Help me help
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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:
\(M=17^{25}+24^4-13^{21}\)
\(=\left(\overline{..7}\right)^{4.6}.\left(\overline{..7}\right)+\left(\overline{..4}\right)^{4.1}-\left(\overline{..3}\right)^{4.5}.\left(\overline{..3}\right)\)
\(\overline{..1}.\overline{..7}+\overline{..6}-\overline{..1}.\overline{..3}\)
\(=\overline{...7}+\overline{..6}-\overline{..3}\)
\(=\overline{...0}\)
\(=>M⋮10\)
B1: a)Dấu hiệu: Điểm ktra môn Toán của 1 nhóm hs
b)Điểm(x) | 7 | 8 | 9 | 10 |
Tần số(n) | 5 | 7 | 5 | 3 | N=20
-Nhận xét: +Có 3 bạn đạt điểm cao nhất là 10 điểm
+Có 5 bạn điểm thấp là 7 điểm
+Có 20 bạn tham gia làm bài
c)AD CT tính số TBC:
\(\dfrac{x_1.n_1+x_2.n_2+...+x_4.n_4}{N}\)
=\(\dfrac{7.5+8.7+9.5+10.3}{20}\)
=8,3
-Mo=8
Bài 4:
a) Xét ΔCAE vuông tại C và ΔDAE vuông tại D có
BE chung
AC=AD(gt)
Do đó: ΔCAE=ΔDAE(Cạnh huyền-cạnh góc vuông)
Suy ra: \(\widehat{CAE}=\widehat{DAE}\)(hai góc tương ứng)
mà tia AE nằm giữa hai tia AC,AB
nên AE là tia phân giác của \(\widehat{CAB}\)
b) Ta có: ΔCAE=ΔDAE(cmt)
nên EC=ED(hai cạnh tương ứng)
Ta có: BC=BD(gt)
nên B nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(1)
Ta có: EC=ED(cmt)
nên E nằm trên đường trung trực của CD(Tính chất đường trung trực của một đoạn thẳng)(2)
Từ (1) và (2) suy ra BE là đường trung trực của CD(đpcm)