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Transformations of quadratic functions

Today, 2020:3030\newlineFigure 66\newlineFigure 66 shows a sketch of the curve \newlineCC with parametric equations\newlinex=8sin⁡2ty=2sin⁡2t+3sin⁡t0≤t≤π2x=8\sin^{2}t \quad y=2\sin 2t+3\sin t \quad0 \leq t \leq \frac{\pi}{2}\newlineThe region \newlineRR, shown shaded in Figure 66, is bounded by \newlineCC, the \newlinexx-axis and the line with equation \newlinex=4x=4\newline(a) Show that the area of \newlineRR is given by\newline∫0a(8−8cos⁡4t+48sin⁡2tcos⁡t)dt\int_{0}^{a}(8-8\cos 4t+48\sin^{2}t \cos t)dt\newlinewhere \newlineaa is a constant to be found.\newline(b) Hence, using algebraic integration, find the exact area of \newlineRR.\newline(44)\newlineNdokumbirawo help apo
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DA PI dLLIC ∠U∠∠ \angle U \angle \angle \newlinepoint\newline1010. The figure below shows three intersecting lines.\newlineWhich statement best describes all the possible measures for the angle labeled y∘ y^{\circ} ?\newlineA. The measure of the angle is x∘ x^{\circ} , and it is between 0∘ 0^{\circ} and 70∘ 70^{\circ} .\newlineB. The measure of the angle is x∘ x^{\circ} , and it is between 110∘ 110^{\circ} and 180∘ 180^{\circ} .\newlineC. The measure of the angle is 110∘+x∘ 110^{\circ}+x^{\circ} , and it is between 0∘ 0^{\circ} and 70∘ 70^{\circ} .\newlineD. The measure of the angle is 110∘+x∘ 110^{\circ}+x^{\circ} , and it is between 110∘ 110^{\circ} and 180∘ 180^{\circ} .
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66) Some people think that ohemisits are more ilkely than other parents to have female children. The Washington State Dept of Heath lists the parents occupations on birth certificates. Between 19801980 and 19901990, 555555 chiliren were bom to lathers who were chemists of these biths, 273273 were girts. During this period, 48.8% 48.8 \% of all births in Washingtion State were girls. Is there evidence at a significance level of 00.0505 , that the proportion of girls bom to chemists is higher than the state proportion?\newline±2⋅.49(1−.49)585p^=173313−7.49±2⋅.021 \begin{array}{l} \pm 2 \cdot \sqrt{\frac{.49(1-.49)}{585}} \quad \hat{p}=\frac{173}{313}-7.49 \\ \pm 2 \cdot .021 \end{array} \newline77) You are told that the population mean score on a recent Calculus test is 5959.55 out of 100100 point You can't believe that and think it should be higher. You take a sample and find that in a sample 118118 students, their mean score was 6565.44 with a standard deviation of 2020.88 . Run a hypothesis te significance level of 00.0303 . Be sure to include your test statistic and p-value and interpret you results.\newlinea. . .0303
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The daily low temperature in Guangzhou, China, varies over time in a periodic way that can be modeled by a trigonometric function.\newlineThe period of change is exactly 11 year. The temperature peaks around July 2626 at 78∘F 78^{\circ} \mathrm{F} , and has its minimum half a year later at 49∘F 49^{\circ} \mathrm{F} . Assuming a year is exactly 365365 days, July 2626 is 206365 \frac{206}{365} of a year after January 11 .\newlineFind the formula of the trigonometric function that models the daily low temperature T T in Guangzhou t t years after January 11,20152015 . Define the function using radians.\newlineT(t)=□ T(t)=\square
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Maya was asked whether the following equation is an identity:\newline(2x+3)(x+1)=2(x+1)2+(x+1) (2 x+3)(x+1)=2(x+1)^{2}+(x+1) \newlineShe performed the following steps:\newline(2x+3)(x+1) (2 x+3)(x+1) \newline↪ Step 1=2x2+2x+3x+3 \stackrel{\text { Step } 1}{\hookrightarrow}=2 x^{2}+2 x+3 x+3 \newline↪ Step 2=2x2+4x+2+x+1 \stackrel{\text { Step } 2}{\hookrightarrow}=2 x^{2}+4 x+2+x+1 \newline↪ Step 3=2(x2+2x+1)+(x+1) \stackrel{\text { Step } 3}{\hookrightarrow}=2\left(x^{2}+2 x+1\right)+(x+1) \newline↪ Step 4=2(x+1)2+(x+1) \stackrel{\text { Step } 4}{\hookrightarrow}=2(x+1)^{2}+(x+1) \newlineFor this reason, Maya stated that the equation is a true identity.\newlineIs Maya correct? If not, in which step did she make a mistake?\newlineChoose 11 answer:\newline(A) Maya is correct.\newline(B) Maya is incorrect. She made a mistake in step 11.\newline(C) Maya is incorrect. She made a mistake in step 22 .\newline(D) Maya is incorrect. She made a mistake in step 44.
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For the rotation 1211∘ 1211^{\circ} , find the coterminal angle from 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} , the quadrant, and the reference angle.\newlineThe coterminal angle is □∘ \square^{\circ} , which lies in Quadrant □ \square, with a reference angle of □∘ \square^{\circ} .
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QUESTION FOUR (88 marks)\newlineThere are four lakeside homes at A, B, C, and D on a man-made quadrilateral lake. There is also a circular road arour the lake that connects the four homes. (As seen in the diagram below). Each of the four families living in the homes argues that their house has the widest angled view of the lake.\newlineIt is known that:\newlinei) Water pipes FA and CE run at a tangent to the circle road.\newlineii) AF A F is parallel BD B D .\newlineiii) Lines BD B D and EF E F both run through the centre of the circle road at O O .\newlineiv) Angle ∠AFO \angle A F O is 47∘ 47^{\circ} \newlinev) Angle ∠CEO \angle \mathrm{CEO} is 32∘ 32^{\circ} \newlineGive geometric reasons for all your working.\newlinea) Find the value of angle ∠FOB \angle F O B \newlineb) Find the value of angle BD B D 00\newlinec) Calculate which house A, B, C, or D has the widest angled view of the lake and the value of that angle.\newlined) Given that length AF is BD B D 11, calculate the distance around the roas from house A to house C.
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A curve is such that dydx=6(2x−1)2 \frac{d y}{d x}=\frac{6}{(2 x-1)^{2}} and P(2,9) P(2,9) is a point on the curve. The normal to the curve at P P meets the y y -axis at Q Q and the x x -axts at R R . Find the coordinates of the midpoint of QR Q R .\newlinedudx=6(x−1)2 grad of normal =−(x−1)269=(−(4−1)26)(2)+ca=−3+c. C=12.  \begin{array}{l} \frac{d u}{d x}=\frac{6}{(x-1)^{2}} \\ \text { grad of normal }=-\frac{(x-1)^{2}}{6} \\ 9=\left(-\frac{(4-1)^{2}}{6}\right)(2)+c \\ a=-3+c \\ \text {. } C=12 \text {. } \\ \end{array} \newlinec=12. ∴ equ of normal ⇒y=(−(2x−1)26)x+12 \begin{array}{l} \quad c=12 \text {. } \\ \therefore \text { equ of normal } \Rightarrow y=\left(-\frac{(2 x-1)^{2}}{6}\right) x+12 \end{array}
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