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andrey2020 [161]
3 years ago
6

An obtuse angle measures between:

Mathematics
2 answers:
Romashka [77]3 years ago
8 0

Answer:

answer is 90 and 180 hope that

velikii [3]3 years ago
5 0
Anything over 90 and bellow 180 degrees is a obtuse. 
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Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.
FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

\rho = \sqrt{x^{2}+y^{2}+z^{2}}

\phi = cos^{-1}\frac{z}{\rho}

For angle θ:

  • If x > 0 and y > 0: \theta = tan^{-1}\frac{y}{x};
  • If x < 0: \theta = \pi + tan^{-1}\frac{y}{x};
  • If x > 0 and y < 0: \theta = 2\pi + tan^{-1}\frac{y}{x};

Calculating:

a) (4,2,-4)

\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}} = 6

\phi = cos^{-1}(\frac{-4}{6})

\phi = cos^{-1}(\frac{-2}{3})

For θ, choose 1st option:

\theta = tan^{-1}(\frac{2}{4})

\theta = tan^{-1}(\frac{1}{2})

b) (0,8,15)

\rho = \sqrt{0^{2}+8^{2}+(15)^{2}} = 17

\phi = cos^{-1}(\frac{15}{17})

\theta = tan^{-1}\frac{y}{x}

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = \frac{\pi}{2}

c) (√2,1,1)

\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}} = 2

\phi = cos^{-1}(\frac{1}{2})

\phi = \frac{\pi}{3}

\theta = tan^{-1}\frac{1}{\sqrt{2} }

d) (−2√3,−2,3)

\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}} = 5

\phi = cos^{-1}(\frac{3}{5})

Since x < 0, use 2nd option:

\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }

\theta = \pi + \frac{\pi}{6}

\theta = \frac{7\pi}{6}

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

r=\sqrt{x^{2}+y^{2}}

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

r=\sqrt{4^{2}+2^{2}} = \sqrt{20}

\theta = tan^{-1}\frac{1}{2}

z = -4

b) (0, 8, 15)

r=\sqrt{0^{2}+8^{2}} = 8

\theta = \frac{\pi}{2}

z = 15

c) (√2,1,1)

r=\sqrt{(\sqrt{2} )^{2}+1^{2}} = \sqrt{3}

\theta = \frac{\pi}{3}

z = 1

d) (−2√3,−2,3)

r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}} = 4

\theta = \frac{7\pi}{6}

z = 3

5 0
3 years ago
division equation where 13 is quotient with the dividend a whole number and a unit fraction is the divisor
lys-0071 [83]

The division equation that has a quotient of 13 is 1/1/13 = 13

<h3>How to determine the equation?</h3>

The given parameters are:

  • Quotient = 13
  • Dividend = Whole number
  • Divisor = Unit fraction i.e. 1/n where n is an integer.

A division equation is represented as:

Dividend/Divisor = Quotient

Substitute 13 for the Quotient

Dividend/Divisor = 13

Recall that:

Unit fraction = 1/n

So, we have:

Dividend/1/n = 13

Let n = 13.

So, we have:

Dividend/1/13 = 13

This gives

13 * Dividend = 13

Divide both sides by 13

Dividend = 1

So, we have:

1/1/13 = 13

Hence, the division equation is 1/1/13 = 13

Read more about division equations at:

brainly.com/question/1622425

#SPJ1

3 0
2 years ago
8<br> 9<br> × i need help<br> 5<br> 8
agasfer [191]

The answer is going to be 5162.

6 0
3 years ago
Read 2 more answers
What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.    

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

  • the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,  

⇒ x⁴ - 81 = 0                      

⇒ (x² +  9)(x² - 9) = 0

⇒ (x² + 9)(x² - 9) = 0

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

#SPJ9

5 0
1 year ago
F
galina1969 [7]
It’s F because I took the test
8 0
3 years ago
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