幾何
三角形、圓、解析幾何與證明題。
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graph each equation. 9) \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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read the passage from sugar changed the world. if you walked down beekm…
A. "Simple enough, but this trade up and down the Atlantic coast was part of a much larger world system."
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question what is the equation of the line that passes through the point…
The equation of the line is \(y=\frac{3}{4}x - 5\) (or in standard form \(3x-4y = 20\))
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read the passage from sugar changed the world. if you walked down beekm…
A. "Simple enough, but this trade up and down the Atlantic coast was part of a much larger world system."
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if these two figures are similar, what is the measure of the missing an…
\(120\)
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if these two shapes are similar, what is the measure of the missing len…
1
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if these two figures are similar, what is the measure of the missing an…
\(120\)
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if these two shapes are similar, what is the measure of the missing len…
1
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the medians of $\\triangle jkl$ are $\\overline{jn}$, $\\overline{kp}$,…
\(QM = 12\), \(JQ = 16\), \(KP = 6\)
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the medians of $\\triangle jkl$ are $\\overline{jn}$, $\\overline{kp}$,…
\(QM = 12\), \(JQ = 16\), \(KP = 6\)
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2. x = 4√5 4. 6. loaded furniture into the back of a moving truck...gro…
s: - Problem 2: \( \boldsymbol{x = 4\sqrt{5}} \) - Problem 4: \( \boldsymbol{x\approx11.65} \) (or exact form \( \sqrt{135.75} \)) - Problem 6: \( \boldsymbol{x\approx14.02} \) (o…
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directions: find the value of x. 1. 2. 3. 4. 5. 6. 7. 8. scott is using…
\( x = \sqrt{149} \) (or approximately \( 12.21 \)) ### Problem 2 (Assuming it's a right triangle, but the diagram is cut off. Let's assume similar to problem 1, but since it's no…
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directions: find the value of x. 1. 2.
\(x = \sqrt{149}\) (or approximately \(12.21\))
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pythagorean theorem maz use the pythagorean theorem to solve for x in e…
For the triangle with legs \( 16 \) and \( x \), hypotenuse \( 24 \), \( x = 8\sqrt{5} \)
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in the figure, there is a right triangle with one leg 9, hypotenuse 27,…
If we leave it in radical form, \(x = 18\sqrt{2}\); if we want a decimal approximation, \(x\approx25.45\) (rounded to two decimal places)
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8. enrichment given that the points a (5,7), b (8, 2) and c (1, 2) are …
5
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14. wheeler is looking at three trees in front of the school. he notice…
The trees form an isosceles triangle.
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which figure is a translation of figure a? figure b figure c figure d
Figure C
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13. $\\triangle cde$ is an isosceles triangle with $\\overline{cd} \\co…
The value of \( x \) is \( 11 \). The lengths of the sides are \( CD = 74 \), \( DE = 74 \), and \( CE = 37 \).
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directions: solve each problem given the information. draw a picture. 1…
The value of \( x \) is \( 6 \), and each side of the equilateral triangle measures \( 83 \) units.
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8. enrichment given that the points a (5,7), b (8, 2) and c (1, 2) are …
5
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find the volume of the rectangular prism. v = l×w×h ? units³
600 units$^3$
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select all the trapezoids.
Top-right quadrilateral, Bottom-left quadrilateral
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Question was provided via image upload.
The side lengths of \( \triangle IJK \) in order from shortest to longest are \( IK \), \( IJ \), \( JK \) (or using the angle - side relationship, the sides opposite the angles \…
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find the longest side of $\\triangle pqr$.
\(PQ\)
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select all the trapezoids.
Top-right quadrilateral, Bottom-left quadrilateral
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find the shortest side of $\\triangle rst$.
