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3 years ago

What is the 8.356 rounded to the nearest hundredth

Mathematics

2 answers:

rodikova [14]3 years ago

5 0

Answer:

8.356 rounded to the nearest hundredth=8.360

KATRIN_1 [288]3 years ago

3 0

Answer:

8.36

Step-by-step explanation:

The hundredth decimal place is the second digit right of the decimal.

There should be 2 decimal places when rounded to the nearest hundredth.

If the third digit is less than 5, round down. If the third digit is 5 or more, round up.

In 8.356, the third digit is 6. 6 is 5 or more so we round the second digit up.

8.356 => 8.36

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A number cube with faces labeled from 1 to 6 will be rolled once.

tresset_1 [31]

Question:

A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling a number greater than 2.

If there is more than one element in the set, separate them with commas.

Answer:

$S = \{1,2,3,4,5,6\}$

$Greater2 = \{3,4,5,6\}$

Step-by-step explanation:

Given

A roll of a 6 sided number cube

Solving (a): The sample space

This implies that we list out all number on the number cube.

So:

$S = \{1,2,3,4,5,6\}$

Solving (b): Outcomes greater than 2

This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.

So:

$Greater2 = \{3,4,5,6\}$

3 0

3 years ago

Need help on this please giving brainlest! :)

masha68 [24]

A is 9,4 B is -9,4 all you do is flip the sign according to the axis

7 0

3 years ago

In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. The same is true for a lower

SCORPION-xisa [38]

Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

$A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0$

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)$ .

So, eigenvalues are $\lambda_1=a,\:\lambda_2=c$

Let's apply it for 3x3 matrix:

$A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0$

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)(\lambda-f)$ .