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zheka24 [161]
3 years ago
12

Find all real and complex roots of the equation below x^4-81=0

Mathematics
1 answer:
Pepsi [2]3 years ago
8 0

x^4-81=0 \\ (x^2-9)(x^2+9)=0\\ (x-3)(x+3)(x^2+9)=0\\ x=3 \vee x=-3 \vee x=3i \vee x=-3i

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My Notes The displacement from equilibrium of an oscillating weight suspended by a spring is given by y(t) = 1 4 cos(5t) where y
ikadub [295]

Answer:

y(0) = 0.25 feet

y(\frac{1}{4}) = 0.0789 \text{ feet}

y(\frac{1}{2}) =-0.2003 \text{ feet}

Step-by-step explanation:

We are given the following information in the question:

The displacement from equilibrium of an oscillating weight suspended by a spring =

y(t) = \displaystyle\frac{1}{4} \cos(5t)

where y is the displacement in feet and t is the time in seconds.

Here, cos is in radians.

1) t = 0

y(0) = \displaystyle\frac{1}{4} \cos(5(0)) = \frac{1}{4} \cos(0) = \frac{1}{4}(1) = \frac{1}{4}

y(0) = 0.25 feet

2) t = \frac{1}{4}

y(\displaystyle\frac{1}{4}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{4})) = \frac{1}{4} \cos(1.25) = \frac{1}{4}(0.31532236) =0.07883059

y(\frac{1}{4}) = 0.0789 \text{ feet}

3) t = \frac{1}{2}

y(\displaystyle\frac{1}{2}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{2})) = \frac{1}{4} \cos(2.5) = \frac{1}{4}(-0.80114362) = -0.200285905

y(\frac{1}{2}) =-0.2003 \text{ feet}

The negative sign indicates the opposite direction of displacement.

3 0
3 years ago
Whats the slope in y=−5−12x.?
jek_recluse [69]

Answer:

-12

Step-by-step explanation:

4 0
3 years ago
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There are 49 dogs signed up for a dog show. There are 36 more small dogs than the large dogs. How many small dogs have signed up
mixer [17]

Answer:

I think it is 41.5...

3 0
2 years ago
Need help with this question please!
Ivenika [448]
The answer is definitely B
5 0
3 years ago
Example 2
PIT_PIT [208]

Answer: (a) 314.2cm², (b) 157.1cm², (c) 78.55cm² (e) 6.77

Step-by-step explanation: (a)  Area of the circle with radius of 10 cm = πr²

                                                                          = 3.142 × 10 × 10

                                                                          = 3.142 × 100

                                                                          = 314.2cm²

The formula                                                       = πr²

(b)  Area of the half of a circle known as semicircle

                                                                         = πr²/2

                                                                         = 3.142 ×10 × 10/2

                                                                         = 3.142 × 50

                                                                         = 157.1cm²

The formula                                                      = πr²/2

(c)  A quarter of a circle is called quadrant

                                                            = πr²/4

                                                            = 3.142 × 10 × 10/4

                                                            = 314.2/4

                                                            = 78.55cm²

The formula is written thus = πr²/4, which implies that the circle is divided into 4 unit

(d) The conjecture about how to determine the area of the sector is

Formula of a sector = ∅/360(πr²)

<u>Information</u>

The arc  cant be 60°, therefore information incomplete.

(e) Area of the sector with the angle AOB of 60° = 24.

To find the radius of the angle, make v the subject of the formula from the formula.

Sector area = πr²∅/360°

equate formula to 24.

Therefore πr²∅/360° = 24

Multiply through by360° to make it a linear expression

It now becomes πr²∅ =24× 360°

                                                     r² = 24  x 360/π × ∅°

                                                     r² = 24 × 360° /3.142 × 60°

                                                     r² = 3,640/188.52

                                                     r² = 45.8

To find r , we take the square root of both side by applying laws of indicies

                                    Therefore r = √45 .8

                                                      r = 6.77

(f)   General formula = ∅°/360° × (πr²)

angle substended at centre by the arc = x°

assuming the radius of the circle = ycm, Therefore,  area of the sector = { ∅°/360° × πy² }

                                                     

                                                   

8 0
3 years ago
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