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2 years ago

What is the value of 25 - 14 +12?

Mathematics

1 answer:

Alenkasestr [34]2 years ago

3 0

Answer:

Step-by-step explanation:

25-14 = 11

11+12=23

that would be your answer

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The function g(x) = x2 is transformed to obtain function h:

Rudik [331]

Answer:

Option C

The graph of g is vertically shifted 5 units down

4 0

2 years ago

If there is a 4 digit code and using the digits 1-4, how many possibilities are there?

timurjin [86]

Answer:

Step-by-step explanation:

1, 2

1, 3

1, 4

2, 1

2, 3

2, 4

3, 1

3, 2

3, 4

4, 1

4, 2

4, 3

1, 2, 3

1, 2, 4

1, 3, 2

1, 3, 4

1, 4, 2

1, 4, 3

2, 1, 3

2, 1, 4

2, 3, 1

2, 3, 4

2, 4, 1

2, 4, 3

3, 1, 2

3, 1, 4

3, 2, 1

3, 2, 4

3, 4, 1

3, 4, 2

4, 1, 2

4, 1, 3

4, 2, 1

4, 2, 3

4, 3, 1

4, 3, 2

1, 2, 3, 4

1, 2, 4, 3

1, 3, 2, 4

1, 3, 4, 2

1, 4, 2, 3

1, 4, 3, 2

2, 1, 3, 4

2, 1, 4, 3

2, 3, 1, 4

2, 3, 4, 1

2, 4, 1, 3

2, 4, 3, 1

3, 1, 2, 4

3, 1, 4, 2

3, 2, 1, 4

3, 2, 4, 1

3, 4, 1, 2

3, 4, 2, 1

4, 1, 2, 3

4, 1, 3, 2

4, 2, 1, 3

4, 2, 3, 1

4, 3, 1, 2

4, 3, 2, 1

3 0

2 years ago

Read 2 more answers

What 3 numbers can be multiplied together to get 180?

hichkok12 [17]

2×6×15=180

Step-by-step explanation:

2×6=12

12×15=180

4 0

3 years ago

Read 2 more answers

What does the letter p in the symbol 4p3 indicates?

GuDViN [60]

Probability is what the letter p indicates

3 0

2 years ago

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