catmg

t is not in the domain when the denominator equals 0. 

When does (t-1)^2 + (t+1)^2 = 0?

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(a+bi)^2 = 2i

a^2 - b^2 + 2abi = 2i

a^2 - b^2 = 0

a^2 = b^2

2abi = 2i

ab = 1

a^2 = b^2 and ab = 1

From ab = 1, we know that either a and b are both positive or both negative, that means we can say that a = b from a^2 = b^2. 

a = b and ab = 1

a = 1, b = 1, or a = -1, b = - 1

(1 + i) or (-1 - i)

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The question isn't showing for me.

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https://web2.0calc.com/questions/quick-question_36

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f(0) = a*0^2 + b*0 + c = 0

0a + 0b + c = 0

c = 0

f(1) = a*1^2 + b*1 + c = 2

a + b + c = 2

a + b = 2

f(2) = a*2^2 + b*2 + c = 5

4a + 2b + c = 5

4a + 2b = 5

2a = 1

a = .5

b = 1.5

f(x) = .5x^2 + 1.5x + 0

f(3) = .5(3)^2 + 1.5(3) + 0 = 9

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For each of them, you want to multiply the numerator and denominator by the conjugate of the denominator.

1/(2 + √3) * (2 - √3)/(2 - √3) = (2 - √3)/(2^2 - sqrt(3)^2) = (2-sqrt(3))/1 = 2-sqrt(3)

Try the other ones. :))

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Yep, a quick way to check is that 80/16 = 5, and 112 and 128 are not divisable by 5, meaning that 16 is the largest factor shared by all 3 numbers. 

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Since it provides us with the numbers, we just have to plug them in. 

V = 1/3(30)(6.5)

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I am going to use the law of cosines since we have the sides. 

sqrt(a^2 + absqrt(3) + b^2)^2 = b^2 + a^2 - 2abcos(A)

a^2 + absqrt(3) + b^2) = b^2 + a^2 - 2abcos(A)

absqrt(3)  = - 2abcos(A)

-sqrt(3)/2 = cos(A)

A = arccos(-sqrt(3)/2) = 2.61799388 radians

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Hello :))

I'm catmg, welcome to the forum. :DD

Good luck on your summer math course, and feel free to ask any questions. 

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