What is the equivalent to 7.4 kilograms

1 answer:

6 0

7.4 kg = 74hg = 740dag = 7400g = 74,000 dg = 740,000 cg = 7,400,000 mg
hectograms dekagrams grams decigrams centigrams milligrams

Now which equivalent were you looking for?

You might be interested in

What is the area of the regular hexagon with a side length of 3 cm and an apothem of 2.6 cm? A. 15.6 cm2 B. 23.4 cm,2 C. 46.8 cm

yarga [219]

Regular hexagons have all equal sides and angles
=((3cmx2.6)/2) x 6
=( 7.8/2) x6
= 3.9 x 6 The answer is B. 23.4cm
= 23.4

7 0

3 years ago

Read 2 more answers

Ax + c = R (solve for x)

sp2606 [1]

Ax + c = R (subtract c from each side)

Ax + c - c = R - c

Ax = R - c (divide A from each side)

Ax/A = (R-c)/A

x = $\frac{R-c}{A}$ <-answer

4 0

3 years ago

Read 2 more answers

Find the mass of the solid paraboloid Dequals={(r,thetaθ,z): 0less than or equals≤zless than or equals≤8181minus−r2, 0less th

Lubov Fominskaja [6]

Answer:

M = 5742π

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

0 ≤ z ≤ 81 - r^2

0 ≤ r ≤ 9

Find:-

Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

$M = \int \int \int {p ( r ,theta, z)} \, dV\\\\$

- We can write the above expression in cylindrical coordinates:

$M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\$

- Perform integration:

$M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\$

$M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] } \, dtheta\\\\M = 5742\pi kg$