
Math You Like 14 hour ago (16:28)
Ta has: abcabc = abc000 + abc
= Abc x 1000 + abc
= Abc. (1000 +1)
= Abc. 1001
= Abc. 7. 11. 13
The abcabc of the abc with 7; 11; 13 => abcabc divide by 7; 11 and 13

Math You Like 15 hour ago (15:50)
Let x be the least satisfied integer and (x > 6)
x leaves a remainder of 6 when divided by 7 and 11, so x  6 is divisible by 7 and 11 or x  6 : 77
=> x  6 = 77
x = 77 + 6
x = 83
So, the answer is 83

Let x be the least satisfied integer (x > 6).
x leaves a remainder of 6 when divided by 7 and 11, so x  6 is divisible by 7 and 11 or \(x6⋮77\)
\(\Rightarrow x6=77\Rightarrow x=83\)
So, the answer is 83

Math You Like 15 hour ago (15:54)
The number of donuts Jon has is:
29 x 51 = 1479 (donuts)
Since 1479 = 123 x 12 + 3, three donuts will be left over when he groups them again
So, the answer is 3

The number of donuts Jon has is : 29 x 51 = 1479 (donuts)
Since 1479 = 123 x 12 + 3, three donuts will be left over when he groups them again
So, the answer is 3

Math You Like 15 hour ago (15:56)
We have:
The last digit the product is the last digit of 1 x 2 x 6
So, the product ends in 2.Then, it is divided by 5 with remainder 2

The last digit of the product is the last digit of 1 x 2 x 6.
So, the product ends in 2. Then, it's divided by 5 with remainder 2

Math You Like 15 hour ago (16:00)
122 x 71 = (11^{2} + 1) . (11 x 6 + 5)
= 11^{3} x 6 + 11^{2} x 5 + 11 x 6 + 5 = 5 (mod11)
So, the answer is : 5

122 x 71 = (11^{2} + 1)(11 x 6 + 5)
= 11^{3} x 6 + 11^{2} x 5 + 11 x 6 + 5 \(\equiv5\left(mod11\right)\)
So, the answer is 5

2^{3000} = (2^{4})^{750} = 16^{750} ends in 6

Math You Like 15 hour ago (15:58)
We have : 2^{3000} = (2^{4})^{750} = 16^{750} ends in 6

We have : P_{ABCDE} = AE + ED + DC + CB + AB
= AE + EQ  DQ + PC  PD + CB + PB + AQ  PQ = 64 units  P_{PDQ}
=> P_{PDQ} = 64  52 = 12 (units) \(\Delta AEQ=\Delta PCB\left(SSS\right)\Rightarrow\widehat{A}=\widehat{P_1}\)
\(\Delta AEQ\infty\Delta PDQ\left(AA\right)\), so the similarity coefficient of 2 triangles is equal to the ratio of their perimeter or :
\(\dfrac{10+10+12}{12}=\dfrac{8}{3}\). The, the ratio of their areas is : \(\left(\dfrac{8}{3}\right)^2=\dfrac{64}{9}\)
Draw \(EH\perp AB\), then the altitude EH of \(\Delta AEQ\) isosceles at E is also the median. So, \(AH=\dfrac{AQ}{2}=\dfrac{12}{2}=6\) (units)
Applying the Pythagorean theorem to \(\Delta AEH\) right at H, we have : \(EH=\sqrt{AE^2AH^2}=\sqrt{10^26^2}=\sqrt{64}=8\)(units)
\(\Rightarrow S_{AEQ}=\dfrac{AQ.EH}{2}=\dfrac{12.8}{2}=48\) (square units)
\(\Rightarrow S_{DPQ}=48:\dfrac{64}{9}=\dfrac{27}{4}\) (square units)
\(\Rightarrow S_{ABCDE}=S_{AEQ}+S_{PCB}S_{DPQ}=48+48\dfrac{27}{4}\)
\(=89.25\) (square units)

Let a be the number of pencils she has \(\left(a\in N;10< a< 100\right)\)
a leaves a remainder of 1 when divided by 3, 4, 5, 6, so a  1 is divisible by 3, 4, 5, 6
LCM(3, 4, 5, 6) = 60
\(\Rightarrow a1\in CM\left(3,4,5,6\right)=\left\{0;60;120;...\right\}\)
Since \(9< a1< 99\) and only 60 is between 9 and 99, a  1 = 60 or a = 61
So, the answer is 61

Math You Like Yesterday at 16:09
a = 20, b = 30
a + 70 + b + 80 + 100
= 20 + 70 + 30 + 80 + 100
= (20 + 80) + (70 + 30) + 100
= 100 + 100 + 100
= 300
Help you solve math selected this answer.

