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Ratio problem

3 days  / 6 times =   x days  / 8 times

3 /  6  =   x /  8

1/ 2   =  x  / 8

8 (1/2)  = x  =  4 times

3x + y  =  15

y = 15 - 3x

So

minimize   x^2 + (15 -3x)^2

minimize  x^2  + 9x^2  - 90x  +  225        

minimize    10x^2  - 90x  +  225         take the derivative and  set to 0

20x  - 90  =  0

x = 90 / 20  = 9/2

x^2  =  81/4

And  y = 15 - 3(9/2)   =  15 - 27/2  =  3/2

y^2  =  9/4

So

min x^2 + y^2  =   (81/4) + (9/4)  =  90 / 4   =  45 / 2  =  22.5

By the change of base theorem

log 625  / log x  - log 25 / log x  =  2

log 25^2 / log x  -  log 25 / log x =   log 100

 [ 2log 25  - log 25 ]  / log x  = log 100

log 25 / log x    = log 100

log x  =  log 25 / log 100

10^(log 25 / log 100)  = x

10^( [log 25 ] / 2) =  x

Draw DB  and   AC 

Since ABCD is a quadrilateral  inscribed in a circle  angles   DAB and DCB are supplementary

And angle  DCB is acute  so  angle  DAB is obtuse

Using the Law of Cosines

DB^2  = 3^2 + 9^2  - 2 (3)(9) cos DCB      (1)

DB^2  =  6^2  +2^2  +2 ( 6)(2) cos DCB    (2)

Subtract (1) from (2)

0  =  -50  +  78 cos DCB

50  = 78 cosDCB

cos DCB  =  50/78 =    25 / 39

And DCB and BCP  are supplementary

So cos BCP =  - 25 /39

arccos (-25/39)  =BCP  

Using the Law of Cosines again   and noting that ADC and ABC are supplementary

AC^2   = 6^2 + 9^2   -2(6)(9)cos ADC     (3)

AC^2  = 2^2  + 3^2  + 2(2)(3)cos ADC    (4)

Subtract  (3) from (4)

0  =   -104  + 120 cos ADC

cos ADC  =  104/120  =  13  /  15

arccos (13/15)= ADC =  PBC

angle BPC =  [ 180  -  angle BCP -  angle PBC]  =   [ arccos(-1)  - arccos(-25/39) -arccos (13/15)]

Then by the Law of Sines

BP  / sin [ BCP]  = 3 / sin [ BPC]

BP  =  sin (arccos (-25/39) ) * 3   / sin [ arccos(-1) - arccos (-25/39)  -arccos (13/15) ]  = 20 / 3

We can use the intersecting chord theorem

Let r be the radius

Let BP  = r + 5

Let AP = r - 5

Let PC  = sqrt (r^2 + 5)

So

PQ * PC   = BP * AP

2sqrt [ r^2 +5^2 ]  =  (r + 5) (r -5)

2sqrt [ r^2 + 5^2 ]  = r^2  - 25          square both sides

4 [ r^2 + 5^2 ] =  r^4  - 50r^2  + 625

4r^2  + 100  = r^4  - 50r^2 + 625

r^4 - 54r^2 + 525  =  0          complete the square on r

r^4  -  54r^2  +  729   =  -525 + 729

(r^2  - 27)^2  =  204        take the square root of both sides

r^2 - 27 =   sqrt (204)

r^2=  sqrt (204)  +  27  

r =  sqrt [  2sqrt 51) + 27 ] ≈ 6.43

Here's the expansion :

100000 x^15 + 50000 x^13 + 10000 x^11 - 25000 x^10 - 1/(32 x^10) + 1000 x^9 - 10000 x^8 + 50 x^7 + 5/(16 x^7) - 1500 x^6 + 2501 x^5 + 25/(8 x^5) - 100 x^4 - 5/(4 x^4) + 750 x^3 - (5 x^2)/2 - 25/x^2 + 75 x + 5/(2 x) - 125

Yep....your answer is  correct!!!

Thanks  !!!

(6/7 - 9/14) : 2/7

(12/14 - 9/14) : 2/7

(3/14) : 2/7

(3/14) : 4/14

3:4

This question, or one very similar, was answered only a day or 2 ago.

Try looking for it.

I fond it, it is not identical but similar enough.

It's approximately five thirds, but being pedantic it's actually 16666666666666667/10000000000000000.

It doesn't cancel so this is the best that can be done.

If you want it to be a recurring decimal then say so.

How about 5/6  = 0.83333...   ?

( 9/11 = 0.81818...  ,  9/10 = 0.90000... )

xy  -2x + 2y + 19   = 0

x ( y - 2)   =  -(2y + 19)

x =     (2y  -19)  / ( 2 - y)

If  y   =  2    then x is  undefined  and no solutions exist

This will form 30 congruent  triangles   

Each triangle  will  have an included angle of   360  / 30   = 12°   

And the sides surrounding this angle = 15

So  the  area  of these 30 triangles  =

30  (1/2) (15)^2  * sin (12°)  =

15^3 * sin (12°)  ≈   701.7 mm^2

2, 4 , 6, 7

We  can  choose  any  3 of 4 =  C(4,3)  =   4

This will form 4 different sets  with 3 elements  each  and permuting these =  3!  = 6

