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ludmilkaskok [199]
3 years ago
13

2x-8=4(x+1). Solve for X please!

Mathematics
2 answers:
julsineya [31]3 years ago
6 0

Answer:

-6 = x

Step-by-step explanation:

2x-8=4(x+1)

Distribute

2x -8 = 4x+4

Subtract 2x

2x-8-2x = 4x-2x+4

-8 = 2x+4

Subtract 4

-8-4 = 2x+4-4

-12 = 2x

Divide by 2

-12/2 = 2x/2

-6 = x

pogonyaev3 years ago
6 0

Answer:

x =  - 6

Step-by-step explanation:

2x - 8 = 4(x + 1) \\ 2x - 8 = 4x + 4 \\ 2x - 4x = 8 + 4 \\  - 2x = 12 \\  \frac{ - 2x}{ - 2}  =  \frac{12}{ - 2}  \\ x =  - 6

hope this helps

brainliest appreciated

good luck! have a nice day!

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Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.
skad [1K]

Answer:

The absolute maximum and minimum is 20\; \text{and} -20.

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

f_x=y-4=0

f_y=x=0

The only solution to this system of equations is the point (0, 4), which lies in the domain.

f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1

\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1

\therefore (0,4) is a saddle point.

Boundary points -  (4,0),  (-4,0), (0,16)

Along boundary  y=16-x^2

   f=x(16-x^2)-4x

=16x-x^3-4x

\Rightarrow f^'=16-3x^2-4=0

\Rightarrow 3x^2=12

\Rightarrow x=\pm2,\;\;y=14

Values of f(x) at these points.

\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}

Therefore, the absolute maximum and minimum is 20\; \text{and} -20.

6 0
3 years ago
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Answer:

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3 years ago
Duik<br>Solve the equation.<br>y + 3 = -y + 9<br><br><br>​
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Step-by-step explanation:

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3 years ago
Read 2 more answers
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 &lt; m &lt; 13 &lt; n, is
tatiyna

Answer:

Hence to get same number of students in each classroom,the sufficient condition is that assign 13n students to each classroom.

Step-by-step explanation:

Given:

There are m classrooms and n be the students

3<m<13<n.

To Find:

Whether it is possible to assign each of n students to one of m classrooms with same no.of students.

Solution:

This problem is related to p/q form  has to be integer in order to get same no of students assigned to the classroom.

As similar as ,n/m ratio

So 1st condition is that,

If it is possible to assign the n/m must be integer and n should be multiple of m,

when we assign 3n students to m classrooms ,we cannot say that 3n/m= integer so that  n is greater than 13 i.e n=14 and m=6

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Otherwise,n=14 and m=7 they will give same number but this condition is not sufficient condition to assign the student.

So 2nd condition is that ,

When we assign 13n students to m classrooms, as 13 is prime number and

3<m<13 which implies the 13n/m to be integer so n and m must be multiple of each other.

Suppose n=20 and m=5 classrooms

then 13*20=260 ,

260/5=52 students in each classroom,

4 0
3 years ago
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