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Naya [18.7K]
4 years ago
10

What decimal is bigger 2.16 or 2.016

Mathematics
2 answers:
Andreyy894 years ago
7 0
2.16 is bigger
!!!!!!!!!!!!!!!!!!!!!!!!!111
levacccp [35]4 years ago
3 0
Salutations!

What decimal is bigger 2.16 or 2.016?

Look at the tenths place, the number 1 is greater than zero. You need to look our for place values where numbers aren't different and compare them. Therefore, 2.16 is greater than 2.016.

Hope I helped (:

Have a great day!
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Help please this question is important!!!!
mestny [16]
A = P(1 + r/n)^nt
p=2000
r=.052
n=4
t=8
answer is 3,023.66
4 0
3 years ago
What Is the solve for n of 2 (8 + 3n) < 34
GREYUIT [131]

Answer:

n<3

Step-by-step explanation:

2(8 +3n) < 34

16 + 6n < 34

6n < 34-16

6n < 18

n < 3

8 0
4 years ago
Please help me understand this problem! It would be greatly appreciated &lt;3 much love
cricket20 [7]

Answer:

all work is shown and pictured

3 0
4 years ago
.75x-20= 620 what does x equal rounded to the nearest 100?
vodomira [7]

To solve this, we’ll have to use inverse operations.

Since addition is the inverse to subtraction, we’ll add 20 to 620.

20 + 620 = 640

From there, since division is the inverse of multiplication, we’ll divide .75 by 640.

640/.75 = 853.33333 repeating. We can round this to 853.33, to round to the nearest hundredth.

Therefore, 853.33 would be your answer! You can check this by solving the equation forwards (853.33 x .75 = ~640, 640 - 20 = 620.)

Hope this helps! :)

4 0
3 years ago
In an arithmetic sequence, a_17 = -40 and a_28 = -73. Please explain how to use this information to write a recursive formula fo
Vinvika [58]

An arithmetic sequence

a_1,a_2,a_3,\ldots,a_n,\ldots

is one in which consecutive terms of the sequence differ by a fixed number, call it <em>d</em>. This means that, given the first term a_1, we can build the sequence by simply adding <em>d</em> :

a_2=a_1+d

a_3=a_2+d

a_4=a_3+d

and so on, the general pattern governed by the recursive rule,

a_n=a_{n-1}+d

We can exploit this rule in order to write any term of the sequence in terms of the first one. For example,

a_3=a_2+d=(a_1+d)+d=a_1+2d

a_4=a_3+d=(a_1+2d)+d=a_1+3d

and so on up to

a_n=a_1+(n-1)d

In this case, we're not given the first term right away, but the 17th. But this isn't a problem; we can use the same exploit to get

a_{18}=a_{17}+d

a_{19}=a_{17}+2d

a_{20}=a_{17}+3d

and so on, up to the next term we know,

a_{28}=a_{17}+11d=-40+11d

(Notice how the subscript of <em>a</em> on the right and the coefficient of <em>d</em> add up to the subscript of <em>a</em> on the left.)

The 28th term is -73, so we can solve for <em>d</em> :

-73=-40+11d\implies -33=11d\implies d=-3

To get the first term of the sequence, we use the rule found above and either of the known values of the sequence. For instance,

a_{17}=a_1+16d\implies-40=a_1-16\cdot3\implies a_1=8

Then the recursive rule for this particular sequence is

\begin{cases}a_1=8\\a_n=a_{n-1}-3&\text{for }n>1\end{cases}

7 0
3 years ago
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