What is the mass, in grams, of 28.52 ml of acetone??

Veronika [31]

Answer:

$159

Step-by-step explanation:

All your doing is adding 45% and a percentage is out of 100 so to put it in decimal form it would be 0.45 added onto 159 which would give you $159.45 rounded to the nearest dollar would be $159

3 0

3 years ago

The sum of two numbers is 8, and their product is 15. What is the largest number?

AfilCa [17]

Hello there!

The sum of two numbers (x + y) is 8 (=8).
Their product (x * y) is 15 (=15).
x + y = 8
x * y = 15

First, we know that the two number must be positive and under 8, as the product is positive.

Let's list out some numbers that are from 1 to 8 that sum up to 8;
1 + 7
2 + 6
3 + 5
4 + 4

Now, we need to multiply the two numbers that add up to 8, and see which makes a product of 15.
1 * 7 = 7, this does not work.
2 * 6 = 12, this does not work.
3 * 5 = 15, this works!
4 * 4 = 16, this does not work.

Since we are looking for the larger number in our pairs, let's pick the largest one out of the correct pair, which is 5.

The largest number is 5.

I hope this helps!

3 0

3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$