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Annette [7]
3 years ago
5

NEED HELP ASAP ON THIS QUESTION

Mathematics
1 answer:
ella [17]3 years ago
8 0

Answer:

AM=AT

Step-by-step explanation:

'

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1 8/9 divided by 1/6
kkurt [141]

11 1/3 is the answer in fraction form but i decimal form it is 11.3 and the exact form is 34/3

4 0
3 years ago
Read 2 more answers
6, 12, 18 and 24 are the first four multiples of 6.<br><br> What are the next 2 multiples?
kondor19780726 [428]

Answer:

30 and 36

Step-by-step explanation:

six times two equals 12

6 * 3 = 18

and six times four equals 24.

to get the next two multiples, You need to multiply 6 by 5, and then 6

6 * 5 = 30 and 6 * 6 = 36

3 0
3 years ago
Hong made 2 identical necklaces, each having beads and a pendant. The total cost of the beads and pendants for both necklaces wa
egoroff_w [7]
B+ p= 10.40
b=2x 2.50
Now using substitution method replace b in first equation with value of second equation basically combing the two equations into one do you end up with one variable to solve .
2 x 2.50 +p =10.40
5 + p=10.40
P=$5.40
b =$5.00
The answer is EACH pendant cost $2.70
(You have to divide 5.40 by two )
6 0
3 years ago
Amelio has 6 more nickels than dimes, and the total value is $2.10. how many nickels does he have?
Nezavi [6.7K]
Amelio have 18 Nickels and 12 dimes!!

2.10−(6×0.05)−(12×0.10)−(12×0.05)= 0
6 0
3 years ago
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
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