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andre [41]
3 years ago
7

POSSIBLE

Mathematics
1 answer:
Levart [38]3 years ago
7 0
I believe the answer is supplementary
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Use the method of lagrange multipliers to find
Yanka [14]

Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y,  λ) = x + y +  λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx =  λ*y + 1

dL/dy =  λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate  λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

(x*x - 196) = 0

x^2 = 196

x = √196 = 14

then the minimum will be:

x + y = x + x = 14 + 14 = 28.

b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y, λ) = x*y +  λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

6 0
3 years ago
-40-8b=-6(1+7b) what’s the answer
Andre45 [30]

\text{Hey there!}

\mathsf{-40-8b=-6(1+7b)}\\\mathsf{SIMPLIFY\ both\ sides\ of\ your\ equation}\\\mathsf{-40-8b=-6(1+7b)}

\mathsf{Distribute\ the\ RIGHT\ side\ of\ the\ equation }\\\mathsf{Distribute [the\ formula\ is: a(b+c)=a(b)+a(c)=ab+ac]}\\\mathsf{-6(1)+(-6)}(7b)\\\mathsf{-6(1)=-6}\\\mathsf{-6(7b)=-42b}\\\mathsf{\rightarrow\ =-6+(-42b)}

\mathsf{-40-8b=-6+(-42b)}\\\mathsf{\rightarrow -8b-40=-42b-6}

\mathsf{ADD\ by\ 42b\ on\ your\ sides}\\\mathsf{-8b-40+42b=-42b-6+42b}\\\mathsf{SOLVE\ above\ correct\ to\ get\ us\ to\ the\ next\ step!}\\\mathsf{\rightarrow 34b-40=-6}

\mathsf{ADD\ by\ 40\ on\ your\ sides}\\\mathsf{34b-40+40=-6+40}\\\mathsf{Cancel\ out\ -40+40\ because\ that\ gives\ the\ result\ of\ 0}\\\mathsf{-6+40=34}\\\mathsf{\rightarrow New\ equation: 34b = 34}

\mathsf{DIVIDE\ both\ of\ your\ sides\ by\ 34}\\\mathsf{\dfrac{34b}{34}=\dfrac{34}{34}}\\\\\mathsf{Cancel\ out: \dfrac{34b}{34}\ because\ that\ gives\ you\ the\ result\ of\ 1b}\\\\\mathsf{Keep:\dfrac{34}{34}\ because\ it\ solves\ for\ b}\\\\\mathsf{\dfrac{34}{34}=34\div34=1}

\boxed{\boxed{\boxed{\boxed{\mathsf{Thus\ your\ answer\ is:\boxed{\boxed{\mathsf{b=1}}}}}}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\dfrac{\frak{LoveYourselfFirst}}{:)}

8 0
3 years ago
X2 + 9x+ c <br><br><br> A. 81/4<br> B. 81/2<br> C. 9/4<br> D. 9/2
VikaD [51]

Answer:

hope it would help you

Step-by-step explanation:

the ans should be A

mark my ans as BRAIN LIST

8 0
3 years ago
Read 2 more answers
PLEASE HELP!! 25 POINTS B bisects segment AC. Find the value of m.
nexus9112 [7]

Answer: theres a answer key in the last slide & its 15 somehow...

6 0
1 year ago
A jet plane has a takeoff speed of vto = 75 m/s and can move along the runway at an average acceleration of 1.3 m/s2. If the len
pishuonlain [190]
Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m

time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s

the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s

Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
6 0
3 years ago
Read 2 more answers
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