1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tju [1.3M]
3 years ago
10

12 minus b=-2 solve for b

Mathematics
2 answers:
ycow [4]3 years ago
5 0
12 - b = -2
12 + 2 = b
14 = b
Vika [28.1K]3 years ago
5 0

Answer:

\:  \:  \:  \:  \: \: 12 - b =  - 2 \\  =  >  - b =  - 2 - 12 \\  =  >  -  b =  - 14 \\   =  >  b = 14

You might be interested in
Select the correct answer.
Nataly [62]

Answer:

D. Part of the solution region includes a negative number of erasers purchased; therefore, not all solutions are viable for the given situation.

Step-by-step explanation:

I got it right on the practice

7 0
3 years ago
A sequence is generated by the formula ax = 3x + 4. What is the value of the seventh term?
lyudmila [28]

Answer:

25

Explanation:

We are given the sequence formula expression;

ax = 3x + 4

You are to get the 7th term. This is gotten by substituting x = 7 into the expression as shown;

a7 = 3(7) + 4

a7 = 21 + 4

a7 = 25

Hence the seventh term of the sequence is 25

5 0
1 year ago
Quiz! Let's see what happens! Fill in the expanded table below (start with standard form)​
djyliett [7]

i put it in coments sence it says it is inapropreait

4 0
3 years ago
Read 2 more answers
How to do the inverse of a 3x3 matrix gaussian elimination.
nata0808 [166]

As an example, let's invert the matrix

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}

We construct the augmented matrix,

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

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 2 by 1/7 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 1 by -1/3 :

\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

So, the inverse of our matrix is

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}

6 0
2 years ago
Pls answer and get branily
Dennis_Churaev [7]

Answer:

Just put 4 basketballs and 5 baseballs in each one

Step-by-step explanation:

If you think about it you will know that each ratio there is the same. So, you do the same thing each time.

All you have to do now is read. It says put 4 basketballs and 5 baseballs.

I hope this helped ;)

6 0
3 years ago
Other questions:
  • A life insurance policy costs $6.88 for every $1,000 of insurance at this rate what is the cost for $17,500 worth of life insura
    7·2 answers
  • Find element a13 of matrix A, if A=
    5·1 answer
  • There are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag. What is the theoretical probability of randomly drawing
    12·2 answers
  • Hello there are 4 questions in this it will look like a lot but it is not and you don't have to answer all of it just look at th
    14·1 answer
  • In order for Josh to graph the solution to the inequality 4x - 2y > 6, which of the following steps does he need to use? Sele
    10·2 answers
  • Enter an equation in point-slope form for the line.
    14·1 answer
  • The perimter of a rectangle is 34 units. Its width is 6.5 units. Write an equation to determine the length (l) if the rectangle
    14·1 answer
  • The quantity five times a number y decreased by twelve. Group of answer choices 5y - 12 12 - 5y 12y - 5 12 + 5y
    7·1 answer
  • Hii, I would really appreciate it if you could help me out and answer this question. So this really isn’t a math question. You s
    9·2 answers
  • PLEASE HELP ME PLEASE!!!!!!!!
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!