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3 years ago

How do you connect the ideas of congruency and rigid motion and how do you prove congruency?

1 answer:

7 0

If you cut a triangle out of a piece of paper, and move that triangle around (sliding, rotating, reflecting) then that triangle will retain its original shape and size. The three movement types mentioned are rigid motions. They do not change the size or shape of the triangle. If two triangles are congruent, then a sequence of one or more rigid motions can be applied to have them line up together.

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Nwhat is the range of data set for 8 1 7 6 14

bazaltina [42]

Answer:

14-1=13 range= 13

Step-by-step explanation:

The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

4 0

2 years ago

What formula should be used to find the circumference of a circle

Inessa05 [86]

The circumference = π x the diameter of the circle (Pi multiplied by the diameter of the circle). Simply divide the circumference by π and you will have the length of the diameter. The diameter is just the radius times two, so divide the diameter by two and you will have the radius of the circle

3 0

3 years ago

Read 2 more answers

Helpppp plzzzzzzzzzzzzz I know it’s supplementary but I can’t find the value of X

Black_prince [1.1K]

Answer:

x = 18

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$

6 0

2 years ago

Home CoworkHelper

9966 [12]

Answer:

See below.

Step-by-step explanation:

PROBLEM A

The formula for the volume of a cylinder is

V = πr²h, where πr² is the area of the base.

V = area of base x height

Choose any four numbers that are less than 10 for the height.
Then, substitute into the formula and isolate "r". 10 = πr²h
Double "r" to find the diameter.
The area of the base is found (bolded) while you solve πr².

*I will use the calculator button for pi (π)

Cylinder A: h = 1

10 = πr²h

10 = πr²(1)

10 = πr² Area of the base is 10m².

r = √ (10/π)

r = 1.78 Radius is 1.78m.

d = 2r = 1.78*2 = 3.56 Diameter is 3.56m.

Cylinder B: h = 2

10 = πr²h

10 = πr²(2)

5 = πr² Area of the base is 5m².

r = √ (5/π)

r = 1.26 Radius is 1.26m.

d = 2r = 1.26*2 = 2.52 Diameter is 2.52m.

Cylinder C: h = 4

10 = πr²h

10 = πr²(4)

2.5 = πr² Area of the base is 2.5m².

r = √ (2.5/π)

r = 0.89 Radius is 0.89m.

d = 2r = 0.89*2 = 1.78 Diameter is 1.78m.

Cylinder D: h = 5

10 = πr²h

10 = πr²(5)

2 = πr² Area of the base is 2m².

r = √ (2/π)

r = 0.80 Radius is 0.8m.

d = 2r = 0.80*2 = 1.6 Diameter is 1.6m.

The answers are reasonable if the answer is close to 10 when you substitute the rounded numbers back into the formula and solve.

There are infinite possibilities because any numbers when used in the formula equates to 10 are possible. The height/radius can be any decimal number, which would make the other dimensions change.

PROBLEM B

Follow similar steps if you know the diameter of a cylinder. Use the same formula V = πr²h.

We need the radius, which is half the diameter. r = d/2 = 10/2 = 5

Choose a random number for volume that divides easily by 25.
Substitute "V" and "r²" (r² = 25).
Isolate "h".
The area of the base is 78.54 cm² every time (A = πr²).

Cylinder A: V = 100

100 = π25h

4 = πh

h = 1.27 The height is 1.27m.

Cylinder B: V = 200

200 = π25h

8 = πh

h = 2.55 The height is 2.55m.

Cylinder C: V = 75

75 = π25h

3 = πh

h = 0.95 The height is 0.95m.

Cylinder D: V = 125

125 = π25h

5 = πh

h = 1.59 The height is 1.59m.

The answers are reasonable if I can substitute two rounded values into the formula and get a number close to the third. The height is also a decimal number which occurs because of pi.

6 0

2 years ago