So

4 * 3! =   24  integers

2sinxcosx   =  sin (2x)cosx + sinxcos(2x) + sin x

2sin x cos x  =  2sinxcos^2x + sinx (cos^2x - sin^2x)  +  sinx

2sinx cos x  = 3sinxcos^2x  - sin ^3 x  +sin x

Since sin x    not  = 0 ,  divide through by sin x

2cosx  = 3cos^2x - sin^2x  + 1

2cosx  = 3cos^2x  - (1 - cos ^2x) + 1

2cos x = 4cos^2 x  

4cos^2x  - 2cos x  =  0

2cos x  ( 2cos x   - 1)  =  0

2cos x  =  0

cos x  = 0

x = 90°   (not acute)

2cos x    -1    = 0

2cos x = 1

cos x =  1/2   

x  =  60° =  theta

arccos (7/10)  ≈  45.57°

arccos (3/10)  ≈ 72.54°

C = ( 180  - 45.57 - 72.54)   =   61.89°

cos 61.89°  ≈  .47

(2 - k)x ^2  +  4x  +  (5k   - 2)

Call the roots  m  and 1/m

The product of the roots  =   (5k - 2)  / (2 - k)

So

m * ( 1 /m)  =  (5k - 2)  / (2 - k)

       1  =   (5k - 2)  / (2 - k)

(2- k)  = ( 5k - 2)

4  = 6k

k = 4 / 6  =  2 / 3

So

(2 - k)x^2  + 4x  + (5k - 2)    = 0

(2 - 2/3)x^2  + 4x + (5 *(2/3) - 2)  = 0

(4/3)x^2 + 4x + 4/3  = 0

4x^2  + 12x  + 4       = 0  

4 ( x^2  + 3x + 1)  = 0

x^2  + 3x + 1  =  0

x^2  + 3x  =  -1

x^2  + 3x + 9/4  =  -1  + 9/4

(x + 3/2)^2  = 5/4

x + 3/2  = sqrt (5) / 2

x =   [sqrt (5)  -  3 ]   / 2  =  m

And    

1/ m  =    2 / [sqrt (5) - 3]

( 6x   +  1 / (2x) )^8

The constant term is

C (8,4)   (6x)^4 * (1/ (2x) )^4   = 

70  (6x)^4  (1/ (2x) )^4   =

70 * 1296x^4 * 1/ (16 x^4)   =

70  *  81   = 

5670

9 /11 ≤ x ≤ 9/10

90 / 110  ≤ x ≤  99 / 110

 Possible  fractions

90  /110  , 91/110 , 92  /110  , 94 /110  , 95  / 110   , 96 /110   , 97/110 , 98/110 , 99/ 110

95  / 110  =     19  /  22     has the smallest denominator

When you mark something up, you add the original price and how much it was marked up. For example, if you marked something up 50%, the new price is (the original price + 50% of the original price).

Applying the same logic here, we have \(23.52 + (1.1 \times 23.52) \approx 49.40\). The 23.52 represents the original price, and the 1.1 x 23.52 represents how much it was marked up. 

The price after the mark up was \($49.40\). We now have to mark it up by \(14.75 \text{%}\). 

Can you do this using the logic I just presented?

Feel free to ask if you need any help!!!

3 numbers have 2 digits (10, 11, 12)

The probability of getting a 2 digit number is \({1 \over 4}\)

We need 2 rolls to satisfy this condition, so the probability of this is \({1 \over 4} ^2\)

We also need 2 rolls to not have 2 digits. The probability of this is \({3 \over 4}^2\)

There are \(4 \choose 2 \) ways to choose which rolls are \(\ge 10 \).

Thus, the probability is:  \(\large{{1 \over 4} \times {1 \over 4} \times {3 \over 4} \times { 3 \over 4} \times {4 \choose 2}} = \color{brown}\boxed{7 \over 128}\)

\(\large\color{brown}\boxed{5 \over 3}\)

Simplify it to: \(4t^2-10t-26\)

The least possible value occurs at: \(\large {-b \over 2a} \)

The quadratic equation is in the form \(ax^2+bx+c\) only difference being that we have t instead of x. 

Plugging in the values, we have the lowest real value occurs at \(\color{brown}\boxed{5 \over 4}\)

Note: I'm not sure if this is right... pls check

Connecting the 4 dots gives us a trapezoid (outlined in green).

This trapezoid can be split into 3 equilateral triangles, each with an area \(4 \sqrt 3\). Thus, the area of the trapezoid is \(12 \sqrt3\)

There are also 2 300 degree sectors, with a total area of \(\large{{300 \over 360} \times 16 \pi \times 2 = {80 \over 3} \pi}\)

We also need to subtract two small parts of the circle (colored in blue on the diagram).

The area of one of these sectors is a 60 degree sector - an equilateral triangle with side length 4. Thus, the area of one of these is: \({8 \over 3} \pi - 4 \sqrt 3 \)

The area of the Venn diagram on top is \(16 \pi - {16 \over 3} \pi + 4\sqrt3\) (\(16 \pi \) is the area of the 2 semi-circles, and \(16- 4\sqrt3\) is the area of the overlap.

 Thus, the area of the shaded region is: \(12 \sqrt 3 + {80 \over 3} \pi - ({16 \over 3} \pi - 8 \sqrt 3) + 16 \pi - ({16 \over 3} \pi - 4 \sqrt 3)\)

This simplifies to: \(\color{brown}\boxed{32 \pi + 24\sqrt3}\)

The area of a parallelogram can be found by multiplying its base (CD) times its height (GE).

CD · GE  =  54 · 18  =  ......