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Over [174]
4 years ago
15

#4 Kerry created a pattern with rectangles. Her first step contained 1

Mathematics
1 answer:
Iteru [2.4K]4 years ago
8 0

Answer:

The constant rate of change is <em>2 rectangles</em>

Step-by-step explanation:

Looking at this question, we can see that what we have is in the form of an arithmetic progression otherwise known as A.P

An arithmetic progression refers to a series of numbers that differ by a constant addition or subtraction of a certain constant number. What this means is that subsequent terms are formed by the addition of a certain constant number to preceding terms.

Now, let's look at the situation in the question;

1, 3 , 5, ?, ?

Looking at the first three terms, we can see that the difference between the first two terms is 2, the difference between the third term and the second term is 2 also. This gives us an ideas of an arithmetic progression, with the first term being 1 and the common difference being 2.

The common difference is what is referred to as the constant term in this questions which is equal to 2

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A rectangle in Quadrant I rotates 270° counterclockwise. In which quadrant will the transformation lie? A) Quadrant I B) Quadran
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4 0
3 years ago
Find the surface area to the nearest hundredth
pochemuha

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226.19

Step-by-step explanation:

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5 0
4 years ago
Read 2 more answers
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]

Answer:

There are 155 ways in which these elements casn occur.

Step-by-step explanation:

We want 3 element subsets whose sum are multiples of 3

1+2+3= 6

1+2+6= 9

1+2+9= 12

1+9+11=21

1+3+5=9

1+4+8=12

1+5+6=12

1+6+8=15

1+7+10=18

1+8+9=18

1+9+11=21

2+3+7=12

2+4+6=12

2+4+9=15

2+5+11=18

2+6+7=15

2+7+9=18

2+8+5=15

2+8+11=21

2+9+10=21

3+6+9= 18

3+9+11=21

3+10+11=24

6+9+10=27

6+8+11=27

6+7+11=24

7+8+9= 24

8+9+10=27

7+9+11=27 .........

We have 11 elements

We need a combination of 3

The combinations can be in the form

even+ even+ odd

odd+odd+odd

even + odd+odd

So there are 3 ways in which these elements can occur

Total number of combinations with  3 elements =11C3= 165

There are 6 odd numbers and 5 even numbers.

Number of subsets with 3 odd numbers = 6C3= 20

Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60

Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75

So 20+60+75=155

There are 155 ways in which this combination can occur

6 0
3 years ago
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