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2 years ago

(01.01 LC)

Mathematics

1 answer:

Scilla [17]2 years ago

3 0

Answer:

Circle

Explanation:

I just took the test and got it right.

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How do you find the area of each triangle?

Pachacha [2.7K]

There's a big triangle with base (2+6) meters and altitude 5 meters.

The formula for the area of a triangle of base B and altitude H is

A = BH/2.

The area of the large triangle is thus A = (8 meters)(5 meters)/2, or
20 square meters.

Use A = BH/2 to find the areas of the other triangles.

6 0

2 years ago

What is the correct value of x? Circle the correct answer.

Andrew [12]

X=5 is the right answer

5 0

2 years ago

Read 2 more answers

3. The figure below is a composite shape that consists of a semicircle and an equilateral triangle (triangle

Ivenika [448]

Answer:

See explanation below

Step-by-step explanation:

Area = Area of triangle + Area of the semi circle

Area of triangle = 1/2bh

Area of triangle = 1/2 *(5/2)(4)

Area of triangle = 1/2(2.5)*4

Area of triangle = 2 * 2.5

Area of triangle = 5cm^2

Area of semicircle = πr²/2

r = 2.5cm

Area of semicircle =(3.14)(2.5)²/2

Area of semicircle = = 3.14 * 4.5

Area of semicircle = 9.8125cm^2

Area pf the figure = 5+9.8125

Area pf the figure = 14.8125cm^2

For the perimeter

Perimeter of the semicircle = 2πr

Perimeter of the semicircle = 2(3.14)(2.5)

Perimeter of the semicircle = 15.7cm

Perimeter of the semicircle = = 15.7 + 5 + 5

Perimeter of the semicircle = 25.7cm

8 0

2 years ago

The diagram below represents a generic right triangle.

Oksi-84 [34.3K]

Bshould be the answer !

4 0

2 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

2 years ago