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skelet666 [1.2K]
3 years ago
10

Helpppppppppp mathhhhhhhhhhhhhhhh

Mathematics
2 answers:
ycow [4]3 years ago
5 0

Answer: ∆TUW ≅ ∆VUW

Step-by-step explanation:

In the given figure we have two triangles ∆TUW  and ∆VUW .

In the given triangles ∆TUW  and ∆VUW , we have

∠TUW  ≅ ∠VUW    [Given]

UW=UW                  [Reflexive]

∠UWT≅∠UWV         [Given]

Therefore, by ASA congruence postulate

∆TUW ≅ ∆VUW  [write corresponding parts in same order.]

  • ASA congruence postulate says that if two angles and the included side of a triangle are congruent or equal to the corresponding parts of another triangle, then the triangles are congruent.

zhenek [66]3 years ago
4 0
1- ASA for all triangle mentioned. The have one side adjacent to 2 angles

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one group of friends has 5 cookies to share between 3 people.a second group of friends has 8 cookies to share between 6 people.w
Paul [167]
Group 1:-
5 cookies
3 people
5/3

Group 2:
8 cookies
6 people
8/6

Group 1 each person gets 5/3 and in group 2 each person gets 8/6.

Which is more?

5/3

5 ÷ 3 ≈ 1.6

8/6

8 ÷ 6 ≈ 1.3

Is 1.3 more or 1.6?

1.6 is more.

5/3 (group 1) receive more. 
8 0
3 years ago
What is the sum of -6 4/5 and 6 4/5 ?
daser333 [38]

Answer:

0.

Step-by-step explanation:

-6 4/5 + 6 4/5 equals zero.

3 0
3 years ago
Read 2 more answers
HELP HELP!!
Goryan [66]

Answer:

0

Step-by-step explanation:

p_1:~~y = x^2+2\\p_2:~~y = 3x^2+2\\ \\ V{p_1} = \Big(-\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) = \Big(-\dfrac{0}{2}, -\dfrac{0^2-4\cdot 2}{4}\Big) = \Big(0,2\Big) \\ \\ Vp_2 = \Big(x_V, -\dfrac{\Delta}{4a}\Big) = \Big(0, -\dfrac{0^2-4\cdot 3 \cdot 2}{4\cdot 3}\Big) = \Big(0,2\Big) \\ \\ \\ \text{The distance is }0,~~\text{Because the vertices are equal.}

7 0
3 years ago
Subtract three-forths from 2 times p
Savatey [412]

Answer:

Hi there!

Your answer is:

2p - 3/4

Hope this helps!

5 0
3 years ago
Solve the system of equations 2x+4y+3z=6 5x+8y+6z=4
MaRussiya [10]

Answer:Solution :

{x,y,z} = {-8,52/7,-18/7}  

 

System of Linear Equations entered :

  [1]    2x + 4y + 3z = 6

  [2]    5x + 8y + 6z = 4

  [3]    4x + 5y + 2z = 0

Solve by Substitution :

// Solve equation [3] for the variable  z  

 

 [3]    2z = -4x - 5y  

 [3]    z = -2x - 5y/2  

// Plug this in for variable  z  in equation [1]

  [1]    2x + 4y + 3•(-2x-5y/2) = 6

  [1]    -4x - 7y/2 = 6

  [1]    -8x - 7y = 12

// Plug this in for variable  z  in equation [2]

  [2]    5x + 8y + 6•(-2x-5y/2) = 4

  [2]    -7x - 7y = 4

// Solve equation [2] for the variable  y  

 

 [2]    7y = -7x - 4

 [2]    y = -x - 4/7

// Plug this in for variable  y  in equation [1]

  [1]    -8x - 7•(-x -4/7) = 12

  [1]    -x = 8

// Solve equation [1] for the variable  x  

  [1]    x = - 8  

// By now we know this much :

   x = -8

   y = -x-4/7

   z = -2x-5y/2

// Use the  x  value to solve for  y  

   y = -(-8)-4/7 = 52/7  

// Use the  x  and  y  values to solve for  z  

 z = -2(-8)-(5/2)(52/7) = -18/7  

Solution :

{x,y,z} = {-8,52/7,-18/7}  

Step-by-step explanation:

4 0
3 years ago
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