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2 years ago

How to find the length of a major arc

1 answer:

8 0

To find arc length, start by dividing the arc's central angle in degrees by 360. Then, multiply that number by the radius of the circle. Finally, multiply that number by 2 × pi to find the arc length.

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Gabriel has 24 balls , his friend gave him the half of balls that he has , whats the half of 24

JulijaS [17]

The answer would be 12

7 0

2 years ago

Read 2 more answers

Factorise 12 raised 2 minus 7x + 1

professor190 [17]

Answer:

(4x - 1)(3x - 1).

Step-by-step explanation:

I think you mean 12x^2 - 7x + 1

= 12x^2 - 4x - 3x + 1

= 4x(3x - 1) -- 1(3x - 1)

= (4x - 1)(3x - 1).

6 0

2 years ago

How do I get the answer

Ronch [10]

Answer:

Step-by-step explanation:

$- |3a + \sqrt{b} |$

By putting the values of a = -4 , b = 9,

$= - |3 \times ( -4) + \sqrt{9} |$

$= - | - 12 + 3|$

$= - | - 9|$

= -(9)

= -9 (Ans)

4 0

2 years ago

2/3 (x - 7 ) = -2 AL y = 5/3 B: 8 1/3 C: 10 D: 4

nika2105 [10]

Answer:

D: 4

Step-by-step explanation:

1. Solve the equation

$\begin{array}{rcl}\dfrac{2}{3}(x - 7) & = & -2\\\\2(x - 7) & = & -6\\x - 7 & = & -3\\x & = & \mathbf{4}\\\end{array}$

2. Check

$\begin{array}{rcl}\dfrac{2}{3}(4 - 7) & = & -2\\\\\dfrac{2}{3}(-3) & = & -2\\\\-2 & = & -2\\\end{array}$

8 0

2 years ago

We select n + 1 diﬀerent integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte

just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3 → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 → N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ → N₃ < N₂

therefore if we continue , we would obtain a number even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

6 0

2 years ago