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Readme [11.4K]
1 year ago
15

LABEL THE CIRCLE BY FILLING THE BOXES WITH THE APPROPRIATE PARTS OF THW CIRCLE.​

Mathematics
1 answer:
N76 [4]1 year ago
4 0

The 1- diameter, 2-inscribed angle, 3-chord, 4-minor arc, 5-central angle, 6-radius, and 7-centre.

<h3>What is a circle?</h3>

It is described as a set of points, where each point is at the same distance from a fixed point (called the centre of a circle)

The radius is the double the diameter.

The longest chord in the circle is diameter.

The angle formed at the centre is known as centeral angle.

Thus, the 1- diameter, 2-inscribed angle, 3-chord, 4-minor arc, 5-central angle, 6-radius, and 7-centre.

Learn more about circle here:

brainly.com/question/11833983

#SPJ1

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Discuss on piazza but do not submit. Divide [-2,2] into (a) 10 subintervals (b) 1000 subintervals of the same length. Then use t
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Answer:

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Step-by-step explanation:

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How do you factor 5x^2b^2 + 14xb^2 -3b^2
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Answer:

5b^2(x-0.2)(x+3)

Step-by-step explanation:

The first thing that I noticed was that all of the terms had a common factor of b^2. You can therefore factor that out first:

5x^2b^2 + 14xb^2 -3b^2= \\\\b^2(5x^2+14x-3)

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Hope this helps!

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8) The total weight of 15 boxes is 45 pounds. How much would 40 boxes weigh?​
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Step-by-step explanation:

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Read 2 more answers
The figure below has a perimeter of 37 feet what is the length in feet of the unknown side
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What is the figure below?
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One positive number is one-fifth of another number. the difference of the two numbers is 84. find the numbers.
seraphim [82]
Answer:  The numbers are:  " 21 " and " 105 " .
___________________________________________________
Explanation:
___________________________________________________
Let "x" be the "one positive number:

Let "y" be the "[an]othyer number".

x = 1/5 (y)
___________________________________________________
Given that the difference of the two number is "84" ;  and that "x" is (1/5) of  "y" ;  we determine that "x" is smaller than "y".

So, y − x = 84 .

Add "x" to each side of this equation; to solve for "y" in terms of "x" ;

y − x + x = 84 + x  ;

 y = 84 + x ;
___________________________________________________
So, we have: 

 x = (1/5) y ;

and:  y = 84 + x  ;

Substitute "(1/5)y" for "x" ;  in  "y = 84 + x " ;  to solve for "y" ;

 y = 84 + [ (1/5)y ]

Subtract  " [ (1/5)y ] " from EACH SIDE of the equation ;

y − [ (1/5)y ] = 84 + [ (1/5)y ] −  [ (1/5)y ]  ;

to get:

  [ (4/5)y ] = 84 ;


       ↔    (4y) / 5 = 84  ;
      
        →  4y = 5 * 84  ;

      Divide EACH SIDE of the equation by "4" ; 
to isolate "y" on one side of the equation; and to solve for "y" ;

           4y / 4 = (5 * 84) / 4 ;

                 y =  5 * (84/4) = 5 * 21 = 105 .

   y = 105 .
___________________________________________________
Now, plug "105" for "y" into:
___________________________________________________
Either:
___________________________________________________
 x = (1/5) y ;

OR:

  y = 84 + x  ;
___________________________________________________
to solve for "x" ;
___________________________________________________
Let us do so in BOTH equations; to see if we get the same value for "x" ; which is a method to "double check" our answer ;
___________________________________________________
Start with:

x = (1/5)y 

    →  (1/5)*(105) = 105 / 5 = 21 ;  x = 21 ; 

___________________________________________________
So, x = 21;  y = 105 .
___________________________________________________
Now, let us see if this values hold true in the other equation:
___________________________________________________
y = 84 + x ;

105 = ? 84 + 21 ?
 
105 = ? 105 ? Yes!
___________________________________________________
The numbers are:  " 21 " and  "105 " .
___________________________________________________

6 0
3 years ago
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