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Mademuasel [1]
4 years ago
6

Landon used a semicircle, a rectangle, and a right triangle to form the figure below. Which is the best estimate of the area of

the figure in square centimeters?

Mathematics
2 answers:
Nuetrik [128]4 years ago
6 0
Let’s start with the rectangle. 6(4) = 24 cm^2.
The triangle is 1/2(bh).
We know the bottom of the triangle is 4 because 10-6=4.
4(4) = 16
16/2 = 8
We’re at 32 square cm so far.
Now the semi circle.
The diameter is 4, so the radius is 2.
A = (pi)r^2
A = (pi)4
A= 4pi
Best estimate could be expressed as 32 + 4pi cm^2, or substituting 3.14 for pi, 44.56 cm^2
Damm [24]4 years ago
4 0

Answer:

38.28cm^2

Step-by-step explanation:

Area of rectangle = 24cm^2

Area of rectangle = 8cm^2

Area of semicircle = 6.28cm^2

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Ganezh [65]

Answer:

first exercise

V = 16 π

second exercise

V= 68π/15

Step-by-step explanation:

Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4

<em>vide</em> picture 1

The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence,  this new red equation is defined by x = (y − 1)2.

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The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.

Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.

(1) Vt = V - Ve

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(2) V=\int\limits^{y_{1} }_{y_{2}}{2*pi*y*f(y)} \, dy

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V = 16 π

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Second part

Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.

<em>vide</em> picture 2

As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.

The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.

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