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Managed bookmarks (589) YouTube Cell Diagram: UT News - The Univ... History Esperanza Rising (5) New Tab Mandatory District... District Student Trai... what's another wor... DeltaMath larr Back to Home Reg Precal cycle 6( will add more) Due: May 14 at 8:00 AM Grade: 35% Adding Vectors Commutatively [Guided] Adding Vectors Numerically Vector Components from Initial / Terminal Points Magnitude from Initial / Terminal Points Multiply Vector By Scalar Graphically Multiply Vector By Scalar Find Vector Magnitude and Direction Double Angle [Sine / Cosine] Angle Sum and Difference Angle Sum and Difference: Tangent Reciprocal Trig Identities Calculator Donovan Kennedy Angle Sum and Difference: Tangent Score: 0//5 Penalty: 1 off Question Show Examples If cos A=(40)/(41) and sin B=(20)/(29) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer Attempt 2 out of 2 (9)/(41) Submit Answer Still Stuck? Copyright Q2024 DeltaMath.com All Rights Reserved. Privacy Policy | Terms of Service Copyrisht E2024 Delta Math.com All Rights Reserved. Privacy Policy / Termis of Service Type here to search

Managed bookmarks\newline(589589) YouTube\newlineCell Diagram:\newlineUT News - The Univ...\newlineHistory\newlineEsperanza Rising (55) New Tab\newlineMandatory District...\newlineDistrict Student Trai...\newlinewhat's another wor...\newlineDeltaMath\newline← \leftarrow Back to Home\newlineReg Precal cycle 66( will add more)\newlineDue: May 1414 at 88:0000 AM\newlineGrade: 35% 35 \% \newlineAdding Vectors Commutatively [Guided]\newlineAdding Vectors Numerically\newlineVector Components from Initial / Terminal Points\newlineMagnitude from Initial / Terminal Points\newlineMultiply Vector By Scalar Graphically\newlineMultiply Vector By Scalar\newlineFind Vector Magnitude and Direction\newlineDouble Angle [Sine / Cosine]\newlineAngle Sum and Difference\newlineAngle Sum and Difference: Tangent\newlineReciprocal Trig Identities\newlineCalculator\newlineDonovan Kennedy\newlineAngle Sum and Difference: Tangent\newlineScore: 0/5 0 / 5 Penalty: 11 off\newlineQuestion\newlineShow Examples\newlineIf cos⁡A=4041 \cos A=\frac{40}{41} and sin⁡B=2029 \sin B=\frac{20}{29} and angles A \mathrm{A} and B \mathrm{B} are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer Attempt 22 out of 22\newline941 \frac{9}{41} \newlineSubmit Answer\newlineStill Stuck?\newlineCopyright Q20242024 DeltaMath.com All Rights Reserved. Privacy Policy | Terms of Service\newlineCopyrisht E20242024 Delta Math.com All Rights Reserved. Privacy Policy / Termis of Service\newlineType here to search

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Q. Managed bookmarks\newline(589589) YouTube\newlineCell Diagram:\newlineUT News - The Univ...\newlineHistory\newlineEsperanza Rising (55) New Tab\newlineMandatory District...\newlineDistrict Student Trai...\newlinewhat's another wor...\newlineDeltaMath\newline← \leftarrow Back to Home\newlineReg Precal cycle 66( will add more)\newlineDue: May 1414 at 88:0000 AM\newlineGrade: 35% 35 \% \newlineAdding Vectors Commutatively [Guided]\newlineAdding Vectors Numerically\newlineVector Components from Initial / Terminal Points\newlineMagnitude from Initial / Terminal Points\newlineMultiply Vector By Scalar Graphically\newlineMultiply Vector By Scalar\newlineFind Vector Magnitude and Direction\newlineDouble Angle [Sine / Cosine]\newlineAngle Sum and Difference\newlineAngle Sum and Difference: Tangent\newlineReciprocal Trig Identities\newlineCalculator\newlineDonovan Kennedy\newlineAngle Sum and Difference: Tangent\newlineScore: 0/5 0 / 5 Penalty: 11 off\newlineQuestion\newlineShow Examples\newlineIf cos⁡A=4041 \cos A=\frac{40}{41} and sin⁡B=2029 \sin B=\frac{20}{29} and angles A \mathrm{A} and B \mathrm{B} are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer Attempt 22 out of 22\newline941 \frac{9}{41} \newlineSubmit Answer\newlineStill Stuck?\newlineCopyright Q20242024 DeltaMath.com All Rights Reserved. Privacy Policy | Terms of Service\newlineCopyrisht E20242024 Delta Math.com All Rights Reserved. Privacy Policy / Termis of Service\newlineType here to search
  1. Calculate sin⁡A\sin A: Calculate sin⁡A\sin A using the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2A + \cos^2A = 1.\newlinesin⁡A=1−cos⁡2A=1−(4041)2=1−16001681=811681=941\sin A = \sqrt{1 - \cos^2A} = \sqrt{1 - (\frac{40}{41})^2} = \sqrt{1 - \frac{1600}{1681}} = \sqrt{\frac{81}{1681}} = \frac{9}{41}.
  2. Calculate cos⁡B\cos B: Calculate cos⁡B\cos B using the Pythagorean identity sin⁡2B+cos⁡2B=1\sin^2B + \cos^2B = 1.\newlinecos⁡B=1−sin⁡2B=1−(2029)2=1−400841=441841=2129\cos B = \sqrt{1 - \sin^2B} = \sqrt{1 - (\frac{20}{29})^2} = \sqrt{1 - \frac{400}{841}} = \sqrt{\frac{441}{841}} = \frac{21}{29}.
  3. Find sin⁡(A+B)\sin(A+B) and cos⁡(A+B)\cos(A+B): Use angle sum identities to find sin⁡(A+B)\sin(A+B) and cos⁡(A+B)\cos(A+B).sin⁡(A+B)=sin⁡A⋅cos⁡B+cos⁡A⋅sin⁡B=941⋅2129+4041⋅2029=1891189+8001189=9891189\sin(A+B) = \sin A \cdot \cos B + \cos A \cdot \sin B = \frac{9}{41}\cdot\frac{21}{29} + \frac{40}{41}\cdot\frac{20}{29} = \frac{189}{1189} + \frac{800}{1189} = \frac{989}{1189}.
  4. Continue with cos⁡(A+B)\cos(A+B): Continue with angle sum identities for cos⁡(A+B)\cos(A+B).cos⁡(A+B)=cos⁡A⋅cos⁡B−sin⁡A⋅sin⁡B=(4041)(2129)−(941)(2029)=8401189−1801189=6601189\cos(A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B = \left(\frac{40}{41}\right)\left(\frac{21}{29}\right) - \left(\frac{9}{41}\right)\left(\frac{20}{29}\right) = \frac{840}{1189} - \frac{180}{1189} = \frac{660}{1189}.
  5. Calculate tan⁡(A+B)\tan(A+B): Calculate tan⁡(A+B)\tan(A+B) using the ratio of sin⁡(A+B)\sin(A+B) to cos⁡(A+B)\cos(A+B).\newlinetan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)=9891189/6601189=989660\tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)} = \frac{989}{1189} / \frac{660}{1189} = \frac{989}{660}.

