Which scenario is a counterexample that disproves the AAA criteria for congruence? Two 45°−45°−90° triangles that share a diagonal to form a square
Two 45 degrees minus 45 degrees minus 90 degrees triangles that share a diagonal to form a square

Two right triangles with hypotenuses that are 6 in and 8 in
Two right triangles with hypotenuses that are 6 in and 8 in

Two isosceles triangles with bases of 2 ft and 7 ft
Two isosceles triangles with bases of 2 ft and 7 ft

Two equilateral triangles with side lengths of 4 cm and 10 cm

the answer is d

step-by-step explanation:

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong