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3 years ago

If hexagon JGHFKI Is congruent to hexagon SRTQUV, which pair of angles must be congruent?

Mathematics

2 answers:

vagabundo [1.1K]3 years ago

6 0

Answer:

j and s

Step-by-step explanation:

i just took the test

Dima020 [189]3 years ago

3 0

Hey there!

If these hexagons are congruent, then each angle is congruent to the one that is in the same position as it is. In this case, I and V are congruent because they are both last. K and U are congruent because they are both second to last.

While looking over the answer options, you should be able to see that Angle J and Angle S is the only answer with angles that match up, because they are both the first angle in the set.

I hope this helps!

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What information does the standard form the equation of a line give you?

Butoxors [25]

I assume you are referencing y=mx+b. The equation gives you the slope of the line (m), and a coordinate (x, y). As well as b, which could be used for many things.

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3 years ago

Point E is the midpoint of AB and point F is the midpoint

GREYUIT [131]

The options are;

1) AB is bisected by CD

2) CD is bisected by AB

3) AE = 1/2 AB

4) EF = 1/2 ED

5) FD= EB

6) CE + EF = FD

Answer:

Options 1, 3 & 6 are correct

Step-by-step explanation:

We are told that Point E is the midpoint of AB. Thus, any line that passes through point E will bisect AB into two equal parts.

The only line passing through point E is line CD.

Thus, we can say that line AB is bisected by pine CD. - - - (1)

Also, since E is midpoint of Line AB, it means that;

AE = EB

Thus, AE = EB = ½AB - - - (2)

Also, we are told that F is the mid-point of CD.

Thus;

CF = FD

Point E lies between C and F.

Thus;

CE + EF = CF

Since CF =FD

Thus;

CE + EF = FD - - - (3)

5 0

2 years ago

Qx+12=13x+P with only one solution

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2 years ago

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Mixed numbers from 0-2 with intervals of 1/3

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0, 1/3, 2/3, 1, 1 1/3, 1 2/3, 2

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2 years ago

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago