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Differentiate these using the product rule.\newlineWrite your answers in fully factorised form with the common factor before the brackets.\newliney=x5(x2+3)y=ex(x5+1)dydx= [2] dydx=e□ ( y=x6(ln⁡x+1)dydx= [2] dydx=ey=e2x(5x+2)y=x5e2xy=5x(x2−2)3dydx=□e□(□) [3] dydx= \begin{array}{l} y=x^{5}\left(x^{2}+3\right) \\ y=e^{x}\left(x^{5}+1\right) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e^{\square} \text { ( } \\ y=x^{6}(\ln x+1) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e \\ y=e^{2 x}(5 x+2) \\ y=x^{5} e^{2 x} \\ y=5 x\left(x^{2}-2\right)^{3} \\ \frac{d y}{d x}=\square e^{\square}(\square) \\ \text { [3] } \frac{d y}{d x}= \\ \end{array}
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IXI.com\newlinePlay Gimkit! - Ent...\newlineadjunct - Google...\newlineroot word bi mea..\newlineAudio Converter:\newlineIXL \mathrm{IXL} \newline−W -\mathrm{W} \newlineMy IXL\newlineLearning\newlineAssessment\newlineAnalytics\newlineAlgebra 1>⋆ 1>\star Y. Write a quadratic function from its x x -intercepts and another point UDD\newlineYou have\newlineWrite the equation of the parabola that passes through the points shown in the table.\newline\begin{tabular}{|c|c|}\newline\hlinex x & y y \\\newline\hline−2-2 & 00 \\\newline\hline 66 & 00 \\\newline\hline 77 & 2727 \\\newline\hline\newline\end{tabular}\newlineWrite your answer in the form y=a(x−p)(x−q) \mathrm{y}=\mathrm{a}(\mathrm{x}-\mathrm{p})(\mathrm{x}-\mathrm{q}) , where a,p \mathrm{a}, \mathrm{p} , and q \mathrm{q} are integers, decimals, or simplified fractions.\newlineSubmit\newlineWork it out\newlineNot feeling ready yet? These can help:\newlineSolve one-step linear equations\newlineLesson: Quadratic equations\newlinePractice in the app
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www-awu.aleks.com\newlineVitalSource Bookshelf: Print Reading for C\'onstruction\newlineA\newlineALEKS - Zachary Zawacki - Learn\newlinePolynomial and Rational Functions\newlineWriting the equation of a rational function given its graph\newlineZachary\newlineThe figure below shows the graph of a rational function ff. It has vertical asymptotes x=−1x=-1 and x=−5x=-5, and horizontal asymptote y=0y=0. The graph has xx-intercept −4-4, and it passes through the point (2,2)(2,2). The equation for f(x)f(x) has one of the five forms shown below. Choose the appropriate form for f(x)f(x), and then write the equation. You can assume that f(x)f(x) is in simplest form.\newline\begin{array}{l} f(x)=\frac{a}{x-b}=\left(\square\right)/\left(\square\right)\ f(x)=\frac{a(x-b)}{x-c}=\left(\square(\square)\right)/\left(\square\right)\ f(x)=\frac{a}{(x-b)(x-c)}=\left(\square\right)/\left(\square(\square)\right)\ f(x)=\frac{a(x-b)}{(x-c)(x-d)}=\left(\square(\square)\right)/\left(\square\right)\ f(x)=\frac{a(x-b)(x-c)}{(x-d)(x-e)}=\left(\square(\square)(\square)\right)/\left(\square\right)\left(\square\right) \end{array}\newlineEspa\~no
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Anastasia tried to solve the differential equation dydx=y2sin⁡(x) \frac{d y}{d x}=y^{2} \sin (x) . This is her work:\newlinedydx=y2sin⁡(x) \frac{d y}{d x}=y^{2} \sin (x) \newlineStep 11: ∫y2dy=∫sin⁡(x)dx \quad \int y^{2} d y=\int \sin (x) d x \newlineStep 22: y33=−cos⁡(x)+C1 \quad \frac{y^{3}}{3}=-\cos (x)+C_{1} \newlineStep 33: y3=−3cos⁡(x)+C \quad y^{3}=-3 \cos (x)+C \newlineStep 44: y=−3cos⁡(x)+C3 \quad y=\sqrt[3]{-3 \cos (x)+C} \newlineIs Anastasia's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Anastasia's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. Anastasia didn't integrate sin⁡(x) \sin (x) correctly.\newline(D) Step 44 is incorrect. The right-hand side of the equation should be −cos⁡(x)3+C \sqrt[3]{-\cos (x)}+C .