\(RS\)
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are these shapes similar? (there are two triangles in the image, with s…
no
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are these shapes similar? k 51 m 51 m i 29 m j u 28 m s 51 m 39 m t yes…
no
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if these two shapes are similar, what is the measure of the missing len…
40
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if these two figures are similar, what is the measure of the missing an…
\(112^\circ\)
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$\\triangle pqr \\sim \\triangle igh$. find the ratio of a side length …
\( \frac{1}{2} \)
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tuvw ~ gdef. find the ratio of a side length in tuvw to its correspondi…
\( 3 \) (or as a fraction \( \frac{3}{1} \), but since it's a whole number, \( 3 \) is appropriate)
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3.1 mm a 1.6 mm what is the length of the missing leg? if necessary, ro…
2.7
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2.4 mi. a 1.9 mi. what is the length of the missing leg? if necessary, …
\( 1.5 \)
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what is the length of the hypotenuse? if necessary, round to the neares…
\(7.3\)
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what is the length of the hypotenuse? if necessary, round to the neares…
\( 3.2 \)
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9 yd. 7 yd. what is the length of the missing leg? if necessary, round …
5.7
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graph each equation. 9) \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(2,0)$, $(-2,0)$, $(0,3)$, and $(0,-3)$, connected by a smooth, oval curve.
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
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 an…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse with vertices at $(0, 3)$, $(0, -3)$, co-vertices at $(2, 0)$, $(-2, 0)$, and a smooth curve connecting these points.
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a circles radius is 3 feet. what is the circles circumference? use 3.14…
18.8 feet
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is an ellipse with x-intercepts at $(-2, 0)$ and $(2, 0)$, y-intercepts at $(0, 3)$ and $(0, -3)$, and a vertical major axis. When plotted on the grid, it is a smooth ov…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is a vertical ellipse with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a smooth oval shape passing through these points on the provid…
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)
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 smo…
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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 \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual drawing involves plotting these points and connecting th…
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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 $(2, 0)$, $(-2, 0)$, drawn smoothly through these points on the provided coordinate grid.
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is a vertical ellipse with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a smooth closed curve passing through these points on the prov…
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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 $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval centered at the origin $(0,0)$.
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is a vertical ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, drawn as a smooth curve connecting thes…
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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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 $(2, 0)$, $(-2, 0)$, forming a vertical-oriented ellipse centered at the origin $(0,0)$.
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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 $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval centered at the origin $(0,0)$.
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graph each equation. 9) \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)
The graph is a vertical ellipse centered at $(0,0)$ with x-intercepts at $(-2,0)$ and $(2,0)$, y-intercepts at $(0,3)$ and $(0,-3)$, forming a smooth closed curve through these po…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertically elongated oval shape passi…
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(2, 0)$, $(-2, 0)$, $(0, 3)$, and $(0, -3)$, connected by a smooth, oval-shaped curve.
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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 $(2, 0)$, $(-2, 0)$, forming a vertical oval shape centered at the origin $(0,0)$.
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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 $(0,0)$ with vertices at $(2, 0)$, $(-2, 0)$, $(0, 3)$, and $(0, -3)$, forming a vertical elongated oval shape passing through these…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
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\\)
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 as a smooth curve passing throug…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) coo…
# Explanation: ## Step1: Identify the ellipse standard form The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2…
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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 $(2, 0)$, $(-2, 0)$, forming a vertical-oriented oval shape passing through these points on the pro…
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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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 $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval centered at the origin $(0,0)$.
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analytical geometry - common core diagnostic test - 1 1. $\triangle abc…
B. \(\frac{AB}{A'B'}=\frac{BC}{B'C'}\) ### Question 2
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is a vertical ellipse centered at the origin $(0,0)$ with vertices at $(2,0)$, $(-2,0)$, $(0,3)$, and $(0,-3)$, forming a smooth oval shape passing through these points.
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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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 $(2, 0)$, $(-2, 0)$, drawn smoothly through these points on the provided coordinate grid.
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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 $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval centered at the origin $(0,0)$.
