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2 years ago

[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = <-2, 12, -3> and w = <-7, 4, -6>

1 answer:

6 0

$\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k$
$-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k$

$\mathbf w=\langle-7,4,-6\rangle=-7\,\vec i+4\,\vec j-6\,\vec k$
$\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k$

$-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}$
$=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k$
$=\dfrac{45}2\,\vec i-\dfrac{27}8\,\vec j-\dfrac{57}2\,\vec k=\left\langle\dfrac{45}2,-\dfrac{27}8,-\dfrac{57}2\right\rangle$

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Calculus 3 help please.

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I assume each path $C$ is oriented positively/counterclockwise.

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$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

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$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

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$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

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and

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Laurie was scuba diving. Each time she dove 5 feet deeper, she would stop and clear her ears. After Laurie had checked her instr

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Answer:

Laurie's maximum depth can be represented using the expressions :

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Step-by-step explanation:

Each depth level at which ears are checked = 5 feets

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Number of times ears are checked = 20/5 = 4 times

This means that for Laurie to have checked her instrument 3 times :

(4 × 3 × 5) = 60 feets

Therefore, Laurie's maximum depth can be expressed as (4 × 3 × 5) = 60 feets

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Yakvenalex [24]

Answer:

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1 ft = 1.894 * 10^-4 miles , so

1 inch = 1.894 * 10^-4 / 12 miles

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= 1.5783* 10^-5 miles ( in scientific form).

Multiplying this by 20 we get

31.567 * 10^-5

= 3.1567 * 10^-4 miles.

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Help meeee!will give brainliest!

meriva

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