Answer:
True, the point (-5, 2) is inside of the circle (lying on it) centered at (0, 7)
Step-by-step explanation:
Hope this helps! Leave any questions or concerns in the comments! byeee <3
Answer:
d = 17
Step-by-step explanation:
The given arithmetic sequence is :
31, 48, 65, 82, ...
We need to find the common difference for this sequence.
First term, a₁ = 31
Second term, a₂ = 48
Common difference = a₂-a₁
= 48-31
= 17
So, the common difference for this arithmetic sequence is equal to 17.
Its is true that C ⊆ D means Every element of C is present in D
According to he question,
Let C = {n ∈ Z | n = 6r – 5 for some integer r}
D = {m ∈ Z | m = 3s + 1 for some integer s}
We have to prove : C ⊆ D
Proof : Let n ∈ C
Then there exists an integer r such that:
n = 6r - 5
Since -5 = -6 + 1
=> n = 6r - 6 + 1
Using distributive property,
=> n = 3(2r - 2) +1
Since , 2 and r are the integers , their product 2r is also an integer and the difference 2r - 2 is also an integer then
Let s = 2r - 2
Then, m = 3r + 1 with r some integer and thus m ∈ D
Since , every element of C is also an element of D
Hence , C ⊆ D proved !
Similarly, you have to prove D ⊆ C
To know more about integers here
brainly.com/question/15276410
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I think this is right maybe
The perimeter of the enlarged garden is 16.56 meters.
<u>Step-by-step explanation:</u>
The garden is in the shape of a rectangle.
- The length of the rectangle = 5.4 meters.
- The width of the rectangle = 1.5 meters.
The Rectangle is enlarged by increasing the length and width by 20%.
<u>To find the enlarged length :</u>
The original length 5.4 is increased by 20%.
⇒ (20/100) × 5.4
⇒ 0.2 × 5.4
⇒ 1.08
- The 20% of the length is 1.08
- The enlarged length is 5.4 + 1.08 = 6.48 meters.
<u>To find the enlarged width :</u>
The original width 1.5 is increased by 20%.
⇒ (20/100) × 1.5
⇒ 0.2 × 1.5
⇒ 0.3
- The 20% of the width is 0.3
- The enlarged width is 1.5 + 0.3 = 1.8 meters.
The perimeter of the enlarged rectangle is found by substituting the enlarged values of length and width in the perimeter formula.
Perimeter of the enlarged rectangle = 2 (length + width)
⇒ 2 (6.48 + 1.8)
⇒ 2 × 8.28
⇒ 16.56 meters.
∴ The perimeter of the enlarged garden is 16.56 meters.
