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mart [117]
3 years ago
7

WILL MARK BRAINIEST What is the conjecture for 3,1,4,1,5 PLEASE HURRY NEED BY 11:00am

Mathematics
1 answer:
Leokris [45]3 years ago
3 0

Answer:

6

Step-by-step explanation:

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Round 17.641 to the nearest whole number
vfiekz [6]
The nearest whole number would be 18 because the 6 is greater than 5. Anything greater than five, would most likely be rounded to the nearest whole number. Hope this helped!


8 0
3 years ago
For the polynomial, (((x^3)+3x+2)/5)+((5x^2)/2)-(x^6)
nirvana33 [79]

Answer:

Hello,

Step-by-step explanation:

(((x^3)+3x+2)/5)+((5x^2)/2)-(x^6)\\\\=\dfrac{x^3}{5} +\dfrac{3*x}{5} +\dfrac{2}{5} +\dfrac{5*x^2}{2} -x^6\\\\\\=-x^6+0x^5+0x^4+\dfrac{x^3}{5} +\dfrac{5*x^2}{2} +\dfrac{3*x}{5}+\dfrac{2}{5}  \\\\\\a)\ \dfrac{3}{5} \\\\b)\ -1\\\\c)\\degree=6\\

4 0
3 years ago
The Cartesian coordinates of a point are given. (a) (−3, 3) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0
irina [24]

Answer:

a) (-3, 3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (3√2, 0.75π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-3√2, 1.75π)

b) (4, 4√3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (8, 0.13π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-8, 1.13π)

Step-by-step explanation:

We know that polar coordinates are related to (x, y) coordinates through

x = r cos θ

y = r sin θ

And r = √[x² + y²]

a) For (-3, 3)

(i) x = -3, y = 3

r = √[x² + y²] = √[(-3)² + (3)²] = √18 = ±3√2

If r > 0, r = 3√2

x = r cos θ

-3 = 3√2 cos θ

cos θ = -3 ÷ 3√2 = -(1/√2)

y = r sin θ

3 = 3√2 sin θ

sin θ = 3 ÷ 3√2 = (1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 0.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (3√2, 0.75π)

(ii) When r < 0, r = -3√2

x = r cos θ

-3 = -3√2 cos θ

cos θ = -3 ÷ -3√2 = (1/√2)

y = r sin θ

3 = -3√2 sin θ

sin θ = 3 ÷ -3√2 = -(1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (-3√2, 1.75π)

b) For (4, 4√3)

(i) x = 4, y = 4√3

r = √[x² + y²] = √[(4)² + (4√3)²] = √64 = ±8

If r > 0, r = 8

x = r cos θ

4 = 8 cos θ

cos θ = 4 ÷ 8 = 0.50

y = r sin θ

4√3 = 8 sin θ

sin θ = 4√3 ÷ 8 = (√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (8, 0.13π)

(ii) When r < 0, r = -8

x = r cos θ

4 = -8 cos θ

cos θ = 4 ÷ -8 = -0.50

y = r sin θ

4√3 = -8 sin θ

sin θ = 4√3 ÷ -8 = -(√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 1.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (-8, 1.13π)

Hope this Helps!!!

8 0
3 years ago
How do I find the answer to this question?
Ierofanga [76]
Answer:
You just did. 
7 0
3 years ago
Which expression has a value of 24?
marin [14]

Answer:

the last one

Step-by-step explanation:

12+(16/4+2)2

12+(4+2)2

12+6*2

12+12

24

5 0
3 years ago
Read 2 more answers
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