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3 years ago

Using the piling method, which of the following can be constructed from polygons alone and discs alone

Mathematics

2 answers:

inna [77]3 years ago

5 0

Answer:

All the mentioned can be constructed with the help of piling method. But, for construction of b and d we need a point as well.

Step-by-step explanation:

a) Yes, a cube can be formed by stacking polygons. If we stack squares one upon another, then the resulting 3-D figure would be a cube.

b) Yes, a cone that does not include the point can be made by stacking disc. If we stack discs such that the discs are arranges in descending order then the resultant 3-D figure would be a cone that does not include a point.

c) Yes, a pyramid can be formed by stacking polygons.

If we stack squares such that they are arranged in descending order of their side length and stacked by a point on the top. Then the resultant figure would be a pyramid that includes a vertex.

f) There are many types of prism. Fro example a rectangular prism can be constructed by simply stacking rectangle polygon on one another.

e) Yes, a cone can be formed by stacking disc.

We need to stack disks arranged in descending order of their radius stacked with a point on the top. The resulting figure would be a cone that includes a point.

f) Yes, a cylinder can be formed by stacking. We need to stack discs of same radius to get a cylinder.

Colt1911 [192]3 years ago

4 0

Answer:

All besides B and D, because a cross section at vertices are nothing more than points, which is neither a polygon nor a disc.

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DedPeter [7]

Answer:

80/89

Step-by-step explanation:

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8 0

3 years ago

Write the inverse of the function: f (x) = 2x+3

34kurt

Answer:

$y = \frac{x - 3}{2}$

Step-by-step explanation:

To find the inverse of a function, simply 'switch' the x and y's and solve for y. $y = 2x + 3$ becomes $x = 2y + 3$ . Now, solving for y, we get $y = \frac{x - 3}{2}$ .

hope this helped! :)

6 0

2 years ago

Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c. What is the distance between the origin and

Elenna [48]

Answer:

The equation is:

(1/a)x + (1/b)y + (1/c)z = 1

Step-by-step explanation:

The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:

<0 - a, b - 0, 0 - 0>

<-a, b, 0> .....................(1)

The direction vector between (0, 0, c) and (0, b, 0) is given as:

<0 - 0, b - 0, 0 - c>

= <0, b, -c> .....................(2)

To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).

Doing this, we have:

<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>

To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:

(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0

(bc)x + (ac)y + (ab)z = (ac)b

Dividing both sides by abc, we have:

(1/a)x + (1/b)y + (1/c)z = 1

3 0

3 years ago

Which of the following expressions is a factor of the polynomial x 2 +3/2x-1

Elodia [21]

Answer:

$\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }$

Step-by-step explanation:

Hello,

let's solve

$x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\$

$\Delta=b^2-4ac=9+4*2*2=9+16=25$

There are two solutions

$x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}$

And

$x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}$

So we can write

$x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

3 years ago

Tammy is selling tickets for the school jazz band concert. Adult tickets cost $5, and student tickets cost $3. If Tammy sells a

N76 [4]

Adult+student=170 ----(1)
5adult+3student=650 ----(2)

(1)*3; 3adult+3student=510 ----(3)
(2)-(3); 2adult = 140
so adult = 70
then student = 170-70 = 100

answer: Tammy sold 70 adult tickets and 100 student tickets

4 0

3 years ago

Read 2 more answers