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3 years ago

Which expression has the smallest value? A. |15| B. |-24| C. |-31| D. |42|

Mathematics

2 answers:

igomit [66]3 years ago

7 0

31 is the answer you want!

Dennis_Churaev [7]3 years ago

6 0

-31 Because it is a negative number
Explaination I’m smart Thank me cause I’m right

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On the photo /picture

asambeis [7]

We need to compute the diagonal of yhe square.

d²=7²+7²=49×2cm²

d=7×sqrt(2)=7×1.41=9.87cm.

The diagonal we just computed is the diameter of the circle that contains the square.

Since d=9.87 < 11, then indeed, the square fits inside the circle, without touching it.

6 0

3 years ago

Is it subtracting that I need to do or is it adding?

alisha [4.7K]

You need to subtract 25 by 18

4 0

3 years ago

What numbers do 9 12 6 3 and 5 go into

KiRa [710]

It wont be a perfect number

6 0

2 years ago

Please help, Will mark brainliest

mihalych1998 [28]

Answer:

y = - 8 x + 2

Step-by-step explanation:

Use any two pairs to find the slope with

slope = (y2-y1)/(x2-x1)

for example: (0, 2), and (1, -6)

slope = (- 6 - 2) / (1 - 0) = - 8

so the equation should look like:

y = -8 x + b

use point (0, 2) to find b:

2 = - 8 (0) + b

b = 2

Then

y = - 8 x + 2

5 0

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.

4 0

3 years ago