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aleksandrvk [35]
3 years ago
12

1/6 x 2/7 in simplest form

Mathematics
2 answers:
gogolik [260]3 years ago
8 0

Answer:

1/21

Step-by-step explanation:

1/6 x 2/7

1x2

6x7

2/42

=1/21

djverab [1.8K]3 years ago
6 0

Answer:

1/6 × 2/7 = 1 / 21

Step-by-step explanation:

by solving you get ,

=》1/6 × 2/7

=》 2 / 42

=》 1 / 21

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The product of a constant factor 15 and a 2-term factor x+4
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6 0
3 years ago
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Lubov Fominskaja [6]
I think its the third one
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4 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

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5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

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3 0
4 years ago
Twenty-six out of 50 teachers drive their cars to school. Approximately what percent of these teacher drive to school
Mars2501 [29]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
What is the domain of y = cos ^-1x?
Bas_tet [7]

Answer:

Step-by-step explanation:

Correct option is

D

[1,(1+π)  

2

]

f(x)=(1+sec  

−1

(x))(1+cos  

−1

(x))

Here the limiting component is cos  

−1

(x), since the domain of cos  

−1

(x) is [−1,1].

Therefore,

f(1)=(1+0)(1+0)

=1

f(−1)=(1+π)(1+π)

=(1+π)  

2

 

Hence range of f(x)=[1,(1+π)  

2

]

8 0
3 years ago
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