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3 years ago

Which is a correct expansion of the following expression?

Mathematics

2 answers:

-BARSIC- [3]3 years ago

8 0

Answer:

A. 18x – 80

Step-by-step explanation:

18(x - 3) - 26

18x - 54 - 26

18x – 80

Sergio [31]3 years ago

5 0

The answer is a. 18x - 80

Because first you will expand the brackets, for this you will multiply everything that is outside the bracket to what’s inside (18 times x) which will give you 18x and (18 times -3) which will give you -54

Now your equation will be

18x -54 - 26

Next you will workout the liked terms with are -54 and -26, since they are both negative just add them together and keep the sign.

To get your final answer

18x -80

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A store receives 2,000 decks of popular trading cards. The number of decks of cards is a function, d, of the

BARSIC [14]

Answer:

-143.4 for 0-5 and -38 for 15-20

Step-by-step explanation:

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3 years ago

Can u help me find the answer to #5 and 6

Arlecino [84]

Answer:

5) x = 24

6) x = 48

Step-by-step explanation:

I'd be glad to help :)

For question 5, we know that 62 and the unknown angle is a straight angle, so 62+y = 180. y = 118. We also know all the angles in a triangle add up to 180, so x+(x+14)+118 = 180. Therefore, 2x+132 = 180. We now know that 2x = 48, so x = 24 for question 5.

For question 6, We know that because the line is divided up, (x+35) can be moved down a line, so (2x+1)+(x+35) = 180, so 3x+36 = 180. We now know that x = 48 for question 6

7 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 $z=x+iy$ .

We are looking for the complex number $z^{-1}=(x+iy)^{-1}$ which in terms of matrices is equivalent to find the matrix

$\left[\begin{array}{cc}x&y\\-y & x\end{array}\right]^{-1}= \dfrac{1}{x^{2}+y^{2}} \left[\begin{array}{ccc}x&-y\\y&x\end{array}\right]$

Hence,

$z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)$

6 0

3 years ago

Indicate the formula for the following conditions.<br><br> C(n,r)

Serjik [45]

The formula for C(n,r), also known as "n choose r" or a combination,
is:
nCr = n! / (r!(n-r)!),
where n is the total number of options and r is number you must choose

The number generated is the total number of possible combinations.

The ! means factorial. For example, 4! = 4 x 3 x 2 x 1

8 0

4 years ago

Read 2 more answers

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dexar [7]

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3 years ago