If a subscriber downloaded 25 applications during a year, the price P is :

\(P=5+1,25x=5+1,25\cdot25=36,25\left(dollars\right)\)

The average cost per application is : \(\dfrac{36,25}{25}=1,45\left(dollars\right)\)

The total heights of 16 girls is : \(63,7\cdot16=1019,2\left(inches\right)\)

The total heights of 16 boys is  : \(65,4\cdot14=915,6\left(inches\right)\)

The mean of the heights of all 30 students is : \(\dfrac{1019,2+915,6}{30}\approx64,5\left(inches\right)\)

The  prime numbers less than 100 have a units digit of 3 are : 3;13;23;43;53;73;83. => There are 7 numbers.

The total prime numbers less than 100 : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

The percent is : \(\dfrac{7}{25}\cdot100\%=28\%\)

1 human fingernail : 3mm/1month

10 human fingernails : 30mm/1month

=> 10 human fingernails : 360mm/1 year.

=> 10 human fingernails : 3600mm/10 years.

So the total meters of Deshawn’s 10 fingernails will grow over the next 10 years is 3600mm=3,6m.

Answer : 3,6 meters.

1 car : First day

2 cars : Second day

4 cars : Third day

So there will be 8 cars : Fourth day.

.... 16 cars : Fifth day.

.... 32 cars : Sixth day.

..... 64 cars : Seventh day.

...... 128 cars : Eighth day.

..... 256 cars : Ninth day.

....... 512 cars : Tenth day.

So it will take him at least 11 days to sell a total of at least 1000 cars.

One digit: 9 (numbers)
Two digits: 9 (numbers)
... 9 Digits: 9 (numbers)

So there are : \(9\cdot9=81\) (numbers)

Answer : There are 81 integers less than one billion are very round numbers.

We have :

1 minute : 155 people gained.

1 day : x people gained.

We have : 1 day = 1440 minutes.

So at the rate 155 people/1 minute, there are \(155\cdot1440=223200\) more people on Earth every day.

One half \(=0,5\)

From \(0,5\rightarrow2,25\)

The number is two-thirds of the distance from 0,5 to 2,25 is \(\left(2,25-0,5\right)\cdot\dfrac{2}{3}=\dfrac{7}{6}\)

So the number is \(\dfrac{7}{6}\left(=1\dfrac{1}{6}\right)\)

Dao Trong Luan If you use the formula \(Radius\cdot2\cdot3,14\) into this math problem you will get a near-right answer.

Your answer is \(\dfrac{\dfrac{24\pi}{3,14}}{2}=\dfrac{24}{2}=12\left(cm\right)\), the right answer of your formula is \(\dfrac{\dfrac{24\pi}{3,14}}{2}=\dfrac{600}{157}\pi=12,00608657\)

So I will answer this math problem again.

Using formula \(\text{circumference}=\text{radius}\cdot2\cdot\pi\)

\(\Leftrightarrow24\pi=\text{radius}\cdot2\cdot\pi\)

\(\Rightarrow\text{radius}=\dfrac{24\pi}{2\pi}=12\left(cm\right)\)

So the radius of the circle is 12 cm.

Following thread we have :

An item with original price : d dollars.

+ In April : \(d\cdot\left(100\%+x\%\right)=d'\)

+ In May : \(\dfrac{d'}{100\%}\cdot\left(100\%-x\%\right)=d''\)

The resulting price is : \(d''=\left(100\%-4\%\right)\cdot d\)

The same with : \(\dfrac{d'}{100\%}\cdot\left(100\%-x\%\right)=96\%\cdot\left(\dfrac{d'}{100\%+x\%}\right)\)

Solve the equation we have x = 20.

So the answer is x = 20.

a) \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

We can see in R.S we have \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=L.S\)

So \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

b) \(a^2+b^2=\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)\)

Start in the R.S we have \(\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)=\dfrac{1}{2}\cdot\left(a^2+2ab+b^2+a^2-2ab+b^2\right)=\dfrac{1}{2}\cdot\left(2a^2+2b^2\right)\)

The same with \(\dfrac{2a^2+2b^2}{2}=a^2+b^2=L.S\)

So \(a^2+b^2=\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)\)

The kangaroo dad and mom have 3 little kangaroo girls.

Each girls has two kangaroo brothers. So 3 little kangaroo girls, each girl also has two kangaroo brothers so we count two kangaroo brothers one time.

So there are \(2+3+2=7\) (members) in the kangaroo family.