\(3\left(2x3\right)x=2\left(3x+1\right)+x\)
\(\Leftrightarrow6x9x=6x+2+x\Leftrightarrow92=6x+x6x+x\)
\(\Leftrightarrow2x=11\Leftrightarrow x=\dfrac{11}{2}\)

The diameter of the circle or the diagonal of the square is :
4 x 2 = 8 (units)
The area of the circle is : \(4^2\pi=16\pi\) (square units)
The area of the square is : \(\dfrac{8^2}{2}=32\) (square units)
The area of the shaded region is :
\(\dfrac{3216\pi}{4}=84\pi\) (square units)

The area of the rectangle is : 100 x 50 = 5000 (m^{2})
The radius of each semicircle is : 50 : 2 = 25 (m)
The total area of 2 semicircles is : \(25^2\pi=625\pi\) (m^{2})
The answer is : \(5000+625\pi\) (m^{2})

Divide the figure into 2 rectangles as shown
The area of the rectangle 2 is : 24 x 12 = 288 (square feet)
The width of the rectangle 1 is : 15  12 = 3 (feet)
The area of the rectangle 1 is : 11 x 3 = 33 (square feet)
This office occupies : 288 + 33 = 321 (square feet)

The area of each removed triangle is : \(\dfrac{2^2}{2}=2\) (square inches)
The area of the square is : 10^{2} = 100 (square inches)
The area of the octagon is : 100  2 x 4 = 92 (square inches)
The answer is : \(\dfrac{92}{100}=92\%\)

The general forms of the satisfied palindromes are \(\overline{bb},\overline{cdc},\overline{effe}\)
The number of 2digit palindromes is : (99  11) : 11 + 1 = 9
There are 9 choices to choose c and 10 choices to choose d, so the number of 3digit palindromes is : 9 x 10 = 90
The only 4digit palindrome is 1001
So, the answer is : 9 + 90 + 1 = 100 (palindromes)

The distance of /4  1 (out of 4 and minus 1) of each number is: 1 (4.11), 2 ( 4.21), 3(4.31).
The least possible score of the distance is: 1.6 = 6
The biggest possible score of the distance is: 3.6 = 18
Therefore, there are: \(186+1=13\left(differentscores\right)\)
ANSWER: 13 different scores.

→இے๖ۣۜBoy™†hïếuGîa 16/09 at 13:38
Let p, n, d, q be the number of pennies, nickels, dimes, quarters we need to make 67 cents respectively (p,n,d,q∈Z+p,n,d,q∈Z+), then p + 5n + 10d + 25q = 67

Let p, n, d, q be the number of pennies, nickels, dimes, quarters we need to make 67 cents respectively (\(p,n,d,q\in Z^+\)), then p + 5n + 10d + 25q = 67
We have :
(q ; d ; n ; p) = (2 ; 1 ; 1 ; 2) ; (1 ; 3 ; 2 ; 2) ; (1 ; 3 ; 1 ; 7) ;
(1 ; 2 ; 4 ; 2) ; (1 ; 2 ; 3 ; 7) ; (1 ; 2 ; 2 ; 12) ; (1 ; 2 ; 1 ; 17)
(1 ; 1 ; 6 ; 2) ; (1 ; 1 ; 5 ; 7) ; (1 ; 1 ; 4 ; 12) ; (1 ; 1 ; 3 ; 17) ;
(1 ; 1 ; 2 ; 22) ; (1 ; 1 ; 1 ; 27)
So, the answer is 13