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Masja [62]
3 years ago
13

Which is true about rigid transformations and circles?

Mathematics
1 answer:
steposvetlana [31]3 years ago
5 0

A rigid transformation is defined to be one in which the pre-image of the object and its new image after the transformation both have the exact same size and shape. So the answer in this question is:

<span>D. a circle's shape is preserved regardless of any rigid transformation</span>

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Peter buys a camper van costing £35000 + 20% VAT. He pays a 50% deposit and then arranges to pay the remaining balance in 20 equ
shutvik [7]

The amount paid each month is 1050.

35,000 plus 20% is 42,000 because 20% of 35,000 is 7,000.

Next, divide 42,000 by 20, and you would get 1050.

3 0
3 years ago
Please help i’ll give brainliest
Setler79 [48]

Answer:

range

Step-by-step explanation:

*JUST AND EDUCATED GUESS* but id say range

7 0
2 years ago
What is the simplified form of x+5/3x+4 + x+4/x+3
alina1380 [7]

x +  \frac{5}{3}x + 4 + x +  \frac{4}{x} + 3 \\ x +  \frac{5x}{3}+ 4 + x +  \frac{4}{x}+ 3 \\ (x +  \frac{5x}{3}  + x) + (4 + 3) +  \frac{4}{x}  \\ x =  \frac{11x}{3}  + 7 +  \frac{4}{x}

3 0
3 years ago
1 (15 Points]. Prove the following statement: is divisible by 3 if only if it is a sum of three consecutive integers. be ddvisbk
Alinara [238K]

Answer:

<em>First.</em> Let us prove that the sum of three consecutive integers is divisible by 3.

Three consecutive integers can be written as k, k+1, k+2. Then, if we denote their sum as n:

n = k+(k+1)+(k+2) = 3k+3 = 3(k+1).

So, n can be written as 3 times another integer, thus n is divisible by 3.

<em>Second. </em>Let us prove that any number divisible by 3 can be written as the sum of three consecutive integers.

Assume that n is divisible by 3. The above proof suggest that we write it as

n=3(k+1)=3k+3=k + k + k +1+2 = k + (k+1) + (k+2).

As k, k+1, k+2 are three consecutive integers, we have completed our goal.

Step-by-step explanation:

4 0
2 years ago
As part of a board game, players choose 5 unique symbols from 9 different symbols to create their secret password. How many diff
frozen [14]

Answer:

15,120

Step-by-step explanation:

For the first symbol, there are 9 options to choose from. Then 8, then 7, and so on. Since each player chooses 5 symbols, they will have a total of 9\cdot 8 \cdot 7 \cdot 6\cdot 5=\boxed{15,120} permutations possible. Since the order of which they choose them matters (as a different order would be a completely different password), it's unnecessary to divide by the number of ways you can rearrange 5 distinct symbols. Therefore, the desired answer is 15,120.

8 0
3 years ago
Read 2 more answers
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