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ElectricPavlov had a great explanation a couple days ago: https://web2.0calc.com/questions/word-prob

https://web2.0calc.com/questions/please-help_66642#r2

Are you sure the problem is right because then DBA and ABD would be the same angle...

We can solve this by finding which range 'm' falls in.

If m is in range 0 < m < 20, clues b and d apply. But this is not possible as no perfect square below 20 is 1 less than a multiple of 3, so m cannot be in range 0 < m < 20

If m is in range 20 < m < 40, clues a, b, and d apply. This is not possible as for m to be a multiple of 7 and be a perfect square, m would have to be m = 49, but that is not less than 40 therefore m cannot be in range 20 < m < 40 either.

Moving on, if m is in range 40 < m < 60, clues a and d apply. As stat5ed earlier, for m to be a perfect square and a multiple of 7, it would need to be 49, and m is in this range and satisfies both clues, therefore m = 49.

We can first choose the color for the middle triangle, which has 3 ways.

Next, we can choose the rest of the on the outside, each of which have 2 colors to choose from.

Therefore we have the equation,

\(3 \cdot 2\cdot2\cdot2 = 24\)

So there are 24 ways he can color it

Hello, apsi! All that in red ink is wrong!

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Now                                                                                              A = 72

 A/ B = 6/5                                                                                    B = 120

B = 72 *5/6                                          Thi is incorrect!!!            C = 72

B = 45                                                                                           D = 60

So no of pupils who failed = 45                                              Total = 324 

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base1 = x     base 2 = 7+x       height = 6

area =  (base1 + base2)/2   * 6       <====== red is median

60 = (x + 7/2 )*6

10 =  median   

median = (13/2    + 13/2+7 )/2 = 10

Let's make the number of 5 coins 'a', number of 10 coins 'b', and the number of 20 coins 'c'. Writing as an equation we get,

\(5a+10b+20c=560\)

\(a+2b+4c=112\)

We also know that she has the same amount of all coins, so we can replace a, b, c, with a common variable of 'x'.

\(x+2x+4x=112\)

\(x = 16\)

Because x  is equal to the amount of each of the coins, she has 16 $20 coins.

f = t = n          five   ten and twenty

5n  +   10n  + 20n  = 560

35n = 560

n = twenty dollar coins = 16

3mn  -   6m  +  7n

3m  ( n  - 2)   +  7 ( n - 2)  =

(n - 2)  ( 3m + 7)

-14     goes in  the blank   

To confirm the fact that 11887 is prime, one needs to check if it is divisible by each of 2, 3, 5, 7,....

What should be the four greatest primes in this list?  

You don't need to check any numbers greater than its square root. The square root of 11887 is 109.027+. 

Since 109 is prime, that would be the largest number.  The next largest primes are 107, 103, & 101.  

Easy

We have  the form

y  = (x - h)^2  +   k     

The  vertex  is  ( h , k)          h = -18      k  =  -15

So       (h, k)   =  (-18 , -15)

sqrt  (10  * 22)    =  sqrt (220 ) =   sqrt  ( 4 * 55)  =    2sqrt (55)   m /s

Jana  gets  the  1st cookie  and the 25th cookie

No. of  people

         3                     possible  each gets  8  and then Jana  gets  the 25th one

         4                     possible each  get  6  and then Jana gets  the 25th one

         5                    not possible  each get  5   and rhen Jana gets the  26th one

         6                     possible each  get  4    and then  Jana gets the  25th one

         7                    not possible  each  get 3  and  then Jana gets  the  22nd one

         8                    possible   each get  3   and Jana gets  the  25th one

         9                    not possible each  get  2   and Jana  gets  the 19th

        10                   not possible  each get 2  and  Jana gets  the  21st

         11                  not possible   each get 2  and Jana gets the  23rd

         12                  possible   each  get 2  and then Jana gets the 25th

        ......

         24                 possible  each get 1  and then Jana gets the  25th

         26                 possible  each  get  1   and  Jana gets the 25th

Number of people possible

3 , 4  , 6  ,  8,  12  ,  24 ,  26

Median =    8

Let  the roots  of  the  first  be  p, q  and   r

Sum of the  roots  =  -1

Sum of the products of the roots, taken two at a time = - 2

Product of all three roots =  3

p + q + r =   -1   

pq + pr  + qr =  -2   ⇒  (pq + pr + qr)^2  =  4  

pqr  = 3   ⇒ (pqr)^2  =  9    ⇒  p^2q^2r^2  =   9

Square  both sides  of  the first equation

(p + q + r)^2   =    1

p^2 + 2 p q + 2 p r + q^2 + 2 q r + r^2  =   1

p^2  + q^2  + r^2  +  2 (pq + pr + qr)  =  1

p^2  + q^2  + r^2  +  2 ( -2)  = 1

p^2  + q^2  + r^2  =   5 

So  b =  -5  and  d  = -9 

And

(pq + pr  +  qr)^2  =   4

p^2 q^2 + 2 p^2 q r + p^2 r^2 + 2 p q^2 r + 2 p q r^2 + q^2 r^2  = 4

(pr)^2  + (pq)^2  + (qr)^2  +  2pqr ( p + q + r)   =   4

(pr)^2  + (pq)^2  + (qr)^2  +  2 (3)  (-1)  =  4

(pr)^2  + (pq)^2  + (qr)^2  -  6   =  4

(pr)^2  + (pq)^2  + (qr)^2  = 10    =    c

(b , c, d)  =  ( -5 , 10 , -9)

