import torch
from PIL import Image
from transformers import AutoModel, AutoTokenizer
model = AutoModel.from_pretrained('openbmb/MiniCPM-Llama3-V-2_5', trust_remote_code=True, torch_dtype=torch.float16)
model = model.to(device='cuda')
tokenizer = AutoTokenizer.from_pretrained('openbmb/MiniCPM-Llama3-V-2_5', trust_remote_code=True)
model.eval()
image = Image.open('xx.jpg').convert('RGB')
question = 'What is in the image?'
msgs = [{'role': 'user', 'content': question}]
res = model.chat(
image=image,
msgs=msgs,
tokenizer=tokenizer,
sampling=True, # if sampling=False, beam_search will be used by default
temperature=0.7,
# system_prompt='' # pass system_prompt if needed
)
print(res)
## if you want to use streaming, please make sure sampling=True and stream=True
## the model.chat will return a generator
res = model.chat(
image=image,
msgs=msgs,
tokenizer=tokenizer,
sampling=True,
temperature=0.7,
stream=True
)
generated_text = ""
for new_text in res:
generated_text += new_text
print(new_text, flush=True, end='')
The image is a radar chart, often referred to as a spider chart or web chart, used to display multivariate data sets. The chart is titled “Performance Comparison of Different Models on NLP Tasks.” It plots the performance of various models across different Natural Language Processing (NLP) tasks. Each task is represented by a point on the outer edge of the chart, and the performance of each model is indicated by its position relative to the origin.
The tasks include:
Object HalBenchmark
TextVQA val
OCRBench
RealWorld QA
DocVQA test
LLABench
Open-Compass
Math-Vista
MMMU val
MMM dev (zh)
The models evaluated are:
Qwen-VL-Chat 9B
Yi-VL-34B
CogVLM-Chat 17B
Gemini Pro
GPT-4V-1106
MiniCPM-Llama3-V2 2.5 8B
Each model’s performance is quantified by a numerical score, with higher scores indicating better performance. The scores for each model across all tasks are connected by lines, forming a polygonal shape within the chart. The overall layout and design suggest a comparative analysis of these models’ capabilities in NLP tasks.
日本語訳
The image is a radar chart, often referred to as a spider chart or web chart, used to display multivariate data sets. The chart is titled “Performance Comparison of Different Models on NLP Tasks.” It plots the performance of various models across different Natural Language Processing (NLP) tasks. Each task is represented by a point on the outer edge of the chart, and the performance of each model is indicated by its position relative to the origin.
The tasks include:
Object HalBenchmark
TextVQA val
OCRBench
RealWorld QA
DocVQA test
LLABench
Open-Compass
Math-Vista
MMMU val
MMM dev (zh)
The models evaluated are:
Qwen-VL-Chat 9B
Yi-VL-34B
CogVLM-Chat 17B
Gemini Pro
GPT-4V-1106
MiniCPM-Llama3-V2 2.5 8B
Each model’s performance is quantified by a numerical score, with higher scores indicating better performance. The scores for each model across all tasks are connected by lines, forming a polygonal shape within the chart. The overall layout and design suggest a comparative analysis of these models’ capabilities in NLP tasks.
To find the volume of the solid, we need to integrate the area of the cross-sections along the y-axis. Since each cross-section is a right triangle with vertices (0, 0), (1, 0), and (1, -1), the area of one cross-section can be calculated using the formula for the area of a triangle: A = (1/2) * base * height. In this case, the base is 1 and the height is -1, so the area of one cross-section is (1/2) * 1 * -1 = -1/2 cubic units.
However, since the y-axis extends from -1 to 1, there are two cross-sections that contribute to the volume of the solid. The first cross-section, when y = -1, will have an area of 0 because the height would be zero. The second cross-section, when y = 1, will have the same area as the first because the height would also be zero.
Therefore, the total volume of the solid is the sum of the areas of these two cross-sections:
Volume = Area of first cross-section + Area of second cross-section Volume = 0 + 0 Volume = 0 cubic units
The volume of the solid enclosed by the parabola x = 1 – y^2 and the y-axis is 0 cubic units.
