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2 years ago

1. Is the square root of 2 between 1.3 and 1.4? How do you know?

Mathematics

2 answers:

KengaRu [80]2 years ago

6 0

Answer:

square root of 2 is equal to 1.414

so,

1) square root of 2 does not lie between 1.3 and 1.4 because 1.414 is greater value than 1.3 and 1.4.

2) square root of 2 does lie between 1.4 and 1.5

we can say it by squaring 1.4 and 1.5

1.4² = 1.96

1.5² = 2.25 where 2 lies between 2.25 and 1.96

3) the square of two lie between 1.4 and 1.5 and the square root of 2 is 1.414

VikaD [51]2 years ago

3 0

Answer:

Step-by-step explanation:

1st . 1.3^2= 1.69

1.4^2= 1.96

It's not bettween 1.3 & 1.4

2. 1.4^2=1.96

1.5^2= 2.25

it's betteen 1.4 & 1.5

3 . 1.4 & 1.5 so it's eqaul 1.4

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3 years ago

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2 years ago

members of a soccer team raised $741.15 to go to a tournament. They rented a bus for $250 and budgeted $25.85 per player for mea

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2 years ago

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3 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 $R_3\rightarrow R_3-2R_2$

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

Apply $R_2\rightarrow R_2-R_1$

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

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.

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3 years ago