Very urgent. brainliest to first correct answer asap.

submarines control how much they float or sink in the ocean by changing the volume of air and water contained in large ballast tanks. when the tanks are completely full of water, the submarine increases its overall mass and sinks down to the bottom. when the tanks are completely fill of air, the submarine reduces its overall mass and floats to the surface. depending on the density if the seawater surrounding the submarine, it will pump seawater in or out of the tanks in order to achieve the same overall density as the sea water and float neutrally in the water. the volume of the submarine never changes. when the tanks are completely full of water, a submarine with a volume of 7.8 x 10^3 m^3 has a total mass of 8 x 10^6 kg. the density of the seawater is 10^3 kg/m^3. to make that submarine float neutrally, and neither float nor sink in the ocean, what volume of water does that submarine need to subtract from its tanks?

a. 8 x 10^6 m^3
b. 2 x 10^2 m^3
c. 8 x 10^3 m^3
d. 2 x 10^5 m^3

B. 2 x 10^2 m^3

Step-by-step explanation:

Given

Volume of submarine when full of water= 7.8 × 10³ m³

Total mass of submarine =8× 10⁶ kg

Density of sea water= 1000 kg/m³

Volume to be occupied by sea water at given mass of  submarine=?

Apply; Density=mass/volume ⇒⇒volume=mass/density

where M=mas, V=volume, D=density

Excess volume= 8×10³ - 7.8× 10³ = 2×10² m³

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