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Area of quadrilaterals and triangles: word problems

Solve the following:\newline(6×5=30)(6\times5=30)\newlinei) Find the domain and range of the function \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}} Page 4747\newline(ii) Prove that If \newlineff is differentiable at a point \newlineaa in Domf\text{Dom} f, then \newlineff is continuous at \newlineaa\newline(ii) Show that polar coordinates \newlineP(3,0)P(3,0) and \newlineQ(−3,π)Q(-3,\pi) represent the same point.\newlineiv) Evaluate \newline∫dxx2+3x+4\int\frac{dx}{x^{2}+3x+4}\newlinev) Find point of Inflection of the curve \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}00 unit 4747 e, \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}11\newline-vi) Find area of the region bounded by \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}22, x-axis, f(x)=1−x2f(x)=\sqrt{1-x^{2}}33, f(x)=1−x2f(x)=\sqrt{1-x^{2}}44\newlineQ.22. Solve the following:\newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}55\newlineH Examine the continuity of \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}66 at \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}77 where \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}88\newlineii) Evaluate \newlinef(x)=1−x2f(x)=\sqrt{1-x^{2}}99\newline(iii) Find the area enclosed by the graph of the circle of radius \newlineff00.\newlineiv) Find reduction formula for \newlineff11 and hence find \newlineff22\newlinev) By using substitution \newlineff33 show that \newlineff44
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QUESTION 33\newlineTABLE 11: THE 20202020/20212021 AND 20212021/20222022 MOGALE CITY WATER TARIFF\newline Mogale City  Locae Municipacity  \frac{\text { Mogale City }}{\text { Locae Municipacity }} \newline\begin{tabular}{|c|c|c|}\newline\hline \begin{tabular}{l} \newlineResidential Water Tariff (Excluding \\\newlineVAT)\newline\end{tabular} & \begin{tabular}{l} \newlineMogale City Approved \\\newlineTariffs July (20202020/20212021)\newline\end{tabular} & \begin{tabular}{|l} \newlineMogale City Approved \\\newlineTariffs July (20212021/20222022) \\\newline\end{tabular} \\\newline\hline Approved Indigents (free 6kl 6 \mathbf{k l} ) & R00 & R00 \\\newline\hline 1−6kl 1-6 \mathrm{kl} & R1717 & \begin{tabular}{l} \newlineR1818 \\\newline\end{tabular} \\\newline\hline 7−15kl 7-15 \mathrm{kl} & R2222 & R2424 \\\newline\hline 16−30kl 16-30 \mathrm{kl} & R28 \mathbf{R} 28 & R2929 \\\newline\hline 31−45kl 31-45 \mathrm{kl} & R3232 & \begin{tabular}{|l|l|}\newline\end{tabular} \\\newline\hline 46−60kl 46-60 \mathrm{kl} & R3535 & R3737 \\\newline\hline 6161 and Above & R3939 & R4141 \\\newline\hline \multicolumn{33}{|c|}{\begin{tabular}{l} \newlineOther consumers: Government Schools, NGOs and Hospitals per kl (Excluding Private \\\newlineSchools \& Private Hospitals)\newline\end{tabular}} \\\newline\hline 1−200kl 1-200 \mathrm{kl} & R2525 & R2626 \\\newline\hline Above 200kl 200 \mathrm{kl} & R4040 & R4343 \\\newline\hline\newline\end{tabular}\newlineSource:\newlinewww.polokwane.gov.za\newline\begin{tabular}{|l|l|r|}\newline\hline 33.22 & Write down the fixed monthly electricity cost for the prepaid system in Rand. & (22) \\\newline\hline 33.33 & \begin{tabular}{l} \newlineIf a household uses 6kl 6 \mathbf{k l} 00 of electricity in a month, how much would they pay \\\newlinefor their electricity consumption, excluding VAT?\newline\end{tabular} & (44) \\\newline\hline & & [99] \\\newline\hline\newline\end{tabular}\newline44\newlineGRAND TOTAL: 5050
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Direct Messages - SEQTA Le\newlineAssessment Master - Exam Wi\newlineassess.scsa.wa.edu.au/engine/index.php/lms/index\newlineSchool Curriculum\newlineand Standards\newlineAuthority\newlineNumeracy\newline() 00:2323:5959 / 00:5050:0000\newline3175162431751624\newline4040 of 4545\newlineBetty uses a diagram, with a scale of 11:2020, to calculate the area of her bathroom.\newlineQuestion 3737\newlineQuestion 3838\newlineQuestion 3939\newlineQuestion 4040\newlineQuestion 4141\newlineQuestion 4242\newlineThe diagram is 15 cm 15 \mathrm{~cm} long and 10 cm 10 \mathrm{~cm} wide. What is the area, in square metres, of Betty's actual bathroom?\newlineQuestion 4343\newlineQuestion 4444\newline150150 square metres\newlineQuestion 4545\newline3030 square metres\newlineConfirmation\newline1010 square metres\newline66 square metres\newlineContent copyright WA School Curriculum and Standards Authority 20242024\newlinea)\newlineAssessment\newlineCopyright ⊕ \oplus SoNET Systems Pty Ltd. 20242024\newlinem\newlineMaster
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e-reader-frontend.macmillanlearning.com/?ro=macmillanlearning.com?rid=7575ee99b8080−04-04aa−4216-4216−867-867e-e99e00b1010f6868e33?req_id=1313f486486c22−619-619b−429-429e... All Bookmarks 1414. Find the area of the shaded region in Figure 1818 enclosed by the circle r=12r=\frac{1}{2} and a petal of the curve r=cos⁡3θr=\cos 3\theta. Hint: Compute the area of both the petal and the region inside the petal and outside the circle. Rogawski et al., Calculus: Early Transcendentals, 44e, (C) 20192019 W. H. Freeman and Company FIGURE 1818 1515. Find the area of the inner loop of the lima\c{c}on with polar equation r=2cos⁡θ−1r=2\cos \theta-1 (Figure 1919). 1616. Find the area of the shaded region in Figure 1919 between the inner and outer loop of the lima\c{c}on r=2cos⁡θ−1r=2\cos \theta-1.