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Baylee tried to solve the differential equation dydx=exy2 \frac{d y}{d x}=\frac{e^{x}}{y^{2}} . This is her work:\newlinedydx=exy2 \frac{d y}{d x}=\frac{e^{x}}{y^{2}} \newlineStep 11: ∫y2dy=∫exdx \quad \int y^{2} d y=\int e^{x} d x \newlineStep 22: y33=ex \quad \frac{y^{3}}{3}=e^{x} \newlineStep 33: y3=3ex \quad y^{3}=3 e^{x} \newlineStep 44: y=3ex3+C \quad y=\sqrt[3]{3 e^{x}}+C \newlineIs Baylee's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Baylee's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. The right-hand side of the equation should be ex+C e^{x}+C .\newline(D) Step 44 is incorrect. The right-hand side of the equation should be 3ex3+C3 \sqrt[3]{3 e^{x}}+\sqrt[3]{C} .
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Daniela tried to solve the differential equation dydx=2x⋅e−y \frac{d y}{d x}=2 x \cdot e^{-y} . This is her work:\newlinedydx=2x⋅e−y \frac{d y}{d x}=2 x \cdot e^{-y} \newlineStep 11: ∫e−ydy=∫2xdx \quad \int e^{-y} d y=\int 2 x d x \newlineStep 22: −e−y=x2+C1 \quad-e^{-y}=x^{2}+C_{1} \newlineStep 33: e−y=−x2+C \quad e^{-y}=-x^{2}+C \newlineStep 44: ln⁡(e−y)=ln⁡(−x2+C) \quad \ln \left(e^{-y}\right)=\ln \left(-x^{2}+C\right) \newlineStep 55: −y=ln⁡(−x2+C) \quad-y=\ln \left(-x^{2}+C\right) \newlineStep 66: y=−ln⁡(−x2+C) \quad y=-\ln \left(-x^{2}+C\right) \newlineIs Daniela's work correct? If not, what is her mistake?\newlineChoose 11 answer:\newline(A) Daniela's work is correct.\newline(B) Step 11 is incorrect. The separation of variables wasn't done correctly.\newline(C) Step 22 is incorrect. Daniela didn't integrate e−y e^{-y} correctly.\newline(D) Step 44 is incorrect. The right-hand side of the equation should be ln⁡(−x2)+C \ln \left(-x^{2}\right)+C .
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The base of a solid is the region enclosed by the graphs of y=ln⁡(x) y=\ln (x) and y=0.1x3−0.5x2 y=0.1 x^{3}-0.5 x^{2} , between x=2 x=2 and x=5 x=5 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 6−x 6-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) ∫25[ln⁡(x)−0.1x3+0.5x2]2dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x \newline(B) ∫25[ln⁡(x)−0.1x3+0.5x2](6−x)dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x \newline(C) ∫25[0.1x3−0.5x2−ln⁡(x)](6−x)dx \int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x \newline(D) ∫25[ln⁡2(x)+(0.1x3+0.5x2)2]dx \int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x
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The base of a solid is the region enclosed by the graphs of y=ln⁡(x) y=\ln (x) and y=0.1x3−0.5x2 y=0.1 x^{3}-0.5 x^{2} , between x=2 x=2 and x=5 x=5 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 6−x 6-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) ∫25[0.1x3−0.5x2−ln⁡(x)](6−x)dx \int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x \newline(B) ∫25[ln⁡(x)−0.1x3+0.5x2](6−x)dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x \newline(C) ∫25[ln⁡2(x)+(0.1x3+0.5x2)2]dx \int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x \newline(D) ∫25[ln⁡(x)−0.1x3+0.5x2]2dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x
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Kendall tried to find all the equations of vertical lines tangent to the curve given by x2+2xy2=25 x^{2}+2 x y^{2}=25 . This is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=−x−y22xy \frac{d y}{d x}=\frac{-x-y^{2}}{2 x y} \newlineStep 22: Forming a system of equations.\newline{x2+2xy2=252xy=0−x−y2≠0 \left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlinex=−5,x=0 x=-5, x=0 , and x=5 x=5 \newlineIs Kendall's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.
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Harold tried to find all the points on the curve given by xy2−x2y=−54 x y^{2}-x^{2} y=-54 where the line tangent to the curve is horizontal. This is his solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=y(2x−y)x(2y−x) \frac{d y}{d x}=\frac{y(2 x-y)}{x(2 y-x)} \newlineStep 22: Forming a system of equations.\newline{xy2−x2y=−54x(2y−x)=0y(2x−y)≠0 \left\{\begin{array}{l} x y^{2}-x^{2} y=-54 \\ x(2 y-x)=0 \\ y(2 x-y) \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newline(6,3) (6,3) \newlineIs Harold's solution correct? If not, at which step did he make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.
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The base of a solid is the region enclosed by the graphs of y=2+x2 y=2+\frac{x}{2} and y=2x2 y=2^{\frac{x}{2}} , between the y y -axis and x=4 x=4 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 4−x 4-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) ∫04[2x2−x2−2](4−x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-2\right](4-x) d x \newline(B) ∫04[2x2+x2+2](4−x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}+\frac{x}{2}+2\right](4-x) d x \newline(C) ∫04[2+x2−2x2](4−x)dx \int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{2}}\right](4-x) d x \newline(D) ∫04[2x2−x2+2](4−x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}+2\right](4-x) d x
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