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Marat540 [252]
3 years ago
15

Determine whether angle-angle-angle (AAA) is a valid means for establishing triangle congruence. If it is a valid criterion, exp

lain why. If it is not valid, use GeoGebra to create a counterexample demonstrating that it doesn’t work and give an explanation. If you construct a counterexample, take a screenshot of your work, save it, and insert the image in the space below.
Mathematics
1 answer:
Igoryamba3 years ago
5 0

AAA is not a valid congruence theorem. This is because we can construct two different triangles, of different shapes, and they both have the same corresponding angles. See the diagram below. Simply form any triangle you want and scale up or down to form a second triangle. The two triangles are similar, but they aren't congruent. They need to be the same size and shape to be congruent.

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It maybe c. I don’t kkow I may be wrong
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3 years ago
Please help me. I am having some trouble with these problems
s344n2d4d5 [400]

Answer:

Step-by-step explanation:

This is a proportion problem. I think you just lost a zero.

12/100 = 9/x                    Cross multiply

12x = 100*9                      Combine the right.

12x = 900                         Divide by 12

12x/12 = 900/12

x = 75 grams

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4 0
3 years ago
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LuckyWell [14K]

Answer:

8

Step-by-step explanation:

we can say that AD is congruent to DC

so your equation for x is: 4x - 1 = 2x

solve this to get x = 0.5

plug x into and equation and multiply you answer by 2 to find the hypotenuse of triangle ABC and DEF

4(0.5) - 1 = 1

hypotenuse: 1 x 2 = 2

since we know x is 0.5, plug this into 4x + 1 to find the length of the leg FE,

4(0.5) + 1 = 3

In the diagram, it shows that the legs of triangle are congruent

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since we know FE is 3, we know that all the other sides are 3 as well

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2 years ago
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natita [175]
You too !! .......... :)
4 0
2 years ago
Read 2 more answers
Write a polynomial of least degree with rational coefficients, a leading coefficient of 1, and zeros at 3 and sqr root 7. Expand
Tresset [83]

Answer:

(x-3)(x^2-7) = x^3 -3x^2 -7x + 21

just change the sign of the root and put x in front of it, then multiply the factors all together

(x-3)(x-sqr7)(x+sqr7)    roots come in conjugate pairs to eliminate irrational coefficients

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Step-by-step explanation:

6 0
3 years ago
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