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jok3333 [9.3K]
2 years ago
9

Need help asap, thanks!

Mathematics
1 answer:
il63 [147K]2 years ago
6 0

Answer:

b option

Step-by-step explanation:

Refer to the attachment.

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>

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a sample of unknown liquid has a volume of 12.0ml and a mass off 6g, what is its density? and how did you do the problem?
maw [93]

Answer:


Step-by-step explanation:

The equation for solving for density is:

Density = mass / volume

Then plug in the values:

Density = 6 g / 12 mL

Divide and don't forget the units!

Density = 0.5 g/mL


5 0
3 years ago
The skating rink rented 218 pairs of skates during the month of April and 3 times that many in May. How many pairs of skates did
Marrrta [24]

Answer: Total number of pairs =872

Step-by-step explanation:

Since we have given that

Number of pairs of  skates during the month of April =218

Number of pairs of skates during the month of May =3 × 218=654

As we know that total means altogether in which we perform the additive operation.

Total pairs of skates the skating rent during April and May = 218+654=872

∴ Total number of pairs =872

7 0
3 years ago
Read 2 more answers
What is the difference &amp; similarities of matter and mass ?
Tamiku [17]
 that differences are that mass is a measurement of the amount of matter something contains  and mass is measured by using a balance comparing a known amount of matter to an unknow of matter
5 0
3 years ago
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9.<br><br><br>How many times greater is the 6 in 6,275 than the 6 in 62.75?​
joja [24]

Answer:100

Step-by-step explanation:

3 0
2 years ago
(4) A spherical balloon is being inflated so that its diameter is increasing at a constant rate of 6 cm/min. How quickly is the
fredd [130]

In terms of its radius r, the volume of the balloon is

V(r)=\dfrac{4\pi}3r^3

The diameter d is twice the radius, so that in terms of its diameters, the balloon's volume is given by

V(d)=\dfrac{4\pi}3\left(\dfrac d2\right)^3=\dfrac\pi6d^3

Differentiate both sides with respect to time t:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi2d^2\dfrac{\mathrm dd}{\mathrm dt}

The diameter increases at a rate of \frac{\mathrm dd}{\mathrm dt}=6\frac{\rm cm}{\rm min}. When the diameter is d=50\,\mathrm{cm}, we have

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi2(50\,\mathrm{cm})^2\left(6\frac{\rm cm}{\rm min}\right)=7500\pi\dfrac{\mathrm{cm}^3}{\rm min}

or about 23,562 cc/min (where cc = cubic centimeters)

5 0
3 years ago
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