximize g(s,t) will be the points that will satisfy the equation ∇f(x, y, z) = λ∇g(x, y, z), therefore:

2 t e^s = λ (2 t), so  e^s = λ           --> 1

t^2 e^s = λ (2 s)                                                --> 2

s^2 + t^2 = 3                                       -->3

 

To solve this problem, note that λ cannot be zero by equation 1 since e^s can never be zero. Therefore plug in equation 1 to 2:

t^2 e^s = (e^s) (2 s)       

t^2 = 2 s                                               --> 4

 

Plug in equation 4 to 3:

s^2 + 2s = 3

 

By completing the square:

s^2 + 2s + 1 = 4

(s + 1)^2 = 4

s + 1 = ±2

s = -3, 1

 

Calculating for t using equation 4:

when s = -3

t^2 = 2(-3)

t = sqrt(-6)

Since t is imaginary, therefore s=-3 is not a solution

 

when s = 1

t^2 = 2(1)

t = sqrt(2) = ±1.414

 

Therefore the maxima and minima points are at:

(1, -1.414) and (1, 1.414)

 

g(1, -1.414)=(-1.414)^2 e^(1) = 5.434

g(1, -1.414)=( 1.414)^2 e^(1) = 5.434

5 0
3 years ago
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