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Mamont248 [21]
3 years ago
13

A circle has a circumference of 450cm

Mathematics
2 answers:
Nastasia [14]3 years ago
7 0

Answer:

288 degrees at the center of the circle

Step-by-step explanation:

Subtends is opposite

90/450=0.2  so the degree of the 90 cm arc is (0.2)(360)=72 degree arc.

It's subtend would be 360-72=288 degree arc.

This is because it is also the major arc of the central angle.

anzhelika [568]3 years ago
3 0

The circumference of the circle is actually the perimeter ( length of the boundary ) of the circle . And a part of the circle which lies between two distinct points on the circumference of the circle is called an arc . If the length of the arc is less than half the circumference , it is called minor arc and remaining portion which is more than half of the circle ( but natural ) is called major arc .

When these two points , which make the arc are joined separately to the centre of circle , these arms make angle at the centre . This is called the angle subtended by the arc at the centre of the circle .

There is a beautiful logical relation exists between arc length and the angle , the arc makes ( subtends ) at the centre of the circle . This relation is as under , the wholle circle subtends an angle of 360 degree at the centre . Half the circumference subtendr 360 / 2 ie 180 degree at the centre . The logical relation becomes Arc Length = Circumference × angle in degrees it ( the arc ) subtends at the centre of the circle / 360 degree . So the answer is very simple :- The Arc Length = 36 × 90 / 360 or 9 units ( may be centimetres or metres or inches , feet , yards , etc ) . Which is definitely length of the minor arc . The length of the major arc ( remaining portion of the circumstance ) is 36 - 9 = 27 units . Hence the required answer of the sum is 9 units .

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The survey shows 300 teens that were surveyed, from ages 15-18, what their first choice would be for an elective in high school.
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79

Step-by-step explanation:

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7 foot is the answer...
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3 years ago
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olga_2 [115]

Answer:

1. Yes

∆RST ~ ∆WSX

by SAS

2. Yes

∆ABC ~ ∆PQR

by SSS

3. Yes

∆STU ~ ∆JPM

by SAS

4. Yes

∆DJK ~ ∆PZR

by SAS

5. Yes

∆RTU ~ ∆STL

by SAS

5. Yes

∆JKL ~ ∆XYW

by SAS

6. No

7. Yes

∆BEF ~ ∆NML

by SAS

8. Yes

∆GHI ~ ∆QRS

by SSS

9. x=22

10. x=12

Step-by-step explanation:

1. RS/WS=ST/SX and m<RST=m<WSX

2. AB/PQ=8/6=4/3

BC/QR=AC/PR=12/9=4/3

AB/PQ=BC/QR=AC/PR

3. ST/JP=10/15=2/3

SU/JM=14/21=2/3

ST/JP=2/3=SU/JM

and m<TSU=70°=m<PJM

4. DK/PR=8/4=2

JK/ZR=18/9=2

DK/PR=2=JK/ZR

and m<DKJ=65°=m<PRZ

5. RT/ST=UT/LT

and m<RTU=m<STL

6. KL/YW=20/18=10/9

JL/XW=36/24=3/2

KL/YW=10/9≠3/2=JL/XW

7. BF/NL=24/16=3/2

BE/NM=39/26=3/2

BF/NL=3/2=BE/NM

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8. GH/QR=32/20=8/5

HI/RS=40/25=8/5

GI/QS=24/15=8/5

GH/QR=HI/RS=GI/QS=8/5

9. x/33=18/27

Simplifying the fraction on the right side of the equation:

x/33=2/3

Solving for x: Multiplying both sides of the equation by 33:

33(x/33)=33(2/3)

x=11(2)

x=22

10. x/16=9/12

Simplifying the fraction on the right side of the equation:

x/16=3/4

Solving for x: Multiplying both sides of the equation by 16:

16(x/16)=16(3/4)

x=4(3)

x=12

4 0
3 years ago
5. Class 7887 has 2:7 math classes daily. How many math classes do they have
ANTONII [103]

Answer:

56.7

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5 0
3 years ago
Work out the formula of the nth term in the following quadratic sequence:<br> 19, 15, 9, 1...
Butoxors [25]

In a quadratic sequence we'll get a linear first difference and a constant second difference.  Let's verify that.

n          1   2   3   4

f(n)       19  15  9   1

1st diff   -4  -6  -8

2nd diff     2   2

We see that we got a constant second difference.  We could just extend that and work back up to get more values.

n          1   2   3   4       5    6      7

f(n)       19  15  9   1      -9   -21   -35

1st diff   -4  -6  -8   -10  -12   -14

2nd diff     2   2    2     2    2

That's just an aside; we're after the general formula.  We have

f(1)=19, f(2)=15, f(3)=9

In general we can assume

f(n) = an²  + bn + c

We get three equations in three unknowns,

19 = a(1²)+b(1)+c = a+b+c

15 = a(2²) + b(2) + c = 4a + 2b + c

9 = a(3²) + b(3) + c =  9a + 3b + c

That's a 3x3 linear system; it's easy to solve directly.  Subtracting pairs,

4 = -3a - b

6 = -5a - b

Subtracting those,

-2 = 2a

a = -1

b = -3a -4 = -1

c = 19-a-b = 21

Answer: f(n) = -n² - n + 21

Check:

f(1) = -1 - 1 + 21 = 19, good

f(2) = -4 - 2 + 21 = 15, good

f(3) = -9 - 3 + 21 = 9, good

f(4) = -16 - 4 + 21 = 1, good

Let's check our extended table, how about

f(7)= -49 - 7 + 21 = -35, good

6 0
3 years ago
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