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3 years ago

What is 48,020,000 expressed in scientific notation?

Mathematics

2 answers:

dem82 [27]3 years ago

7 0

A is the correct answer

Bad White [126]3 years ago

6 0

Answer:

im pretty sure it is answer A. 4.802 x 10^7

Step-by-Step explanation

The purpose of scientific notation is for scientists to write very large, or very small, numbers with ease.

Calculating scientific notation for a positive integer is simple, as it always follows this notation:

a x 10b

Follow the steps below to see how 48,020,000 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: 48,020,000

New Number: 4.8020000

Step 2

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

New Number: 4 . 8 0 2 0 0 0 0

Decimal Count: 1 2 3 4 5 6 7

There are 7 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 = 4.802 (Please notice any zeroes on the end have been removed)

b = 7

Now the whole thing:

4.802 x 107

Step 4

Check your work:

107 = 10,000,000 x 4.802 = 48,020,000

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$