請問一下，這邊為什麼會這樣子寫？

72V + 100 in 2.1 362 Lob Rth 12v Rth= [(10+2)/16]+1=5 vth=24+(-15)=9V 当R:52 Pmax=(7)x5 =405(W) 102 12 a 652 3A R 22 103A "open" M 22 36R 101 1 6 -Vth = 72X To+6+2 = 24(V) m 62 22 13A √th = 3x 12+6x6+3X1 -ob = 15(V) b

初會 想問c小題的分錄要怎麼寫

A NT$1,200. It was properly recorded as a purchase in 2022 but was counted with 2021's ending inventory. 尚未 A NT$3,400 shipment of goods to a customer on January 3 was recorded as a sale in 2022 but was not included in the December 31, 2021, ending inventory balance. The goods cost d. The company had goods costing NT$8,000 on consignment with a customer, and NT$6,000 of NT$2,300. TTC61 800

請問這題正六邊形面積式子怎麼來的

*J (3976) ( +1²) = (a+b)". 18x32 (a+b)² 數學A考科 > 0<a+b≤2+6=" 9 zab3 ab=2 TO 5 2uth, 397b 2345 ab33√3 9.93 30+5=18 atb 1953 = "2 30²= b² Ab=53a α=53,6=3 2 P2ab請記得在答題卷簽名欄位以正楷簽全名 10. 如右圖,六角柱 ABCDEFGHIJKL 之上底面ABCDEF與下底面 GHIJKL 為兩個全等的正六邊形,側面為六片全等的矩形,且上、下底面的正六邊 Ap 5 形與側面矩形皆垂直。設AB=a, AG=b, AI=32,請選出正確的選 項。 30+ $827-18 1 341 (1)3a²+b=18 (2)矩形 ABHG 的面積有最大值為3-03 CAB ab=356號 3矩形ABHG的面積有最大值時,六角柱 ABCDEF-GHIJKL 的體積為27~3 +za+bb = 2x256=456 (4)矩形 ABHG的周長有最大值為2~62(a+b)+ 512G+b) (5)矩形ABHG的周長有最大值時,六角柱 ABCDEFGHIJKL的表面積為 19x9x+axa h (3,1) 第 4 第共 頁頁 7 F E D \B L 32 a H a I 9√√3+54 2 a+- 202 -=20 x² = 30

求解DE🙏🏻

(BIC 設 deg f(x)≥4,若f(x)除以(x-1)餘式為3x+2,除以(x+2)餘式為5x-3,請選出正確的選項。 D (A)x-1除f(x) 餘式為5 (C)(x-1)(x+2)除f(x)餘式為6x-1 4-5-6 (B)x+2除f(x)餘式為-13 (x-174(x+2)+0xbxte a+b+c=5, 42-2b+c=13 (D)(x-1)2(x+2)除f(x) 餘式為-x2+5x+130-35--18. (E) (x-1)² (x+2)² f(x) (x²+5x+1 = (x-179 +3x+2 +11x=5 = (x-1)x+2) of torty f(x)= 5-2015=13 4654 +(x+2)+5x-3 11-27--13 +-513 3α=18 ☆02. 已知a、b、c均為實數,若二次函數f(x)=ax²+bx+c滿足f(7)=f(-7)、f(3)<0、f(5)>0,請選出正確的

