Aspacecraft can attain a stable orbit 300 kilometers above earth if it reaches a velocity of 7.7 kilometers per second. the formula for a rocket's maximum velocity v in kilometers per second is vequalsminus0.0098tplusc ln upper r, where t is the firing time in seconds, c is the velocity of the exhaust in kilometers per second, and r is the ratio of the mass of the rocket filled with fuel to the mass of the rocket without fuel. find the velocity of a spacecraft whose booster rocket has a mass ratio of 20, an exhaust velocity of 2.1 km/s, and a firing time of 15 s. can the spacecraft achieve a stable orbit 300 km above earth?

18.8 cubic inches.

step-by-step explanation:

volume of a cone = πr²h

r = 1.5 inches

h = 8 inches

pi = 3.14

volume of our cone = × 3.14 × 1.5² × 8 = 18.84 square inches

or equals to 18.8 square inches rounded up to nearest tenth.

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)