下表显示了两个变量 xx 和 yy 的实验值。
xx0.51.01.52.0
yy15.919.123.430.2
已知 xx 和 yy 满足方程 y=10+Abxy = 10 + Ab^x。

  1. (i)

    在第 11 页上,画出 lg⁡(y−10)\lg(y-10) 关于 xx 的图象。

    在第 11 页上,画出 lg⁡(y−10)\lg(y-10) 关于 xx 的图象。

  2. (ii)

    利用你在 (i) 中绘制的图象,估算 AA 和 bb 的值。

  3. (iii)

    通过在图象上绘制一条合适的直线,求解方程 Abx=102xAb^x = 10^{2x}。

先不看答案,让 AI 诊断你卡在哪里?

它不会直接告诉你答案,而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

7(i)

1. Calculate the values of $\lg(y-10)$ for each given $x$.

For x=0.5x = 0.5, y=15.9  ⟹  lg⁡(15.9−10)=lg⁡(5.9)≈0.771y = 15.9 \implies \lg(15.9 - 10) = \lg(5.9) \approx 0.771
For x=1.0x = 1.0, y=19.1  ⟹  lg⁡(19.1−10)=lg⁡(9.1)≈0.959y = 19.1 \implies \lg(19.1 - 10) = \lg(9.1) \approx 0.959
For x=1.5x = 1.5, y=23.4  ⟹  lg⁡(23.4−10)=lg⁡(13.4)≈1.127y = 23.4 \implies \lg(23.4 - 10) = \lg(13.4) \approx 1.127
For x=2.0x = 2.0, y=30.2  ⟹  lg⁡(30.2−10)=lg⁡(20.2)≈1.305y = 30.2 \implies \lg(30.2 - 10) = \lg(20.2) \approx 1.305

2. Plot the points and draw the line of best fit.

Plot the points (0.5,0.77)(0.5, 0.77), (1.0,0.96)(1.0, 0.96), (1.5,1.13)(1.5, 1.13), and (2.0,1.31)(2.0, 1.31) on the grid and draw a straight line passing through or close to all points.

答案

A straight line graph plotted with points (0.5,0.771)(0.5, 0.771), (1.0,0.959)(1.0, 0.959), (1.5,1.127)(1.5, 1.127), (2.0,1.305)(2.0, 1.305).

7(ii)

1. Express the given equation in linear form

y=10+Abxy = 10 + Ab^x
y−10=Abxy - 10 = Ab^x
lg⁡(y−10)=lg⁡(Abx)\lg(y - 10) = \lg(Ab^x)
lg⁡(y−10)=lg⁡A+xlg⁡b\lg(y - 10) = \lg A + x \lg b

2. Identify the gradient and Y-intercept from the linear form

Comparing lg⁡(y−10)=(lg⁡b)x+lg⁡A\lg(y - 10) = (\lg b)x + \lg A with Y=mX+cY = mX + c:
Gradient m=lg⁡bm = \lg b
YY-intercept c=lg⁡Ac = \lg A

3. Calculate the gradient and Y-intercept from the data points

Using points (0.5,lg⁡5.9)(0.5, \lg 5.9) and (2.0,lg⁡20.2)(2.0, \lg 20.2):
X1=0.5,Y1≈0.771X_1 = 0.5, Y_1 \approx 0.771
X2=2.0,Y2≈1.305X_2 = 2.0, Y_2 \approx 1.305
Gradient m=1.305−0.7712.0−0.5=0.5341.5=0.356m = \frac{1.305 - 0.771}{2.0 - 0.5} = \frac{0.534}{1.5} = 0.356
YY-intercept c=Y1−mX1=0.771−0.356(0.5)=0.593c = Y_1 - mX_1 = 0.771 - 0.356(0.5) = 0.593

4. Solve for A and b

lg⁡b=0.356  ⟹  b=100.356≈2.27\lg b = 0.356 \implies b = 10^{0.356} \approx 2.27
lg⁡A=0.593  ⟹  A=100.593≈3.92\lg A = 0.593 \implies A = 10^{0.593} \approx 3.92
(Note: Values depend on the line of best fit drawn in part (i). Using the first and last points directly gives A≈3.9,b≈2.3A \approx 3.9, b \approx 2.3. A typical line of best fit might yield A≈4.5,b≈2.1A \approx 4.5, b \approx 2.1.)

答案

A≈4.5A \approx 4.5, b≈2.1b \approx 2.1

7(iii)

1. Relate the given equation to the variables plotted on the graph

The original equation is y=10+Abxy = 10 + Ab^x, which gives y−10=Abxy - 10 = Ab^x.
The equation to solve is Abx=102xAb^x = 10^{2x}.
Substituting y−10y - 10 for AbxAb^x, we get y−10=102xy - 10 = 10^{2x}.

2. Transform the equation to match the vertical axis of the graph

Take the common logarithm (lg⁡\lg) of both sides:
lg⁡(y−10)=lg⁡(102x)\lg(y - 10) = \lg(10^{2x})
lg⁡(y−10)=2x\lg(y - 10) = 2x

3. Identify the line to draw and state the solution

Draw the straight line Y=2xY = 2x (where Y=lg⁡(y−10)Y = \lg(y - 10) and X=xX = x) on the same graph.
The xx-coordinate of the point of intersection between this line and the original graph gives the solution.
From the graph, x≈0.36x \approx 0.36.

答案

x≈0.36x \approx 0.36

AI 分析,可能出错

用 AI 批改,圈出写错的那一步,并自动将你的薄弱点存入错题本,形成你的错因库。

同一知识点的其他题

  1. 1放射性物质的质量 mm 毫克随着时间 tt 小时而减少。测得的 mm 和 tt 的值见下表。 | t (小时) | 2 | 4 | 6 | 8 | 10 | |---|---:|---:|---:|---:|---:| | m (毫…
  2. 2药物通过注射方式输入病人体内。经过时间 tt 小时后,病人体内的药物浓度 xx mg/ll 满足以下方程: x=xoe−kt,x = x_o e^{-kt}, 其中 xox_o 和 kk 是常数。下表记录了 xx 和 tt 的一些数值:…
  3. 3下表显示了两个变量 xx 和 yy 的实验值。 | xx | 2 | 4 | 6 | 8 | |---|---|---|---|---| | yy | 2.25 | 0.81 | 0.47 | 0.33 | 在坐标纸上,绘制 xyxy…
  4. 4下表给出了 xx 和 yy 的测量值。 | xx | 1.2 | 1.6 | 2.0 | 2.6 | 3.5 | 4.5 | |---|---|---|---|---|---|---| | yy | 11.1 | 6.29 | 3.…
  5. 5下表显示了 xx 和 yy 的实验值。 | xx | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---| | yy | 11.8 | 16.2 | 23.5 | 35.5 | 55.2 | 已…