Condition \(x\ge4\)

a) When m = 1 we have equation : 

\(\sqrt{x-3-2\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=1\)

\(\Leftrightarrow\sqrt{\left(x-4\right)-2\sqrt{x-4}+1}+\sqrt{4-4\sqrt{x-4}+x-4}=1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}-1\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-4}-1\right|+\left|2-\sqrt{x-4}\right|=1\)

Using inequality \(\left|a+b\right|\le\left|a\right|+\left|b\right|\) we have :

L.S \(\ge\left|\sqrt{x-4}-1+2-\sqrt{x-4}=1\right|\)

So the given equation is the same with 

\(\left(\sqrt{x-4}-1\right)\left(2-\sqrt{x-4}\right)\ge0\)

\(\Leftrightarrow1\le\sqrt{x-4}\le2\Leftrightarrow5\le x\le8\)

b) The equation has given is the same with \(\left|\sqrt{x-4}-1\right|+\left|2-\sqrt{x-4}\right|=m\)

Using checklist (Trans : Bang Xet Dau) : You make the list by your own.

\(+\) If \(0\le\sqrt{x-4}\le1\Leftrightarrow4\le x\le5\)

The equation becomes : \(3-2\sqrt{x-4}=m\Leftrightarrow\sqrt{x-4}=\dfrac{3-m}{2}\)

The equation has solution \(\Leftrightarrow0\le\dfrac{3-m}{2}\le1\Leftrightarrow1\le m\le3\)

+ If \(1< \sqrt{x-4}< 2\Leftrightarrow5< x< 8\)

The equation becomes \(1=m\)

All cases \(m\ne1\) the equation has no solution.

+ If \(\sqrt{x-4}\ge2\Leftrightarrow x\ge8\)

The equation becomes \(2\sqrt{x-4}-3=m\Leftrightarrow\sqrt{x-4}=\dfrac{m+3}{2}\)

The equation has solution \(\Leftrightarrow\dfrac{m+3}{2}\ge2\Leftrightarrow m\ge1\)

Answer :

- If \(m\ge1\) the equation has solution.

- If \(m< 1\) the equation has no solution.

(If you don't understand anything in my answer please inbox me :))

We have to prove \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

Let it be : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(1).

We have (1) <=> \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\), true with all a,b > 0.

Done ! 

Using Bunhiakovski's inequaility :

\(\left(1\cdot\sqrt{4a+1}+1\cdot\sqrt{4b+1}\right)^2\le\left(1^2+1^2\right)\cdot\left(4a+1+4b+1\right)=2\cdot6=12\)

=> \(\sqrt{4a+1}+\sqrt{4b+1}\le\sqrt{12}\)

The "=" happens when a = b = \(\dfrac{1}{2}\)

The last 6 more people must have you Help you solve math :))

I've search some newest users, I can't understand if a question the thread just have the order "Calculate" in hundred kind ( 250 + 100 + ... = ?) you have more than 5 ers in each question.

I've also find some new users and find out that more than 70% new users are your own accounts, cause when I click in that accounts I will find a line that "............. d an answer of a question" when I click in that question, I always find out that you're in that question.

You also for another accounts to low down the doubt that the accounts are yours.

These ways of cheating cannot pass my eyes dude :)) I've catched red-handed a lot of cheating ways that you've never think once in your life.

P/s : Just for the weekly prize but you do by cheating, the knowledge won't stay long with your brain.

Henry and John started walking from the same point.

Henry went 1 km north, 2 km west, 4 km south and finally 1 km west.

John went 1 km east, 4 km south and 4 km west.

When comparing we give the conclusion that the final walk of John must be (B) 1 km north so that he could reach the same point as Henry.

17

We call that interger is \(\overline{abc}\).

Cause \(a\cdot b\cdot c=135\) so c must be 5.

Next \(a\cdot b=27\) so (a;b) could be (1;27),(27;1),(3;9),(9;3).

So (a;b) just could be (3;9),(9;3).

So \(\overline{abc}=395\) or \(\overline{abc}=935\)

So the sum of the digit is \(a+b+c=3+9+5=17\)

(The picture is not all the same with thread, just take an example)

We know that AF = 35, AC = 12, BD = 11, CE = 12 and DF = 16.

So \(AC+CE=AB+BC+CD+DE=12+12=24\)

\(\Rightarrow EF=AF-\left(AB+BC+CD+DE\right)=35-24=11\)

Then : \(BD+DF=BC+CD+DE+EF=11+16=27\)

\(\Rightarrow AB=AF-\left(BC+CD+DE+EF\right)=35-27=8\)

So the distance BE is \(BE=35-AB-EF=35-11-8=16\)

Call the number of marbles is a.

Following thread we have : \(\left(a-2\right)⋮3\) and \(\left(a-2\right)⋮5\)

So we take an example that \(32:3=10\left(remainder-2\right)\)and \(32:5=6\left(remainder-2\right)\)

So at least she needs 13 more marbles so that 32 + 13 = 55 marbles which is divisible for 3 and 5 => there won't be any left when she arranges them in groups of 3 and in groups of 5.

So the answer is 13.