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3 years ago

Angles 1 and 2 are called?

Mathematics

2 answers:

Slav-nsk [51]3 years ago

8 0

3 main angles: Right; a right angle is 90 degrees, Obtuse; blunt more then 90 and less then 180, Acute: Less then 90.

Nitella [24]3 years ago

7 0

Angles larger than a right angle and smaller than a straight angle (between 90° and 180°) are called obtuse angles ("obtuse" meaning "blunt"). An angle equal to 12 turn (180° or π radians) is called a straight angle.

You might be interested in

Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder X^2+y^2=1

hichkok12 [17]

Answer:

$\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}$

Step-by-step explanation:

Given that:

The surface area (S.A) $z = x^2 +y^2$

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

$A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA$

where :

D = cylinder $x^2 +y^2 =1$

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, $\dfrac{\partial z}{\partial x} = 2x$ and $\dfrac{\partial z}{\partial y} = 2y$

Therefore;

$S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA$

$= \iint_D \sqrt{1+4(x^2+y^2)} \ dA$

$= \int^{2 \pi}_{0} \int^{1}_{0} \sqrt{1+4r^2} \ r \ dr \d \theta$

$= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0$

$= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]$

$\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}$

6 0

3 years ago

Find the domain and range of each relation. Then determine if the relation is a function. D={-2, -1, 3, 5}, R={-7 ,0 ,4} Functio

poizon [28]

Answer:

$D = \{-2,-1,3,5\}$ -- Domain

$R = \{-7,0,4\}$ -- Range

It is a function

Step-by-step explanation:

Given

$D = \{-2,-1,3,5\}$

$R = \{-7,0,4\}$

Required

State the domain and range

Determine if the relation is a function

From the question, we have the domain and range as:

$D = \{-2,-1,3,5\}$ -- Domain

$R = \{-7,0,4\}$ -- Range

Next, is to determine if the relation is a function or not.

Yes, it is a function.

The number of elements in the domain is 4

The number of elements in the range is 3

When the domain has more elements than the range, this is called a many-to-one function, and it is a valid type of function.

5 0

3 years ago

A quadilateral ABCD is drawn to circumcribe a circle . prove that AB+CD=AD+BC

lora16 [44]

Answer:

Helllllp

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Factor this problem ? 125 +27y^3

koban [17]

Answer:

(5 + 3y)(25 - 15y + 9y²)

Step-by-step explanation:

This is a sum of cubes and factors in general as

a³ + b³ = (a + b)(a² - ab + b²), thus

125 + 27y³

= 5³ + (3y)³ with a = 5 and b = 3y

= (5 + 3y)(5² - 5(3y) + (3y)² )

= (5 + 3y)(25 - 15y + 9y²)

6 0

3 years ago

Which two tables represent the same function?

topjm [15]

Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

(x2 , y2) is $\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ $\frac{y-8}{x-4}=\frac{7-8}{6-4}$

∴ $\frac{y-8}{x-4}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ $\frac{y-5}{x-4}=\frac{4-5}{6-4}$

∴ $\frac{y-5}{x-4}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ $\frac{y-8}{x-2}=\frac{5-8}{8-2}$

∴ $\frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

∴ $\frac{y-10}{x-2}=\frac{14-10}{6-2}$

∴ $\frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1$

- By using cross multiplication

∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

∴ -8 = x - y ⇒ switch the two sides

∴ x - y = -8 ⇒ (4)

# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)

∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ $\frac{y-9}{x-2}=\frac{6-9}{8-2}$

∴ $\frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 9) = -1(x - 2) ⇒ simplify

∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

∴ The 1st and the 5th tables represent the same function

3 0

3 years ago