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3 years ago

Write 843 in scientific notation.

Mathematics

1 answer:

yarga [219]3 years ago

7 0

a x 10b

Follow the steps below to see how 843 is written in scientific notation.

Step 1

To find a, take the number and move a decimal place to the right one position.

Original Number: 843New Number: 8.43Step 2

Now, to find b, count how many places to the right of the decimal.

New Number:8.43Decimal Count:12

There are 2 places to the right of the decimal place.

Step 3

Building upon what we know above, we can now reconstruct the number into scientific notation.

Remember, the notation is: a x 10b

a = 8.43

b = 2

Now the whole thing:

8.43 x 102

Step 4

Check your work:

102 = 100 x 8.43 = 843

More Scientific Notation Examples

8418428448458.41 x 1028.42 x 1028.44 x 1028.45 x 102

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The outside of a picture frame measures 14 in and 20 in 196in^2 of the picture shows find the thickness of the frame

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The described picture frame can be visualized into two separate parts. The first area is equal to the area using the outermost dimensions for the length and width.
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Area = (20 in) x (14 in)
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We are given that the area is only equal to 192 in². We subtract this value from the computed area.
Difference = 280 in² - 192 in²
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The area,
88 = (20 - 2x)(14 - 2x)

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3 years ago

How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

4 0

2 years ago

Write 843 in scientific notation.​

Write 843 in scientific notation.