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nignag [31]
4 years ago
7

Which pairs of quadrilaterals can be shown to be congruent using rigid motions?

Mathematics
2 answers:
Mnenie [13.5K]4 years ago
7 0

Answer:- 1. Congruent .

2. Not Congruent .

3. Not Congruent .

4. Not Congruent .

Explanation:-

A basic rigid transformation is a transformation of the figure that does not affect the size of the figure . The size of figure doesn't reduce or get enlarge. There are three basic rigid transformations:-reflections, rotations, and translations.

1. Reflection:- A reflection is a transformation that maps every point of a figure in the plane to point of image of figure, across a line of reflection .

2.Rotation:-A rotation of some degrees is a transformation which rotate a figure about a fixed point called the center of rotation.

3.Translation:-A translation is a transformation of a figure that moves every point of the figure a fixed distance in a particular direction.

Now,

1. Quadrilateral 1 and quadrilateral 2 are congruent by reflection.

2. Quadrilateral 3 and quadrilateral 4 are not congruent by using any rigid transformation.

3.Quadrilateral 1 and quadrilateral 4 are not congruent by using any rigid transformation.(as 4 is not a reflected image of 1)

4.Quadrilateral 2 and quadrilateral 3  are not congruent by using any rigid transformation.


OverLord2011 [107]4 years ago
6 0
Answers:
True
Not True
Not True
True
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Five times the larger number is 11 more than three time the smaller number. Also three times the smaller is 16 less four times t
alisha [4.7K]
Let the larger number be w and the smaller number be k;
5w-11=3k
3k+16=4w
Reorganizing the equations;
5w-3k=11
4w-3k=16
Subtracting equation 2 from equation 1:
w=-5
Replacing value of w in equation 1;
5(-5)-11=3k
-25-11=3k
3k=-36
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The numbers are -5 and -12
3 0
3 years ago
The amount of toothpaste in a tube is normally distributed with a mean of 6.5 ounces and a standard
Anarel [89]

Answer:

a) 266 tubes ,  TC_r = $53.2

b) 266 tubes ,  T.Loss = $13.30

Step-by-step explanation:

Given:

- The sample size of tubes n = 1,000 tubes

- The mean of the sample u = 6.5 oz

- The standard deviation of the sample s.d = 0.8 oz

- Cost of manufacturing a tube C_t = 50 cents

- Cost of refilling a tube C_r = 20 cents

- Profit loss per tube Loss = 5 cents

Find:

a). How many tubes will be found to contain less than 6 ounces? In that case, what will be the total cost of the  refill?

b) How many tubes will be found to contain more than 7 ounces? In that case, what will be the amount of  profit lost?

Solution:

- First we will compute the probability of tube containing less than 6 oz.

- Declaring X : The amount of toothpaste.

Where,                         X ~ N ( 6.5 , 0.8 )

- We need to compute P ( X < 6 oz )?

Compute the Z-score value:

                  P ( X < 6 oz ) =  P ( Z < (6 - 6.5) / 0.8 ) = P ( Z < -0.625 )

Use the Z table to find the probability:

                               P ( X < 6 oz ) = P ( Z < -0.625 ) = 0.266

- The probability that it lies below 6 ounces. The total sample size is n = 1000.

       The number of tubes with X < 6 ounces = 1000* P ( X < 6 oz )

                                                                           = 1000*0.266 = 266 tubes.

- The total cost of refill:

                            TC_r = C_f*(number of tubes with X < 6)

                            TC_r = 20*266 = 5320 cents = $53.2

- We need to compute P ( X > 7 oz )?

Compute the Z-score value:

                  P ( X > 7 oz ) =  P ( Z > (7 - 6.5) / 0.8 ) = P ( Z < 0.625 )

Use the Z table to find the probability:

                               P ( X > 7 oz ) = P ( Z > 0.625 ) = 0.266

- The probability that it lies above 7 ounces. The total sample size is n = 1000.

       The number of tubes with X > 7 ounces = 1000* P ( X > 7 oz )

                                                                           = 1000*0.266 = 266 tubes.

- The total cost of refill:

                            T.Loss = Loss*(number of tubes with X > 7)

                            T.Loss = 5*266 = 1330 cents = $13.30

5 0
3 years ago
For normal distribution with mean μ = 400 and standard deviation σ = 100 find probability that x is greater than 619: p(x &gt; 6
denis23 [38]
A suitable calculator shows p(x > 619) ≈ 1.43%.

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3 years ago
A student draws two parabolas both parabolas cross the x axis at (-4,0) and (6,0) the y intercept of the first parabolas is (0,-
faust18 [17]

Answer:

The positive difference between the a values is 0.5

Step-by-step explanation:

we know that

Both parabolas cross the x axis at (-4,0) and (6,0)

so

The general equation is

y=a(x+4)(x-6)

<em>Find the value of a in the first parabola</em>

The y-intercept is (0,-12)

so

For x=0, y=-12

substitute

y=a(x+4)(x-6)

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a=0.5

<em>Find the value of a in the second parabola</em>

The y-intercept is (0,-24)

so

For x=0, y=-24

substitute

y=a(x+4)(x-6)

-24=a(0+4)(0-6)

-24=a(-24)

a=1

<em>Find the positive difference between the a values for the two functions</em>

so

1-0.5=0.5

5 0
3 years ago
Read 2 more answers
About how many times would u you expect to get an odd number in 200 trials-I need the answer urgent
ale4655 [162]

Answer:

50

Step-by-step explanation:

4 0
3 years ago
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