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katrin2010 [14]
2 years ago
13

6. Triangle ABC is rotated 90 degrees counterclockwise and then translated 3 units up to form

Mathematics
1 answer:
joja [24]2 years ago
7 0

Given:

Consider triangle ABC is rotated 90 degrees counterclockwise and then translated 3 units up to form triangle A'B'C'.

To find:

The transformation that can be used to map each point (x,y) on Triangle ABC to its corresponding point on Triangle A'B'C'.

Solution:

If a figure rotated 90 degrees counterclockwise, then

(x,y)\to (-y,x)

P(x,y)\to P_1(-y,x)

If a figure translated 3 units up, then

(x,y)\to (x,y+3)

P_1(-y,x)\to P'(-y,x+3)

If  triangle ABC is rotated 90 degrees counterclockwise and then translated 3 units up to form triangle A'B'C', then the rule of transformation is

(x,y)\to (-y,x+3)

Therefore, the required rule of transformation is (x,y)\to (-y,x+3).

Note: Option are not in proper form.

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A company produces boxes of dvds of a rate at 52 boxes per hour. they begin to produce boxes when they first open for the day an
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Answer:

284 boxes

Step-by-step explanation:

7 0
2 years ago
If the farmer has 234 feet of fencing, what are the dimensions of the region which enclose the maximal area?
marissa [1.9K]

Answer: 58 ft × 58 ft  

Step-by-step explanation:

Let the length of the region = x feet

And, the width of the region = y feet

Since, the perimeter of the region = 234 feet ( Given )

⇒ 2(x+y) = 234

⇒ x+y = 117

⇒ y = 117 - x

Again the area of the region, A = xy

⇒ A(x) = x(117-x)

⇒ A(x) = 117x - x^2

By differentiating the above equation with respect to x,

⇒ A'(x) = 117 - 2x  

For maxima or minima,

A'(x) = 0

\implies 117 - 2x = 0

\implies -2x = -117

\implies x = 58.5

Again differentiating equation A'(x) with respect to x,

We get, A''(x) = -2

Hence, For x = 58.5 A''(x) = negative

⇒ For x = 58.5 feet the area A(x) is maximum,

⇒ The length of the region having maximum area = x = 58.5 feet

And, the width of the region having maximum area = y = 117-x= 117 - 58.5=58.5 feet,

⇒ The dimension of the region having the maximum area = 58.5 ft × 58.5 ft

8 0
2 years ago
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A hospital takes record of any birth that occurs there every day. On one day, the hospital reports that 35 of the 62 babies born
andre [41]

Answer:

Step-by-step explanation:

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2 years ago
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Sunny earns \$12$12dollar sign, 12 per hour delivering cakes. She worked for xxx hours this week. Unfortunately, she was charged
dezoksy [38]

Answer:

12x - 15 dollars

Step-by-step explanation:

Sunny earns $12 per hour for delivering cakes.

She worked for x hours this week.

Unfortunately, she was charged $15 for a late delivery on Tuesday

She was supposed to earn $12 × x = $12x this week

But she was charged $15 for late delivery on Tuesday

So her net earning this week is; $12x - $15

7 0
2 years ago
Assume there are 365 days in a year.
MissTica

Answer:

1) The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.

Step-by-step explanation:

Given : Assume there are 365 days in a year.

To find : 1) What is the probability that ten students in a class have different birthdays?

2) What is the probability that among ten students in a class, at least two of them share a birthday?

Solution :

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

Total outcome = 365

1) Probability that ten students in a class have different birthdays is

The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883

The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday

P(2 born on same day) = 1- P( 2 not born on same day)

\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]

\text{P(2 born on same day) }=1-[\frac{364}{365}]

\text{P(2 born on same day) }=0.002

The probability that among ten students in a class, at least two of them share a birthday is 0.002.

6 0
2 years ago
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