All categories

4 years ago

Can somebody please help me

Mathematics

1 answer:

yKpoI14uk [10]4 years ago

4 0

For #1 the answer would be ; " y = - x - 4 "
( you just have to plug in the given numbers to the slope intercept form )

You might be interested in

Kayla earns a weekly salary of $290 plus a 5.5% commission on sales at a gift shop. How much would she make in a week if she sol

Mama L [17]

$\$290+5.5\%\cdot\$5799=\$290+0.055\cdot\$5799\approx\$608.95$

$Ans.\ Kayla\ make\ in\ a\ week\ about\ \$608.95$

6 0

3 years ago

Read 2 more answers

What is the value of 12⁰?

MatroZZZ [7]

Heyo! ;D

So, I think I may of heard about this topic in the past...are you working with degrees of circles? Plane angles? If so, you can use this formula to find your answer:

12° × $\pi$ ÷ 180 = <u>0.2094(rad)</u>

I'm not entirely sure, however I hope this somewhat helped! If so, please lmk! Tysm and good luck!

8 0

3 years ago

Read 2 more answers

Sam has a strip of wood that is 3 4 foot long and 1 6 foot wide. He wants to use the wood to patch a damaged area on his desk. H

MArishka [77]

Answer:

1 / 8

Step-by-step explanation:

I'm assuming that is meant to be 3/4 and 1/6...

Area = Length * Width

A = 3/4 * 1/6

A = 3 / 24

simplify the fraction to get your answer

6 0

3 years ago

Read 2 more answers

Which fraction has a repeating decimal as its decimal expansion?

Kobotan [32]

Answer:

Step-by-step explanation:

a=0.15789474

b=0.1875

c=0.27272727

d=0.375

7 0

3 years ago

Read 2 more answers

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