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Alexandra [31]
3 years ago
6

Mr Richards science class is conducting a variety of expirements with projectiles and falling objects DROP DOWN ANSWERS ARE GROU

PS A-D

Mathematics
1 answer:
ASHA 777 [7]3 years ago
7 0
First remember the following kinematic equations:
 a = g
 vf = g * t + vo
 rf = (1/2) * g * t ^ 2 + vo * t + ro
 g: gravity
 t: time
 vf: final speed
 Vo: initial speed
 For this case what we should do is to choose the equation that best suits each statement.
 We have then:
 GroupA: h (t) = - 4.9 * t ^ 2 + 19 * t
 Group B: h (t) = - 16 * t ^ 2 + 50 * t
 Group C: h (t) = - 4.9 * t ^ 2 +19
 Group D: h (t) = - 16 * t ^ 2 +50
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In Cartesian coordinates, the region is given by -3\le x\le3, -\sqrt{9-x^2}\le y\le\sqrt{9-x^2}, and -\sqrt{16-x^2-y^2}\le z\le\sqrt{16-x^2-y^2}. Converting to cylindrical coordinates, using

\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}

we get a Jacobian determinant of r, and the region is given in cylindrical coordinates by 0\le\theta\le2\pi, 0\le r\le3, and -\sqrt{16-r^2}\le z\le\sqrt{16-r^2}.

The volume is then

\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=-\sqrt{16-r^2}}^{z=\sqrt{16-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\dfrac{4(64-7\sqrt7)\pi}3
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Given: Line A R bisects ∠BAC; AB = AC Triangle A B C is shown. Point R is at the middle of the triangle. Lines are drawn from po
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Answer:

It is SAS

Step-by-step explanation:

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We changed the form from y=x to y=x+c. In this case, we use 1/2 for the and constant C value which is -12.

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