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tia_tia [17]
3 years ago
7

20 points! Identify the solids that form the compos

Mathematics
1 answer:
Troyanec [42]3 years ago
3 0

The composite shape is made up of a cube with a side length of 5 inches and a cylinder with a radius of 2 inches and a height of 4 inches.

The composite solid's surface area is 225.4 square inches.

Step-by-step explanation:

Step 1:

The given composite shape is made up of a cube with a side length of 5 inches and a cylinder with a radius of 2 inches and a height of 4 inches.

The surface area of the composite shape is given by summing the individual surface areas.

The composite shape's surface area = The cube's surface area + the cylinder's surface area.

Step 2:

Any cube's surface area is calculated by multiplying 6 with the square of the side length (a^2).

The cube's surface area = 6a^2 = 6(5^{2} ) = 150 square inches.

Step 3:

Any cylinder's surface area is calculated with the following formula;

The cylinder's surface area = 2 \pi r h+2 \pi r^{2} = 2 \pi (2) (4)+2 \pi (2)^{2} = 75.398 square inches

Step 4:

The composite shape's surface area = The cube's surface area + the cylinder's surface area.

The composite shape's surface area = 150 + 75.398 = 225.398 square inches. Rounding this off, we get the area as 225.4 square inches.

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g The tangent plane to z=f(x,y) at the point (1,2) is z=5x+2y−10. (a) Find fx(1,2) and fy(1,2). fx(1,2)= Number fy(1,2)= Number
murzikaleks [220]

Answer:

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Approximation at points (1.1,1.9) is 0.7

Step-by-step explanation:

Given:

Tangent plane to  a surface z=5x+2y-10 as the function at point (1,2)

To find :

f(x,y) at (1,2)

partial derivatives of function w.r.t. (x and y) and value of that function at given points.

Solution:(refer the attachment also)

Now we know that

the equation of tangent plane at given points to the surface is given by,

f(x1,y1,z1) and z=f(x,y)

z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)

here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.

now

taking partial derivative w.r.t. x we get

Fx(x1`,y1)=\frac{d}{dx} (5x+2y-10)

=5.

Then w.r.t y we get

Fy(x1,y1)=

\frac{d}{dy}(5x+2y-10)

=2.

The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.

Using the Linearization or linear approximation we get

L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)

=-1+5(x-1)+2(y-2)

=5x+2y-10

Approximation at F(1.1,1.9)

=5(1.1)+2(1.9)-10

=5.5+3.8-10

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Approximation at points (1.1,1.9) is 0.7

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Answer:

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