求兩題詳解

6. 密碼學是一種將信息隱藏的方法,隨 1 都應用到 7. 小 批 了數學、電子科技及工程學等現代科學。 密碼學的應用相當廣泛,包括New 卡、電腦密碼、手機網路、銀行自動取款機與電子商務、 在密碼學中,直接可以看到的內容為「明碼」,對明碼進行相關處理後得到 碼學來保密相關資料,讓我們的生活既便利又安全。 ( 寫這26個英文。 26 這26個正整數,其對應方式如下表。 h f d i j 5 6 7 8 9 10 11 k e 1 C IT 的內容為「密碼」 已知有一種密碼,將a、b、c….…(不論大寫或小寫) 依序對應到1、2、3、…、 字母 序號 2 3 字母 序號 18 20 b_2 12 13 到一a-1-n-H t u y WS X 1 s_9 y Z 24 | 25 26 21 22 23 15 16 17 1 當明碼對應 當明碼對應的序號為奇數時,密碼所對應的序號為y=-x+x; 2 2 序號x為偶數時,密碼所對應的序號為y= x+ 13。若依上述的對應方式, MAECAM (S-2) (8.2-) (I-SA 求: (1)明碼「aengdb」所譯成的密碼為, (2)若有一明碼所譯成的密碼為「love」,則該明碼為。 (ED) - 0.2-8 (s.e-A A STREET 10 (A)=- A:,三、A(C an A(4.2.) (1.5-). ( AOA ^(-x: ||| 24

最小範圍如何求。給我建議或回覆。我一定會回覆～謝謝某某。不會沒禮貌。🥺

bat 0 New 詳答參見詳解本P2 (2)設d=1,b=2,實數x、y的範圍為3<x<5,-4<y<-2,求 xa + y) 的最 大 大可能範圍為 0< | xu 0< | xa+b|≤13 由3<x<5 且|a|=1,3S|x|<5;由-4<y<-2且|b|=2,4<b=8 若x=5,y=-4且u 與同向,則| xu + b 有最大值為5+8 = 13 若x=4"y=-2且 ca 與b 反向,則a+b|有最小值為0 故所求為0 xa + b < 13 最範圍如何求 X

請問這兩題要怎麼算 log好難🤯

- 3. (1) L'ula=12345006.in<logan+1 tip new n= (2)已Il a=0.00000123456,若n<logu<n+1,其中n為整數,則n= 4 若 loga=1+logb=(-2)+logc,則: (1)c是a的 倍。 (2)b是c的 倍

【三角函數】 求解釋，謝謝🙏

19 , TEAABC #, EFI BC =10(1-сos A), #AABC #3>NEW+BILD 3, H cosA = (化為最 簡分數) 10-10 COSA BA 男生暗

求第二題的作法 我只會往右往上的算法 可以往下的要怎麼算😰😰

B 一例題7 加法原理與乘法原理 如右圖,從4點到B點依下列行進方式,各有幾種走法: (1)只能向上或向右。(5分) (2)只能向上、向下或向右。(5分) 1 news Now

（MI證明題） 可以請問一下倒數第四行到第三行是怎麼移項的？？ 那個55是怎麼來的

/ 6+*# *31) < an> 3 - Hikoi aa = 2*** 2+ 3+ new nel s'a, Q1+ 2+ 3+1 na2 #4, Az+2+ 3 = 539+ \\x 49 1) .4 6.81] 7-11 6n-3 3 bK-3 3 ski + 3“|內衣 12 A On - 2 1 - 3 11 AIRES 1 1 - 1 of a 11 Angin ben- koz, ak= 2 +3=1m (mother tongue) 引言 n-HILD, 6(kH) - 3 (k+1)- Akty*3 (6K-3)+6 7 3 1742 * Tak = 26k-3 + L 6 66-3 2 2 26k 3 + 3 x 3 * 9+(2** 131") +55° 244] [IM [ 11,5 - 1x (30+ x 2 ) 191名人,才有8]或正 t? HMI, 1378: An = 2* * 359 to 1913,52), WHEN THE P. 2

根本看不懂啦＼(￣▽￣;)／

Your report should follow these guidelines, although you may choose how you present it: How to Write a Mathematics Report In writing your report, remember that you are writing up a mathematical story and so, like all good stories, it will need a beginning, a middle and an end. More formally, the main components of this writing style are: Introduction, Formulating the Problem, Solving the problem, Discussion of Results, and Conclusion. We will now consider some of the detail in each of these aspects. Introduction This is the beginning of the story. Give a brief explanation of what the problem is about what the goals of the report are and what will be presented. Assume that your reader does not know what the problem is about or how to solve it. Formulating the problem Translate the situation into a maths problem. Explain how you will begin to solve the problem and break it into simpler stages. Discuss any assumptions made. What quantities are variables and which values are fixed? You may use sub-headings if they assist you. Solving the Problem Show any calculations and mathematical reasoning that you use. (Assume that your reader does not know much maths). Do not show the same types of calculations repetitively. Just give one or two examples of a calculation and then put the rest of the results in a table. Use diagrams or graphs if they assist you. Make general remarks about what you observe in your calculation results and, possibly, why. You may want to criticise your work and go on to improve it in the next section. Explain what you will do next and why. Discussion of Results - Evaluate and Verify Summarise your results if necessary and refer to your mathematical reasoning. Justify procedures used. Interpret your results. First, are they reasonable or does something not look right and need further investigation or checking? Is there a decision to be made? Here is where you should present the decision-making process. Evaluate the strengths and limitations of your solutions. Conclusion Summarise your findings. Refer to the problem outlined in the introduction. Make sure that you answer the question that was asked. Make recommendations. No new material should be presented here.

高一素養題 求解🥺拜託了🙏 真的不知道要怎麼下筆

3. 坡度:以百分數表路面縱向傾斜之程度,亦即在路線中心線方向上,兩點間高程差與其平距比值之百 分數。例如:兩點相距100公尺,其高程差為2.5 公尺,則該兩點間坡度為2.5%,上坡為正,下坡為 負。為適應設計行車速率、地形區別以及慢車行駛等因素,據實驗所得最大縱坡度如表所示。於開挖 路段之坡度設計,為免除路面積水,規定必需設置最小之縱坡,一般以0.5%至0.35%為宜。 設計行車速率 最大縱坡度(%) (公里/時) 平原區 丘陵區 山嶺區 120 3 4 100 3 4 6 80 4 5 7 5 60 7 9

求解C、D、E🙏

一!一上下了進度請載 平合與協倪關僚 下烈各敘述何者正確? (多選) ~(A)過直線 / 外一點 ,恰有一直線平行於已知直線 / @@過平面<外一點 2x怡有一直線平行於已知平面人, 3字 (0過空間中任意一點 5,怡有一直柱王直已知直禁和 、/(D)過空間中任意一點 5,恰有一平面垂直已知直線 / 全 過空間中任意一點 >, 恰有一平面垂直已知平面 F。 As 滲 (了一玉 (2 (2 如 2 2 亡 ANS: (A 0D)

B怎麼算？看解答還是不會

6 為 0.657 pg63 的尾數相同 ,小數點後第 4得8分,答錯不倒拓。未3 77,7C8) 二3ZEHIIOE EE 為0f \ 單位,得圖形 G。# y New讓 之中點 \為 。