Answer:
length = 200 m
width = 400 m
Step-by-step explanation:
Let the length of the plaing area is L and the width of the playing area is W.
Length of fencing around three sides = 2 L + W = 800
W = 800 - 2L ..... (1)
Let A is the area of playing area
A = L x W
A = L (800 - 2L)
A = 800 L - 2L²
Differentiate with respect to L.
dA/dL = 800 - 4 L
It is equal to zero for maxima and minima
800 - 4 L = 0
L = 200 m
W = 800 - 2 x 200 = 400 m
So, the area is maximum if the length is 200 m and the width is 400 m.
The solutions of the equations of the situation can be:
z = 1 , y = 5, x = 4
z= 2 , y = 3, x = 5
z=3 , y = 1, x = 6
z = 0 , y = 7, x =3
The question can be expressed as a equation
6 x + 8 y + 10 z = 84
also, x + y + z = 10
⇒ x = 10- y - z
Putting it in first equation,
6(10 - y - z ) +8y + 10z = 84
⇒ 60 +2y + 4z = 84
⇒2(y + 2z ) = 14
⇒ y + 2z = 7
Now putting
z = 1 , we get y = 5,
z= 2 , y = 3
z=3 , y = 1
z = 0 , y = 7
So, only 4 possible solutions.
Therefore, there can only be 4 possible solutions for the equations.
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Answer:
Step-by-step explanation:
Consider the given expression is
![\ln (x\sqrt[3]{x^2+1})](https://tex.z-dn.net/?f=%5Cln%20%28x%5Csqrt%5B3%5D%7Bx%5E2%2B1%7D%29)
We need to rewrite the expression as a sum,difference,or multiple of logarithms.
![[\because \sqrt[n]{x}=x^{\frac{1}{n}}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5Bn%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5D)
Using the properties of logarithm we get
![[\because \ln (ab)=\ln a+\ln b]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20%28ab%29%3D%5Cln%20a%2B%5Cln%20b%5D)
![[\because \ln (a^b)=b\ln a]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cln%20%28a%5Eb%29%3Db%5Cln%20a%5D)
Therefore, the simplified form of the given expression is .
Given that <span>Jack rakes the yard in 5 hours and Jill rakes the yeard in 8 hours. Let the amount of time that it takes them to rake the yard together be t, then:
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