Grade 6

Sentence 1: Thanksgiving is always the last Thursday of November

Sentence 1: Thanksgiving is always the last Thursday of November

câu 1 là đúng

(D)8 or (E)9

(D)8 or (E)9

I chose the answer : E

hjh.hj.kjh,jhb,bjv,

I choose D because :
a<b => a-b<0 which means a-b is negative so I is true.

0<a<b => ab>0 => \(\dfrac{1}{ab}\) is positive so II is true.

\(\dfrac{1}{b}\) -\(\dfrac{1}{a}\)=\(\dfrac{a-b}{ab}\) but a-b<0 and ab>0 => \(\dfrac{a-b}{ab}\) is negative so III is false

We have :  6x6 , 36x6 , ...

So the result is 6

You should post the question in English

mk rất khó để viết bài toán bằng tiếng anh bởi vì mk ko bt nhiều về tiếng anh du mk giới

Because he threw 8 darts at the dartboard => The score must be even. (Odd + Odd = Even)

The most score that he could throw is: \(8.9=72\left(points\right)\)

The least score that he can throw is: \(8.3=24\left(points\right)\)

Satisfy all this operation, only \(C.42\) could be his score.

So the answer is C

Because he threw 8 darts at the dartboard => The score must be even. (Odd + Odd = Even)

The most score that he could throw is: 8.9=72(points)

The least score that he can throw is: 8.3=24(points)

Satisfy all this operation, only C.42

 could be his score.

So the answer is C

Let A,D,H be the age of each person respectively.We have :

\(\circledast A+D+H=49;\circledast D=3H;\circledast A+5=3\left(D+5\right)\)

\(\Rightarrow H=\dfrac{1}{3}D;A=3D+10\)

\(\Rightarrow\left(3D+10\right)+D+\dfrac{1}{3}D=49\Rightarrow\dfrac{13}{3}D=39\Rightarrow D=9\)

\(\Rightarrow H=3;A=37\Rightarrow A.D.H=37.9.3=999\)

So,the product of their ages is 999

Let A,D,H be the age of each person respectively.We have :

⊛A+D+H=49;⊛D=3H;⊛A+5=3(D+5)

⇒H=13D;A=3D+10

⇒(3D+10)+D+13D=49⇒133D=39⇒D=9

⇒H=3;A=37⇒A.D.H=37.9.3=999

So,the product of their ages is 999

Because this topic is not to say that the faces should be different so that their sum is the smallest, then the remaining 3 [can be said to be x, y, z] smallest and satisfy:
  - 12*x = 9*y = 6*z
=> x = y = z = 0 is the smallest
So the sum is: 0 + 0 + 0 + 6 + 9 + 12 = 27
Answer: 27

Because this topic is not to say that the faces should be different so that their sum is the smallest, then the remaining 3 [can be said to be x, y, z] smallest and satisfy:
  - 12*x = 9*y = 6*z
=> x = y = z = 0 is the smallest
So the sum is: 0 + 0 + 0 + 6 + 9 + 12 = 27
Answer: 27

The answer is 901.

From 1 to 9 we have sum : 45.

From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.

Next from 20 to 29 we have sum 65.

................. 30 to 39 we have sum 75.

................. 40 to 49 we have sum 85.

................. 50 to 59 we have sum 95.

................. 60 to 69 we have sum 105.

................. 70 to 79 we have sum 115.

................. 80 to 89 we have sum 125.

................. 90 to 99 we have sum 135.

And the last 100 we have sum 1 + 0 + 0 = 1.

So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.

The answer is 901.

From 1 to 9 we have sum : 45.

From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.

Next from 20 to 29 we have sum 65.

................. 30 to 39 we have sum 75.

................. 40 to 49 we have sum 85.

................. 50 to 59 we have sum 95.

................. 60 to 69 we have sum 105.

................. 70 to 79 we have sum 115.

................. 80 to 89 we have sum 125.

................. 90 to 99 we have sum 135.

And the last 100 we have sum 1 + 0 + 0 = 1.

So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.

From 1 to 9 the sum of the individual digits are: \(1+2+3+4+5+6+7+8+9=45\)

From 10 to 19: the tens digit all have the same digit, the ones digit have the rules from 1 to 9.

So as the same with 20-99.

The number 100 has the individual digits sum: \(1+0+0=1\)

She has obtained the sum: \(\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+\left[\left(1+2+3+4+5+6+7+8+9\right).10\right]+1\)

\(=450+450+1=901\)

So she obtained the sum: \(901\)

It's a good problem :((

So we have the time is \(\dfrac{2}{60}\left(hour\right)=\dfrac{1}{30}\)

So we have the speed to get to each intersection right on the first time is : \(\dfrac{3,5}{x}=\dfrac{1}{30}\Rightarrow x=105\) (kilometres per hour).

So he can travel to get to each intersection at the third time as it just changes to green at the speed of \(\dfrac{105}{3}=35\) (kilometres per hour)

That's what I think :))

It's a good problem :((

So we have the time is 260(hour)=130

So we have the speed to get to each intersection right on the first time is : 3,5x=130⇒x=105

 (kilometres per hour).

So he can travel to get to each intersection at the third time as it just changes to green at the speed of 1053=35

 (kilometres per hour)

That's what I think :))

We put the numbers 1-2 ; 3-4 and 5-6 opposite each other. So then 2-3 and 4-5 will have the same side. So he will lose at least 2 points.

So the answer is: \(C:2points.\)

ohhh.....! I'm sorry , the last row will be :

So , there are 120 kilometres do you live from the station . 

Let {V1=4V2=20

 (km/h)  

We got this :

S=V.t

 (with t is the time , S is the length of roads and V is the speed)

⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩t1=SV1t2=SV2

From the topic :

t1=t2+24

  (minute)

⇔SV1=SV2+24

⇔S4=S20+24

⇔S4=S+24.2020

<=> 20S = 4S + 4.20.24

<=> 16S = 4.20.24

<=> S = 120 (km)

So , there are 20 kilometres do you live from the station .

Let \(\left\{{}\begin{matrix}V_1=4\\V_2=20\end{matrix}\right.\) (km/h)  

We got this :

\(S=V.t\) (with t is the time , S is the length of roads and V is the speed)

\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{S}{V_1}\\t_2=\dfrac{S}{V_2}\end{matrix}\right.\)

From the topic :

\(t_1=t_2+24\)  (minute)

\(\Leftrightarrow\dfrac{S}{V_1}=\dfrac{S}{V_2}+24\)

\(\Leftrightarrow\dfrac{S}{4}=\dfrac{S}{20}+24\)

\(\Leftrightarrow\dfrac{S}{4}=\dfrac{S+24.20}{20}\)

<=> 20S = 4S + 4.20.24

<=> 16S = 4.20.24

<=> S = 120 (km)

So , there are 20 kilometres do you live from the station . 

No isosceles triangles . Because 2 < 3 < 7 < 11

There is no equal edge 

Convert : 1/2 hour = 30 minutes

The number of minutes it takes Maya to finish her homework questions is : \(30:\dfrac{1}{3}=90\)

The number of minutes it takes Mây to finish the remaining questions is : 90 - 30 = 60

So,the answer is 60 minutes

Convert : 1/2 hour = 30 minutes

The number of minutes it takes Maya to finish her homework questions is : 30:13=90

The number of minutes it takes Mây to finish the remaining questions is : 90 - 30 = 60

So,the answer is 60 minutes