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kupik [55]
3 years ago
9

The equation 2(5+7)=(2x 5) +(2 x 7) illustrates what property

Mathematics
1 answer:
Marina86 [1]3 years ago
5 0

Answer:

Distrubutive property.

                                  <u><em>Thank You</em></u>

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Call an integer $n$ oddly powerful if there exist positive integers $a$ and $b$, where $b&gt;1$, $b$ is odd, and $a^b = n$. How
azamat

Answer:

There are 16 oddly powerful integers less than 2010

Step-by-step explanation:

∵ b is an odd integer

∵ b > 1

∴ The first value of b is 3

∵ a is an integer

- We can use a = 1, 2, 3, ..........

∵ a^{b}=n

∵ n < 2010

- Let a = 1, 2, ............... 12 because 12³ is greatest integer  < 2010

∵ 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216, 7³ = 343,

   8³ = 512, 9³ = 729, 10³ = 1000, 11³ = 1331, 12³ = 1728

∴ There are 12 oddly powerful integers with b = 3

Now the second value of b is 5

1^{5}=1 but we took 1 before so we will start with 2

∵ 2^{5}=32, 3^{5}=243, 4^{5}=1024

- 4^{5} is the greatest integer < 2010

∴ There are 3 oddly powerful integers with b = 5

Now the third value of b is 7

∵ 2^{7}=128

- 2^{7} is the greatest integer < 2010

∴ There is 1 oddly powerful integers with b = 7

Now the fourth value of b is 9

∵ 2^{9}=512

- 2^{9} is the greatest integer < 2010

- But we used 512 before

∴ There is no oddly powerful integers with b = 9

- 9 is the greatest value of b which makes a^{b}

∵ 12 + 3 + 1 = 16

∴ There are 16 oddly powerful integers less than 2010

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Use the quadratic formula to solve x² + 7x + 8 = 0. What are the solutions to the equation? Round irrational solutions to the ne
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