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andrezito [222]
3 years ago
13

You need at least 15 pencils or markers. You want to spend at most $14 on pencils and markers. Pencils p are $0.85 each and mark

ers m are $1.45 each. Which system of inequalities models the situation?
Mathematics
1 answer:
S_A_V [24]3 years ago
6 0

Answer:

.85p + 1.45m <= 14

Step-by-step explanation:

You don't want to spend more than 14, but you can spend 14. So the inequality must be less than or equal to. To find the combination of pencils and markers you must multiply their price times the number of pencils. Sorry if this explanation isn't amazing, I know how to do work but not explain my reasoning. :)

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aliya0001 [1]

Answer:

Step-by-step explanation:

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8 0
2 years ago
Y=6x – 11<br> -2x – 3y =-7
Eva8 [605]

Answer:

x=2   and   y=1

Step-by-step explanation:

1). Plug the y value into the second equation. It will look like the following afte you plug in what y equals:

              -2x-3(6x-11)=-7

2). Then, you distribute the "-3" to all the terms inside the parenthesis.

              -2x-18x+33=7

3). Then, you subtract the positive 33 from both sides to try to get the x alone on one side.

              -2x-18x=-40

4.) Then, you combine all of the terms containing an x.

              -20x=-40

5). Finally, you divide both sides by -20 and the outcome is x=2.

6).  Hold up...we ain't done yet..

7).  Next, you have to plug the x back into the either equation to get the outcome of what y will be, but for this purpose I will plug it back into the first equation.

              y=6(2)-11

8). I will then distribute the 6 into the all the terms in the parenthesis and then it will look like this.

              y= 12-11

9). Do the math...and the answer is y=1.

4 0
3 years ago
Give an example of a 2x2 matrix without any real eigenvalues:___________
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Answer:

Step-by-step explanation:

An eigenvalue of n × n is a function of a scalar \lambda  considering that there is a solution (i.e. nontrivial) to an eigenvector x of Ax =  

Suppose the matrix A = \left[\begin{array}{cc}-1&-1\\2&1\\ \end{array}\right]

Thus, the equation of the determinant (A - \lambda1) = 0

This implies that:

\left[\begin{array}{cc}-1-\lambda &-1\\2&1- \lambda\\ \end{array}\right] =0

-(1 - \lambda^2 ) + 2 = 0

-1 + \lambda ^2 + 2= 0

\lambda^2 +1 =0

Hence, the eigenvalues of the equation are \mathtt{\lambda = i , -i}

Also, the eigenvalues can be said to be complex numbers.

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3 years ago
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Answer:

Step-by-step explanation:

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Answer:

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