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4 years ago

Solve (t - 3)2 = 6. The arrow is at a height of 48 ft after approximately _ sand after _s

Mathematics

2 answers:

Karo-lina-s [1.5K]4 years ago

4 0

Answer: 0.55 & 5.45

Step-by-step explanation:

Slav-nsk [51]4 years ago

3 0

Answer:

The arrow is at a height of 48 ft after approximately 0.55 s and after 5.45 s

Step-by-step explanation:

The following information is missing:

The height of an arrow shot upward can be given by the formula s = v0*t - 16*t², where v0 is the initial velocity and t is time.How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/s?

If the arrow is at a height of 48 ft and its initial velocity is 96 ft/s, then:

48 = 96*t - 16*t²

16*t² - 96*t + 48 = 0

16*(t² - 6*t + 3) = 0

t² - 6*t + 3 = 0

t² - 6*t + 3 + 6 = 0 + 6

t² - 6*t + 9 = 6

(t - 3)² = 6

t - 3 = √6

t - 3 = 2.45; t = 2.45 + 3; t = 5.45

t - 3 = -2.45; t = -2.45 + 3; t = 0.55

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The denominator of a rational number is greater than its numerator by 5. If 5 is subtracted from the numerator and 4 is added to

nasty-shy [4]

Answer:

7/12.

Step-by-step explanation:

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x-5=⅛x+1⅛

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thus the initial fraction is x/x+5= 7/12

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3 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

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