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OleMash [197]
2 years ago
15

What is 6.42 written as a fraction in lowest terms?

Mathematics
1 answer:
IrinaVladis [17]2 years ago
6 0

Answer:

315/50

Step-by-step explanation:

trust me bro!

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5 2c 4 8 1 0

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2 years ago
The answer to this question. i’m confused and don’t understand it.
Georgia [21]

Answer:

5,120 bacterias

Step-by-step explanation:

Given the amount of bacteria after x hours modelled by the equation:

f(x) = 20(2)^x/12

Note that x is in hours not days

To calculate the number of bacteria that will be contained within the culture after 4 days, we will first convert 4days into hours since x is in hours.

24hours = 1day

xhours = 4days

x = 24×4

x = 96hours

Substituting x = 96 into the equation

f(96) = 20(2)^96/12

f(96) = 20(2)^8

f(96) = 20×2^8

f(96) = 20×256

f(96) = 5,120

The number of bacteria that will be contained after 4days is 5,120bacterias

3 0
2 years ago
The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge
Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

5x+4y+5z=-1

x+y+2z=1

2x+y-z=-3

Firs we find determinant of system of equations

Let a matrix A=\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right] and B=\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]

\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}

\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply R_1\rightarrow R_1-4R_2 and R_3\rightarrow R_3-2R_2

\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]

ApplyR_2\rightarrow R_2-R_1

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]

Apply R_3\rightarrow R_3+R_2

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]

Apply R_3\rightarrow- \frac{1}{2} and R_2\rightarrow R_2-R_3

\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Apply R_1\rightarrow R_1-R_3

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0
3 years ago
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