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If you're comfortable Cat, what grade are you in?

I'm in 7th btw.

The rectangular base has an area \(20 \times 15 = 300\)

The triangles have an area \(20 \times 8 \div 2 = 80\)

Thus the area is \(300 + 80 = \color{brown}\boxed{380}\)

k has to be negative, so the answer is \(\color{brown}\boxed{1 - 4 \sqrt7}\)

Here is the graph: https://www.desmos.com/calculator/fwa8nyw6it

It's nice to see you too. :))

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I don't this so.

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well glad you are here today

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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It has to be negative, so, unfortunately, you're not right. But, good try tho!

That is correct  cat Long time u havent been on

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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"Given that m - 2 is a positive integer that divides 3m^2 - 2m + 11, find the sum of all such values of m."

For the last term to be an integer we must have m - 2 = 1, so m = 3;  or  m - 2 = 19, so m = 21

Hence sum = 3 + 21 = 24

The probability is 3/14.

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