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m_a_m_a [10]
3 years ago
10

8x + 1 = -8 - X What is X?

Mathematics
2 answers:
worty [1.4K]3 years ago
6 0

Answer:

x=-1

Step-by-step explanation:

8x+1=−8−x

8x+1=−8+−x

8x+1=−x−8

8x+1+x=−x−8+x

9x+1=−8

9x+1−1=−8−1

9x=−9

9x9=−99

x=−1

Alexxx [7]3 years ago
5 0

Answer:

<u><em>X = -1</em></u>

<u><em>Step-by-step explanation:</em></u>

  • <u><em>Add x to both sides.</em></u>
  • <u><em>8x + 1 + x = -8</em></u>
  • <u><em>Combine 8x and x to get 9x.</em></u>
  • <u><em>9x + 1 = -8</em></u>
  • <u><em>Subtract 1 from both sides</em></u>
  • <u><em>9x = -9</em></u>
  • <u><em>Divide both sides by 9 </em></u>
  • <u><em>x = -9/9</em></u>
  • <u><em>x = -1</em></u>

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What is the value of the expression All of 3.9 multiplied by 10 to the power 5 over all of 1.3 multiplied by 10 to the power 2?
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4 0
3 years ago
1. Gavin and Seiji both worked hard over the summer. Together, they earned a total of $425. Gavin earned $25 more than Seiji. Ho
Gre4nikov [31]
Let x = the amount that Seini earned
Then x + 25 = the amount that Gavin
The amount Gavin earned would also = 425 - x (your directions asked you to have 2 equations.)
Together they earned 425:
x + (x + 25) = 425
2x + 25 = 425
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Now divide both sides by 2 to find x.

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8 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
I cant see the answer what is it
JulsSmile [24]
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8 0
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