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GrogVix [38]
3 years ago
6

Consider an experiment in which the letters "A"

Mathematics
1 answer:
spin [16.1K]3 years ago
6 0

0.2 iis the answer you are loooking for when you say pal is not formed

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Find function domain <br> f(x) = sqrt( 2sin x - 1 )
vekshin1

Answer:

{x ∈ ℝ : x ≥ π/6 +2πn and x ≤ π/6 + 2πn and n ∈ ℤ}

Step-by-step explanation:

sinx can run from -1 to +1

2sinx can run from -2 to +2

2sinx -1 can run from -3 to +1

However, the square root is imaginary when x < 0. So, the condition is

2sinx -1 ≥ 0

2sinx ≥ 1

sinx ≥ ½

x ≥ π/6 (30°)

So, in the interval [0, 2π], π/6 ≤ x ≤ 5π/6

However, the sine is a cyclic function and repeats itself every 2π.

Over all real numbers, the condition is (π/6 +2πn) ≤ x ≤ (5π/6 + 2πn).

The domain is then

{x ∈ ℝ : x ≥ π/6 +2πn and x ≤ π/6 + 2πn and n ∈ ℤ}

5 0
3 years ago
finding the distance from (–6, 2) to the origin. Use (0, 0) as (x1, y1). d = StartRoot (x 2 minus x 1) squared + (y 2 minus y 1)
Naddik [55]

Answer:

2√10

Step-by-step explanation:

Given the coordinates (-6, 2) and (0,0)

We are to find the distance between the coordinates. Using the distance formula;

d = √(x2-x1)²+(y2-y1)²

d = √(0-2)²+(0+6)²

d = √(-2)²+(6)²

d = √4+36

d = √40

d = √4*10

d = 2√10

Hence the required distance is 2√10

4 0
3 years ago
You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

6 0
3 years ago
Mean average of this question plz.
stira [4]
 for this answer
 I got 32
7 0
3 years ago
Evaluate the given algebraic expressions replacing x with -2 and y with 1<br><br> <img src="https://tex.z-dn.net/?f=3x%5E%7B2%7D
Savatey [412]

Answer: -6

Step-by-step explanation: -2^1 = -2

                                              3 x -2 = -6

6 0
2 years ago
Read 2 more answers
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