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Nataliya [291]
3 years ago
10

What’s the answer to this algebra question

Mathematics
1 answer:
natka813 [3]3 years ago
4 0

Answer:

x=9, x=7

Aka x= 9,7

Step-by-step explanation:

Oh goodie, factoring.

I used to be very good at this. My memory is a LITTLE rusty but let's see if we can bust out some penzoil and fix my aching joints (yes I just stole the quote from that terrible show, sue me), shall we?

1. x^2 - 11x + 68 = 5x +5

2. Chuck all the x variables to the left: x^2 - 16x + 68 = 5

3. Get rid of that pesky loser, 5, by subtracting from both sides of the equation (Note that it's important to remember LIKE TERMS, don't try subtracting 5 from 16x, you'll hurt yourself): x^2 - 16x + 68 - 5 = 5 - 5

New equation: x^2 - 16x + 63 = zilch (0)

4. Use AC method to factor: x2+bx+c. AC method definition: a pair of numbers whose product is c and whose sum is b: (x-9)(x-7)=0

5. Solve for x.

Booyah.

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Call an integer $n$ oddly powerful if there exist positive integers $a$ and $b$, where $b>1$, $b$ 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

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∵ 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,

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∴ 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

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∴ There are 3 oddly powerful integers with b = 5

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Now the fourth value of b is 9

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∴ There are 16 oddly powerful integers less than 2010

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Step-by-step explanation:

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