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3 years ago

What is 8% of 101.49

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

8.1192

you need to multiply the decimal to the grand number:

0.08 x 101.49 = 8.1192

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If (-1, y) lies on the graph of y=2^2x then y=

zubka84 [21]

Answer:

y is 1/4 when x is -1 if I did interpret your equation right. Please see that I did.

I believe the equation to be $y=2^{2x}$ . If you meant $y=2^2x$ , please let me know.

Thanks kindly.

Step-by-step explanation:

If (-1,y) lies on the graph of $y=2^{2x}$ , then by substitution we have:

$y=2^{2\cdot -1}$ .

We just need to simplify to determine y:

$y=2^{2 \cdot -1}$

Multiply the 2 and -1:

$y=2^{-2}$

Use reciprocal rule for exponents to get rid of the negative:

$y=\frac{1}{2^2}$

$2^2$ just means $2(2)$ :

$y=\frac{1}{2(2)}$

$y=\frac{1}{4}$ .

8 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

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