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3 years ago

What is 4,952 rounded to the nearest hundred?

Mathematics

2 answers:

Juliette [100K]3 years ago

6 0

Answer:

4,952 rounded to the nearest hundred is:

5,000

Step-by-step explanation:

1. Look at the hundreds place.

2. Look one to the right.

3. If the number is 5 or more add one to the hundreds place and make the rest of the numbers to the right o's. But if it is below five, leave the hundreds place alone and make the rest of the numbers o's. But in this case, the hundreds place is a nine. So you move to the next digit. Which is 4, and make it a five.

4. 5,000

Hope this helps you out : D

Blizzard [7]3 years ago

3 0

Answer:

5000

Step-by-step explanation:

We'd normally just round up the 5, but the digit in the hundreds place is a 9, which rounds up to 10, so changes the number to 5000.

im really bad at explaining stuff

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One number is 6 less than another. Their sum is 14. Find the numbers.

umka21 [38]

Answer:

4 and 10

Step-by-step explanation:

10-4=6

and 10 plus 4 equals 14

so the numbers are 4 and 10

6 0

3 years ago

A = ???? 4 −2

irinina [24]

Answer:

1. The matrix A isn't the inverse of matrix B.

2. |B|=12, |A|=12

Step-by-step explanation:

1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.

We have:

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

and

$B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]$

A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.

Then if A is the inverse of B:

$B.A=I$

Where,

$I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Observation:

If you have two matrices:

$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]$

Now:

$B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]$