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just olya [345]
2 years ago
8

A figure is made up of a square and a rectangle. The area of the square is 64 cm². The breadth of the rectangle is the same leng

th of the square. If the total area of the figure is 232 cm², what is the perimeter of the whole figure? ​
Mathematics
1 answer:
antoniya [11.8K]2 years ago
6 0

Answer:

90 cm

Step-by-step explanation:

Let the side of the square be S

Area of the square = S² = 64

So, S = √64 = 8 cm

Let L be the length of the rectangle and B be the breadth

We know B = S = 8

Area of rectangle = BL = 8L

Total area = Area of square + area of rectangle

Total area   = 64 + 8L  = 232  (given)

Subtracting 64 from both sides yields

8L = 232-64 = 168 or L = 168/8 = 21

Perimeter of square = 4S = 4 x 8 = 32

Perimeter of rectangle = 2(B + L) = 2(8 + 21) = 2 x 29 = 58

Total perimeter = 32 + 58 = 90 cm

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Answer:

    Step-by-step explanation:

D

4 0
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
PLEASE HELP I WILL MARK BEAINLIST
Vlada [557]

Answer:

5981.25 m²

Step-by-step explanation:

1. Divide the court into two sections. One section being the semicircle, the other being the rectangle.

2. Multiply 100 by 50 to get the area of the rectangle, which is 5000.

3. Divide 50 by 2 to get the radius of the semicircle, which is 25.

4. Find the formula used for area of a semicircle A = \frac{1}{2}\pir²

5. Insert 25 into where the "r" is in the equation: A = \frac{1}{2}\pi(25²), resulting in about 981.25. (I replaced \pi with 3.14)

6. Add the area of the rectangle with the area of the semicircle, 5000 + 981.25 = 5981.25.

7. The final answer is 5981.25 m².

6 0
3 years ago
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Answer:

Step-by-step explanation:

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substitute : -cos^2 ( sin^2 )

substitute  :  -(1-sin^2 ) ( sin^2)

multiply : -sin^2 + sin^4 or sin^4 - sin^2

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Answer:

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Step-by-step explanation:

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