All categories

2 years ago

FRANK has a paper clip.it has a mass of 9g and a volume of 3cm3.what is its density?

Mathematics

1 answer:

Pachacha [2.7K]2 years ago

7 0

Answer:

Step-by-step explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{9}{3} = 3 \\$

We have the final answer as

Hope this helps you

You might be interested in

Pleasee help<br><br> 15 points

Nostrana [21]

Answer:

Step-by-step explanation:

4 0

2 years ago

A store sells 3 categories of kitchen items: cups, bowls, and spoons. There are 5 types of cups, 4 types of bowls, and 2 types o

Slav-nsk [51]

Answer:

40.

Step-by-step explanation:

That is 5 * 4 * 2

= 40 combinations of a set of bowl, cup and spoon.

6 0

3 years ago

Please I NEED HELP HURRY The quotient of six and a number subtracted from 100 Question A 100 - 6/x Question B 6/x - 100 Question

jasenka [17]

For this case we have that the quotient of 6 and a number, can be expressed as:

$\frac {6} {x}$

Where the variable "x" represents the incognito number.

Now we have that expression is subtracted from 100. Now, we can write the following:

$100- \frac {6} {x}$

ANswer:

Option A

4 0

3 years ago

Answer a and b ASAP

melomori [17]

Answer:

The area of the rectangle is 48cm2

6 0

2 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago