Which values are solutions to the inequality -3x - 4 < 2? Check all of the boxes that apply.

Mathematics

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

-2

Step-by-step explanation:

You would add 4 to 2 inorder to get 6, then divide by 3 to get -2

You might be interested in

A driver training school calculates the cost(C) of training for a course using the formula C=60+16.5h, where h is the number of

Crazy boy [7]

C = 60 +16.5h
C = 60 +16.5(10)
C = $225.00

5 0

3 years ago

Solve −2x2 +3x − 9 = 0.

Dmitry_Shevchenko [17]

I assume you are asking to solve:

-2x^2 + 3x - 9 = 0

So first, this equation does not have real roots.

It only has 2 complex roots.

The roots are:

3/4 * (1-i√7) and 3/4 * (1+i√7)

8 0

3 years ago

A complex number, represented by z = x + iy, may also be visualized as a 2 by 2 matrix

Marat540 [252]

Answer:

Step-by-step explanation:

A) Suppose that we have the complex numbers

$z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}$

Remember that to sum complex numbers, we sum the real parts of the two numbers to get the real part and the imaginary parts of the two numbers to get the imaginary part. Hence,

$z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})$

On the other hand, if we sum the matrix visualizations of $z \quad \text{and} \quad \tilde{z}$ we get

$\left[\begin{array}{cc}x &y\\-y&x\end{array}\right] + \left[\begin{array}{cc}\tilde{x}&\tilde{y}\\ -\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x + \tilde{x}& y + \tilde{y}\\-(y+\tilde{y})&x+\tilde{x}\end{array}\right]$

which is the matrix visualization of $z + \tilde{z}$ .

To multiply two complex numbers, we use the distributive law to multiplly and then separete the real part from the imaginary part

$z \cdot \tilde{z}= (x + iy) \cdot (\tilde{x} + i \tilde{y})=(x \tilde{x} + i x \tilde{y} + i \tilde{x} y - y\tilde{y} ) = (x\tilde{x}-y\yilde{y})+i(x\tilde{y}+\tilde{x}y)$

Again, if we multiply the matrix visualizations of $z \quad \text{and} \quad \tilde{z}$ we get

$\left[\begin{array}{cc}x&y\\-y&x\end{array}\right]\left[\begin{array}{cc}\tilde{x}&\tilde{y}\\-\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x\tilde{x}-y\tilde{y}&x\tilde{y}+y\tilde{x}\\-y\tilde{x}-x\tilde{y}&x\tilde{x}-y\tilde{y}\end{array}\right]$

which is the matrix viasualization of $z\cdot\tilde{z}.$

B) Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number $z=x+iy$ .