More problems from Transformations of functions

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What does the transformation f(x)↦f(x)−6 f(x) \mapsto f(x) - 6 do to the graph of f(x) f(x) ?\newlineChoices:\newline(A) translates it 6 units down\text{translates it } 6 \text{ units down}\newline(B) translates it 6 units right\text{translates it } 6 \text{ units right}\newline(C) translates it 6 units left\text{translates it } 6 \text{ units left}\newline(D) translates it 6 units up\text{translates it } 6 \text{ units up}
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Find g(x)g(x), where g(x)g(x) is the reflection across the yy-axis of f(x)=−4x+1f(x) = -4x + 1.\newlineChoices:\newline[g(x) = 4x - 1]\text{[g(x) = 4x - 1]}\newline[g(x) = -4x + 1]\text{[g(x) = -4x + 1]}\newline[g(x) = 4x + 1]\text{[g(x) = 4x + 1]}\newline[g(x) = -4x - 1]\text{[g(x) = -4x - 1]}
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Posted 2 months ago

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What kind of transformation converts the graph of f(x)=−8(x−4)2+6 f(x) = -8(x - 4)^2 + 6 into the graph of g(x)=−8(x+6)2+6 g(x) = -8(x + 6)^2 + 6 ?\newlineChoices:\newline[A]translation 10 units left\text{[A]translation 10 units left}\newline[B]translation 10 units right\text{[B]translation 10 units right}\newline[C]translation 10 units up\text{[C]translation 10 units up}\newline[D]translation 10 units down\text{[D]translation 10 units down}
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Posted 2 months ago

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Find g(x)g(x), where g(x)g(x) is the reflection across the x-axis of f(x)=−6(x−7)2+2f(x) = -6(x - 7)^2 + 2.\newlineChoices:\newline(A) g(x)=6(x−7)2−2g(x) = 6(x - 7)^2 - 2\newline(B) g(x)=−6(x+7)2+2g(x) = -6(x + 7)^2 + 2\newline(C) g(x)=−6(x−7)2−2g(x) = -6(x - 7)^2 - 2\newline(D) g(x)=6(x+7)2−2g(x) = 6(x + 7)^2 - 2
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Posted 2 months ago

Question
lim⁡x→π4cos⁡(x)=? \lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=? \newlineChoose 11 answer:\newline(A) 12 \frac{1}{2} \newline(B) 11\newline(C) 22 \frac{\sqrt{2}}{2} \newline(D) The limit doesn't exist.
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Posted 3 months ago

Question
Tess tried to solve the differential equation dydx=2xy−6x \frac{d y}{d x}=2 x y-6 x . This is her work:\newlinedydx=2xy−6x \frac{d y}{d x}=2 x y-6 x \newlineStep 11: dydx=(y−3)2x\quad \frac{d y}{d x}=(y-3) 2 x \newlineStep 22: ∫1y−3dy=∫2xdx \quad \int \frac{1}{y-3} d y=\int 2 x d x \newlineStep 33: ln⁡∣y−3∣=x2+C1 \quad \ln |y-3|=x^{2}+C_{1} \newlineStep 44: eln⁡∣y−3∣=ex2+C1 \quad e^{\ln |y-3|}=e^{x^{2}}+C_{1} \newlineStep 55: ∣y−3∣=ex2+C1 \quad|y-3|=e^{x^{2}}+C_{1} \newlineStep 66: y−3=±ex2+C2 \quad y-3= \pm e^{x^{2}}+C_{2} \newlineStep 77: y=±ex2+C \quad y= \pm e^{x^{2}}+C \newlineIs Tess's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Tess's work is correct.\newline(B) Step 11 is incorrect. We're not allowed to factor an expression when solving differential equations.\newline(C) Step 44 is incorrect. The right-hand side of the equation should be ex2+C1 e^{x^{2}+C_{1}} .\newline(D) Step 77 is incorrect. Tess didn't account for adding 33 to the right side.
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Posted 1 month ago

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Posted 2 months ago

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