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The portion of the  tub  filled by the coldwater faucet  every minute  = 1/17

The portion of the  tub  filled by  both faucets  in one minute  = 1/7

So....the portion of the  tub filled by  the  hot water faucet  in one minute   =

1/7  - 1/17 =

[17 - 7 ] / [7 * 17]  =

[10 ] / [ 119]     (1)

To find out how long (in minutes) the hot water faucet alone takes to  fil the tub.....take the reciprocal of (1)   =

119  /10    =

11.9  min

(- 4, 3√5)  =  ( -4, √45)  =  (x , y)

cos B  = x  / r

r  =  √ [(-4)^2  + (√45)^2 ]  = √ [ 16 + 45 ] =  √61

So

cos B  = -4/√61  =  -4√61  / 61

We have  a  30  - 60  - 90   right triangle

The  side opposite the 30°   angle    =   15 / √ 3 =     15√3 / 3  =  5√3

The  side opposite the  90° angle  =  2 * 5√3  =   10√3

Just as Hypotenuisance  said  !!!!

This is a geometric series  with   r =  (-1/4)

The sum  is  given by

48  [ 1  -  (-1/4)^6  ] / [ 1 - (-1/4) ]  =

48 [ 1  - (-1/4)^6]  / [ 5/4]  =

48 (4/5) [ 1  - (-1/4)^6 ]  =  

2457 / 64  = 

38.390625

A simple google search could've helped you quicker but:

7t+119/17t = 1

7t + 119 = 17t            | *17t

119 = 17t - 7t = 10t

t = 11.9

Thanks for the A2A!

The answer is \(10\sqrt{3}\).

Cheers!

Oh wait, I got it.

Turn them into modular arithmetic expressions such as \(a\equiv 5\bmod{9}\)  and  \(b\equiv 33\bmod{36}\).

Then, convert them into mod 3 and mod 9 expressions in order to get the units and tens digits of the base 3 equivalents of \(a\) and \(b\).

\(a\equiv 2\bmod{3}\) --> The first digit of a in base 3 is 2.

\(a\equiv 5\bmod{9}\) --> The value of the two last digits of a in base 3 is 5. Therefore, the second digit of a in base 3 is 1.

\(b\equiv 0\bmod{3}\) --> The first digit of b in base 3 is 0.

\(b\equiv 6\bmod{9}\) --> The value of the two last digits of b in base 3 is 6. Therefore, the second digit of b in base 3 is 2.

So, we end up with 12 as the last two digits of a and 20 as the last two digits of b (both in base 3).

To subtract, we do 2 - 0 = 2 for the units digit. Since the second digit of a is less than the second digit of b, we carry over 3 from the third digit, which we must assume has a value. Then we do 4 - 2 = 2 for the second digit.

The last two digits of a - b in base 3 is 22.

We have   2^4  =  16  possible codes

Codes with  three 0s

0001

0010

0100

1000

So....the probability  is    4  / 16   =   1  / 4

Let  x  be the amount  of the  60%  solution  in L

Let  30 - x  be the amount of the 20%  solution in  L

So

.60x  + .20(30 - x)   = .50(30)    simplify

.60x  + 6  - .20x  = 15

.40x  +  6   = 15     subtract 6 from both sides

.40x  = 9     divide both sides by  .40

x  =  22.5  =   amount of  60% solution (in L)

30  - x =  7.5  =  amount of  20% solution ( in L )

Awesom! Thank you for checking them for me

Perimeter   =  2  ( L + W)  =    2  ( 5√8  + 2√8)  =  2  ( 7√8)  =  14√8   in

Thanks, Isabelle   !!!!

Glad to see you back.....

See the following :

Let  AD   be the bisector....and this sets up the following relationship 

 AC / AB   = CD / BD

12/ 5  = CD/ BD

Thus  there are  (12 +5)  =  17 equal parts  to BC  so  CD  is (12/17) *13  =  156/17

And angle DAC  = 45°

And the sin  of angle BCA  = 5/13

So....using the Law of Sines

CD  / sin DAC  =  AD  /sin (BCA)

156/17 / sin (45)  = AD  / (5/13)

(156/17) (5/13) √2  = AD  =  60√2  / 17  ≈  4.99

c + 2c + (2c - 1.50)= 12.25
5c - 1.5 = 12.25 then 5c =13.75
c= 2.75
She spent $2.75 on oranges, 2(2.75) 􏰂 $5.50 on blueberries, and 5.50 􏰀-1.50= 􏰂 $4.00 on
   strawberries.

I'm assuming  that the side length is    6√6

The area  =   (side)^2  =  (6√6)^2   =  (6)^2  * (√6)^2  =  36 * 6   =  216

You can get  a basic idea here :

https://www.desmos.com/calculator/4p7ywck2dr

If we  reflect  f(x)  about the  y axis,  we get  f(-x)

If we   move this  4 units to  the right, we have

f(-(x - 4))   =  f(-x + 4)   =  g(x)

How many times could light travel around the circumference of the Earth in 8.6 second? Let the circumference of the Earth be 24,000 miles.   

Light travels at 186,000 miles per second 

Divide 186,000 by 24,000 and the quotient is 7.75 

That's how many times light would go around the world in 1 second 

Multiply 7.75 by 8.6 and the product is 66.65 

That's how many times light would go around the world in 8.6 seconds  66.65 times 

_.

See  the   following  :

Since  PX  is an angle bisectior....we have  the following relationnship  :

PR  / PQ  = XR /XQ    so

10/9  = XR / XQ

And  XR  + XQ  =   QR   =17

So

XR  = (10/19)*17  =   171/19    =  9

And using the Law of Cosines

9^2   = 17^2  + 10^2  - 2(17*10)*cos (PRQ)

81  - 289  - 100  =  -340*cos *PRQ)

-308 / -340  = cos (PRQ)   = 77/85

So....  sin (PRQ)  =  sqrt ( 1 - (cos (PRQ)^2 )   =  sqrt ( 1  - (77/85)^2 )   =    36/85

And by the Law of Sines

XR /sin (90)  =  XY / sin (PRQ)

9 / 1  =  XY / (36/85)

9 (36/85)  =  XY   =   324 / 85   ≈  3.81

cool

0.6x*0.2(1-x)=0.5
Factor out

0.6x*0.2-0.2x=0.5

Move x to one side

0.6x-0.2x=0.5-0.2

0.4x=0.3

x=0.3/0.4
x=0.75

So we need 0.75 of 60% solution and 0.25 (i.e 1-0.75) of 20% solution

You need to know distance from X to michigan to calculate anything of value.

Good job! And, thank you

CPhill: The question ends by saying "The sum of the mother's age and the square root of that of the daughter's is 77 years". Find their ages

Let mother's age 10 years from now = M
Daughter's age 10 years from now =M - 21
[M - 10] + sqrt[M - 31] = 77 [sum of their ages today], solve for M
M =80 years - mother's age 10 years from now
80 - 21 =59 years - daughter's age 10 years from now.
80 - 10 =70 years - mother's age now
59 - 10 =49 years - daughter's age now.

So that: [70 + sqrt(49) ] = 77 years.

The  amount  of paint  is

1(2  1/2)   + 2(4)  + 1(5  1/2)  =

5/2  +  8   + 11/2 =

16/2  + 8

8 +  8

16

So....each can would contain   16/2  =   8 pints