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3 years ago

Find the following quotient. (5−4i)/(5+4i) Answer in the form (a+bi)/c .

Mathematics

1 answer:

Umnica [9.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

$\frac{5-4i}{5+4i}=\frac{(5-4i)*(5-4i)}{(5+4i)*(5-4i)}\\\\$

Now numerator is in the form (a - b)² & denominator in the form (a+b)(a - b)

(a -b)² = a² - 2ab + b²

(a + b)(a-b) = a² - b²

$\frac{(5-4i)^{2}}{(5)^{2}-(4i)^{2}}=\frac{5^{2}-2*5*4i+(4i)^{2}}{25-16i^{2}}\\\\=\frac{25-40i+16i^{2}}{25-16*(-1)}\\\\=\frac{25-40i+16*(-1)}{25+16}\\\\=\frac{25-40i-16}{41}\\\\=\frac{9-40i}{41}$

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What is the coefficient of the variable in the expression 6-4x-8x+2

Daniel [21]

I think the trick is to add the two middle terms together. (And the 2 end terms).

8 - 12x

The variable is x. The coefficient is - 12.

8 is a constant.

The coefficient always goes with a variable.

3x^2 + 2x + 5

Nothing can be combined. The leading coefficient is 3 (which goes with the greatest power of the variable x^2 in this case) and the other coefficient is 2. 5 is a constant.

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3 years ago

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Answer:

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2 years ago

5/6x - 4 = 11 what does x equal?

dezoksy [38]

Answer:

Step-by-step explanation:

(5/6)x – 4 = 11

(5/6)x = 11 + 4

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7 0

2 years ago

Read 2 more answers

Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta

kherson [118]

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}$

which follows from $x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x$ .

$\dfrac{\mathrm dy}{\mathrm dt}$ is then a function of $x$ ; denote this function by $f(x)$ . Then by the product rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Then the ODE in terms of $t$ is

$\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0$

The characteristic equation

$r^2+8r-20=(r+10)(r-2)=0$

has two roots at $r=-10$ and $r=2$ , so the characteristic solution is

$y_c(t)=C_1e^{-10t}+C_2e^{2t}$

Solving in terms of $x$ gives

$y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}$

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3 years ago

Solve the equation 5+7 | 2x-1 | = -44

dlinn [17]

The easy part is isolating the absolute-value term:

5 + 7 |2x - 1| = -44

7 |2x - 1| = -49

|2x - 1| = -7

Remember that the absolute value function returns a positive number that you can think of as the "size" of that number, or the positive distance between that number and zero. If x is a positive number, its absolute value is the same number, |x| = x. But if x is negative, then the absolute value returns its negative, |x| = -x, which makes it positive. (If x = 0, you can use either result, since -0 is still 0.)

The important thing to take from this is that there are 2 cases to consider: is the expression in the absolute value positive, or is it negative?

• If 2x - 1 > 0, then |2x - 1| = 2x - 1. Then the equation becomes

2x - 1 = -7

2x = -6

x = -3

• If 2x - 1 < 0, then |2x - 1| = - (2x - 1) = 1 - 2x. Then

1 - 2x = -7

-2x = -8

x = 4

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2 years ago