Shenhed finished his math assament in 3/8

1 answer:

4 0

Math assignment = 3/8 hours
History assignment = 1/4 hours

We can not compare this as both fractions do not have alike denominators.

We can fix this problem easily by turning 3/8 and 1/4 into alike denominator fractions.

We can turn 1/4 into 2/8 as it is equivalent fractions.

1 x 2 = 2
4 x 2 = 8

(Whatever you do to the denominator, you have to do the same to the numerator)

Once we get this, we can see that Shen spent 2/8 hours on his homework while he spent 3/8 hours on his math assignment.

3/8 - 2/8 = 1/8

1/8 = 0.125

Shen spent 1/8 of an hour (or 0.125 of an hour) more on his math assignment than his history assignment.

You might be interested in

James wants to invest $12,000. He can invest the money at 7.2% simple interest for 30 yr or he can invest at 6.8% with interest

Sever21 [200]

Over time, compound interest at any rate will outperform simple interest. When the rates are nearly equal to start with, compound interest will be greater in very short order. Here, it takes less than 1 year for compound interest to give a larger account balance.

In 30 years, the simple interest will be

... I = P·r·t = 12,000·0.07·30 = 25,200

In 30 years, the compound interest will be

... I = P·(e^(rt) -1) = 12,000·(e^(.068·30) -1) ≈ 80,287.31

_____

6.8% compounded continuously results in more total interest

4 0

2 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$