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3 years ago

The number of terms between 200 and 800 which are divisible by 7 is

Mathematics

1 answer:

NikAS [45]3 years ago

7 0

Answer:

Step-by-step explanation:

Mathematically, the numbers that are divisible by 7 here forms an arithmetic sequence, with 203 being the first term

Now, we need the last term

The last term before 800 is 798

Now, the nth term of a sequence (arithmetic) can be represented as;

Tn = a + (n-1)d

Kindly note that the numbers have 7 as their common difference

Thus;

798 = 203 + (n-1)7

798-203 = 7n - 7

7n = 798 -203 + 7

7n = 602

n = 602/7

n = 86

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What is 100/7 divided by 4 1/3 simplified if you answer thank you and happy holidays

SVETLANKA909090 [29]

Answer: 300/91

Step-by-step explanation:

1. Convert 4 1/3 to an improper fraction

13/3

2. Divide, using the keep, change, flip method to divide fractions

- Keep the first fraction the same (100/7)

- Change the division sign to a multiplication sign

- Flip the second fraction (13/3 -> 3/13)

3. Plug-in and evaluate

100/7 * 3/13

300/91

If you are looking for the solution as a mixed number, it is 3 27/91

4. Simplify, if applicable. This fraction cannot be simplified any further.

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3 years ago

Find the circumference of circle P in terms of pi , P = 5ft

ra1l [238]

If by P=5ft you mean the radius is 5 then c = 2pi r or c = 10pi or approx 31.4 ft

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4 years ago

Read 2 more answers

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

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Bogdan [553]

The answer is three (3)

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3 years ago

What is h(x) = - 8x + 360

grandymaker [24]

h(x)=-8x+360

-360

h (x)= -8x

sorry if that is incorrect

5 0

3 years ago