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soldier1979 [14.2K]
2 years ago
10

What is the maximum number of times two planes can intersect? What is the minimum number of times they can intersect?

Mathematics
1 answer:
nordsb [41]2 years ago
8 0
The minimum number of times two planes can intersect is zero, because if the planes are parallel to each other, they will never intersect. we can take the example of floor and roof, both are parallel to each other and no intersection. If the two planes are in same plane the maximum number of times they can intersect is infinite for example if we consider a line we can intersect at infinite points.
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Solve 7x- c= k for x. <br>A. X = 7(k+C) <br>B. x = 7(K-C) <br>C. X = k+c/7 <br>D. X= k-c/7​
antoniya [11.8K]

Answer:

C

Step-by-step explanation:

you add c to both side which will then make it 7x=k+c then you would divide 7 from both sides leaving you with x=k+c/7. Except 7 will be under the k and c.

7 0
1 year ago
Which term of the AP. 8 , -4 , -16 , -28 ,......... is -880 ?​
denis-greek [22]

Answer:

75th term

Step-by-step explanation:

hope it is well understood

6 0
1 year ago
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Yves bought 420 tropical fish for a museum display. He bought 6 times as many parrotfish as angelfish. How many of each type of
ohaa [14]
The answer would be d , because you have to add how many of each fish to equal up to the sum of 420 . which is x y . so the first equation is x + y = 420 . then it says he bought 6 times as many parrotfish as he did angelfish so you would have to multiply x which represents angelfish by 6 . to get the second equation of y = 6x .
5 0
1 year ago
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A park has a 333 meter (\text{m})(m)left parenthesis, start text, m, end text, right parenthesis tall tether ball pole and a 6.8
dybincka [34]

Answer:option B & D are correct

Step-by-step explanation:

Given options are  

                                             A              B          C             D           E        

Tether ball pole shadow      1.35         1.8        3.75        0.6       2

Flagpole shadow                  3.4          4.08      8.35       1.36      4.8

Ratio of corresponding height & shadows  should be same

3 meter tall tether ball pole and a 6.8 tall flagpole

=> Ratio  = 6.8/3    = 68/30  = 34/15  

A    - 3.4/1.35 =   340/135     ≠ 34/15( 306/135)

B  -  4.08/1.8  = 408/180   =  34/15  = 34/15  ( Correct option)

C    8.35/3.75  = 835/375  =  167/75   ≠ 34/15( 170/75)

D  1.36/0.6  = 136/60   =  34/15  = 34/15   ( Correct option)

E  4.8/2  = 48/20   =  12/5   = 36/15    ≠ 34/15

option B & D are correct

6 0
1 year ago
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
1 year ago
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