Find the product of 28 and 4. ONLY type in the number.

Mathematics

1 answer:

Jet001 [13]3 years ago

8 0

Answer: 112

Step-by-step explanation:

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A piece of aluminium has a volume of 6.0cm^3 and a density of 2.70 g\cm^3 what is its mass????

8090 [49]

Density = Mass / volume is the original equation.

If we change it to solve for mass:

Mass = (Density)(volume)

Plug in the numbers:

M = (2.70 g/cm^3)(6.0cm^3) = 16.2 g

Hope that helps! :)

7 0

3 years ago

Please help:)<br><br> what is the mean height when looking at a histogram? <br> how do i find it?

kari74 [83]

Answer:

For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. Each of these products corresponds to the sum of all values falling within each bar. Summing all products gives us the total sum of all values, and dividing it by the number of observations yields the mean.

Step-by-step explanation:

I hope this helps! :)

5 0

3 years ago

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. Ifpo

steposvetlana [31]

The two curves $y=x$ and $y=x^{1/15}$ intersect at $x=0$ and $x=1$ , with $x^{1/15}\ge x$ over $0\le x\le1$ .

Washer method

$\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}$

Shell method

We have $y=x^{1/15}\implies x=y^{15}$ . The curves $x=y$ and $x=y^{15}$ intersect at $y=0$ and $y=1$ , with $y\ge y^{15}$ over $0\le y\le1$ .

$\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}$