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inysia [295]
3 years ago
8

Is this Positive or Negative (3,528)(−17,438)(0)(−5,280)

Mathematics
2 answers:
adell [148]3 years ago
6 0

Answer:

positive

Step-by-step explanation:

since you're multiplying by 0, the answer will be zero, and zero is always positive

Mekhanik [1.2K]3 years ago
4 0

Answer:

Positive

Step-by-step explanation:

The negative and negative cancel each other out, and positive times positive is positive.

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The points (-4, 3) and (4, r) lie on a line with slope 1/2<br> Find the missing coordinate r.
melisa1 [442]

Answer:

7

Step-by-step explanation:

8 units separate -4 and 4

8 divided by 2 (which is the x of slope) is four

Four times 1 (which is y of slope) equals four

3+4 equals seven

(4,7)

7 0
3 years ago
Determine which equation is written in function form:
Mila [183]

<em>Greetings from Brasil... </em>

A 1st degree function can be written in two ways:

F(X) = AX + B

or

Y = AX + B

note that F(X) = Y

So, looking at the alternatives in the question, only Y = 7X/3 - 8 is equivalent to Y = AX + B, where:

A = 7/3

B = - 8

then

\large{\boldsymbol{\mathbf{Y = \frac{7X}{3}-8\Leftrightarrow Y=AX+B}}}   <em>a function of the 1st degree</em>

answer:

<h2>Y = (7/3)X - 8</h2>
7 0
3 years ago
What eigen value for this matix <br> (1 -2)<br> (-2 0)
natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

  1. Compute the matrix A' = A-\lambda I, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
  2. Compute the determinant of A'
  3. Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

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

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

Finally, we have

\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}

So, the two eigenvalues are

\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}

5 0
3 years ago
Read 2 more answers
AAAAA
mr_godi [17]

Answer:

600

Step-by-step explanation:

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How much money do I spend if 75%off for an item that's 1.23
stira [4]
If an item cost 1.23 and you get 75% off the item costs $0.92
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