help me
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24 - 16(x - 1/2) = 23
=> 16(x - 1/2) = 24 - 23
=> 16(x - 1/2) = 1
=> x - 1/2 = 1/16
=> x = 1/16 + 1/2
=> x = 9/16
\(24-16(x-\frac{1}{2})=23\)
\(16(x-\frac{1}{2})=24-23\)
\(16(x-\frac{1}{2})=1\)
\(x-\frac{1}{2}=\frac{1}{16}\)
\(x=\frac{1}{16}+\frac{1}{2}\)
\(x=\frac{9}{16}\)
Vậy số thực x cần tìm là \(\frac{9}{16}\)
Chúc bạn hok tốt ~
I am sorry, I can’t help you.” Peter said to me.
A. Peter promised to help me. B. Peter refused to help me.
C. Peter asked me for help. D. I couldn’t help Peter.
To make Peter surprised, we _______ and when he comes, we ________.
A. are going to hide / will jump out and shout B. will hide / are jumping out and shouting
C. are hiding / are going to jump out and shout D. are hiding / are jumping out and shouting
It’s kind of you to help me wash the dishes after the party.
A. Washing the dishes is kind of you to help me.
B. You are so kind when you help me wash the dishes.
C. To help me wash the dishes after the party you are kind.
D. It’s your kind to help me with the dishes after the party.
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So sánh:\(10^{10}\) và \(48.50^5\)
Ta có:
\(10^{10}=10^{2.5}=\left(10^2\right)^5=100^5=\left(2.50\right)^5=2^5.50^5=32.50^5\)
Vì \(32.50^5< 48.50^5\)
\(\Rightarrow10^{10}< 48.50^5\)
Our Greener Would will able to have a clean air. It has many green trees around the Earth. Air around the world will not be able to pollute the air. People will not be affected by breathing. So, I love it so much.
Hơi ngắn
Our green world will be able to have a clean air. It has many green trees around the Earth. Air around the world will not be able to pollute the air. People will not be affected by breathing. So, I love it so much.
3:
a: 5^n luôn có chữ số tận cùng là 5 với mọi n là số tự nhiên
=>5^100 có chữ số tận cùng là 5
b: \(2^{4k}\) có chữ số tận cùng là 6 với mọi k là số tự nhiên
mà 100=4*25
nên 2^100 có chữ số tận cùng là 6
c: 2023 chia 2 dư 1
mà \(9^{2k+1}\) luôn có chữ số tận cùng là 9
nên \(9^{2023}\) có chữ số tận cùng là 9
d: 2023 chia 4 dư 3
\(7^{4k+3}\left(k\in N\right)\) luôn có chữ số tận cùng là 3
Do đó: \(7^{2023}\) có chữ số tận cùng là 3
Quy luật:
+) các số có c/s tận cg là 0,1,5,6 nâng lên lũy thừa bậc nào (≠0) thì c/s tận cg vẫn là nó.
+) các số có tận cg là 2,4,8 nâng lên lt bậc 4n(n≠0) thì đều có c.s tận cg là 6.
+)các số có c/s tận cg là 3,7,9 nâng lên lt bậc 4n(n≠0) thì đều có c/s tận cg là 1.
+) số có tận cg là 3 khi nâng lên lũy thừa bậc 4n+3 sẽ có tận cùng là 7
+) số có tận cg là 7 khi nâng lên lũy thừa bậc 4n+3 sẽ có tận cùng là 3
+) số có tận cg là 2 khi nâng lên lũy thừa bậc 4n+3 sẽ có tận cùng là 8
+) số có tận cg là 8 khi nâng lên lũy thừa bậc 4n+3 sẽ có tận cùng là 2
+) số có c/s tận cg là 0,1,4,5,6,9 khi nâng lên lũy thừa bậc 4n+3 thì c/s tận cg là chính nó
Bài 3: áp dụng quy luật bên trên
\(a.5^{100}=\overline{..5}\)
\(b.2^{100}=2^{4.25}=\overline{..6}\)
\(c.9^{2023}=\overline{..9}\)
\(d.7^{2023}=7^{4.505+3}=\overline{...3}\)
Bài 4:
\(A=17^{2008}-11^{2008}-3^{2008}\)
\(=\left(\overline{...7}\right)^{4.502}-\left(\overline{..1}\right)^{2008}-\left(\overline{..3}\right)^{4.502}\)
\(=\overline{..1}-\overline{...1}-\overline{...1}\)
\(=\overline{..9}\)
Bài 5:
\(M=17^{25}+24^4-13^{21}\)
\(=\left(\overline{..7}\right)^{4.6}.\left(\overline{..7}\right)+\left(\overline{..4}\right)^{4.1}-\left(\overline{..3}\right)^{4.5}.\left(\overline{..3}\right)\)
\(\overline{..1}.\overline{..7}+\overline{..6}-\overline{..1}.\overline{..3}\)
\(=\overline{...7}+\overline{..6}-\overline{..3}\)
\(=\overline{...0}\)
\(=>M⋮10\)
- Gen 1: Có g = 3/2. A = 600. 3/2 = 900 nu.
Mà A1 = T2 = 225 nu => T1 = A2 = 600 - 225 = 375 nu.
G1 = X2 = 475 nu => X1 = G2 = 900 -475 = 425 nu.
- Gen 2 có: Tổng nu = 2A + 2G = 600 + 900 = 1500. và 2A = 2/3. 3G
=> A = 375 và G = 375.
Mà A1 = T2 = 180 nu => T1 = A2 = 375 - 180 = 195 nu.
G1 = X2 = 200 nu => X1 = G2 = 375 - 200 = 175 nu.
* ý 2 và 3 chưa đủ điều kiện để giải:
- Cấu trúc của gen gồm 3 vùng điều hòa, mã hóa và kết thúc.
- mARn chỉ được tổng hợp dựa trên trình tự nu của vùng mã hóa => Chiều dài của mARN sơ khai = đoạn mã hóa của gen và < chiều dài của gen.
- Ở sinh vật nhân thực, sau phiên mã, mARN sơ khai còn phải trải qua quá trình trưởng thành, cắt bỏ các đoạn ko mã hóa aa (intron) và tái tổ hợp các đoạn exon để hình thành nên các loại mARN hoàn chỉnh làm khuôn cho dịch mã.
* Lưu ý:
- Không sử dụng khái niệm ribonucleotit để chỉ đơn phân của ARN. Thực tế đơn phân của ARN hay ADN đều gọi chung là nucleotit do cấu trúc hóa học và chức năng tương tự nhau.
- Không sử dụng khái niệm sao mã để chỉ quá trình phiên mã (tổng hợp ARN) vì khái niệm này ko chỉ được bản chất của quá trình tổng hợp ARN. Thực tế, hiện tượng nhân đôi ADN nếu gọi là sao mã cũng không sai.
(Mong có người đọc được và suy nghĩ!!!!!!!!!!!)