幾何
三角形、円、座標幾何、証明問題。
-
estimate the x- and y-intercepts from the graph. write each intercept a…
$(0, 0)$
-
estimate the x- and y-intercepts from the graph. graph of a circle on a…
x-intercept(s): $(-4, 0)$, $(4, 0)$ y-intercept(s): $(0, -1)$, $(0, 9)$
-
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…
-
graph this inequality: x ≤ 1 plot points on the boundary line. select t…
# Explanation: ## Step1: Identify the boundary line The inequality is \( x \leq 1 \). The boundary line is \( x = 1 \), which is a vertical line. Since the inequality is "less tha…
-
the medians of $\\triangle pqr$ are $\\overline{pt}$, $\\overline{qu}$,…
\(PV = 18\), \(QV = 24\), \(RS = 33\)
-
the medians of $\\triangle def$ are $\\overline{dk}$, $\\overline{el}$,…
\(DK = 18\), \(MJ = 3\), \(ML = 8\)
-
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…
-
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…
-
pythagorean theorem maz use the pythagorean theorem to solve for x in e…
For the triangle with legs \( 16 \) and \( x \), hypotenuse \( 24 \), \( x = 8\sqrt{5} \)
-
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)
-
14. wheeler is looking at three trees in front of the school. he notice…
The trees form an isosceles triangle.
-
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 \( CE = 37 \).
-
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.
-
8. enrichment given that the points a (5,7), b (8, 2) and c (1, 2) are …
500 tickets of \$40 and 354 tickets of \$60 were sold. ### Question 11 Solution:
-
find the volume of the rectangular prism. v = l×w×h ? units³
600 units$^3$
-
find the graph of this system of linear inequalities. \\(\\begin{cases}…
The middle graph (with blue region for \(y \leq -2\) and orange region for \(y < 2x - 1\), overlapping purple region as the solution set)
-
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 \…
-
select all the trapezoids.
1. Top-left blue figure 2. Top-right orange figure 3. Bottom-right blue figure
-
select all the trapezoids.
The trapezoids are the first (purple), second (green parallelogram - like), and fourth (orange) figures. The third (green irregular) figure is not a trapezoid.
-
select all the trapezoids.
All four figures (the blue - bordered, purple - bordered, orange - bordered, and green - bordered quadrilaterals) are trapezoids as each has at least one pair of parallel sides.
-
are these shapes similar? (there are two triangles in the image, with s…
no
-
are these shapes similar? k 51 m 51 m i 29 m j u 28 m s 51 m 39 m t yes…
no
-
if these two shapes are similar, what is the measure of the missing len…
40
-
if these two figures are similar, what is the measure of the missing an…
\(112^\circ\)
-
$\\triangle pqr \\sim \\triangle igh$. find the ratio of a side length …
\( \frac{1}{2} \)
-
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)
-
3.1 mm a 1.6 mm what is the length of the missing leg? if necessary, ro…
2.7
-
9 yd. 7 yd. what is the length of the missing leg? if necessary, round …
5.7
-
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…
-
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.
-
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…
-
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.
-
a circles radius is 3 feet. what is the circles circumference? use 3.14…
18.8 feet
-
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…
-
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…
-
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…
-
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…
-
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.
-
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…
-
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 $(0,0)$.
-
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…
-
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…
-
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 $(0,0)$.
-
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 $(0,0)$.
-
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…
-
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…
-
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…
-
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…
-
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.
-
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 $(0,0)$.
-
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…
-
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…
-
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…
-
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…
-
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…
-
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…
-
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 $(0,0)$.
-
analytical geometry - common core diagnostic test - 1 1. $\triangle abc…
B. \(\frac{AB}{A'B'}=\frac{BC}{B'C'}\) ### Question 2
-
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.
-
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…
-
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…
-
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.
-
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 $(0,0)$.
-
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…
-
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…
-
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. $m\angle Q' = 67.5^\circ$
-
21. plot points (11, 11) (16, 16) (16, 6). label this figure #21. 22. t…
24. $(-11,11)$, $(-16,16)$, $(-16,6)$ 25. $(11,11)$, $(16,16)$, $(6,16)$ 26. $(-11,-11)$, $(-16,-16)$, $(-6,-16)$ 27. $(11,-11)$, $(16,-16)$, $(6,-16)$ 28. $(-11,11)$, $(-16,16)$,…
-
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)…
-
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)…
-
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)…
-
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…
-
challenge 1 the vertices of pentagon vwxyz are v(4,5), w(6,9), x(8,7), …
A.
-
challenge the vertices of pentagon vwxyz are v(4,5), w(0,5), x(6,7), y(…
Correct graph: A Distance between $V$ and $V'$: $\approx10.2$ units
-
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…
-
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)…
-
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 \((0,3)\), \((0, - 3)\), \(…
-
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。
-
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…
-
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…
-
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 (…
-
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…
-
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. …
-
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…
-
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…
-
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…
-
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…
-
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 oval shape passing through t…
-
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 $(0, -3)$, forming a vertically elongated oval shape passing through the…
-
read the sentences below, filling in the blanks with items from this li…
1. The formula for finding the volume of a **cone** is $V = \frac{1}{3}\pi r^2 h$. 2. The distance from the center to the edge of the circular base of a cone is called the **radiu…
-
1. a cone has a volume of 200π cm³ and a base radius of 10 cm. what is …
1. c. 6 cm 2. b. It explains why a cone's volume is one-third that of a cylinder. 3. c. It quadruples 4. d. The cone's volume is one-third that of the cylinder. 5. d. $3\sqrt{5}$ …
-
8. the volume of a cylinder is directly proportional to which of the fo…
8. b. Base area and height 9. a. $36\pi$ $\text{cm}^2$ 10. c. 10 meters 11. d. Their volumes are equal 12. b. 50.24 inches 13. c. 10 meters 14. c. 5 meters
-
1 what is the circumference of a circle with a diameter of 14 meters? a…
1. a. 43.96 meters 2. c. 5 meters 3. c. 15.92 meters 4. a. The area of a two-dimensional slice of a solid at a given height 5. c. Circumference divided by $\pi$ 6. d. $45\pi$ cm³ …
-
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 connecting these points.
-
23. if the point n(2, 7) is rotated clockwise about the origin, it will…
- \( 90^\circ \): \( N'(7, -2) \) - \( 180^\circ \): \( N'(-2, -7) \) - \( 270^\circ \): \( N'(-7, 2) \) - \( 360^\circ \): \( N'(2, 7) \) ### Question 24
-
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…
-
graph each equation. 9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)
The graph is an ellipse with vertices at $(0, 3)$, $(0, -3)$ and co-vertices at $(2, 0)$, $(-2, 0)$, forming a smooth, symmetric oval centered at the origin.
-
21. the sum of jeff’s and his grandfather’s ages is 80 years. jeff’s gr…
Jeff = 13 Grandfather = 67 ### Question 22
-
read the sentences below, filling in the blanks with items from this li…
1. The ratio of the circumference to the diameter of any circle is known as **pi** 2. The approximate value of π (pi) is **3.14** 3. A circle's diameter is twice the length of its…