To find the volume of the solid, we need to integrate the area of the equilateral triangles formed by the parabola x = 1 – y^2 and the y-axis, from y = -1 to y = 1. The height of each triangle is given by the y-coordinate of the parabola, which is y. The base of each triangle is the distance from the y-axis to the parabola, which is x = 1 – y^2. The area of each triangle is (1/2) * base * height = (1/2) * (1 – y^2) * y. To find the volume, we integrate this area from y = -1 to y = 1:
Volume = ∫[-1, 1] (1/2) * (1 – y^2) * y dy
= (1/2) * ∫[-1, 1] (y – y^3) dy
= (1/2) * [ (1/2) * y^2 – (1/4) * y^4 ] evaluated from -1 to 1
(a) The base of a solid is the region enclosed by the parabola x = 1 – y^2 and the y-axis. Each cross-section perpendicular to the y-axis is an equilateral triangle, as shown in the diagram.
Find the volume of the solid.
Solve the questions in the image.
To find the volume of the solid, we need to integrate the area of the equilateral triangles formed by the parabola x = 1 – y^2 and the y-axis, from y = -1 to y = 1. The height of each triangle is given by the y-coordinate of the parabola, which is y. The base of each triangle is the distance from the y-axis to the parabola, which is x = 1 – y^2. The area of each triangle is (1/2) * base * height = (1/2) * (1 – y^2) * y. To find the volume, we integrate this area from y = -1 to y = 1:
Volume = ∫[-1, 1] (1/2) * (1 – y^2) * y dy
= (1/2) * ∫[-1, 1] (y – y^3) dy
= (1/2) * [ (1/2) * y^2 – (1/4) * y^4 ] evaluated from -1 to 1
この立体の体積を求めるには、x = 1 – y^2 の放物線と y 軸で形成される正三角形の面積を、y = -1 から y = 1 まで積分する必要があります。各三角形の高さは放物線の y 座標、つまり y で与えられます。各三角形の底辺は y 軸から放物線までの距離、つまり x = 1 – y^2 です。各三角形の面積は、(1/2) * 底辺 * 高さ = (1/2) * (1 – y^2) * y です。体積を求めるには、この面積を y = -1 から y = 1 まで積分します。 体積 = ∫-1, 1 * (1 – y^2) * y dy = (1/2) * ∫[-1, 1] (y – y^3) dy = (1/2) * [ (1/2) * y^2 – (1/4) * y^4 ] -1 から 1 までの評価値 = (1/2) * [(1/2) * 1^2 – (1/4) * 1^4 – (1/2) * (-1)^2 + (1/4) * (-1)^4] = (1/2) * [1/2 – 1/4 – 1/2 + 1/4] = (1/2) * [0] 立体の体積は0です。 (a) 立体底面は、放物線 x = 1 – y^2 と y 軸で囲まれた領域です。y 軸に垂直な各断面は、図のように正三角形です。 この立体の体積を求めなさい。 画像の問題を解いてください。 この立体の体積を求めるには、y = -1 から y = 1 までの範囲において、x = 1 – y^2 の放物線と y 軸で形成される正三角形の面積を積分する必要があります。各三角形の高さは放物線の y 座標、つまり y で与えられます。各三角形の底辺は y 軸から放物線までの距離、つまり x = 1 – y^2 です。各三角形の面積は、(1/2) * 底辺 * 高さ = (1/2) * (1 – y^2) * y です。体積を求めるには、この面積を y = -1 から y = 1 まで積分します。 体積 = ∫-1, 1 * (1 – y^2) * y dy = (1/2) * ∫[-1, 1] (y – y^3) dy = (1/2) * [ (1/2) * y^2 – (1/4) * y^4 ] -1 から 1 までの評価値 = (1/2) * [(1/2) * 1^2 – (1/4) * 1^4 – (1/2) * (-1)^2 + (1/4) * (-1)^4] = (1/2) * [1/2 – 1/4 – 1/2 + 1/4] = (1/2) * [0] この立体の体積は 0 です。
GPT-4oの結果はこちら
To find the volume of the solid whose base is the region enclosed by the parabola ( x = 1 – y^2 ) and the ( y )-axis, and whose cross-sections perpendicular to the ( y )-axis are equilateral triangles, we follow these steps:
Identify the bounds of integration:
The parabola ( x = 1 – y^2 ) intersects the ( y )-axis at ( y = -1 ) and ( y = 1 ).
Determine the side length of the equilateral triangle cross-section:
For a fixed ( y ), the length of the side of the equilateral triangle is the distance across the parabola, from ( x = 0 ) to ( x = 1 – y^2 ), which is ( 1 – y^2 ).
