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    Calculating masses of unknown quantities of reactants.

    Homework Statement Calculate the number of moles and mass of BaCl2 and NaCl in the original mixture. Homework Equations We prepared a solution of 0.35M Na2SO4. We then obtained an unknown mixture of BaCl2•2H2O and NaCl, weighed 1 g and added it to 200 mL of water and 10 mL HCl. Finally we...
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    Evaluating an Inverse Trigonometric Function

    I apologize for being so confused, independent study is not being kind to me. Thank you very much, your help is immensely appreciated.
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    Evaluating an Inverse Trigonometric Function

    Alright. So if cos(90°-x)=sinx, which in my case is 70°, I still do not understand why Tanya then encourages me to use sin^-1(sinθ)=θ. How is cos70° to be replaced by sinθ? By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°. Would my equation then be accurately...
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    Evaluating an Inverse Trigonometric Function

    Ah a trig identity I see. So cos(90°-70°)=sinθ cos(20°)=sinθ 0.94=sinθ sin^-1(sinθ)=θ sin^-1(0.94)=θ 0.94=θ
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    Evaluating an Inverse Trigonometric Function

    Could you possibly explain the cos(90°-θ)=sinθ?
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    Evaluating an Inverse Trigonometric Function

    Homework Statement Evaluate sin^-1(cos70°) Homework Equations The Attempt at a Solution sin^-1(cos70°)=θ sinθ=cos70° sinθ=1/2 sinθ=∏/3
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    Derivative of Trigonometric Functions

    Oh I guess I could change the term in the denominator to 4/3.
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    Derivative of Trigonometric Functions

    Can anyone tell me if I am on the right track please?
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    Derivative of Trigonometric Functions

    ((2/√3)(√3/3)+((2/√3)(1/3))-(8/3^(3/2)))/(4/3)+((2√3)/3) =(2/9+2/(3√3)-8/(3^3/2))/((2(2√3+1))/3√3)
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    Derivative of Trigonometric Functions

    Yes, I was simply confusing myself. As far as my current work, can it be simplified any more?
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    Derivative of Trigonometric Functions

    Is this as far as it goes? Can it be simplified any more?
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    Derivative of Trigonometric Functions

    =((1+tanx)(secxtanx)-(secx)(sec^2x))/(1+tanx)2 =(secxtanx+secxtan^2x-sec^3x)/(1+tanx)2 Is this looking any closer to a correct answer?
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    Derivative of Trigonometric Functions

    Right, they are not. Hahaha. Don't mind my red cheeks.
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    Derivative of Trigonometric Functions

    Of course I have. 1/cos(x) is the derivative of sec(x) and (sin(x)/cos(x)) is the derivative of (1+tan(x)) are they not?
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    Derivative of Trigonometric Functions

    Homework Statement d/dx(sec(x)/1+tan(x) Evaluate at x=∏/6 Homework Equations The Attempt at a Solution ((1/cos(x))(1+tan(x))-(sec(x))(sin(x)/cos(x)))/(1+tan(x))^2 ((1/cos(x))-(sec(x))(sin(x)/cos(x)))/(1+tan(x)) I used the quotient rule and reduced what I could. Have I done this...
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    Write the polar form of a complex number in the form of a+ib

    Finding sine and cosine of 390° or 30° is not difficult. Converting sin390°=0.5 into 1/2 is simple as well. However, when it comes to converting cos390°=0.866 into √3/2, I am at a loss.
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    Write the polar form of a complex number in the form of a+ib

    I understand how to convert between radians and degrees. What I am unsure of is how to get from the radians to those fractions. When I write the exam, I am not allowed a scientific calculator either, so I definitely need to know this. Also, Sammy, while this may be basic to you, it is not...
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    Write the polar form of a complex number in the form of a+ib

    Homework Statement 4{cos(13∏/6)+isin(13∏/6)} = 4((√3/2)+(i/2)) = 2√3+2i Homework Equations The Attempt at a Solution This is an example from my textbook. The part which I do not understand is how to convert the cos and sin of radians into those fractions. Any help is greatly appreciated.
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    Prove whether or not a transformation is one to one.

    So if T(u) = T(v) then 0 = T(u)-T(v) = T(u-v) so u-v is in kerT and u-v=0 so u=v? But only if the kerT=0.
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    Prove whether or not a transformation is one to one.

    Homework Statement Let T: R3 -> M22 by T\begin{bmatrix} a \\ b \\ c \label{T} \end{bmatrix} = \begin{bmatrix} a-b & b-c \\ a+b & b+c\\ \end{bmatrix} Is this transformation one-to-one? Homework Equations The Attempt at a Solution I am not really certain...
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    Use exponential notation to form a+ib

    I don't know what I did there. Does it instead behave the same as ∏/6? Making my answer 2√3+2i?
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    Use exponential notation to form a+ib

    z1z2 = 2*2ei11∏/6+i∏/3 = 4ei13∏/6 4{cos13∏/6+isin13∏/6} Trig confuses me, so let's see. 4{-cos∏/6-isin∏/6} 4(-√3/2)-(i/2) -2√3-2i
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    Use exponential notation to form a+ib

    z1 = 2{cos11∏/6+isin11∏/6} = 2ei11∏/6
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    Use exponential notation to form a+ib

    z1 = √2{cos(-∏/4)+isin(-∏/4)} = √2ei(-∏/4) z2 = 2{cos∏/3+isin∏/3) = 2ei∏/3 z1z2 = 2√2ei(-∏/4)+i∏/3 =2√2ei∏/12 So far so good?
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    Use exponential notation to form a+ib

    Well, the problem states "Use exponential notation to write (√3-i)(1+i√3) in the form a+ib." Have I been wasting my time?
  26. T

    Use exponential notation to form a+ib

    That is what I was asking and what I thought. Thank you. Now let's see if I can wrestle this into exponential form.
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