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Kitty [74]
3 years ago
11

Find the recurrence relation satisfied by rn, where rn is the number of regions that a plane is divided into by n lines, if no t

wo of the lines are parallel and no three of the lines go through the same point.

Mathematics
1 answer:
joja [24]3 years ago
5 0
1. Take a look at the pictures attached.

2. 

i) the first line divides the plane into 2 regions 

ii) the second line adds 2 more regions so we have 4 in total.

iii) the third line adds 3 more regions, so 4+3=7 regions

iv) the fourth line adds 4 more regions.

so the n^{th} line adds n more regions to the ones created by the previous n-1 lines.

3. 

r_1=2

r_2=2+2=4

r_3=4+3=7

r_4=7+4=11

r_n=r_n_-_1+n

So the recurrence relation is

r_1=2
r_n=r_n_-_1+n

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85

Step-by-step explanation:

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PLSSS HELP ME WITH MATH
labwork [276]

Step-by-step explanation:

always remember, equations and inequalities are like a scale with 2 pans (left and right).

a variable is like a placeholder, an empty spot on the pans. and the expressions on both sides are kind of the conditions for the things to put in these empty spots so that the status of the scale stays unchanged.

for an equation the scale has to stay balanced all the time.

and in an inequality one side is heavier that the other (it normally does not matter how much heavier) with "<" or ">" signs, or the scale can be also balanced with "<=" or ">=" signs. just the other side must never be heavier.

A

so, in our case here

2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = 4. no other value of x allows the scale to be balanced.

2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = 4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. 5 ?

2×5 - 5 >= 3

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = 5 is also a valid solution.

what else ?

let's simplify the inequality

2x - 5 >= 3

2x >= 8

x >= 4

so, really every value of x that is greater or equal to 4 is a valid solution for the inequality.

B

-2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = -4. no other value of x allows the scale to be balanced.

-2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = -4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. -5 ?

-2×-5 - 5 >= 3

remember, -×- = +

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = -5 is also a valid solution.

what else ?

let's simplify the inequality

-2x - 5 >= 3

-2x >= 8

x <= -4

a multiplication or division by a negative value flips the inequality sign, because such an operation makes a light weight heavy and a heavy weight light.

so, really every value of x that is less or equal to -4 is a valid solution for the inequality.

3 0
2 years ago
At a yearly basketball tournament, 64 different teams compete. After each round of the tournament, one-fourth of the teams remai
artcher [175]

Answer:

64*(1/4)^x

Step-by-step explanation:

64 is the starting number because it is given, and x represents how many rounds there are. 1/4 represents how much of the teams remain after each round.

5 0
3 years ago
A-b
Lynna [10]

Answer:

\boxed{\sf d. \ \frac{a}{b}  =  \frac{10}{7}}

Step-by-step explanation:

\sf \implies \dfrac{a - b}{b}  =  \dfrac{3}{7}  \\  \\  \sf \implies  \frac{a}{b}  -  \frac{b}{b}  =  \frac{3}{7}  \\  \\  \sf \implies  \frac{a}{b}  - 1 =  \frac{3}{7}  \\  \\  \sf \implies  \frac{a}{b}  - 1 + 1 =  \frac{3}{7}  + 1 \\  \\  \sf \implies  \frac{a}{b}  =  \frac{3}{7}  + 1 \\  \\  \sf \implies  \frac{a}{b}  =  \frac{3}{7}  +  \frac{7}{7}  \\  \\  \sf \implies  \frac{a}{b}  =  \frac{3 + 7}{7}  \\  \\  \sf \implies  \frac{a}{b}  =  \frac{10}{7}

6 0
3 years ago
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