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Tanzania [10]
3 years ago
12

Find the value of A A: 18 B: 21 C: 20 D: 17

Mathematics
2 answers:
statuscvo [17]3 years ago
8 0

These are vertical angles so they are equal so

6a +  11  = 2a + 83  

Solve for A

= 18

-  Prodixy ☕

AnnZ [28]3 years ago
6 0

These are vertical angles so they are equal so

6a +  11  = 2a + 83  

now solve for a

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If 1/64=4^2s-1 times 16^2s+2 what is the value of s?
galina1969 [7]
S=-1 see attachment for explanation

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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.

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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)

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3 years ago
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grigory [225]
Linear would be the answer
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3 years ago
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What is the solution 4x+6<18
mash [69]

Answer:

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add and subtract 6 from both sides

4x + 6 - 6 < 18 -6

4x < 12

x< 12/4

X < 3

4 0
3 years ago
The minimum of the graph of a quadratic function is located at (–1, 2). The point (2, 20) is also on the parabola. Which functio
ss7ja [257]
The min or max of a parabola/quadratic function is the vertex

for
y=a(x-h)²+k
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so
vertex/min is at (-1,2)
h=-1
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y=a(x-(-1))²+2
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given, (2,20) is on the graph

20=a(2+1)²+2
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y=2(x+1)²+2 is da equation

3rd one
f(x)=2(x+1)²+2
5 0
3 years ago
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