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tatiyna
3 years ago
15

What is v^4-1 simplified

Mathematics
2 answers:
serg [7]3 years ago
4 0
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
v^4-1\implies v^4-1^4\implies (v^2)^2-(1^2)^2\implies \underline{(v^2-1^2)}(v^2+1^2)
\\\\\\
\underline{(v-1)(v+1)}(v^2+1^2)
lina2011 [118]3 years ago
3 0
The answer is:
(v-1)(v+1)(v^2+1)
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HELP WITH THESE QUESTIONS PART 2.
polet [3.4K]

Answer: (a) IV (b) positive (c) \frac{\pi}{4} (d) C (e) \sqrt{2}

<u>Step-by-step explanation:</u>

a) \frac{15\pi}{4} - \frac{8\pi}{4} = \frac{7\pi}{4}, which is located in Quadrant IV <em>per the Unit Circle.  </em>

b) sec is \frac{1}{cos}. Cos is the x-coordinate.  The x-coordinate inQuadrant IV is positive.

c) the reference angle is the angle from \frac{7\pi}{4} to 2π = \frac{\pi}{4}

d) since the angle is below the x-axis and the reference angle is\frac{\pi}{4}, then the angle is equal to -sec(\frac{\pi}{4})

e) sec = \frac{1}{cos}   ⇒   sec \frac{7\pi}{4} = \frac{2}{\sqrt{2} } = \frac{2}{\sqrt{2} }(\frac{\sqrt{2}} {\sqrt{2}}) = \frac{2\sqrt{2} }{2} = \sqrt{2}

**********************************************************

Answer: (a) \frac{3\pi}{2} (b) (0, -1) (c) A (d) -1

<u>Step-by-step explanation:</u>

a) \frac{11\pi}{2} - \frac{4\pi}{2} = \frac{7\pi}{2} - \frac{11\pi}{2} = \frac{3\pi}{2}

b) \frac{3\pi}{2} is on the Unit Circle at (0, -1)

c) sin = \frac{opposite}{hypotenuse} which equals \frac{y}{r} on the Unit Circle.

d) sin is the y-coordinate. sin (\frac{3\pi}{2}) = -1

6 0
3 years ago
A LOOOOOOOOT OF POINTS AND THANK YOU AND BRAINLIEST!!!!!!!!
Anarel [89]

Answer:

Step-by-step explanation:

vertex is 1 units left.

y is always positive.

y=|x+1|

5 0
3 years ago
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Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.
tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: \frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3} square units

(22)

Similar to (21)

Area = 216\sqrt{3} square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

altitude=\frac{\sqrt{3}}{2}*side

side = \frac{36}{\sqrt{3}}

Hence, area of the hexagon will be: 648\sqrt{3} square units

(24)

Given is the inradius of an equilateral triangle.

Inradius = \frac{\sqrt{3}}{6}*side

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:

Side = 16 units

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4 0
3 years ago
WHAT DOES THIS MEAN I DONT UNDERSTAND
Art [367]

Answer:

the goal is to try and get x alone. For the 1st problem, x=3.

Step-by-step explanation:

3x + 1 = 10      Subtract 1 both sides to cancel out the 1 on the left side. 10-1 = 9

3x = 9             To get x alone here, Divide 3 on both sides. 9/3 = 3

x = 3                Now x is alone, which means x = 3.

This is right because if you plug x with 3, the sum = 10.

(3*3) + 1 =10

9 + 1 = 10

So, that's how you know 3 is the right answer.

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2 years ago
At the start of 2014 Mike's car was worth ?12000.The value of the car decreased by 30% every year.Work out the valure of his car
Contact [7]

Answer:

$1200

Step-by-step explanation:

12,000*(-90%)

12,000*(.1)=1200

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3 years ago
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