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3 years ago

Find the distance using the modulus of the difference between z1 = -8 + 3i and z2 = 7 - 4i.

Mathematics

1 answer:

geniusboy [140]3 years ago

3 0

Answer:

$d = 16.55$

Step-by-step explanation:

Given

$z_1 = -8 + 3i$

$z_2 = 7 - 4i$

Required

The distance between them

First, calculate the difference (d) between z1 and z2

$d = z_1 - z_2$

$d = -8 + 3i - [7 - 4i]$

Open bracket

$d = -8 + 3i - 7 + 4i$

Collect like terms

$d = -8 - 7+ 3i + 4i$

$d = -15+ 7i$

Using modulus, we have:

$d = \sqrt{R^2 + I^2}$

Where

R = Real Part = -15

I = Imaginary Part = 7

So, we have:

$d = \sqrt{(-15)^2 + 7^2}$

$d = \sqrt{225 + 49}$

$d = \sqrt{274}$

$d = 16.55$

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help me ;-; this question is hard. if possible, can you try explaining how to do it too after answering? that would be blessing

Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

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3 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
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6 0

4 years ago

Bryan and Jadyn had barbeque potato chips and soda at a football party. Bryan ate 4 oz of chips and drank 1 cup of soda for a to

Inga [223]

Answer:

1 oz of chips has 140 mg of sodium, and 1 cup of soda has 40 mg of sodium.

Step-by-step explanation:

This question is solved by a system of equations.

I am going to say that:

x is the amount of sodium in 1oz of chips.

y is the amount of sodium in 1 cup of soda.

Bryan ate 4 oz of chips and drank 1 cup of soda for a total of 600mg of sodium.