Géométrie
Triangles, cercles, géométrie analytique et démonstrations.
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graph each equation. 9) $\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1$ coordin…
Ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), plotted on the given coordinate grid and connected with a smooth curve.…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
Ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$
An ellipse centered at the origin with vertices at \((0, 3)\) and \((0, -3)\), co - vertices at \((2, 0)\) and \((-2, 0)\)
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graph each equation. 9) \\( \\frac{x^2}{4} + \\frac{y^2}{9} = 1 \\) coo…
Ellipse centered at the origin with vertices at \((0, 3)\) and \((0, -3)\), co - vertices at \((2, 0)\) and \((-2, 0)\)
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
Ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) $\\frac{x^2}{4} + \\frac{y^2}{9} = 1$ graph wit…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). To draw it, plot these four points and…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\), plotted as described above. (To actually draw it, connect the plo…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
Ellipse centered at the origin with \( x \)-intercepts \( (\pm2,0) \) and \( y \)-intercepts \( (0,\pm3) \)
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graph each equation. 9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$
The graph is an ellipse centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and it is drawn by connecting these points smoo…
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graph each equation. 9) $\\frac{x^2}{4} + \\frac{y^2}{9} = 1$
The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\). To sketch it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \…
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graph each equation. 9) $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), and it is drawn by connecting these points with a smooth curve. (The actu…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
An ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) coord…
Plot the ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
An ellipse centered at the origin with vertices at \((0, 3)\) and \((0, -3)\), co - vertices at \((2, 0)\) and \((-2, 0)\)
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
Ellipse centered at the origin with \( x\)-intercepts \((\pm2,0)\) and \( y\)-intercepts \((0,\pm3)\)
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\).
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graph each equation. 9) $\\frac{x^2}{4} + \\frac{y^2}{9} = 1$
Plot the ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) $\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1$ coordin…
An ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\)
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), symmetric about the \(x\) - axis and \(y\) - axis. (The actual graph is…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
Ellipse centered at the origin with vertices at \((0, 3)\), \((0, -3)\), co - vertices at \((2, 0)\), \((-2, 0)\)
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn by connecting these points smoo…
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graph each equation. 9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$
An ellipse centered at the origin with \(x\) - intercepts \((\pm2,0)\) and \(y\) - intercepts \((0,\pm3)\)
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graph each equation. 9) $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
# Explanation: ## Step1: Identify the conic section type The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{…
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graph each equation. 9) $\\frac{x^2}{4} + \\frac{y^2}{9} = 1$
An ellipse centered at the origin with vertices at \((0, 3)\) and \((0, -3)\), co - vertices at \((2, 0)\) and \((-2, 0)\)
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), plotted and connected smoothly as described. (To present the final visual…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) coord…
The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), sketched by plotting these points and drawing a smooth, symmetric curve t…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) coo…
The graph is an ellipse centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connecting these po…
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graph each equation. 9) \\( \\frac{x^2}{4} + \\frac{y^2}{9} = 1 \\) gra…
The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) (and the ellipse is drawn through these points as described in the steps a…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\): 1. Recognize it as an ellipse with \(a = 3\) (along \(y\) - axis) and \(b = 2\) (along \(x\) - axis). 2. Plot vertices \((0,3)\), \…
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第一题 已知：△abc中，ab = bc，d是ac的中点，过d作de⊥bc于e，连接ae，取de中点f，连接bf。求证：ae⊥bf 简证：rt…
We have proved that \(AE\perp BF\) by using the properties of similar triangles and cyclic quadrilaterals. The key steps are proving the similarity of right - angled triangles, us…
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problem 25 a semicircle is inscribed in an isosceles triangle with base…
\(\frac{120}{17}\) (or \(120/17\))
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin with vertices at \((0, \pm 3)\) and co - vertices at \((\pm 2,0)\), drawn by connecting these points smoothly. (The actual drawing w…
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graph the following system of inequalities on the coordinate plane and …
# Explanation: ## Step1: 分析第一个不等式\(y > 2x - 3\) 首先，将其视为直线方程\(y = 2x - 3\)。该直线的斜率\(m = 2\)，截距\(b = -3\)。由于不等式是\(y > 2x - 3\)，所以直线应该画成虚线（因为不包含等号），然后在直线上方的区域进行阴影标记（因为\(y\)大于直线上的点）。 #…
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graph $y = \\frac{4}{5}x - 7$.
