Answer:
Im thinking 12 im so sorry if its wrong!
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
Permutation :
nPr = n! ÷ (n - r)!
nCr = n! ÷ (n-r)!r!
11 objects taken 3 at a time :
11P3 = 11! ÷ (11 - 3)!
11P3 = 11! / 8!
11P3 = (11 * 10 * 9 )
11P3 = 990
Dividing by 3!
3! = 3*2*1 = 6
= 990 / 6
= 165
For combination :
11C3 = 11! ÷ 8!3!
11C3 = (11 * 10 * 9) / (3 * 2 * 1)
11C3 = 990 / 6
11C3
Answer:
∠ 1,3,5,7 = 32°
∠2,4,6,8,= 148°
Step-by-step explanation:
From the figure attached,
AB and CD are two parallel lines and another transverse line is intersecting these line at two distinct points.
Since, m∠1 = 32°,
∠1 and ∠4 are supplementary angles [Linear pair of angles]
m∠1 + m∠4 = 180°
32° + m∠4 = 180°
m∠4 = 180° - 32°
m∠4 = 148°
therefore,
m∠1,3,5,7 = 32°
m∠2,4,6,8,= 148°
You just have to reverse the operations
1. -48-8= -56
2. -56/3= -18.66
ANSWER = -18.66
Answer:
The polar coordinates are as follow:
a. (6,2π)
b. (18, π/3)
c. (2√2 , 3π/4)
d. (2, 5π /6)
Step-by-step explanation:
To convert the rectangular coordinates into polar coordinates, we need to calculate r, θ .
To calculate r, we use Pythagorean theorem:
r =
---- (1)
To calculate the θ, first we will find out the θ
' using the inverse of cosine as it is easy to calculate.
So, θ
' =
cos
⁻¹ (x/r)
If y ≥ 0 then θ = ∅
If y < 0 then θ = 2
π − ∅
For a. (6,0)
Sol:
Using the formula in equation (1). we get the value of r as:
r = 
r = 6
And ∅ =
cos
⁻¹ (x/r)
∅ =
cos
⁻¹ (6/6)
∅ =cos
⁻¹ (1) = 2π
As If y ≥ 0 then θ = ∅
So ∅ = 2π
The polar coordinates are (6,2π)
For a. (9,9/
)
Sol:
r = 9 + 3(3) = 18
and ∅ =
cos
⁻¹ (x/r)
∅ =
cos
⁻¹ (9/18)
∅ = cos
⁻¹ (1/2) = π/3
As If y ≥ 0 then θ = ∅
then θ = π/3
The polar coordinates are (18, π/3)
For (-2,2)
Sol:
r =√( (-2)²+(2)² )
r = 2 √2
and ∅ =
cos
⁻¹ (x/r)
∅ =
cos
⁻¹ (-2/ 2 √2)
∅ = 3π/4
As If y ≥ 0 then θ = ∅
then
θ = 3π/4
The polar coordinates are (2√2 , 3π/4)
For (-√3, 1)
Sol:
r = √ ((-√3)² + 1²)
r = 2
and ∅ =
cos
⁻¹ (x/r)
∅ =
cos
⁻¹ ( -√3/2)
∅ = 5π /6
As If y ≥ 0 then θ = ∅
So θ = 5π /6
The polar coordinates are (2, 5π /6)