A technical machinist is asked to build a cubical steel tank that will hold 410 L of water. Calculate in meters the smallest possible inside length of the tank. Round your answer to the nearest .001 .

0.743 m

Step-by-step explanation:

Given that :

Volume of cubical tank = 410 L

Volume of a cube = L³

First we convert 410 L into meter cube

1 L = 10^-3 m³

1L = 0.001 m³

410 L = (410 * 0.001) m³

410 L = 0.410 m³

0.410 = l^3

Cube root of both sides

(0.410)^1/3 = l

0.7428958 = l

l = 0.7428958 m

Hence, smallest possible inside length of the cube would be : 0.743m

e) none of the above because the measure of angle cba is 101°

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}