Answer:
V = ∫∫∫rdrdθdz integrating from z = 2 to z = 4, r = 0 to √(16 - z²) and θ = 0 to 2π
Step-by-step explanation:
Since we have the radius of the sphere R = 4, we have R² = r² + z² where r = radius of cylinder in z-plane and z = height² of cylinder.
So, r = √(R² - z²)
r = √(4² - z²)
r = √(16 - z²)
Since the region is above the plane z = 2, we integrate z from z = 2 to z = R = 4
Our volume integral in cylindrical coordinates is thus
V = ∫∫∫rdrdθdz integrating from z = 2 to z = 4, r = 0 to √(16 - z²) and θ = 0 to 2π
Answer:
She would have saved 14 dollars more
Step-by-step explanation:
Hope this helps and plz mark brainliest :)
Answer:
option D is true.
Step-by-step explanation:
The right-angled triangle is shown.
From the right-angled triangle,
The angle Ф = 60°
We know that the trigonometric ratio
tan Ф = opposite / adjacent
Thus,
tan 60 = 4 / n
√3 = 4/n
n = 4/√3
Thus,
n = 4/√3
= (4 × √3) / (√3 × √3)
= 4√3 / 3
Thus,
n = 4√3 / 3
Using Pythagorean theorem
m = √n²+4²
Thus,
Therefore, option D is true.
If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
I used 3.14 for pi to solve easier.
6. 6.28
7. 21.98
8. 94.2
