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Bogdan [553]
3 years ago
15

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h

We substitute the values to obtain;

Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm

Area=\frac{1}{2}(21.6cm)\times 6cm

Area=64.8cm^2

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

C=2\pi r

where r=3.5ft is the radius of the circular table cloth.

We substitute this value and \pi=3.14 into the formula to get;

C=2(3.14)(3.5)ft

C=21.98ft

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

C=\pi r

where r=12.5cm is the radius of the semicircle.

C=3.14\times 12.5cm

C=39.25cm

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh

Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24

Area=345.3125+300

Area=645.3125cm^2

Area\approx645cm^2

QUESTION 6

The area of the parallelogram is given by the formula;

A=bh

From the diagram, b=y\:in.,h=3in..

Given, A=23.7 inches squared.

We substitute the values to obtain;

23.7=3y

y=\frac{23.7}{3}

y=7.9in.

QUESTION 7

The area of a rectangle is given by the formula;

Area=l\times b

The area of the bigger rectangle =9\times15=135in^2

The area of the smaller rectangle =8\times 13=104in^2

The area of the  shaded region =135-104=31in^2

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Answer:

x - 8y - z = 1

Step-by-step explanation:

Data provided according to the question is as follows

f(x,y) = z = ln(x - 8y)

Now the equation for the tangent plane to the surface

For z = f (x,y) at the point P (x_0,y_0,z_0) is

z - z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\\

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f_x(x,y) = \frac{1}{x-8y} \\\\f_y(x,y) = \frac{8}{x-8y} \\\\P(x_0,y_0,z_0) = (9,1,0)\\\\f_z(9,1,0) = (\frac{1}{x-8y})_^{(9,1,0)}

\\\\=\frac{1}{9-8}

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f_y(9,1,0)=(\frac{8}{x-8y})_{(9,1,0)}\\\\ = -\frac{8}{9 - 8}

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So, the tangent equation is

z - 0 = 1\times (x - 9) -8\times (y - 1)

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z = x - 9 - 8y + 8

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