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2 years ago

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Zan has created this rule for generating sequences of whole numbers. If a number is 25 or less, double the number. If a number i

s more than 25, subtract 12 from it. For example, if Zan starts with 10, she gets the sequence 10, 20, 40, 28, 16,... If the third number in Zan's sequence is 36, what is the sum of the four distinct numbers that could have been the first number in her sequence

1 answer:

enot [183]2 years ago

7 0

Answer:

123

Step-by-step explanation:

Given that:

Method to generate the sequence:

If a number is less than or equal to 25, then the number is doubled.

And if the number is more than 25, then 12 is subtracted from it.

Here, we are given a sequence with its third number is 36.

Now, let us have a look by making the number 36 as half of its value.

$\frac{36}{2}$ = 18

This could have been the second number in the.

According to this, the first number will be $\frac{18}{2} = \bold{9}$

Other option for the first term can be 18 + 12 = 30

By second method, the second method can be:

36 + 12 = 48

By this method, First number 48 + 12 = 60

Other option for first number can be $\frac{48}{2 } = \bold{24}$

Therefore, sum of four options of first term:

9 + 30 + 60 + 24 = <em>123</em>

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Step-by-step explanation:

g(x) = x² + 2x+4

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =

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Answer:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

$\bigtriangledown f=\lambda \bigtriangledown g$

for some scalar $\lambda$ called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

The gradient of f is

$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

We have

$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

Let's compute the partial derivatives

$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

The Lagrange condition leads to

$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

Operating and simplifying

$1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11$

Replacing the value of $\lambda$ in the two first equations, we get

$1=11+2x\mu\\2=-11 +2y\mu$

From the first equation

$\displaystyle 2\mu=\frac{-10}{x}$

Replacing into the second

$\displaystyle 13=y\frac{-10}{x}$

Or, equivalently

$13x=-10y$

Squaring

$169x^2=100y^2$

To solve, we use the restriction h

$x^2 + y^2 = 1$

Multiplying by 100

$100x^2 + 100y^2 = 100$

Replacing the above condition

$100x^2 + 169x^2 = 100$

Solving for x

$\displaystyle x=\pm \frac{10}{\sqrt{269}}$

We compute the values of y by solving

$13x=-10y$

$\displaystyle y=-\frac{13x}{10}$

For

$\displaystyle x= \frac{10}{\sqrt{269}}$

$\displaystyle y= -\frac{13}{\sqrt{269}}$

And for

$\displaystyle x= -\frac{10}{\sqrt{269}}$

$\displaystyle y= \frac{13}{\sqrt{269}}$

Finally, we get z using the other restriction

$x - y + z = 1$

Or:

$z = 1-x+y$

The first solution yields to

$\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

And the second solution gives us

$\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Complete first solution:

$\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

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Answer:

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Step-by-step explanation:

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