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3 years ago

X^4-x^2-42=0 need to solve equation and find solution set

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

5 0

That's a quadratic equation in $x^2$ , which means we get an extra square root and $\pm$ at the end.

$x^4 - x^2 - 42 = 0$

We factor,

$(x^2 - 7)(x^2+6) = 0$

$x^2 = 7 \quad \textrm{or} \quad x^2 = -6$

$x = \pm \sqrt{7} \quad \textrm{or} \quad x = \pm i \sqrt{6}$

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A manufacturing company is expected to pay a dividend of br. 1.25 per share at the end of the year (D1=br.1.25). The stock sells

Mama L [17]

Answer:

the equilibrium expected growth rate is 6.65%

Step by step Explanation:

We were given stock sold per share of $32.50

Dividend per share =$1.25

Required Return rate = 10.5%

Then we can calculate Percentage of Dividend for share as;

dividend of br. 1.25 per share at the end of the year (D1=br.1.25)

= 1.25×100= 125

Let the dividend percentage = y

stock sold per share × y= 125

125= 32.50y

y = 125/32.50

y= 3.85

y= 3.85*100%

Then the Dividend percentage = 3.85%

Growth rate=(required rate of return -Dividend percentage)

= 10.5 - 3.85 = 6.65

Therefore, the equilibrium expected growth rate is 6.65%

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4 years ago

3. 8.2(D) Which list shows the numbers in ascending order?

krek1111 [17]

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7 0

3 years ago

Read 2 more answers

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

$u\neq 0_{R2}$

$v\neq 0_{R2}$

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to $R^{2x2}$ using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

$A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]$

We used the first vector ''u'' as the first column of the matrix A

We used the second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

$Det(A)=6-p$

We need this determinant to be different to zero

$6-p\neq 0$

$p\neq 6$

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that $p\neq 6$

We can write : p ∈ IR - {6}

Notice that is $p=6$ ⇒

$u=(3,6)$

$v=(1,2)$

If we write $3v=3(1,2)=(3,6)=u$ , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.