Solve all that

Creato il: 9 maggio 2025

Risposto usando O3 da Chat01

Domanda

IMG_5295.jpegIMG_5295.jpeg
IMG_5294.pngIMG_5294.png
IMG_5293.jpegIMG_5293.jpeg
IMG_5292.jpegIMG_5292.jpeg

Solve all that

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from PIL import Image import matplotlib.pyplot as plt img = Image.open('/mnt/data/IMG_5295.jpeg') plt.imshow(img); plt.axis('off') (-0.5, 1169.5, 1463.5, -0.5)

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plt.figure(figsize=(4,8)) plt.imshow(img.crop((580,100,720,350))); plt.axis('off') (-0.5, 139.5, 249.5, -0.5)

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plt.imshow(img.crop((480,80,730,380))); plt.axis('off') (-0.5, 249.5, 299.5, -0.5)

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plt.imshow(img.crop((350,0,760,450))); plt.axis('off') (-0.5, 409.5, 449.5, -0.5)

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plt.imshow(img.crop((40,320,340,580))); plt.axis('off') (-0.5, 299.5, 259.5, -0.5)

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plt.imshow(img.crop((120,400,500,780))); plt.axis('off') (-0.5, 379.5, 379.5, -0.5)

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img2 = Image.open('/mnt/data/IMG_5294.png') plt.imshow(img2); plt.axis('off') (-0.5, 1169.5, 2531.5, -0.5)

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plt.imshow(img2.crop((0,0,460,780))); plt.axis('off') (-0.5, 459.5, 779.5, -0.5)

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plt.imshow(img2.crop((0,0,640,1000))); plt.axis('off') (-0.5, 639.5, 999.5, -0.5)

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plt.imshow(img2.crop((0,900,640,1700))); plt.axis('off') (-0.5, 639.5, 799.5, -0.5)

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plt.imshow(img2.crop((0,850,640,1200))); plt.axis('off') (-0.5, 639.5, 349.5, -0.5)

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plt.imshow(img2.crop((0,1150,640,2000))); plt.axis('off') (-0.5, 639.5, 849.5, -0.5)

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plt.imshow(img2.crop((0,1700,640,2600))); plt.axis('off') (-0.5, 639.5, 899.5, -0.5)

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img3 = Image.open('/mnt/data/IMG_5293.jpeg') plt.imshow(img3); plt.axis('off') (-0.5, 1169.5, 1579.5, -0.5)

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img4 = Image.open('/mnt/data/IMG_5292.jpeg') plt.imshow(img4); plt.axis('off') (-0.5, 1169.5, 2410.5, -0.5)

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plt.imshow(img2.crop((360,60,620,380))); plt.axis('off') (-0.5, 259.5, 319.5, -0.5)

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plt.imshow(img2.crop((340,80,480,340))); plt.axis('off') (-0.5, 139.5, 259.5, -0.5)

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plt.imshow(img2.crop((440,60,680,380))); plt.axis('off') (-0.5, 239.5, 319.5, -0.5)

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plt.imshow(img2.crop((450,50,720,530))); plt.axis('off') (-0.5, 269.5, 479.5, -0.5)

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plt.figure(figsize=(2,4)) plt.imshow(img2.crop((620,0,1000,780))); plt.axis('off') (-0.5, 379.5, 779.5, -0.5)

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import math, pprint, json, os, sys 105.29957264870546

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plt.imshow(img3.crop((80,410,240,700))); plt.axis('off') (-0.5, 159.5, 289.5, -0.5)

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plt.imshow(img3.crop((450,700,700,1000))); plt.axis('off') (-0.5, 249.5, 299.5, -0.5)

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plt.imshow(img3.crop((450,830,730,1250))); plt.axis('off') (-0.5, 279.5, 419.5, -0.5)

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plt.imshow(img3.crop((450,850,1000,1400))); plt.axis('off') (-0.5, 549.5, 549.5, -0.5)

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plt.imshow(img3.crop((190,950,440,1400))); plt.axis('off') (-0.5, 249.5, 449.5, -0.5)

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plt.imshow(img3.crop((190,1150,440,1550))); plt.axis('off') (-0.5, 249.5, 399.5, -0.5)

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plt.imshow(img3.crop((80,880,450,1240))); plt.axis('off') (-0.5, 369.5, 359.5, -0.5)

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plt.imshow(img3.crop((80,650,450,880))); plt.axis('off') (-0.5, 369.5, 229.5, -0.5)

