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11Alexandr11 [23.1K]
1 year ago
15

Edge proof would be amazing and I will give brainliest for it!

Mathematics
1 answer:
inna [77]1 year ago
7 0

Answer: The answer is B.

Step-by-step explanation: The scale on the y-axis could be changed to 100–120.

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115. Sami opens an account and deposits $100 into it at the end of each month. The account earns 2% per year compounded monthly.
givi [52]

Step-by-step explanation:

Geometric series.

Month 3.

      100(1+\frac{0.02}{12})^2 + 100(1+\frac{0.02}{12})+100

Month 4.

100( 1 + \frac{0.02}{12})^3 + 100( 1 + \frac{0.02}{12})^2+100(1+\frac{0.02}{12})+100

Month 5.

100(1 + \frac{0.02}{12})^4 + 100(1 + \frac{0.02}{12})^3 + 100(1 + \frac{0.02}{12})^2 +100( 1 + \frac{0.02}{12}) + 100

4 0
3 years ago
Color of car. a. Qualitative/Ordinal b. Qualitative/Nominal c. Quantitative/Discrete d. Quantitative/Continuous
Ymorist [56]
Colors are Qualitative and Nominal: Qualitative, because there is no number involved for "color" and Nominal, because their is no natural ordering for color
3 0
3 years ago
Choose all the fractions that are in simplest form.<br> 4/26<br> 8/21<br> 6/19<br> 9/10<br> 10/22
dimulka [17.4K]
4/26 can be simplified by 2. So, this is not the answer.
8/21 cannot be simplified. So, it is one of the answers.
6/19 cannot be simplified. So, it is one of the answers.
9/10 cannot be simplified. So, it is one of the answers.
10/22 can be simplified by 2. So, this is not the answer.

So, the fractions in simplest forms are 8/21, 6/19, and 9/10.
3 0
3 years ago
James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo
Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒  for n = 2 , y = 4       ..... (2)

⇒y  = 2^2  =  2^n

Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

⇒  There are total 128 equal parts.

Lastly,  if the paper is folded ELEVEN  times  ⇒  n = 11

So, y = 2^ n = 2^{11} = 2048

⇒  There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

8 0
3 years ago
Someone plz help me giving brainliest!
Arada [10]

Answer:

the estimate is : 39 7/32

the answer is :   46  9/35

4 0
3 years ago
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