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3 years ago

Work out the perimeter of this quarter circle. Take Pi to be 3.142 and write down all the digits given by your calculator.

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

3 0

Answer:

Perimeter of the quarter circle will be 24.997 cm.

Step-by-step explanation:

Picture shown is a sector of a circle having an angle inscribed = 90° at the center

Formula to get the circumference of the sector is,

Circumference of the sector = $\frac{\theta^\circ}{360}(2\pi r)$

C = $\frac{90^\circ}{360^\circ}(2\pi)(7)$

C = $\frac{7\pi }{2}$

C = 10.997 cm

Now perimeter of the figure = Sum of lengths of all sides

P = 2r + C

P = 7 + 7 + 10.997

= 24.997 cm

Therefore, Perimeter of the quarter circle will be 24.997 cm.

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A ball is thrown at an initial height of 7 feet with an initial upward velocity at 27 ft/s. The balls height h (in feet) after t

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Answer:

The height of ball is 17 ft at t=0.55 and t=1.14.

Step-by-step explanation:

The general projectile motion is defined as

$y=-16t^2+vt+y_0$

Where, v is initial velocity and y₀ is initial height.

It is given that the initial height is 7 and the initial upward velocity is 27.

Substitute v=27 and y₀=7 in the above equation to find the model for height of the ball.

$h(t)=-16t^2+27t+7$

The height of ball is 17 ft. Put h(t)=17.

$17=-16t^2+27t+7$

$0=-16t^2+27t-10$

On solving this equation using graphing calculator we get

$t=0.549,1.139$

$t\approx 0.55,1.14$

Therefore the height of ball is 17 ft at t=0.55 and t=1.14.

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3 years ago

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mrs_skeptik [129]

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Answer:

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Step-by-step explanation:

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3 years ago

Compute the matrix of partial derivatives of the following functions.

s344n2d4d5 [400]

For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$

(c)

$Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}$

(d)

$Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}$

5 0

3 years ago