3 years ago

What is the density of a block with a mass of 420 g and a length of 2cm, 3cm wide, and 1cm long

Mathematics

1 answer:

Basile [38]3 years ago

4 0

Answer:

Step-by-step explanation:

Find the volume first.

V= l * w * h

l = 2 cm

w = 3 cm

h = 1 cm

V = 2 * 3 * 1 = 6 cm^3

Density = mass / volume

Density = 420 grams / 6 cm^3

Density = 70 grams / cc^3

Kind of large. Check your givens.

You might be interested in

The market equilibrium point for a product is reached when 12000 units are produced and sold at $27 per unit. The manufacturer w

mamaluj [8]

If the equilibrium is such that only 12000 units are sold for $27, then the total earnings from the given scenario is $324,000. The supply equation would then be,
supply: 324000 = 6p ; p = 324000/6 = 54000
demand: 324000 = 69p ; p = 324000/69 = 4695.65 ≈ 4696

8 0

3 years ago

Help please i want help please

Rasek [7]

Don't listen but i'm gonna go with the third one

3 0

3 years ago

Solve for z w(-4+z)=Mz+17

bogdanovich [222]

Answer:

z = $\frac{17+4w}{w-M}$

Step-by-step explanation:

6 0

3 years ago

Will give 25 points and brainliest answer if correct

Norma-Jean [14]

Let $d$ be the common difference between terms in the sequence $\{a_n\}_{n\ge1}$ . Then

$a_{19}=-58$
$a_{20}=a_{19}+d$
$a_{21}=a_{19}+2d$
$\implies-164=-58+2d\implies d=-53$

The first term in the sequence, $a_1$ , satisfies

$a_2=a_1+d$
$a_3=a_1+2d$
$a_4=a_1+3d$
...
$a_{19}=a_1+18d$
$\implies-58=a_1+18(-53)\implies a_1=896$

So the explicit formula for the $n$ -th term in the sequence is

$a_n=a_1+(n-1)d\implies a_n=896-53(n-1)$

5 0

3 years ago

Which transformation is happening in the image below

s2008m [1.1K]

Answer:

B.) Reflection over y=-3

Step-by-step explanation:

8 0

3 years ago

What is the density of a block with a mass of 420 g and a length of 2cm, 3cm wide, and 1cm long​

What is the density of a block with a mass of 420 g and a length of 2cm, 3cm wide, and 1cm long