Recent content by thatguythere
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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...- thatguythere
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- quantities
- Replies: 1
- Forum: Biology and Chemistry Homework Help
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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.- thatguythere
- Post #11
- Forum: Precalculus Mathematics Homework Help
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Evaluating an Inverse Trigonometric Function
sin^-1(sin(20°))=20°=∏/9- thatguythere
- Post #9
- Forum: Precalculus Mathematics Homework Help
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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...- thatguythere
- Post #7
- Forum: Precalculus Mathematics Homework Help
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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- thatguythere
- Post #5
- Forum: Precalculus Mathematics Homework Help
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Evaluating an Inverse Trigonometric Function
Could you possibly explain the cos(90°-θ)=sinθ?- thatguythere
- Post #4
- Forum: Precalculus Mathematics Homework Help
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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- thatguythere
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- Function Inverse Trigonometric
- Replies: 14
- Forum: Precalculus Mathematics Homework Help
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Derivative of Trigonometric Functions
Bump.- thatguythere
- Post #17
- Forum: Calculus and Beyond Homework Help
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Derivative of Trigonometric Functions
Oh I guess I could change the term in the denominator to 4/3.- thatguythere
- Post #16
- Forum: Calculus and Beyond Homework Help
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Derivative of Trigonometric Functions
Let's try this- thatguythere
- Post #15
- Forum: Calculus and Beyond Homework Help
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Derivative of Trigonometric Functions
Can anyone tell me if I am on the right track please?- thatguythere
- Post #13
- Forum: Calculus and Beyond Homework Help
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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)- thatguythere
- Post #12
- Forum: Calculus and Beyond Homework Help
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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?- thatguythere
- Post #10
- Forum: Calculus and Beyond Homework Help
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Derivative of Trigonometric Functions
Is this as far as it goes? Can it be simplified any more?- thatguythere
- Post #8
- Forum: Calculus and Beyond Homework Help
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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?- thatguythere
- Post #6
- Forum: Calculus and Beyond Homework Help