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slamgirl [31]
3 years ago
12

Find the coordinate of K' after a Glide reflection of the triangle: translation 4 units up and 4 units right then a reflection a

cross the line y = 1.
Part 3a​

Mathematics
1 answer:
monitta3 years ago
4 0

Answer:

  a.  (-1, 3)

Step-by-step explanation:

The "glide" 4 units right and up is the transformation ...

  (x, y) ⇒ (x+4, y+4)

The reflection across y=1 is the transformation ...

  (x, y) ⇒ (x, 2-y)

Combined, you have the transformation ...

  (x, y) ⇒ (x +4, 2-(y +4)) = (x +4, -2-y)

Then, ...

  K(-5, -5) ⇒ K'(-5 +4, -2-(-5)) = K'(-1, 3)

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11.3x = 15 - 2x<br> Help me
schepotkina [342]

Answer:  see below

Step-by-step explanation:

11.3x=15-2x

We move all terms to the left:

11.3x-(15-2x)=0

We add all the numbers together, and all the variables

11.3x-(-2x+15)=0

We get rid of parentheses

11.3x+2x-15=0

We add all the numbers together, and all the variables

13.3x-15=0

We move all terms containing x to the left, all other terms to the right

13.3x=15

x=15/13.3

x=1+1/7.8235294117647

   

7 0
3 years ago
Read 2 more answers
Suppose that an airline quotes a flight time of 128 minutes between two cities. Furthermore, suppose that historical flight reco
ANTONII [103]

Answer:

(a) The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

(b) The value of P (129 ≤ X ≤ 146) is 0.3462.

(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

Step-by-step explanation:

The random variable <em>X</em> is defined as the flight time between the two cities.

Since the random variable <em>X</em> denotes time interval, the random variable <em>X</em> is continuous.

(a)

The random variable <em>X</em> is Uniformly distributed with parameters <em>a</em> = 10 minutes and <em>b</em> = 154 minutes.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

(b)

Compute the value of P (129 ≤ X ≤ 146) as follows:

Apply continuity correction:

P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)

                           = P (128.50 < X < 146.50)

                           =\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx

                           =\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx

                           =\frac{1}{52}\times (146.50-128.50)

                           =0.3462

Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.

(c)

It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. <em>X</em> ≥ 128 + 3 = 131.

Compute the value of P (X ≥ 131) as follows:

Apply continuity correction:

P (X ≥ 131) = P (X > 131 + 0.50)

                 = P (X > 131.50)

                 =\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx

                 =\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx

                 =\frac{1}{52}\times (154-131.50)

                 =0.4327

Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

6 0
3 years ago
BC is included between?
hammer [34]

Answer:

bc is included between ab and ac

Step-by-step explanation:

8 0
2 years ago
What is the maximum value of the picture below?
Cloud [144]

Answer:

y = 4

Step-by-step explanation:

The maximum value occurs at the vertex

vertex = (2, 4 )

when x = 2 the maximum value is y = 4

4 0
2 years ago
Is it possible for a system of linear equations to have exactly one solution?
Elanso [62]
No, it's not possible
8 0
3 years ago
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