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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 $(0,0)$ with vertices at $(2, 0)$, $(-2, 0)$, $(0, 3)$, and $(0, -3)$, forming a vertical elongated oval shape passing through these…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
The graph is a vertical ellipse with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, drawn as a smooth curve connecting these points on the provided coordi…
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triangle qrs
a. The image $\triangle Q'R'S'$ has vertices $Q'(5,5)$, $R'(9,9)$, $S'(7,11)$ (translated 2 units right and 2 units up, matching the given graph). b. $m\angle Q' = 67.5^\circ$
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11,-11)$, $(16,-6)$, $(16,-16)$ 23. $(-11,-11)$, $(-16,-6)$, $(-16,-16)$ 24. $(-11,11)$, $(-16,16)$, $(-16,6)$ 25. $(11,11)$, $(16,16)$, $(6,16)$ 26. $(-11,-11)$, $(-16,-16)…
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11,-11)$, $(16,-6)$, $(16,-16)$ 23. $(-11,-11)$, $(-16,-6)$, $(-16,-16)$ 24. $(-11,11)$, $(-16,16)$, $(-16,6)$ 25. $(11,11)$, $(16,16)$, $(6,16)$ 26. $(-11,-11)$, $(-16,-16)…
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11, -11)$, $(16, -6)$, $(16, -16)$ 23. $(-11, -11)$, $(-16, -6)$, $(-16, -16)$ 24. $(-11, 11)$, $(-16, 16)$, $(-16, 6)$ 25. $(11, 11)$, $(16, 16)$, $(6, 16)$ 26. $(-11, -11)…
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11, -11)$, $(16, -6)$, $(16, -16)$ 23. $(-11, -11)$, $(-16, -6)$, $(-16, -16)$ 24. $(-11, 11)$, $(-16, 6)$, $(-16, 16)$ 25. $(11, 11)$, $(6, 16)$, $(16, 16)$ 26. $(-11, -11)…
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11,-11)$, $(16,-6)$, $(16,-16)$ 23. $(-11,-11)$, $(-16,-6)$, $(-16,-16)$ 24. $(-11,11)$, $(-16,16)$, $(-16,6)$ 25. $(11,11)$, $(16,16)$, $(6,16)$ 26. $(-11,-11)$, $(-16,-16)…
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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 \((\pm2,0)\), and the ellipse is drawn by connecting these points smoothly. (To a…
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challenge 1 the vertices of pentagon vwxyz are v(4,5), w(6,9), x(8,7), …
A.
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challenge the vertices of pentagon vwxyz are v(4,5), w(6,5), x(6,7), y(…
A. The correct graph is Option A, and the distance between V and V' is $\sqrt{104}$ units (or ~10.2 units)
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11,-11)$ $(16,-6)$ $(16,-16)$ 23. $(-11,-11)$ $(-16,-6)$ $(-16,-16)$ 24. $(-11,11)$ $(-16,16)$ $(-16,6)$ 25. $(11,11)$ $(16,16)$ $(6,16)$ 26. $(-11,-11)$ $(-16,-16)$ $(-6,-1…
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21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
22. $(11, -11)$, $(16, -6)$, $(16, -16)$ 23. $(-11, -11)$, $(-16, -6)$, $(-16, -16)$ 24. $(-11, 11)$, $(-16, 6)$, $(-16, 16)$ 25. $(11, 11)$, $(6, 16)$, $(16, 16)$ 26. $(-11, -11)…
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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 \((\pm2,0)\), formed by connecting the plotted points \((0,3)\), \((0, - 3)\), \(…
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graph each equation. 9) \\(\\frac{x^2}{4} + \\frac{y^2}{9} = 1\\)
该方程表示的是长轴在y轴上的椭圆，顶点为$(0,3)$、$(0,-3)$、$(2,0)$、$(-2,0)$，平滑连接这些点即可得到对应的椭圆图像。
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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 $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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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 $(0,0)$, with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passin…
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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 $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
-
read the passage from sugar changed the world. if you walked down beekm…
"Simple enough, but this trade up and down the Atlantic coast was part of a much larger world system."
-
graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
1. Plot the vertical vertices at $(0, 3)$ and $(0, -3)$, and horizontal co-vertices at $(2, 0)$ and $(-2, 0)$ on the provided coordinate grid. 2. Draw a smooth, symmetrical oval (…
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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 $(2, 0)$, $(-2, 0)$, drawn as a smooth closed curve connecting these points on the provided coordin…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\) a c…
To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\): 1. Recognize it as a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis) and \(b=2\) (semi - minor axis). 2. …
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual graph is a smooth curve connecting these points as descr…
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graph each equation. 9) \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse centered at the origin $(0,0)$ with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a vertical elongated oval shape passing…
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graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
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 in a smooth, c…