To graph \( y=\frac{4}{5}x - 7 \): 1. Plot the y - intercept at \( (0,-7) \). 2. Use the slope \( \frac{4}{5} \) to find another point: from \( (0,-7) \), move 5 units right and 4…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse with center at \((0,0)\), vertices at \((0,\pm3)\), co - vertices at \((\pm2,0)\), and is drawn by connecting these points with a smooth curve. (The actual…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connecting these poi…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), and a smooth curve connecting these points symmetric abou…
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph w…
To graph \(\boldsymbol{\frac{x^2}{4}+\frac{y^2}{9}=1}\): 1. Recognize it as a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis, along \(y\) - axis) and \(b…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) coo…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). To graph it, plot these four points an…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\) graph…
To graph \(\frac{x^2}{4}+\frac{y^2}{9}=1\): 1. Recognize it is a vertical ellipse centered at \((0,0)\) with \(a = 3\), \(b = 2\). 2. Plot vertices \((0,3)\), \((0,-3)\) and co - …
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph w…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual drawing should be a smooth…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connectin…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\): 1. Recognize it is an ellipse with \(a = 3\) (semi - major axis along \(y\) - axis) and \(b = 2\) (semi - minor axis along \(x\) - …
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\): 1. Recognize it is an ellipse with major axis along the \(y\) - axis, \(a = 3\), \(b = 2\). 2. Plot the vertices \((0,3…
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph w…
The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (plotted and connected to form the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
# Explanation: ## Step1: Identify the conic section type The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse, \(\frac{x^{2}}{b^{2}}+\frac{y^…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse with vertices at \((0,3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual drawing involves plotting these points and drawing a smooth c…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\). The ellipse is drawn by connecting th…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) coo…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\). To draw it, plot these points and …
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin \((0,0)\), with vertices at \((0, 3)\) and \((0, - 3)\), and co - vertices at \((2, 0)\) and \((-2, 0)\) (plotted and connected in a…
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read this excerpt from the world on turtles back. the birds of the sea …
B. felt compassion for the woman because she was scared
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) coordin…
The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn by connecting these points smoothly. (To a…
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graph each equation. 9) $\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1$ graph w…
The graph is an ellipse centered at \((0,0)\), with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), and a smooth curve connecting these points (…
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph w…
The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (plotted on the given coordinate grid as described in the steps). To…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) gra…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and a smooth curve passing through t…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) coo…
The graph is an ellipse centered at the origin \((0,0)\) with vertices \((0, 3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\) (represented by the smooth curve passing thro…
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graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph w…
The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual graph is a smooth curve p…
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graph each equation. 9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$
The graph is an ellipse centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((- 2,0)\), drawn through these points. (To actually graph…
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draw the line of reflection that reflects $\\triangle abc$ onto $\\tria…
The line of reflection is the x - axis ( \( y = 0 \) ). To draw it, draw a horizontal line passing through the origin (where the x - axis and y - axis intersect) on the coordinate…
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1. we can map $\\triangle abc$ using a sequence of rigid transformation…
A. We mapped one figure onto the other using rigid transformations. ### 2)
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this square - based oblique pyramid has a volume of $125\\ m^3$. what i…
\( 15 \)
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below is the proof that $de = \\frac{1}{2}cb$. the proof is divided int…
- Step 12: \( \frac{CA}{DA} \) (or \( \frac{BA}{EA} \), but using the ratio from Part A, \( \frac{CA}{DA} \)) - Step 13: \( 2 \) - Step 14: \( \frac{DE}{2} \) (the expression to m…
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complete the proof that $\\triangle abd \\sim \\triangle cbe$. (there i…
1. Statement: $\boldsymbol{\angle A \cong \angle C}$; Reason: All right angles are congruent. 2. Statement: $\boldsymbol{\angle ABD \cong \angle CBE}$; Reason: Given. 3. Criterion…
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in the right triangle shown, $m\\angle k = 60^{\\circ}$ and $kl = 2$. h…
B. \( 2\sqrt{3} \)
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triangle $\triangle abc$ is rotated $135^{circ}$ about point $s$ to cre…
10.9
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the circle centered at point a has a radius of length ab. the circles c…
Step 4: \(\boldsymbol{\triangle ABE\cong\triangle ADE}\) Step 5: \(\boldsymbol{\angle DAE\cong\angle BAE}\) (by CPCTC from \(\triangle ABE\cong\triangle ADE\))
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the circle centered at point a has a radius of length ab. the circles c…
Step 2: \( BE = DE \) Step 4: \( \triangle BAE \cong \triangle DAE \)
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fill in the blanks in pengs solution. if we perform a $180^{\\circ}$ ro…
The measure of angle \(x\) is \(120^\circ\), and the relevant ray mappings are: Ray \(\overrightarrow{OI}\) maps onto ray \(\overrightarrow{OH}\), Ray \(\overrightarrow{OJ}\) maps…
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peng was asked to find the measure of angle ( x ) and explain his reaso…
- Ray \( \overrightarrow{OI} \) maps onto \( \boldsymbol{\overrightarrow{OH}} \). - Ray \( \boldsymbol{\overrightarrow{OJ}} \) maps onto ray \( \overrightarrow{OK} \). - Therefore…
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triangle abc has the following vertices: • a(1,9) • b(11,−7) • c(−9,3) …