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plt.imshow(img3.crop((350,700,730,1100))); plt.axis('off') (-0.5, 379.5, 399.5, -0.5)

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vol3 = (1/3)*3.14*(3.1**2)*5.9 vol3 59.34495333333334

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vol4 = (1/3)*3.14*(12**2)*9 vol4 1356.48

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plt.imshow(img4.crop((0,300,640,600))); plt.axis('off') (-0.5, 639.5, 299.5, -0.5)

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plt.imshow(img4.crop((300,60,700,350))); plt.axis('off') (-0.5, 399.5, 289.5, -0.5)

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plt.imshow(img4.crop((40,810,480,1400))); plt.axis('off') (-0.5, 439.5, 589.5, -0.5)

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plt.imshow(img4.crop((40,1200,480,1700))); plt.axis('off') (-0.5, 439.5, 499.5, -0.5)

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plt.imshow(img4.crop((40,1600,480,2200))); plt.axis('off') (-0.5, 439.5, 599.5, -0.5)

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plt.imshow(img4.crop((40,2200,480,3000))); plt.axis('off') (-0.5, 439.5, 799.5, -0.5)

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plt.imshow(img4.crop((40,1100,440,1600))); plt.axis('off') (-0.5, 399.5, 499.5, -0.5)

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plt.imshow(img4.crop((40,1400,440,2200))); plt.axis('off') (-0.5, 399.5, 799.5, -0.5)

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x,y=sp.symbols('x y') sol=sp.solve([-3*x+5*y-21,2*x-3*y+13],[x,y]) sol {x: -2, y: 3}

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vol13=3.14*(3.4**2)*1.2 vol13 43.55808

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plt.imshow(img4.crop((380,1150,900,1600))); plt.axis('off') (-0.5, 519.5, 449.5, -0.5)

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plt.imshow(img4.crop((600,1150,1000,1600))); plt.axis('off') (-0.5, 399.5, 449.5, -0.5)

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vol8 = (1/3)*3.14*(16.4**2)*12.9 vol8 3631.49792

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vol5=(1/3)*3.14*(9**2)*15 vol5 1271.7

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3.14*1215 3815.1000000000004

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straight = math.sqrt(78**2 + 13**2) straight 79.07591289387686

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plt.imshow(img4.crop((150,1600,440,2000))); plt.axis('off') (-0.5, 289.5, 399.5, -0.5)

Pensiero

For problem 8, I computed various radius options and focused on adjusting the dimensions for an accurate diagonal. By testing different radius and width combinations, I arrived at a solution where r=27 and w=36, which gives an exact diagonal of 45. The other combinations either didn’t reach 45 or weren't the most suitable. Now, I'm finalizing test prep with the radius choices for verification, possibly providing a more typical example.

Thought for 11m 1s

Risposta

Here are complete step-by-step solutions (rounded as the book asks) for every numbered exercise that appears in the four photos you shared.


Page 386

#WorkAnswer
11 AHeight of a right-circular cone is found from h=s2r2h=\sqrt{s^{2}-r^{2}}.<br>• Cone A: h=172102=18913.75 cmh=\sqrt{17^{2}-10^{2}}=\sqrt{189}\approx13.75\text{ cm}<br>• Cone B: h=182122=18013.42 cmh=\sqrt{18^{2}-12^{2}}=\sqrt{180}\approx13.42\text{ cm}Cone A is taller.
11 B13.7513.42=0.33 cm13.75-13.42=0.33\text{ cm}0.33 cm
12In the triangular gate brace, x=382242=86829.5 in.x=\sqrt{38^{2}-24^{2}}=\sqrt{868}\approx29.5\text{ in.}29.5 in.
13Space diagonal of a cube d=s3=233.5 md=s\sqrt3=2\sqrt3\approx3.5\text{ m}3.5 m

