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3 years ago

How can you write the expression with a rationalized denominator? √3-√6/√3+√6

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

The answer to your question is -3+2√2

kozerog [31]3 years ago

5 0

The answer is √3-√2+√6 = in decimal form 2.76732698

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3 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒ $x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

8 0

3 years ago

Read 2 more answers

What are the powers of 4 through 1000

Ahat [919]

Answer:

{4^0, 4^1, 4^2, 4^3, 4^4}, or, with a bit of rounding off,

{4^0, 4^1, 4^2, 4^3, 4^4, 4^5} = {1, 4, 16, 64, 256, 1024}

Step-by-step explanation:

I wanted to determine the largest power of 4 that does not exceed 1000. Here's what I did:

4^x = 1000. Taking the ln of both sides, we get x ln 4 = ln 1000.

6.9078

Solving for x: x = --------------- = 4.98

1.3863

Thus, the answer to this question is {4^0, 4^1, 4^2, 4^3, 4^4}. If you're willing to talk in terms of approximations, then there'd be one more power of 4 in this set: 4^5. Note that 4^5 = 1024, whidh is just a bit over 1000.

7 0

3 years ago