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san4es73 [151]
3 years ago
10

WILL GIVE 100 POINTS AND BRAINLIEST! Prove that the square of the second number out of three consecutive odd numbers is four gre

ater than the product of the first and the third numbers.
Mathematics
2 answers:
vodomira [7]3 years ago
6 0

Pick 3 consecutive odd numbers:

3, 5, 7

Square the 2nd number: 5 ^ 2 = 25

Multiply the first and 3rd:

3 x 7 = 21

25-21 = 4

This is true.

Try another set of numbers:

9, 11, 13

11^2 = 121

9 x 13 = 117

121-117 = 4

Again it’s true.

Ad libitum [116K]3 years ago
4 0

Answer:

Let's choose the three odd consecutive numbers 1, 3, and 5.

3^2=9

1•5=5

9-5=4

Let's try the same thing, but with the numbers 101, 103, and 105.

103^2=10609

101•105=10605

10609-10605=4

So, yes, the square of the second number out of three consecutive odd numbers is four greater than the product of the first and the third numbers.

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I am very confused, if somebody could help me it would mean alot. :)
Yuliya22 [10]

Answer:

3n+7=16

Step-by-step explanation:

3(3)+7=16

9+7=16

16=16

The second one is wrong(x=18)

8 0
3 years ago
Is this correct ? My answers 10 btw. NO BEGINNERS , & whoever answers first gets a thanks :)
borishaifa [10]
The answer would be 54
6 0
3 years ago
Can anyone help solve this?
Mashcka [7]

Answer:

12.2

Step-by-step explanation

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5 0
3 years ago
Explain how to find 27% of 16 using multiplication by
just olya [345]

Answer:

4.32

Step-by-step explanation:

You can multiply 16 by 0.27 to get the answer.

16*0.27=4.32

You can check your work with estimation because 27% is about 1/4 of the number, and just by looking at the number, 4 is 1/4 of the number.

6 0
3 years ago
Read 2 more answers
Evaluate g(3) if g(x)=2x+2
klemol [59]
Plug in 3 for x so
2(3)+2
6+2=
8
The evaluation is 8
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