16x^2  +  56x  +  36

Let  m = 4,  n  =  7

Note :  56  =   2 (4)(7)

(4x + 7)^2   =  (4x + 7) ( 4x + 7)  =  16x^2 + 56x  +  49

mn   =  4 * 7    = 28

Let  r  be  the common ratio

So

18/16  =  r^2

9/8   = r^2

3 / (2sqrt 2)  =  r

So

b =   18 ( 3   /  (2sqrt 2) )  =     27 / sqrt 2   =   27 sqrt (2)  /2   =  13.5*sqrt (2)

2010a + 2014b = 2018         (1)

2012a + 2016b = 2020         (2)

Subtract:  (2) - (1)

2a + 2b = 2  or

a + b = 1                             (3)

so b = 1-a, put this into (1)

2010a +2014(1 - a) = 2018    Simplify:

-4a = 4

a = -1

put this back into (3) to find b, then find a - b.

A B C D E F G H I J

A+B/2= *integer*

A+B+C/3= *integer*

A+B+C+D/4= *integer*

yadda yadda yadda

Because of divisibility rules, J must be 0 due to math being math.

Such is the same for 5, which means that E is 5.

This goes on for quite a bit and frankly it's very time consuming for me to type, and for you to read.

The answer to this annoying question is 3816547290

3/1=3

38/2=19

381/3=127

3816/4=954

38165/5=7633

381654/6=63609

3816547/7=545221

38165472/8=4770684

381654729/9=42406081

3816547290/10=3816547290

Yay!

The  radius  of  25  will be  the slant height of the  cone

The  circumference of the  cone's  base =   2pi  (25)   *  32 / 36  =  400pi  /  9

The radius  of  the cone  =    (400pi / 9)  / (2pi)   =  400/18 =  200/9

Using  the Pythagorean Theoren, the height of  the  cone =   sqrt  ( 25^2  -  (200/9)^2  )   =  (25./9)sqrt 17

Volume of cone   =  (1/3)pi  *[ 200/9]^2 *  [ (25/9)sqrt 17 ]  ≈  5922.8 units^3

Thx so much 😆😀

Let  x =  the  equal sides  and  y  =  the remaining side

We  have  the  feasible region defined  by

2x + y  =  30

2x  >  y

x + y > x

See  the  graph here :  https://www.desmos.com/calculator/brjqinmutc

6 triangles

8 - 8 - 14

9 - 9 - 12

11 - 11  - 8

12 - 12 - 6

13 - 13 - 4

14 - 14 - 2

The exterior angle in all polygons is 360

If we divide 360 by 18, we would get 20

Yay!

xy  - z    =  15

xz  - y    =  15

yz  - x =     15

Subtract  the  2nd equation from the 1st

xy - xz  + y - z   =   0

x (y -z)  + 1(y -z)   =  0

(x + 1)  ( y - z)  =  0                          since x is positive, divide both sides  by  (x + 1)

y - z   =   0

y  = z

Subtract  the  3rd  equation from the  2nd

xz - yz  + x - y =  0

z (x - y)  + 1(x - y)   = 0

(z + 1) ( x - y)   = 0                  again, z is positive,  divide both sides by (z + 1)

x - y   = 0

x = y

So   x = y =  z

In the 1st equation  we  have  that

x^2  - x =  15

x^2 - x  + 1/4   =  15  + 1/4

(x - 1/2)^2  = 61/4

x - 1/2  =   sqrt (61) / 2

x =  ( 1 + sqrt (61) ) / 2

So

xyz =    [  ( 1 + sqrt (61) ) /2  ] ^3    =    23  +  8sqrt (61)

For every value of a, the graph of each of these three equations is a line:
\(ax + y = 1 \\ x + ay = 1\\ x + y = 9\)
For what value of a do all three of these lines intersect at the same point?

My attempt:

\(\begin{array}{|lrcll|} \hline (1): & ax + y &=& 1 \\ (2): & x + ay &=& 1 \\ (3): & x + y &=& 9 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & ax + y &=& 1 \\ & y &=& 1-ax \\\\ (3): & x + y &=& 9 \\ & x+1-ax &=& 9 \\ & x(1-a) &=& 8 \\ & \mathbf{x} &=& \mathbf{\dfrac{8}{1-a}} \\\\ & y &=& 9-x \\ & \mathbf{y} &=& \mathbf{9-\dfrac{8}{1-a}} \qquad (4) \\\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3): & x + y &=& 9 \\ & x &=& 9-y \\\\ (2): & x + ay &=& 1 \\ & 9-y + ay &=& 1 \\ & y(1-a) &=& 8 \\ & \mathbf{y} &=& \mathbf{\dfrac{8}{1-a}} \qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (5)=(4):& y = \dfrac{8}{1-a} &=& 9-\dfrac{8}{1-a} \\\\ & \dfrac{8}{1-a} &=& 9-\dfrac{8}{1-a} \\\\ & \dfrac{2*8}{1-a} &=& 9 \\\\ & \dfrac{16}{1-a} &=& 9 \\\\ & 16 &=& 9(1-a) \\ & 16 &=& 9-9a \\ & 9a &=& 9-16 \\ & 9a &=& -7 \\\\ & \mathbf{a} &=& \mathbf{ -\dfrac{7}{9} } \\ \hline \end{array}\)