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Alborosie
2 years ago
7

Please help me i only have 4 minutes to turn this in u will get branliest for life!!!!!!!!!!!!!!!!!!!!11

Mathematics
1 answer:
zavuch27 [327]2 years ago
8 0
Ok the answer is c!!!!!


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Classify the triangle by its angles and sides
Vlad [161]
This is an Acute Isosceles Triangle. 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
What is the perimeter of a trapezoid having side lengths of 20 ft, 20 ft, 26 ft and 18 ft?
Irina18 [472]
The perimeter would be 84
8 0
3 years ago
This is a graph of y<2x+5 what is true about point A on the graph
Ilya [14]

Answer: "Point A is a solution because it is in the shaded area

Step-by-step explanation:

When we have an inequality like the one shown in the graph, the shaded area represents the possible solutions to the inequality.

If we have a dashed line, then the points in the line are not solutions (this is the case of the symbols > and < )

If the line is solid, then the points of the line are solutions (this is the case of the symbols ≤ and ≥)

In the graph we can see that point A is inside the shaded area, then point A is a solution.

Then the correct option is:

"Point A is a solution because it is in the shaded area"

3 0
2 years ago
X2 - 4x + 4 = 0 solve following quadratic equations graphically and a table
stepan [7]

Let's create a table, and use arbitrary values of x between -3 and 3 (if the solution is not in this interval, we can use another one later). If we call:

f(x)=x^2-4x+4

We want to find the x's for with f(x) = 0

This is the table:

And now we can evaluate f(x) in each value of x to complete the table:

\begin{gathered} f(-3)=(-3)^2-4\cdot(-3)+4=9+12+4=25 \\ f(-2)=(-2)^2-4\cdot(-2)+4=4+8+4=16 \\ f(-1)=(-1)^2-4\cdot(-1)+4=1+4+4=9 \\ f(0)=0^2-4\cdot0+4=4 \\ f(1)=1^2-4\cdot1+4=1-4+4=1 \\ f(2)=2^2-4\cdot2+4=4-8+4=0 \\ f(3)=3^2-4\cdot3+4=9-12+4=1 \end{gathered}

The table is:

If we use this values and plot them in the cartesian plane.

We get:

And now if we plot a line that connects the points, we get the graph of the quadratic equation:

Since the problem ask us to find the value of x for which f(x) = 0, we can see both in the table and in the graph that this value is x = 2

6 0
11 months ago
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