求解 41題E為何對

300 60 +60 51m ↓ wkw/hr bo kw 40. ~42. 題為題組 5 bob 360 360=6 Flomin 沁 西元前3世紀時,埃拉托斯特尼擔任亞力山 1360x60 360x 70 x 5=300 To w/miAA hr' kw 卓圖書館的館長。某天他在埃及南部的塞伊 度电 亞力山卓 8.00km. 阿斯旺 [Syene,現今稱為阿斯旺(Aswan),如圖16〕, 注意到夏至中午陽光直射當地一口井時,沒有任 何陰影,說明太陽位於頭頂;同時夏至中午在位 埃及北部的亞力山卓(Alexandria),由垂直地 面桿投射出的陰影,計算出此地的太陽位於與垂 直線偏向 7°12'。埃拉托斯特尼得知阿斯旺與亞力 山卓兩地相距 800公里,就能計算出地球的大小。(註:1°=60',1'稱為1弧分或1角分) 深井 12 12 圖 16 27R= F 7 505 40. 依據題幹敘述,太陽位於與垂直線偏向7°12'相當於360°的多少比例,以及地球半徑約 多少公里?(應選2項) 2xRx50 = 800 36 5. 4020000 1 (A) 7°12'是360°的 R=800x50 Ms 1 36 3600 5 (B) 7°12'是360°的 50 30 元 40 50 1 26400 (C)7°12'是360°的 地球半徑約為6400 公里 YO = Su -50 100.11 (E)地球半徑約為7200公里 41. 依據題幹敘述,下列有關埃拉托斯特尼的觀察與基本假設,哪些正確?(應選3項) (A)亞力山卓應位於北回歸線上 B LE ← (B)阿斯旺應位於北回歸線上 (C)埃拉托斯特尼假設太陽位於無窮遠處,即陽光照射在阿斯旺及亞力山卓為平行光 ( 埃拉托斯特尼尚未考慮陽光通過大氣層產生明顯的折射現象,所以此測量有巨大誤 (E)埃拉托斯特假設亞力山卓在阿斯旺的正北方夏至中午陽光直射(或近乎直射) ㄇ射角乎O

為什麼是a

ABCD 28.圖(十三)為荷蘭藝術家蒙德里安在西元 1930年所創作的作品《紅藍黃的構 成》,他以形狀、色彩的平衡表達對宇 宙和諧的追求。若此畫作展示在展覽問 內,單獨使用紅光、藍光、綠光照射此 畫作,則哪一色光照射此畫作時,畫作 所呈現的黑色面積最大2(已知黃色物 體能反射紅光與綠光) 白色 白色十 藍色- (A)紅光 (B)藍光 (C)綠光 黑色線條 圖(十三) 白色 ABC (D)三種色光照射時所呈現的黑色面積相同 請閱讀下列敘述後,回答29~30題 如圖(十四),某款智能玻璃能依環境條件自動調整進入室內的光 量,氣溫高或室外光線較強時能讓光線進入室內的量減少,避免室內溫度 升高;氣溫低或室外光線較弱時則能讓光線進入室內的量增加,使室內明 亮並有助於室內升溫。其原理是在兩層玻璃中夾一層「高分子液晶」,此 | 液晶在不同溫度下會改變液晶分子的排列方式,因此可以改變光線的通過 量,圖(十五)為兩種溫度下,智能玻璃內液晶分子的排列示意圖。 智能玻璃一 升溫 升溫 口 降溫 降溫 小於25℃時 25~32℃之間 大於32℃時 透明 逐漸霧化 完全霧化 圖(十四) 光線 光線 -玻璃. -液晶分子 hash ·玻璃. 圖(十五) 29. 圖(十四)中,溫度升高時無法看見玻璃後方植物的原因,下列何才 理? (A)照射至植物的光線反射後,無法經玻璃折射傳遞至眼睛 (B)照射至玻璃的光線反射後,無法經玻璃折射傳遞至眼睛 (C)照射至玻璃的光線反射後,無法經玻璃反射傳遞至眼睛 (D)玻璃無法反射照射至植物的光線

為何是c

請閱讀下列敘述後 回答21~22題 甲~丙為三種在不同介質中傳遞的橫波,且均沿著X軸向右傳播。當時 間為時,此三個波的部分波形如圖(八)所示,已知圖中O、P分別為波上的 兩個質點,且OP=100cm ° 平衡①100 位置 甲 乙 平衡 O 平衡! 位置 位置 圖(八) 丙 X 21. 若甲波由時間t(初始時間)開始傳播,則當波上的O點再次回到平衡位置 7 時,波傳遞的距離為何? (A) 25cm (C) 100cm (B)50cm (D) 200cm

