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IgorLugansk [536]
2 years ago
13

Square OABC is drawn on a centimetre grid. O is (0,0) A is (2,0) B is (2,2) C is (0,2) Write down how many invariant points ther

e are on the perimeter of the square when OABC is rotated 90 degrees clockwise, centre (2,0).

Mathematics
1 answer:
Nina [5.8K]2 years ago
7 0

Answer:

One invariant point;

Point A = (2, 0)

Step-by-step explanation:

The coordinates of the square vertices are;

O = (0, 0)

A = (2, 0)

B = (2, 2)

C = (0, 2)

Therefore, we have by 90° clockwise rotation;

O' = (2, 2)

A' = (2, 0)

B' = (4, 0)

C' = (2, 4)

Therefore, since only A' (2, 0) = A (2, 0), we have only one invariant point on the perimeter of the square when it is rotated 90° about the center (2, 0) which is the point A = (2, 0).

You might be interested in
How many positive integers between 1000 and 9999 inclusive
bekas [8.4K]

Answer:

Step-by-step explanation:

a.

first number is  1000-1+9=1008

9)1000(1

    9

-------

     10

       9

    -----

       10

         9

       ----

         1

       ----

last number is 9999

9| 9999

  ---------

    1111 |0

    --------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

  ---------

  |4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total  numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

  ---------

  |  142-6

1000-6+7=1001

7 | 9999

  |---------

    1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

        70

       ----

        300

        280

        ------

           20

           ---

1000-20+35=1015

35)9999(285

     70

    ----

     299

     280

     -----

        199

        175

        ----

          24

         ----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

7 0
3 years ago
Find the distance from P to l. Line l contains points (-4,2) and (3,-5).Point P has coordinates (1,2).
Annette [7]
Hello,

A=(-4,2)
B=(3,-5)
AB≡y-2=(x+4)(-7)/7==>y=-x-2

Slope of the perpendicular: 1

y-2=(x-1)1==>y=x+1 is the equation of the perpendicular.

Intersection: y=x+1 and y=-x-2 ==>(-3/2,-1/2)=Q

|PQ|²=(-1/2-2)²+(1+3/2)²=25/2
|PQ|=5√2/2≈3.5355339059....



7 0
3 years ago
Write an equation of a line parallel to y=1/2x+3 that passes through (2,5)
schepotkina [342]
Y= 1/2x +4
hopes this helps!
7 0
3 years ago
The radius of a circle is 9 M what is the circle's circumference use 3.14 for pi​
MissTica

Answer:

56.52 is the circumference. If the radias is 9 and the diameter is then 18.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
The function h(x)=x^2+3 and g(x)=x^2-6. if g(x)=h(x)+k, what us the value of k
lutik1710 [3]

Answer:

The value of k is -9

Step-by-step explanation:

Subsitute

g(x)=h(x)+k

(x^2-6)=(x^2+3)+k

x^2-6=x^2+3+k

subtract x^2+3 from both sides (or add -x^2-3 to both sides)

-9=k

k=-9

3 0
2 years ago
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