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2 years ago

12

you are going around texas and oklahoma with your family for diffrent types of shoes clothing s ect but you have a 10000$ buying

limit not including gass and food how much cloths and shoes can you buy

1 answer:

user100 [1]2 years ago

4 0

Depends on what the cost of the clothing is- not enough info.

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1x+5=10 pls do it for me i'll give you brainly

STatiana [176]

1x+5=10

1x-5=10

x=5 that is the answer to the question hoped it helped.

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3 years ago

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FrozenT [24]

Is between 0 and 1 and is less than 1

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1 year ago

Order from greatest to least:show your work❗️<br><br> 12,-4,8,-3,0

Aneli [31]

The answer would be 12, 8, 0, -3, -4.

The answer is that because 12 is a positive number so that's higher than negative, 8 goes second because its positive and higher than the others, 0 is third because negatives numbers are LOWER than 0, not higher, -3 is next because higher negative numbers are always lower, not greater than smaller numbers because -8 is LOWER than -4 since its on the left of a number line.

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3 years ago

Inez has 27 basketball cards and 45 baseball cards in her card collection. What time is the ratio of Inez’s baseball cards to he

lisov135 [29]

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2 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

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

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3 years ago