To find an object’s density (D), first measure its mass (m) and volume (V). Then use the following equation:

Mathematics

1 answer:

Mariana [72]3 years ago

7 0

Answer:

C. 20 cm^3

Step-by-step explanation:

Using the equation D=m/v, you would plug in the given mass (227 g) into m and the given density of lead (11.35 g/cm^3). Solving for v=m/D, we find the answer to be 20 cm^3.

Also, you could cancel out all the answers except C because it is the only one in units of volume (1 cm^3 = 1 mL).

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Last Monday local mail carriers delivered 1,344 pieces of mail. Of all the deliveries they made, 1 / 6 were magazines, 1 / 12 we

Brut [27]

Answer: 1008 letters

Step-by-step explanation:

From the question, last Monday, a local mail carriers delivered 1,344 pieces of mail and of all the deliveries they made, 1 / 6 were magazines, 1 / 12 were packages, and the rest were letters. The fraction of letters will be:

= 1 - (1/6 + 1/12)

= 1 - (3/12)

= 9/12 = 3/4

We will now multiply 3/4 by 1344 to get the number of letters delivered.

= 3/4 × 1344

= 1008 letters

6 0

3 years ago

2 x 6/7 <br><br> SEND HELP!!!!!!!!!!!

Allushta [10]

Answer: C. 2 6/7

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8 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is