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GuDViN [60]
3 years ago
13

No link answers or a n answer just to get points please you will be reported I ry need help

Mathematics
1 answer:
mart [117]3 years ago
4 0
The correct answer is A
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I need help on this fast pls help me
andreev551 [17]

Answer:

Choice 2) 5x+15

Step-by-step explanation:

6 0
3 years ago
The first five terms of a sequence are 2, 5/3, 3/2, 7/5 and 4/3. what is the explicit formula?
katen-ka-za [31]

Answer:

  an = (n+3)/(n+1)

Step-by-step explanation:

The sequence can be written as ...

  4/2, 5/3, 6/4, 7/5, 8/6, ...

where the first numerator is n+3 and the first denominator is n+1. Each numerator and denominator increase by 1 from term to term, so the explicit formula can be written

  an = (n+3)/(n+1)

4 0
3 years ago
One metre of ribbon costs $0.45. If Tabitha buys 225 metres of ribbon, how much will she have to pay? (Please help me! I have to
tatuchka [14]

Answer: $101.25

Step-by-step explanation:

Since we are given the information that one metre of ribbon costs $0.45 and that Tabitha buys 225 metres of ribbon, the amount that she will have to pay will be gotten by multiplying the cost of one meter of ribbon by the total meters that she bought. This will be:

= $0.45 × 225

= $101.25

Therefore, the cost is $101.25

6 0
3 years ago
When you divide a number by a decimal less than 1, is the quotient less than or greater than the original number?
mote1985 [20]

Answer:

Step-by-step explanation:

7 0
3 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 =
Taya2010 [7]

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

5 0
3 years ago
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