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adelina 88 [10]
3 years ago
13

Which angles are neither obtuse angles nor acute angles?

Mathematics
2 answers:
mr Goodwill [35]3 years ago
8 0

Answer:

Acute angles are less than 90

obtuse is more than 90

right angles are exactly 90

Step-by-step explanation:

mylen [45]3 years ago
4 0

Answer:

First option: 90 degrees

Step-by-step explanation:

Obtuse is greater than 90.

Acute is less than 90.

Right is 90.

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Give the properties for the equation x 2 - 3 y 2 - 8x + 12y + 16 = 0 Center (4, -4) (4, 2) (4, -2)
Sergio [31]
Start by completing the square.

x^2-8x-3y^2+12y=-16
x^2-8x+16-3(y^2-4y+4)=-16+16-12=-12

Notice the addition of those constants allows us to factor these expressions and because we did it to both sides it’s completely legit.

(x-4)^2-3(y-2)^2=-12

However, it seems we get a negative radius, which is not allowed. I think you entered the problem incorrectly.
6 0
3 years ago
I’ll mark brainliest
Naddik [55]

Answer:

A

Step-by-step explanation:

ok so pretend the 2.00 was just 2 so whats 8 divided by 2? 4 right so if we test that witht he other problems its the same so its A

Hope this helps!!!!     :D

4 0
3 years ago
Does anyone know how to do this ?
AysviL [449]

9514 1404 393

Answer:

  y = 21

Step-by-step explanation:

In straightforward fashion, we can find the midpoints of the two segments, and set the distance between them equal to 10.

  M = midpoint of AB

  M = (A+B)/2 = ((-1, 0) +(-3, 6))/2 = (-4, 6)/2 = (-2, 3)

  N = midpoint of CD

  N = (C+D)/2 = ((3, 1) +(5, y))/2 = (8, 1+y)/2 = (4, (1+y)/2)

Then the distance MN is given by the distance formula.

  |N-M| = 10 = √((4 -(-2))² +((1+y)/2 -3)²) . . . . . distance formula*

  100 = 36 +((1+y)/2 -3)² . . . . . square both sides

  64 = ((1 +y)/2 -3)² . . . . . . . . . subtract 36

  8 = (1 +y)/2 -3 . . . . . . . . . . . . (positive) square root

  11 = 1+y/2 . . . . . . . . add 3

  22 = 1+y . . . . . . . . . multiply by 2

  21 = y . . . . . . . . . . . subtract 1

_____

* The distance formula is an application of the Pythagorean theorem. Basically, the differences of x- and y-coordinates are taken to be the legs of a right triangle. They length of the hypotenuse is the distance between the two points.

  d² = (x2 -x1)² +(y2 -y1)² . . . . . . . from the Pythagorean theorem

  d = √((x2 -x1)² +(y2 -y1)²) . . . . . the usual distance formula

8 0
3 years ago
Select the correct answer.<br> Which function has the same graph as x + y = 11?
djyliett [7]

The function that have the same graph as x + y = 11 is the function 3x + 3y = 33

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depends on other variable while a dependent variable is a variable that depends on other variable.

Given the function, x + y = 11. To get another function with the same graph, multiply by a constant:

(x + y) * 3 = 11 * 3

3x + 3y = 33

The function that have the same graph as x + y = 11 is the function 3x + 3y = 33

Find out more on equation at: brainly.com/question/2972832

#SPJ1

3 0
1 year ago
A boy is playing a ball in a garden surrounded by a wall 2.5 m high and kicks the ball vertically up from a height of 0.4 m with
kow [346]

Answer:

2.5 sec

Step-by-step explanation:

Height of wall = 2.5 m

initial speed of ball = 14 m/s

height from which ball is kicked = 0.4 m

we calculate the speed of the ball at the height that matches the wall first

height that matches wall = 2.5 - 0.4 = 2.1 m

using  =  + 2as

where a = acceleration due to gravity = -9.81 m/s^2 (negative in upwards movement)

=  + 2(-9.81 x 2.1)

= 196 - 41.202

= 154.8

v =  = 12.44 m/s

this is the velocity of the ball at exactly the point where the wall ends.

At the maximum height, the speed of the ball becomes zero

therefore,

u = 12.44 m/s

v = 0 m/s

a = -9.81 m/s^2

t = ?

using V = U + at

0 = 12.44 - 9.81t

-12.44 = -9.81

t = -12.44/-9.81

t = 1.27 s

the maximum height the ball reaches will be gotten with

=  + 2as

a = -9.81 m/s^2

0 =  + 2(-9.81s)

0 = 196 - 19.62s

s = -196/-19.62 = 9.99 m. This the maximum height reached by the ball.

height from maximum height to height of ball = 9.99 - 2.5 = 7.49 m

we calculate for the time taken for the ball to travel down this height

a = 9.81 m/s^2 (positive in downwards movement)

u = 0

s = 7.49 m

using s = ut + a

7.49 = (0 x t) +  (9.81 x  )

7.49 = 0 + 4.9

 = 7.49/4.9 = 1.53

t =  = 1.23 sec

Total time spent above wall = 1.27 s + 1.23 s = 2.5 sec

3 0
3 years ago
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