D. No, because \( \triangle ABC \) doesn't have a pair of perpendicular sides.
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• $overleftrightarrow{tv}$ is parallel to $overleftrightarrow{wx}$. • t…
- First blank: 6 - Second blank: 8 - Third blank: 36.87
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find the values of $\\theta$ if $0 \\leq \\theta \\leq \\pi$. $(\\cot \…
\(\theta = \frac{3\pi}{4}\), so the numerator is \(3\) and the denominator is \(4\).
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Question was provided via image upload.
(a): \( \boxed{168} \) litres ### Part (b)
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sara is considering charging her electric car. if she goes to the charg…
6
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consider the following pyramids, whose bases are a square and an equila…
A. The base areas are not the same.
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use the law of sines to find the indicated side length in this triangle…
\(11.08\)
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quadrilateral l is the image of quadrilateral l under a dilation. what …
A. A
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in the diagram below, $overrightarrow{qr}$ is perpendicular to $overlin…
\(24.00\)
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triangle $\\triangle abc$ is the result of dilating $\\triangle abc$ ab…
s: - \( \overline{AB} \) and \( \overline{A'B'} \) are on the same line: True - \( \overline{AC} \) and \( \overline{A'C'} \) are on distinct parallel lines: False
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△ijk is a translation of △ijk. write the translation rule. (x, y) ↦ (x …
\( (x, y) \to (x + 9, y + 13) \) (so the blanks are 9 and 13)
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here is a cube: blue cube image what two of the following shapes of cro…
A. Triangle C. Rectangle
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a circle is centered at o(0, 0) and has a radius of $2\\sqrt{3}$. where…
C. Outside the circle
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angle a is circumscribed about circle o, what is the measure of ∠a? □° …
\( 88^\circ \)
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draw the image of quadrilateral (and triangle?) under a translation by …
The image of \( \triangle ABC \) is obtained by transforming each vertex \( (x,y) \) to \( (x + 1,y + 6) \), getting \( A'(2,2) \), \( B'(-6,4) \), \( C'(-4,7) \), and then drawin…
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which rotation of the plane can we use to prove angles u and v are cong…
A. A \(180^\circ\) rotation about \(O\) maps ray \(\overrightarrow{OM}\) onto \(\overrightarrow{OP}\) and vice versa, and the same for rays \(\overrightarrow{ON}\) and \(\overrigh…
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the following triangular prism has a volume of 10 cubic units and a hei…
A. \(\frac{10}{3}\) cubic units
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rectangle j has an area of 2 square units. amira drew a scaled version …
4
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segment \\(\\overline{ab}\\) is the result of rotating \\(\\overline{ab…
C. $\overline{AB}$ and $\overline{A'B'}$ have the same lengths.
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vera wants to prove that any rectangle is also a parallelogram. rectang…
B. For quadrilateral \(ABCD\), if \(m\angle A = m\angle B = m\angle C = m\angle D = 90^\circ\), then \(\overline{AB} \parallel \overline{DC}\) and \(\overline{AD} \parallel \overl…
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in the right triangle shown, df = ef = 3. how long is de? choose 1 answ…
A. \( 3\sqrt{2} \)
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△elm is rotated $-90^{circ}$ about the origin. draw the image of this r…
To draw the image of \(\triangle ELM\) after a \(-90^{\circ}\) rotation about the origin: 1. Find the coordinates of the vertices of \(\triangle ELM\) (e.g., \( E=(0,3) \), \( L=(…
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which triangles are similar to △abc? choose 1 answer: a △def only b △gh…
C. Both
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the triangle shown below has an area of 24 units². find the missing len…
\( 12 \)
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what is the magnitude of \\(\\vec{v}\\)?\\(\\vec{v}\\) points to (8, -2…
\(8.25\)
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use the law of cosines to find the indicated angle in this triangle. 4 …
\(101.54^{\circ}\)
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how many boxes will fit along the width of the shipping crate? | | stor…
16
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each bottle in this package is 5.75 inches tall and 1.5 inches wide. ho…
0.25
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a sequence of transformations is described below: - a rotation about a …
C. Both angle measures and segment lengths
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find the value of x in the triangle shown below. x = box° (there is a t…
$\boxed{65}$ (approximate, considering rounding)