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3 years ago

6×11=66,89×101=8989,706×1001=706706.next two sequence are???

Mathematics

2 answers:

lions [1.4K]3 years ago

8 0

The solution would be like this for this specific problem:

2nd factor product goes like this: 11, 101, 1001, so the next 2nd factor would be 10001.

1st factor can be solved thru using a polynomial function.

(n) = 267n² – 718n + 457

FirstFactor(4) = 1857

So, the 4th product is 1857 * 10001 = 18571857

REY [17]3 years ago

3 0

Answer:

4th= 7, 067, 766, 706

5th=706, 783, 783, 366, 706

Step-by-step explanation:

To get the next two numbers on the sequence you just have to continue the pattern that you are given in the example, in this case you are adding one zero between the ones at the precedent number, so the next ones would be:

4th- 10001

5th- 100001

WHen you multiply 706706 by 10001 you get 7, 067, 766, 706

Then you have to multiply that number by 100001 and you get:

7, 067, 766, 706 x 100001= 706, 783, 783, 366, 706

And that´s how yo calculate the next two in sequence.

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3 years ago

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I hope this helps you

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3 years ago

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faltersainse [42]

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Step-by-step explanation:

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3 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$