Page 384 – On Your Own

#WorkAnswer
3 ASidewalk: 78+13=91 m78+13=91\text{ m}; grass shortcut: 782+13279.1 m\sqrt{78^{2}+13^{2}}\approx79.1\text{ m}Sidewalk route
3 B9179.111.9 m91-79.1\approx11.9\text{ m}11.9 m
4Space diagonal =362+242+962105.3 in.=\sqrt{36^{2}+24^{2}+96^{2}}\approx105.3\text{ in.}105.3 in.
5Slant height x=102+42=11610.77 mx=\sqrt{10^{2}+4^{2}}=\sqrt{116}\approx10.77\text{ m}10.77 m
6Need any l,wl,w with l2+w245\sqrt{l^{2}+w^{2}}\approx45. Two convenient pairs:<br>27.0 cm×36.0 cm27.0\text{ cm}\times36.0\text{ cm} (exact 3-4-5 triple)<br>17.3 cm×41.5 cm17.3\text{ cm}\times41.5\text{ cm} (rounded)Any two correct pairs; sample shown.
7Face diagonal of 24 in×16 in24\text{ in}\times16\text{ in} rectangle: x=242+16228.8 in.x=\sqrt{24^{2}+16^{2}}\approx28.8\text{ in.}28.8 in.
8In the cone, r=s2h2=902402=650080.6 mmr=\sqrt{s^{2}-h^{2}}=\sqrt{90^{2}-40^{2}}=\sqrt{6500}\approx80.6\text{ mm}80.6 mm
9Space diagonal =542+122+40268.3 cm=\sqrt{54^{2}+12^{2}+40^{2}}\approx68.3\text{ cm}68.3 cm

Pages 444–445 – Find Volume of Cones

Use V=13πr2hV=\dfrac{1}{3}\pi r^{2}h with π3.14\pi\approx3.14 where asked.

#WorkVolume (nearest 0.1)
1A cylinder with same rr and hh as a cone has 3 × the cone’s volume.<br>Given cone fountain V=2.6 m3V=2.6\text{ m}^3 ⇒ cylinder V=7.8 m3V=7.8\text{ m}^3.7.8 m^3
2r=2.0×103 mm,  h=6.0×103 mmr=2.0\times10^{3}\text{ mm},\;h=6.0\times10^{3}\text{ mm}<br>V=13πr2h=13π(2.0×103)2(6.0×103)=8.0×109π mm3V=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\pi(2.0\times10^{3})^{2}(6.0\times10^{3})=8.0\times10^{9}\pi\text{ mm}^38.0×109π mm38.0\times10^{9}\pi\text{ mm}^3
3r=3.1 cm,  h=5.9 cmr=3.1\text{ cm},\;h=5.9\text{ cm}59.3 cm^3
4r=12 in,  h=9 inr=12\text{ in},\;h=9\text{ in}1 356.5 in^3
5r=9 ft,  h=15 ftr=9\text{ ft},\;h=15\text{ ft}1 271.7 ft^3
6r=5 m,  h=3 mr=5\text{ m},\;h=3\text{ m}78.5 m^3
7With diameter dd (so r=d2r=\dfrac{d}{2}):<br>V=13π(d2)2h=πd2h12V=\dfrac{1}{3}\pi\left(\dfrac{d}{2}\right)^{2}h=\boxed{\dfrac{\pi d^{2}h}{12}}

Page 445 – Test Prep

#Solution
8Substitute into V=13πr2hV=\frac{1}{3}\pi r^{2}h: V=13π(16.4)2(12.9) cm3\displaystyle V=\frac{1}{3}\pi(16.4)^{2}(12.9)\text{ cm}^{3} (≈ 3 631.5 cm^3). The four boxes should contain 1, 3, 16.4, 12.9 in that order.
9For r=3 ft,h=5 ftr=3\text{ ft},h=5\text{ ft}:<br>• Cone volume =15π ft3=15\pi\text{ ft}^3<br>• Cylinder volume =45π ft3=45\pi\text{ ft}^3<br>True statements are the ones saying: volume of the cone is 15π15\pi; volume of the cylinder is 45π45\pi; the cylinder’s volume is 3 times that of the cone; the ratio (V_{\text{cone}}:V_{\text{cyl}}=1:3.
10Smallest V=13πr2hV=\dfrac{1}{3}\pi r^{2}h comes from the smallest r2hr^{2}h. Computing each choice shows radius 1 m, height 15 m gives the least volume.

Page 445 – Spiral Review

#WorkAnswer
11Slope 100503010=2.5\dfrac{100-50}{30-10}=2.5.The line rises about 2.5 quarts of iced coffee for every 1 °F increase in temperature.
12Solve {3x+5y=212x3y=13\begin{cases}-3x+5y=21\\[2pt]2x-3y=-13\end{cases}x=2,  y=3x=-2,\;y=3.(2,3)(-2,\,3)
13Cylinder: V=πr2h=3.14(3.4)2(1.2)43.6 m3V=\pi r^{2}h=3.14(3.4)^{2}(1.2)\approx43.6\text{ m}^3.43.6 m^3

Feel free to let me know if you’d like any extra explanation or alternative methods for any of these problems!

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