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lilavasa [31]
3 years ago
6

In the ellipse, if c=6cm, and b=8cm, what is the length of the major axis?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

Option B). 10 cm

Step-by-step explanation:

Distance between center and the focus of an ellipse is given by the formula,

c² = a² - b²

where, a = length of the major axis

b = length of the minor axis

and c = distance between focus and center

If c = 6 cm, b = 8 cm,

6² = a² - 8²

a² = 6² + 8²

a² = 36 + 64

a² = 100

a = 10 cm

Therefore, length of the major axis is 10 cm.

Option (B) is the answer.

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The series must be decreasing, and the limit as the n-th term goes to infinity should be 0.

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What is the difference between 12/8 and 3/4?
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Answer:

(12/8) and (3/4) = 2 1/4 or 2.25 decimal thus: A:

Step-by-step explanation:

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3 years ago
Brainliest for you if you answer!
eduard

<u>To find the constant of proportionality:</u>

  ⇒ must define constant of proportionality

    ⇒ <em>Constant of Proportionality:  ratio of two proportional values at a </em>

       constant value

       ⇒ (in this case) two proportional values are x and y

<u>The ratio between x and y is equal to</u>

  \frac{y}{x} =\frac{66}{1.2}=55

<u>Answer:</u> <u>55</u> or <u>Choice 1</u>

<u></u>

Hope that helps!

7 0
2 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
Kurt deposited money into his bank account every month. The amount of money he deposited in the first 5 months of 2014 is listed
Irina-Kira [14]

Answer:

that is the answer

Step-by-step explanation:

3 1
3 years ago
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