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Archy [21]
3 years ago
10

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

plain why. If it isn’t valid, use GeoGebra to create a counterexample demonstrating that it doesn’t work and give an explanation. If you construct a counterexample, include a screenshot of your work in your answer.
Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0
Can yu pls send us some picture i cant understand the question and then i will comment an answer thx...
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Jobisdone [24]
-1/6 The x is coefficient is negative and is 1/6
5 0
3 years ago
01.01) Rewrite the expression with rational exponents as a radical expression. 7 times x to the two thirds power cube root of th
jeka94

Answer:

(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2

Step-by-step explanation:

Given the expression (7x)^{\frac{2}{3}, to express this as a radical expressions, we'd apply the rule/law of indices that deals with converting expressions that has rational exponents into radical expressions.

The rule of indices to apply is: b^{\frac{m}{n}} = (\sqrt[n]{b})^m

To apply this to the expression, (7x)^{\frac{2}{3}, the denominator of the fraction of the exponent would determine the root, that is, cube root in this case. The numerator of the exponent would then determine the exponent of the radical expressions.

Thus:

(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2

6 0
3 years ago
Help please, i need this done asap thanks in advance
alina1380 [7]

Answer:

<AOC

<AOB

<BOC

Step-by-step explanation:

The central angles are formed at point O, which is the center of the circle.

All the central angles are marked in the attachment.

The central angles are

: m\angle AOC. This is a right angle and it is 180 degrees.

:m\angle AOB

:m\angle BOC

7 0
3 years ago
5 for $49.95 or 3 for $34.95?​
balu736 [363]

Answer:

I'd love to help but we need more details

Step-by-step explanation:

I'll edit my answer to solve if you can give more information about the problem, we don't know the context and what to solve

7 0
3 years ago
Read 2 more answers
Which equation has the solutions X=1 +-SQRT 5?
Vesna [10]

Answer:

D is correct. x^2-2x-4=0

Step-by-step explanation:

We are given the root of the equation

x=1+\pm \sqrt{5}

If we are given solution of the equation then find equation using formula.

If a and b are the solution of equation then equation would be (x-a)(x-b)=0

Here, a=1+\sqrt{5} , b=1-\sqrt{5}

Equation form would be (x-1-\sqrt{5})(x-1+\sqrt{5})=0

Now we simplify the above equation to get correct option.

x^2-x+x\sqrt{5}-x+1-\sqrt{5}-x\sqrt{5}+\sqrt{5}-5=0

x^2-2x-4=0

So, D is correct.  x^2-2x-4=0

8 0
4 years ago
Read 2 more answers
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