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Files\newlineQuestion\newlineReader\newlineArea bou\newlineWorked\newlineByteLear\newlinepolar coo\newlineArc Leng\newlineMail - Br\newlineMy Office\newlinee-reader-frontend.macmillanlearning.com/?ro=macmillanlearning.com?rid=7575ee99b8080−04-04aa−4216-4216−867-867e-e99e00b1010f6868e33?req_id=1313f486486c22−619-619b−429-429e...\newlineNew Chrome available :\newlineAll Bookmarks\newline1414. Find the area of the shaded region in Figure 1818 enclosed by the circle r=12 r=\frac{1}{2} and a petal of the curve r=cos⁡3θ r=\cos 3 \theta . Hint: Compute the area of both the petal and the region inside the petal and outside the circle.\newlineRogawski et al., Calculus: Early Transcendentals, 4e 4 \mathrm{e} , (C) 20192019 W. H. Freeman and Company\newlineFIGURE 1818\newline1515. Find the area of the inner loop of the limaçon with polar equation r=2cos⁡θ−1 r=2 \cos \theta-1 (Figure 1919).\newline1616. Find the area of the shaded region in Figure 1919 between the inner and outer loop of the limaçon r=2cos⁡θ−1 r=2 \cos \theta-1 .
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Math Al\newlineBest Al Homework Hel\newlineLogging in\newlineA\newlineALEKS - Jameson Colle\newlineByteLearn\newlinehttps://Www-awu.aleks.com/alekscgi/y/lsl.exe/1010_u-IgNslkasNW88D...\newlineImport favorites\newlineMVNU Students Ho...\newlineA\newlineALEKS - Jameson C...\newlineTrigonometric Identities and Equations\newlineSum and difference identities: Problem type 33\newlineJameson\newlineEspañol\newlineUse the information given below to find cos⁡(α−β) \cos (\alpha-\beta) .\newlinecos⁡α=35 \cos \alpha=\frac{3}{5} , with α \alpha in quadrant I\newlinetan⁡β=34 \tan \beta=\frac{3}{4} , with β \beta in quadrant III\newlineGive the exact answer, not a decimal approximation.\newlinecos⁡(α−β)= \cos (\alpha-\beta)= \newline□ \square \newline□ \square \newline□ \square \newline□ \sqrt{\square} \newline[Aa\newline□ \square \newlineExplanation\newlineCheck\newlineC 20242024 McGraw Hill LLC. All Rights Reserved.\newlineTerms of Use\newlinePrivacy Center\newlineAccessibility\newlineType here to search
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My Apps\newlineEnglish II - O\newlinePresentation\newline(S) Home|Scho\newlineWorld Hist: S\newlineWhat to Brinc\newlineTeen Driver E\newlineTest 12 A 12 \mathrm{~A} on\newlineByteLearn\newlineapp.formative.com/formatives/66265406186626540618fa6565b99a5151fb7575e\newlineNew Chrome available\newlineYouTube\newlineHome - Roblox\newlineWatch TV Online, St...\newlineMaps\newlinePrettyLittleThing\newline3838:0606\newline11\newline22\newline33\newline44\newline55\newline66\newline77\newline88\newline99\newline1010\newline1111\newline1212\newline1313\newline77\newline33. Find the area of the shaded sector.\newlined=19ft \mathrm{d}=19 \mathrm{ft} \newline746.63ft2 746.63 \mathrm{ft}^{2} \newline186.66ft2 186.66 \mathrm{ft}^{2} \newline228.0ft2 228.0 \mathrm{ft}^{2}
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1010. Wesley is building a patio shaped like a trapezoid. The dimensions of the patio are shown below. What is the area of the patio in square feet?\newlineEnter the correct answer in the box. Not all answers will be used.\newline1,125ft2 1,125 \mathrm{ft}^{2} \newline240ft2 240 \mathrm{ft}^{2} \newline22,250250\newline260tt2 260 \mathrm{tt}^{2} \newlineA= A=
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