想問第十題的選項2是黃色的紙那樣算嗎？

* 35 數學B考科 請記得在答題卷簽名欄位以正楷簽全名 z 10.設P(a,b)為兩函數y=logx和y=sinx 在 為實數,請選出正確的選項。 (10<a+b<3<a<l, o<b<l (2)(bar)為函數y= (3) aπ 振=1 第 4 共7 頁 頁頁 [02]內之交點,其中為圓周率,a、b (2)b=logan P (ar, b) 0.301 0.301 10 圖形上一點 -b 為函數y=log x 圖形上一點 2 元 3,14 y=10gx For (4)(z,b)為函數y=sinx以原點為中心水平伸縮a倍後圖形上一點 (5)(an,b)為函數y=sinx 向左平移單位後圖形上一點 (11. 筊是利用兩個半月形的木頭筊杯,透過將筊杯 預測事情。根據筊杯的正反可分成以下三種狀況 二種為兩正面,又名「笑杯」、第三種為兩反面 出聖杯表「成功」,無杯表「失敗」就不再擲 或「失敗」才停止。若每一個筊杯正、反面出現 "ar = (1/1) b logar = log + = b to 頁

為什麼[A(org)]=n[An(org)]成立 (第三張圖equation 7.9) 感謝

Experiment 7 Experiment 7 The partition of Organic acid between Water and Organic solvent Objectives Understand the partition of a solute between two immiscible solvents. Introduction A chemical analysis that is performed primarily with the aid of volumetric glassware (e.g., pipets, burets, volumetric flasks) is called a volumetric analysis. For a volumetric analysis procedure, a known quantity or a carefully measured amount of one substance reacts with a to-be-determined amount of another substance with the reaction occurring in aqueous solution. The volumes of all solutions are carefully measured with volumetric glassware. The known amount of the substance for an analysis is generally measured and available in two ways: 1. As a primary standard: An accurate mass (and thus, moles) of a solid substance is measured on a balance, dissolved in water, and then reacted with the substance being analyzed. 2. As a standard solution: A measured number of moles of substance is present in a measured volume of solution - a solution of known concentration, generally expressed as the molar concentration (or molarity) of the substance. A measured volume of the standard solution then reacts with the substance being analyzed. The reaction of the known substance with the substance to be analyzed, occurring in aqueous solution, is generally conducted by a titration procedure. The titration procedure required a buret to dispense a liquid, called the titrant, into a flask containing the analyte. A reaction is complete when stoichiometric amounts of the reacting substances are combined. In a titration this is the stoichiometric point. In this experiment the stoichiometric point for the acid-base titration is detected using a phenolphthalein indicator. Phenolphthalein is colorless in an acidic solution but pink in a basic solution. The point in the titration at which the phenolphthalein changes color is called the endpoint of the indicator. Indicators are selected so that the stoichiometric point in the titration coincides (at approximately the same pH) with the endpoint of the indicator.

初會ㄉ銀行調節表問題🥲 請問b,c小題要怎麼寫啊？我快居居了

1 Balance b/d " 7 B Man " 16 A Sun " 28 M Bai " 30 K Ng 2,379 The bank account of C Chan's business for June 2009 and the bank statement for the same month are as follows: 2009 Jun Cash at bank $ 2009 Jun 5 D Wong " 158 12 J Yau " 93 16 B Shek " 307 29 Orange Club " 624 30 Balance c/d 3,561 $ 150 433 88 57 2,833 3,561 Bank Statement Date 2009 Jun Details Dr Cr Balance $ $ $ 1 Balance b/d " 8 Cheque " 11 Cheque " 17 Cheque " 17 Cheque " 18 Cheque " 29 Cheque " 29 Standing order - Union Credit " 30 Credit transfer - RS Ltd " 30 Bank charges 2,379 Cr 158 2,537 Cr 150 2,387 Cr 93 2,480 Cr 433 2,047 Cr 88 1,959 Cr 307 2,266 Cr 44 44 2,222 Cr 70 70 90 90 2,312 Cr 2,242 Cr (a) Cross out all the items that appear in both cash book and the bank statement. (b) Prepare a bank reconciliation statement as at 30 June 2009. Start with the balance as per cash book (c) Update the bank account of C Chan's business and prepare a bank reconciliation statement as at 30 June 2009. (-24 (b) Bank Reconciliation Statement as at 30 June 2009 Balance per cash book add credit transfer less: Standing order Bank charges. Balance per bank statement less: check 3447 $ 90 Total 3,5615 40 (44) (70) (114) $3537 $2242 (624) (57)