catmg

Yep, in person school has been quite busy. :((

I wish I could be on more.

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It is always nice to show some work alongside an an answer. :))

That way, people can learn and share ideas. 

y = 2x^2 + kx + 6

y = -x - 8

-x - 8 =  2x^2 + kx + 6

0 = 2x^2 + (k+1)x + 14

There is one solution when the discriminant is 0.

(k+1)^2 - 4(2)(14) = 0

(k+1)^2 = 112

k+1 = +/- 4sqrt(7)

k = -1 +/- 4sqrt(7)

I did not seem to get 16. 

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a = x - 1/x

a^2 = x^2 - 2 + 1/x^2

x^2 + 1/x^2 = 171

a^2 = 171 - 2

a^2 = 169

\(a = \pm 13\)

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sin(2x) = 2/3 

2sin(x)cos(x) = 2/3

sin(x)cos(x) = 1/3

sin^2(x)cos^2(x) = 1/9

3sin^2(x)cos^2(x) = 1/3

sin^2(x) + cos^2(x) = 1

sin^6(x) + 3sin^4(x)cos^2(x) + 3sin^2(x)cos^4(x) + cos^6(x) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(sin^2(x) + cos^2(x)) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(1) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x) = 1

sin^6(x) + cos^6(x) + 1/3 = 1

sin^6(x) + cos^6(x) = 2/3

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If there are zeroes at x = 2 and x = -4, the quadratic must be in the form of a(x - 2)(x + 4)

Plugging in x = 3, a(3 - 2)(3 + 4) = 6, solve for a and expand the quadratic. :))

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1/3 log_4(8) + 3/2 log_4(36) - 3 log_(4^4) (81)

1/3 log_4(8) + 3/2 log_4(36) - 3/4 log_4(81)

log_4(8^(1/3)) + log_4(36^(3/2)) - log_4(81^(3/4))

log_4(2) + log_4(216) - log_4(27)

log_4(2*216/27)

log_4(16)

2

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In the equation 2y = f(2x)/2 + 2, try plugging x as 4.5 and see what happens.

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Your logic makes perfect sense to me. :))

I think that if 10^(n+1) + 1 contains at least 1 perfect square factor, then this statement can be disproven. 

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HAPPY ANNIVERSARYYYYY :)))

Thank you for being such an amazing person on web2.0calc.

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I'm not sure I understand this question properly. 

If n = 4, then z = 3. 

So there must exist an interger a such at 4 = a^6. 

I don't think an integer a exists.

Am I interpreting the problem wrong?

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