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3 years ago

{0.23,19%1/5} Order each set of numbers from least to greatest show work

1 answer:

4 0

First you have to change them all to the same form, so you have to pick one out of them, so for instance you could pick percent which would be; 23%, 19%, and 20%, so from least to greatest it would be; 19%, 20%, and 23%. Hope this helps!

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Determine if the given ordered pair is a solution to the equation. Show work full.

sleet_krkn [62]

Answer:

The ordered pair is not a solution of the equation

Step-by-step explanation:

we know that

If a ordered pair is a solution of a linear equation, then the ordered pair must satisfy the linear equation

we have

$y=-3x-13$

Substitute the value of x and the value of y of the given ordered pair in the linear equation and analyze the results

For x=-1, y=15

$15=-3(-1)-13$

$15=3-13$

$15=-10$ ----> is not true

the ordered pair not satisfy the equation

therefore

The ordered pair is not a solution of the equation

5 0

3 years ago

After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 lb. What was Lisa's original weight?

AlekseyPX

Answer:

175lb

Step-by-step explanation:

Let's say 100% - original weight

and 12% - 21lb

weight = (100*21)/12=175lb

6 0

3 years ago

Read 2 more answers

Please help me ASAP Please

Romashka [77]

Answer:

I think it’s 20 or 26 sorry if I get it worng :(

5 0

3 years ago

The area of the Indian Ocean is 39% of the area of which ocean shown in the table

ValentinkaMS [17]

Is there a way that you could show me the table so I can help you please?

8 0

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

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

4 0

3 years ago