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lapo4ka [179]
3 years ago
10

15 over 18 simplified

Mathematics
2 answers:
Sphinxa [80]3 years ago
8 0

Answer:

5/6

Step-by-step explanation:

svlad2 [7]3 years ago
3 0

Answer:

5/6

0.83333333

Step-by-step explanation:

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Which values represent the first two terms of the sequence a1=2 and an=-4(an-1)^2?
cricket20 [7]
<span>a1= (which is the first term.)
 to find the 2nd term you plug in 2 for (an) in the formula given to you. =-4(2-1)^2 *remember an represents the term u are looking for *
 now you start to solve it
 -4(2-1)^2
 -4 (1)^2 .
-4^2 = 16
 so your first two terms are 2 and 16 .</span>
4 0
4 years ago
PLEASE HELP ME!!Write the quotient as a mixed number.<br> 46 divided by 9 = 5 R1
Zigmanuir [339]

Answer:

5\frac{1}{9}

Step-by-step explanation:

When dividing 46 by 9 quotient is 5 and remainder is 46 - 45 = 1

5\frac{1}{9}

5 0
3 years ago
Can you help me with 12-3(x+1)=30
inysia [295]
Greetings!

12-3(x+1)=30
Distribute the Parenthesis.
12-3x-3=30
Combine like terms.
9-3x=30
Add -9 to both sides.
(15-3x)+(-9)=(30)+(-9)
-3x=21
Divide both sides by -3.
(-3x)/-3=(21)/-3
The Answer Is:
x=-7

Hope this helps.
-Benjamin
7 0
4 years ago
If you take a number multiply by 7 then takeaway 1. You get the same as if you took the number multiply by 5 then takeaway 8. Wh
muminat
7x - 1 = 5x - 8
Minus 5x on each side
2x -1 = -8
Plus 1 on each side
2x = -7
Divide by 2 on each side
x = -7/2

The number is most likely -7/2
7 0
3 years ago
Read 2 more answers
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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