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Verdich [7]
3 years ago
13

Point A’ is the image of point A under a rotation about Q. Determine the angles of rotation

Mathematics
2 answers:
Fynjy0 [20]3 years ago
7 0

Answer:

ok but wheres point A and where's Q

Step-by-step explanation:

Finger [1]3 years ago
7 0

Answer: 90 degrees clockwise & 270 counterclockwise

Step-by-step explanation: did it on khan academy

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Are 5p+5c and 5(p+c) equivalent
Flura [38]
Yes they are equivilent because of the distributive property
which is
a(b+c)=ab+ac
or in this case
5(p+c)=5p+5c
if you don't believe it *cough* *cough* *Brainiac01* *cough* *cough*
then subsitute values
lets sayp=4 and c=3
5(4)+5(3)=5(4+3)
20+15=5(7)
35=35
tada


so the answer is yes
6 0
3 years ago
Use the GCF and the distributive property to find the sum of 22+33
hram777 [196]

Answer:

22 + 33 = 55 the GCF of 22 and 33 is 11

Step-by-step explanation:

5 0
3 years ago
Solve. Check that your answer is reasonable. Paula's painting has a perimeter of 1.22 meters. She wants to put ribbon around the
iren2701 [21]

Answer:

5 pieces

Step-by-step explanation:

Given:

Paula's painting has a perimeter of 1.22 meters.

She wants to put ribbon around the edge.

If the ribbon comes in pieces that are 25 centimeters long.

Question asked:

How many pieces of ribbon does she need to go all the way around her painting?

Solution:

Perimeter of painting = 1.22 meters

First of all we will convert it into centimeters:-

As we know:

1 meter = 100 centimeters

1.22 meters = 100 \times 1.22 = 122 centimeters

Now, as given that ribbon comes in pieces of 25 cm long and total we should have 122 cm to go all the way around the painting.

Number of pieces of ribbon will be needed = \frac{122}{25} =4.88\

Thus, she need 5 pieces of 25 cm long ribbon to go all the way around the painting.

<u>Note</u>:-This is by default understood that few length of 5th ribbon will be left over.

3 0
3 years ago
If Mary traveled 200 miles on foot, then traveled 200 miles on bike then traveled 200 miles by car how long did it take her to g
Serjik [45]

Step-by-step explanation:

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast, how far, or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance, rate, and time. To understand the difference among these, think about the last time you drove somewhere.

The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time

d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

You drove 25 miles—that's the distance.

You drove an average of 50 mph—that's the rate.

The drive took you 30 minutes, or 0.5 hours—that's the time.

According to the formula, if we multiply the rate and time, the product should be our distance.

And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25, which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance, rate, or time. Let's try another simple problem.

7 0
3 years ago
Given m || n, find the value of x and<br> y.
Ann [662]

Answer:

x = 41 , y = 139

Step-by-step explanation:

x and 41° are alternate exterior angles and are congruent , so

x = 41

x and y are a linear pair and sum to 180° , that is

x + y = 180

41 + y = 180 ( subtract 41 from both sides )

y = 139

3 0
2 years ago
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