All categories

4 years ago

What is the solution to x4 - 12x2+10>0

Mathematics

1 answer:

alexira [117]4 years ago

5 0

Answer:

(-√(6-√26) < x < √(6-√26)) ∪ (x < -√(6 +√26)) ∪ (√(6 +√26) < x)

Step-by-step explanation:

Using x^2 = z, the equation can be rewritten as ...

z^2 -12z +10 > 0

(z -6)^2 -26 > 0

|z -6| > √26

This resolves to two equations.

This one ...

x^2 -6 < -√26 . . . . substitute x^2 for z

|x| < √(6-√26) . . . . add 6, take the square root; use √a^2 = |a|

-√(6-√26) < x < √(6-√26)

and this one ...

x^2 -6 > √26

|x| > √(6 +√26)

x < -√(6 +√26) ∪ √(6 +√26) < x

You might be interested in

8 and 3/15 - 4 and 3/15

Iteru [2.4K]

Answer:

Step-by-step explanation:

8 3/15 - 4 3/15 the 3/15 cancel each other out so then there is only 8-4 which is 4

8 0

3 years ago

Read 2 more answers

115 is what percent of 200?

Valentin [98]

57.5 is the answer. Because think of 200 as 100% and 15 out of a hundered is ______. Right.

4 0

4 years ago

What are the values of m and n in the matrix addition below?

Aleksandr [31]

ANSWER

The correct answer is $m=45,n=12$ .

<u>EXPLANATION</u>

We were given the matrix equation;

$\left[\begin{array}{cc}n-1&6\\-19&m+3\end{array}\right] +\left[\begin{array}{cc}-1&0\\16&-8\end{array}\right] =\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

We must first simplify the Left Hand Side of the equation by adding corresponding entries.

$\left[\begin{array}{cc}n-1+-1&6+0\\-19+16&m+3-8\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

That is;

$\left[\begin{array}{cc}n-2&6\\-3&m-5\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

Since the two matrices are equal, their corresponding entries are also equal. we equate corresponding entries and solve for m and n.

This implies that;

$n-2=10$

We got this equation from row one-column one entry of both matrices.

$n=12$

Also, the row three-column three entries of both matrices will give us the equation;

$m-5=40$

$m=45$

Hence the correct answer is $m=45,n=12$ .

The correct option is option 2

6 0

3 years ago

Determine the values of X, Y and Z on the following number line:<br> HH<br> X -11 Y 9 Z

Dmitrij [34]

-11 y 17 with z and x on the line

3 0

3 years ago

Assume V and W are finite-dimensional vector spaces and T is a linear transformation from V to W, T: Upper V right arrow Upper

scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

8 0

3 years ago