Calculate the area of the equilateral triangle:
The area ( A ) of an equilateral triangle with side length ( s ) is given by: [ A = \frac{\sqrt{3}}{4} s^2 ]
In this case, ( s = 1 – y^2 ), so the area of the cross-section at height ( y ) is: [ A(y) = \frac{\sqrt{3}}{4} (1 – y^2)^2 ]
Integrate the area to find the volume:
The volume ( V ) of the solid is given by integrating the area of the cross-sections from ( y = -1 ) to ( y = 1 ): [ V = \int_{-1}^{1} A(y) \, dy = \int_{-1}^{1} \frac{\sqrt{3}}{4} (1 – y^2)^2 \, dy ]
Thus, [ V = \frac{\sqrt{3}}{4} \int_{-1}^{1} (1 – 2y^2 + y^4) \, dy ]
Evaluate the integral:
Split the integral into separate terms: [ V = \frac{\sqrt{3}}{4} \left( \int_{-1}^{1} 1 \, dy – 2 \int_{-1}^{1} y^2 \, dy + \int_{-1}^{1} y^4 \, dy \right) ]
Calculate each term separately: [ \int_{-1}^{1} 1 \, dy = 2 ] [ \int_{-1}^{1} y^2 \, dy = \frac{2}{3} ] [ \int_{-1}^{1} y^4 \, dy = \frac{2}{5} ]
Combine the results: [ V = \frac{\sqrt{3}}{4} \left( 2 – 2 \cdot \frac{2}{3} + \frac{2}{5} \right) ] [ V = \frac{\sqrt{3}}{4} \left( 2 – \frac{4}{3} + \frac{2}{5} \right) ] [ V = \frac{\sqrt{3}}{4} \left( \frac{30}{15} – \frac{20}{15} + \frac{6}{15} \right) ] [ V = \frac{\sqrt{3}}{4} \cdot \frac{16}{15} ] [ V = \frac{4\sqrt{3}}{15} ]
So, the volume of the solid is ( \boxed{\frac{4\sqrt{3}}{15}} ).
放物線 ( x = 1 – y^2 ) と ( y ) 軸で囲まれた領域を底面とし、各 ( y ) 軸に垂直な断面が正三角形である立体の体積を求めるための手順は次のとおりです。
積分の範囲を特定する:
放物線 ( x = 1 – y^2 ) は ( y ) 軸と ( y = -1 ) と ( y = 1 ) で交わる。
正三角形の断面の辺の長さを決定する:
固定された ( y ) の場合、正三角形の辺の長さは放物線 ( x = 0 ) から ( x = 1 – y^2 ) までの距離であり、これは ( 1 – y^2 ) である。
正三角形の面積を計算する:
辺の長さが ( s ) の正三角形の面積 ( A ) は次の式で表される： [ A = \frac{\sqrt{3}}{4} s^2 ]
この場合、( s = 1 – y^2 ) であるため、高さ ( y ) における断面の面積は： [ A(y) = \frac{\sqrt{3}}{4} (1 – y^2)^2 ]
面積を積分して体積を求める:
立体の体積 ( V ) は、( y = -1 ) から ( y = 1 ) までの断面の面積を積分することによって求められる： [ V = \int_{-1}^{1} A(y) \, dy = \int_{-1}^{1} \frac{\sqrt{3}}{4} (1 – y^2)^2 \, dy ]
積分を簡略化する:
被積分関数を展開する： [ (1 – y^2)^2 = 1 – 2y^2 + y^4 ]
したがって、 [ V = \frac{\sqrt{3}}{4} \int_{-1}^{1} (1 – 2y^2 + y^4) \, dy ]
積分を評価する:
積分を個々の項に分割する： [ V = \frac{\sqrt{3}}{4} \left( \int_{-1}^{1} 1 \, dy – 2 \int_{-1}^{1} y^2 \, dy + \int_{-1}^{1} y^4 \, dy \right) ]
各項を別々に計算する： [ \int_{-1}^{1} 1 \, dy = 2 ] [ \int_{-1}^{1} y^2 \, dy = \frac{2}{3} ] [ \int_{-1}^{1} y^4 \, dy = \frac{2}{5} ]
The text in the image is in Chinese characters. The OCR text provided does not match the content in the image, so I cannot use it to extract text. However, I can provide a translation of the visible text in the image:
企業No. 04
注文日 2023/12/12
業者 OX(株)
業者No 210A
品目證号
數量 20,000
単位 0.865
金額 17300
入購予定日 2023/12/15
086465332A 6.350
単位 1.255
単位 7,170
入購予定日 2023/12/26
32804d14088 3300
単位 2.800
単位 9,240
入購予定日 2023/12/20
08633v63244 570
數量 14.058
単位 7,170
入購予定日 2023/2/13
合計 41/880
The image also contains a handwritten note at the bottom with the number ’41/880′.
