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VMariaS [17]
3 years ago
6

Which statement is true about the value of |-14|?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0
It is the distance between -14 and 0 on the number line
You might be interested in
Jody practiced a piano piece for 500 seconds bill practiced a piano piece for 8 minutes who practiced longer explain
Mrac [35]

Answer:

Jody

Step-by-step explanation:

You can either convert 500 seconds to minutes or 8 minutes to seconds to compare. I'll do both.

There is 60 seconds in a minute. To find the total minutes of 500 seconds, divide 500 by 60 → 8.3, which means that Jody practiced for 8.3 minutes.

To find the total seconds of 8 minutes, multiply 60 seconds per minute by 8 minutes → 60 * 8 = 480 seconds which means that Bill practiced for 480 seconds.

Now you can compare. Jody practiced for 500 seconds, or 8.3 minutes. Bill practiced for 480 seconds, or 8 minutes. Jody practiced longer.

7 0
3 years ago
One-third of a number decreased by six equals eight. What is the number?
love history [14]

Answer:

42

Step-by-step explanation:

8+6=14

14×3=42

6 0
3 years ago
I don't know how to solve this. help please?
iragen [17]

Answer:

51   74

Step-by-step explanation:

x = smaller

x +23 = larger

added together   (x   + x+23)    equal to 28 less than 3x      = 3x-28

x    + x+23  = 3x-28      subtract 2x from both sides of the equation

   23 = x -28      add 28 to both sides

 51 = x      then the larger   =  51 + 23 = 74

8 0
2 years ago
Read 2 more answers
Answer this please thanks
densk [106]

Answer:

z=5, m<b=66 degrees

Step-by-step explanation:

8z+74+2z+56=180

10z+130=180

10z=50

z=5

3 0
3 years ago
Read 2 more answers
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
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