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3 years ago

Write in expanded form 3 ways 23.56

Mathematics

1 answer:

AveGali [126]3 years ago

4 0

Answer:

23+ .5 + .06

20+3+.56

10+10+ 3.56

Step-by-step explanation:

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Please help!! Consider the equation x/5 -2=11, where x represents the

ira [324]

Answer:

x=65

Each NBA team can have a maximum of 15 players, 13 of which can be active each game.

65/5=13 so it's safe to say tgat 5 team can be crated

Step-by-step explanation:

x/5-2=11-->transposing x/5=11+2---->x=13×5--->x=65

verification 65/5-2=11--->13-2=11

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2 years ago

How much money, in dollars, is available on his card after he takes 0<br> rides?

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Answer:

However much he had on his card in the first place.

Step-by-step explanation:

Say he had $500 on his card. He took 0 rides ( no rides ) so he doesn't lose any money. Leaving him with his starting amount, $500.

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2 years ago

Please answer In(5x-3)=4

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Answer:

$\huge\boxed{Answer\hookleftarrow}$

$(5x - 3) = 4 \\ 5x - 3 = 4 \\ 5x = 4 + 3 \\ 5x = 7 \\ x = \frac{7}{5} \\ x = 1.4$

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# ꧁❣ RainbowSalt2²2² ࿐

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2 years ago

12 Units to the right of zero on a number line

vredina [299]

Answer:

Step-by-step explanation:

0 + 12 = 12

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2 years ago

Read 2 more answers

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.

4 0

3 years ago