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Convert $\frac{a}{b} = \frac{a}{d}$ to $\frac{a}{b} = \frac{c}{d}$

We have :

$\frac{a}{b} = \frac{c}{d} \Leftrightarrow \frac{2 a}{2 b} = \frac{3 c}{3 d}$

Apply the same sequence properties , we got :

$\frac{2 a}{2 b} = \frac{3 c}{3 d} = \frac{2 a - 3 c}{2 b - 3 d} = \frac{2 a + 3 c}{2 b + 3 d}$

$\Rightarrow \frac{2 a - 3 c}{2 b - 3 d} = \frac{2 a + 3 c}{2 b + 3 d}$

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There are $9$ digits could be the ones digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9.$

If the ones digit is $0$ then there are: $9 - 0 = 9 (n u m b e r s)$ satisfy the question.

If the ones digit is $1$ then there are: $9 - 1 = 8 (n u m b e r s)$ satisfy the question.

....

If the ones digit is $9$ then there are: $9 - 9 = 0 (n u m b e r s)$ satisfy the question.

So there are total: $9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 45 (n u m b e r s)$ satisfy the question

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Point B has the numbers of students it receives :

40 * 25% = 40 * 1/4 = 10 ( students )

So the result of this problem is 10 students.

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1.2+2.4+3.6 / 2.4 + 4.8 + 6.12

=1.2 + 2.4 + 3.6 / 4 * 1.2 + 2.4 * 4 + 2*3.2.6

=1.2 + 2.4 + 3.6 / 4 * ( 1.2 + 2.4 + 3.6 )

=1 / 4.

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11+22+33+44+...+999

=11(1+2+...+111)

=11+111+1/2x111/2

=34188

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a) The term number of the sum s is

( 101 - 1 ) : 3 + 1 = 34 (number)

b) number 50 is

( 50 - 1 ) . 3 + 1 = 148

c) total S is

( 1 + 100 ) . 34 : 2 = 1717

answer: a) 34 number

b) 148

c) 1717

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There are three squares can be draw.

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We have: $1 c m = 10 m m$

=> $20 c m = 200 m m$

So the answer is $200 m m$ (200 millimeter).

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There is total: $(100 - 1) + 1 = 100 (n u m b e r s)$

The sum of all numbers is: $(1 + 100) .100 : 2 = 5050$

The average of all numbers is: $5050 : 100 = 50, 5$

So the answer is: $50, 5$

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The least possible number is: $1000$

The biggest possible number is: $9998$

There are: $(9998 - 1000) : 2 + 1 = 4500 (n u m b e r s)$

So there are $4500$ 4-digit-even numbers.

Thanks !

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The sum of 5 numbers will be:

18 x 5=90

The sum of 5 numbers increases more than the old amount by:

1+2+3+4+5=15

Total of 5 numbers increased;

90+15=105

Average of 5 numbers increased:

105:5=21

=>The correct answer is 21.

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mình s đầu tiên

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Arrcoding to the question, we have the ratio of the salt and fresh water in Protaras is 7:193

=> The ratio of the salt and the sea water = 7:(193+7) = 7:200

First, we exchange: $1000 h g = 100 k g$ (sea water).

Because the ratio of the salt and sea water is 7:200

=> $7 k g$ salt are in $200 k g$ sea water.

$200 k g$ sea water are more than $100 k g$ sea water: $200 : 100 = 2 (t i m e s)$

In $100 k g$ sea water there is: $7 : 2 = 3, 5 (k g)$ salt.

(This question has no answer because there is no 3,5 kg salt.

If there is an answer, you must change $1000 h g$ sea water -> $1000 k g$ sea water or change the questions into: How many hectograms of salt are there in 1000hg of sea water)

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We have: $x . y = 14$ and $y . z = 10$ and $z . x = 35$

<=> $x . y . y . z . z . x = 14.10.35$

<=> $x^{2} . y^{2} . z^{2} = 4900$

<=> ${(x . y . z)}^{2} = 70^{2}$

=> $x . y . z = 70$ <1>

From the context and <1>

=> $x = (x . y . z) : (y . z) = 70 : 10 = 7$

=> $y = (x . y) : x = 14 : 7 = 2$

=> $z = (z . x) : x = 35 : 7 = 5$

So $x = 7; y = 2; z = 5$

$x = 7; y = 2; z = 5$

So the answer of the problem is C.14 .

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a,Ta tách số trên thành các số từ 1 đến 53.

Từ 1 đến 9 có 9 số có 9 chữ số ,vậy cần phải có 41 chữ số nữa.

41:2=20(dư 1)

Ta đi tìm số thứ 20 của dãy số có 2 chữ số.

Gọi chữ số thứ 20 của dãy số có 2 chữ số là a, ta có:

(a-10):1+1=20

(a-10):1=19

a-10=19

a=19+10

a=29

Vậy số thứ 20 của dãy số có 2 chữ số là:29

Mà chữ số tiếp theo của số 9 là số 3 vậy nên chữ số thứ 50 của số A là chữ số 3 của số 30.

b,Từ 1 đến 9 có 1 chữ số 2.

Từ 10 đến 19 có 1 chữ số 2.

Từ 20 đến 29 có 11 chữ số 2.

Từ 30 đến 39 có 1 chữ số 2.

Từ 40 đến 49 có 1 chữ số 2.

50 53 có 1 chữ số 2.

Vậy số A có 1+1+11+1+1+1=17(chữ số 2).

c,Bạn tự tính đi.

Sorry because I not good answering the question in English.Thanks!

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20+20=40

mình rồi

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100*5=500

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I don't understand meaning of the post

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$Δ A B C$ has AB = AC,so it isosceles at A $\Rightarrow \hat{C_{1}} = \hat{A B C} = 75^{0}$

The quadrilateral ABCD has :

$\hat{A B C} + \hat{B C D} + \hat{A D C} + \hat{B A D} = 360^{0}$

$\Rightarrow \hat{B C D} = 360^{0} - 80^{0} - 75^{0} - 65^{0} = 140^{0}$

$\Rightarrow \hat{C_{2}} = \hat{B C D} - \hat{C_{1}} = 140^{0} - 75^{0} = 65^{0}$

$Δ A C D$ has $\hat{C_{2}} = \hat{A D C} = 65^{0}$ ,so the triangular $Δ A C D$ isosceles at A

$\Rightarrow A C = A D$ but $A B = A C \Rightarrow A B = A D$

$\Rightarrow Δ A B D$ isosceles at A $\Rightarrow \hat{D_{1}} = \frac{180^{0} - \hat{B A D}}{2} = \frac{100^{0}}{2} = 50^{0}$

$\Rightarrow \hat{D_{2}} = \hat{B D C} - \hat{D_{1}} = 65^{0} - 50^{0} = 15^{0}$

So the answer is B

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2014 2016 = (2014 2) 1008

= (...6)1008 = ....6