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Debora [2.8K]
3 years ago
10

HELP ASAP IM BEING TIMEDD

Mathematics
2 answers:
MrRissso [65]3 years ago
5 0

Answer:

53/100

Step-by-step explanation:

First, we convert the fraction to a decimal number by dividing the numerator by the denominator:

8 / 15 = 0.533

There are two parts to the decimal number above:

Integer Part: 0

Fractional Part: 533

Now, we will make the Fractional Part just two digits (nearest hundredth) by using our rounding rules.*

In this case, Rule I applies, so 8/15 (or 0.533) rounded to the nearest hundredth in decimal format is:

0.53

 

Next, we will make 8/15 rounded to the nearest hundredth in fraction format. Since you can divide our decimal format answer above by 1 and keep the same value, you can make it like this:

0.53 = 0.53/1

Then, we multiply the numerator and denominator by 100 to get rid of the decimal point:

(0.53 x 100) / (1 x 100) = 53/100

That's it. 8/15 rounded to the nearest hundredth is displayed below (simplified if necessary):

53/100

Masteriza [31]3 years ago
4 0

Answer:

53.33% is the correct answer

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900 yards

Step-by-step explanation:

540/3=180

180x5=900

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(12x + y + z = 26
mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}26\\17\\23\end{array}\right].

Step 3:

The linear system B is represented as

\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}23\\17\\26\end{array}\right].

Step 4:

The linear system C is represented as

\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\11\\12\end{array}\right].

Step 5:

The linear system D is represented as

\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}12\\11\\4\end{array}\right].

Step 6:

Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.

4 0
3 years ago
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