There are \(\dfrac{100-2}{2}+1=50\) (even number) so the sum of 50 even numbers that is 50.

There are \(\dfrac{99-1}{2}+1=50\)(odd number) so the sum of 50 odd numbers that is -50.

So the sum of the expression \(50+\left(-50\right)=0\)

\(4x^2+3xy-y^2\)

= \(3x^2+3xy+x^2-y^2\)

= \(3x\cdot\left(x+y\right)+\left(x-y\right)\cdot\left(x+y\right)\)

= \(\left(x+y\right)\cdot\left(3x+x-y\right)\)

= \(\left(x+y\right)\cdot\left(4x-y\right)\)

We have : \(AH^2=BH\cdot CH\Leftrightarrow3^2=BH\cdot6\Rightarrow BH=\dfrac{9}{6}=\dfrac{3}{2}\)

So \(BC=BH+CH=6+\dfrac{3}{2}=7,5\)

=> \(AC^2=7,5\cdot6\Rightarrow AC=3\sqrt{5}\)

So the area of the triangle ABC is 

\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot5\cdot3\sqrt{5}=\dfrac{15\sqrt{5}}{2}\)

\(S=1+3+3^2+...+3^{50}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{50}+3^{51}\)

\(\Rightarrow3S-S=2S=\left(3+3^2+3^3+...+3^{50}+3^{51}\right)-\left(1+3+3^2+...+3^{50}\right)=3^{51}-1\)

So \(S=\dfrac{3^{51}-1}{2}\)

This is my quesion in Hoc24.vn, cheater :))

Câu hỏi của Đức Minh - Toán lớp 9 | Học trực tuyến

You can't take it easily :))

Call that number is \(\overline{ab}\)

Following the thread we have :

\(\left\{{}\begin{matrix}a+b=\dfrac{\overline{ab}}{6}\\ab+25=\overline{ba}\end{matrix}\right.\)

Now we just calculate a and b then we'll find the answer :)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{10a+b}{6}\\ab+25=10b+a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6a+6b=10a+b\\ab+25=10b+a\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5b=4a\\ab+25=10b+a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{4a}{5}\\a\cdot\dfrac{4a}{5}+25=10\cdot\dfrac{4a}{5}+a\end{matrix}\right.\)

Solve the second expression we have a = 5.

So \(b=\dfrac{4a}{5}=\dfrac{4\cdot5}{5}=4\)

So the number \(\overline{ab}\) is 54.

Cause \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

So we get an example that \(\left\{{}\begin{matrix}x=1\\y=2\\z=-\dfrac{6}{5}\end{matrix}\right.\)(cause \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{-\dfrac{6}{5}}=0\))

Okay now \(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{2\cdot-\dfrac{6}{5}}{1^2+2\cdot2\cdot\left(-\dfrac{6}{5}\right)}+\dfrac{1\cdot-\dfrac{6}{5}}{2^2+2\cdot1\cdot-\dfrac{6}{5}}+\dfrac{1\cdot2}{\left(-\dfrac{6}{5}\right)^2+2\cdot1\cdot2}\)

\(A=\dfrac{161}{646}\)

MATH Error :v

Kaya Renger There is no mistake in the thread.

You can find the answer in here : Câu hỏi của Đức Minh - Toán lớp 9 | Học trực tuyến

The perimeter of the original photo is \(2\cdot\left(8+10\right)=36\left(inches\right)\)

The area of the original photo is \(8\cdot10=80\left(inches^2\right)\)

After reduced in size the new perimeter will equal \(\dfrac{27}{36}=\dfrac{3}{4}\) the original perimeter. So each measurements of the new photo will equal \(\dfrac{3}{4}\) the original measurements of the original photo.

The area of the new photo will equal \(\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\) the area of the original photo.

So the area of the new photo will be \(80\cdot\dfrac{9}{16}=45\left(inches^2\right)\)

P/s mình nói tiếng việt tí nhé :))

Cho hỏi bạn Toàn kia học lớp mấy thế, bài này ko biết làm thì xuống học lại lớp 1 đi bạn :)) đừng hỏi những điều mà ai cũng biết :v

Hay là bạn "Help you solve math" tự hỏi tự trả lời thế ?

\(75+34+43+25+66+57\)

\(=\left(75+25\right)+\left(43+57\right)+\left(34+66\right)\)

\(=100+100+100=300\)

\(75+12+43+25+88+50+57\)

\(=\left(75+25\right)+\left(12+88\right)+\left(43+57\right)+50\)

\(=100+100+100+50=350\)

So we start :

The first person can pick any number from 1 to 10.

The second person has a chance of 9/10 to pick different number.

The third person has a chance of 8/10 to pick different number.

The fourth person has a chance of 7/10 to pick different number.

The probability that four people pick different numbers is \(1\cdot\dfrac{9}{10}\cdot\dfrac{8}{10}\cdot\dfrac{7}{10}=\dfrac{63}{125}\)

The probability that at least two of the people pick the same integer is \(1-\dfrac{63}{125}=0,496=49,6\%\)

In 32c coin we have 7 combinations :

32+16+2

32+8+8+2

32+8+4+4+2

32+8+4+2+2+1+1

32+16+1+1

32+8+8+1+1

32+8+4+4+1+1

In 16c coin we have 5 combinations :

16+16+8+8+2

16+16+8+4+4+2

16+16+8+4+2+2+1+1

16+16+8+8+1+1

16+16+8+4+4+1+1

There are 7+5=12 combinations that have a combined value of 50¢.

A survey showed that of 120 secondary school students: 52 of them have a dog, 31 of them have a cat and 19 of them have both.
How many of the 120 students have neither a dog nor a cat? 
Answer: 56 students.

The students have neither a dog or a cat is \(120-52-31+19=56\) students.

I found the number that STAR present is \(\overline{STAR}=8712\)

Cause \(RATS\cdot4=STAR=2178\cdot4=8712\)

So the value of S+T+A+R is 8 + 7 + 1 + 2 = 18.

The chance of winning the draw of bag A is 1/10.

The chance of winning the draw of bag B is \(1-\left(\dfrac{99}{100}\right)^{20}\)

The difference is : \(\left(\dfrac{1}{10}-\left(1-\left(\dfrac{99}{100}\right)^{20}\right)\right)=-\dfrac{9}{10}+\dfrac{99}{100}^{20}=0,082\)

We have \(x^2-2x+1=\left(x-1\right)^2\)

We know that x have 11 values, we apply each values into \(\left(x-1\right)^2\) to calculate, we recognize that only 7 elements that 0,1,4,9,16,25,36 are in set.

The list that 2, 0, 1 and 4 are used to create is :

The sum of all numbers is : 45108.

So the arithmetic mean of all these integers is \(\dfrac{45108}{16}=2506\)