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stira [4]
3 years ago
15

Solve for x in the equation.

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
5 0
x^2 + 20x + 100 = 36

(x + 10)^2 = 36

x + 10 = \pm \sqrt{36}

x + 10 = 6   or   x + 10 = -6

x = -4   or   = -16
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.The endpoints of AB are A(1,4) and B(6,-1).
Brums [2.3K]

Answer:

5√2

Step-by-step explanation:

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4 0
3 years ago
Help Please my we dobnt know
erica [24]

Answer:

its B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A rectangle has a width of 4 inches and a length of 6 inches. A similar rectangle has a width ​
qaws [65]

Answer:

Step-by-step explanation:

6 times 4 is 24 so the answer is 24

3 0
3 years ago
Jermaine has just painted a large mural. It is 8 1/2 feet wide. The wall he wants to place it on is 11 3/4 feet wide. About how
PolarNik [594]
First turn 1/2 to 2/4 so you have a common denominator.
Next turn the mixed numbers into improper fractions:
11 3/4 --> 47/4 and 8 2/4 --> 34/4
Subtract:
47/4 - 34/4 = 13/4
Divide:
13/4 = 3.25
3.25/2 --> 1.625 or 1 5/8
Therefore, he will have to place it 1 5/8 feet from each side

Hope this helps!
6 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
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