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So, first let's analyze the question. 

Note that when Dirk finishes, Edith has ran 14km and Foley has run 9km. 

Well, this means Foley runs \(9/14\) as fast as Edith. 

So when Edith runs the entire 24km, Foley has ran \(24 (9/14) ≈ 15.43 km\)

Which means that Foley is \(24 - 15.43 ≈ 8.57 km \) behind Edith. 

So 8.57 km is our answer!

Thanks! :)

So, right now there are \(5\) workers. They need to get the job completed 20 days earlier.

Since each additional worker adds on \(4\) days, we simply have

\(\frac{\text{Number Of Days}}{\text{Days Saved Per Worker}}\). Thus, we can plug in \(20/4 = 5\) workers.

So our answer is 5 workers. 

Thanks! :)

We can write an equation to solve this problem. We have

\((3p+12)(t-7)=2tp\)

Now, factoring the left side, we get

\(3pt-21p+12t-84=2tp\\ 12t+pt-21p=84\\ pt=21p-12t+84\\ pt=7(3p+12)-12t\\ pt+12t=7(3p+12)\\ t(p+12)=7(3p+12)\\ t/7=\frac{3p+12}{p+12}\)

We could test some values here, since no time was given. t has to be less than 24 and a multiple of 7. 

When t is 7, p is 0. 

When t is 14, p is 12.

When t is 21, p is undefined. 

Thanks! :) 

First off, we need to put a in terms of b. This helps us identify a range. 

We get

\(a+b=1\\ a=1-b\)

Now, plugging that into the expression, we get

\(\frac{1}{a}+\frac{1}{b}= \frac{1}{1-b}+\frac{1}{b}=\frac{b+1-b}{b-b^2}=\frac{1}{b-b^2}\)

Now, we can easily find the set of all real numbers. We have

\(\{ R>1\} |\{a,b\} \subset R\) where \(1> a,b > 0\)

Thanks! :)

First, let's add up the two numbers. We get \(3578 + 2238 = 5816\)

Now, we want to convert 5816 to base 8. 

First, we have to do \(5816/8\). We get a quotient of 727 and a remainder of 0.

Next, we have \(727/8\). There's a remainder of 7 but a quotient of 90. 

Doing \(90/8\) has a quotient of 11 and a remainder of 2.

\(11/8 = 1 R 3\)

\(1/8=0R1\)

So for our final answer, we get \(13270_8\)

Thanks! :)

2*12^5 + 10*12^4 + 3 *12^3 + 1* 12^2 +  1*12  + 2   = 710366

11* 12^4 + 7 * 12^3 + 5 * 12^2 + 1 * 12 + 2  = 240926

710366 - 240926  = 469440

Convert back t base 12

469440  / 12  = 39120 R 0

39120 / 12 = 3260 R 0

3260 / 12 = 271 R 8

271 / 12  = 22 R 7

22 R 12  = 1 R 10

1 /12  =  0 R 1

1A7800₁₂

11₂  = 3

110₂ = 6

1000111₂ = 71

101011₂  =  43     

111₂  =   7

10111₂  =  23   

 1100101₂ = 101

3 + 6 + 71 + 43 + 7 + 23 + 101   = 254

Convert back to base 2

254 / 2  =  127  R 0

127  / 2  = 93 R 1

63 / 2  = 31 R 1

31 / 2 =   15 R 1

15 / 2  = 7 R 1

7 / 2 = 3 R 1

3 / 2 = 1 R 1

1 / 2 = 0 R 1

11111110₂

5a = 57

a = 11.4

4(11.4) + 5b = 63

45.6 + 5b = 63

5b = 17.4

b = 3.48

3(11.4) + 4(3.48) + 5c = 34

34.2 + 13.92 + 5c = 34

48.12 + 5c = 34

5c = -14.12

c = -2.824

2(11.4) + 3(3.48) + 4(-2.824) + 5d = 15

22.8 + 10.44 - 11.296 + 5d = 15

21.944 + 5d = 15

5d = -6.944

d = -1.3888

11.4 + 2(3.48) + 3(-2.824) + 4(-1.3888) + 5e = 47

11.4 + 6.96 - 8.472 - 5.5552 + 5e = 47

4.3328 + 5e = 47

5e = 42.6672

e = 8.53344

(a, b, c, d, e) = (11.4, 3.48, -2.824, -1.3888, 8.53344)

35A811 =   3 * 11^5  + 5 * 11^4 + 10 * 11^3  + 8 * 11^2  + 1 * 11 + 1   = 570648

A0411  = 10* 11^4  + 0 * 11^3  + 4*11^2 + 1 * 11  + 1   = 146906

570648 + 146906  =  717554 base 10

Convert  back to base 11

717554  / 11  = 65232 R 2

65232 /11 = 5930 R 2

5930  /11  = 539 R 1

539 / 11 = 49 R 0

49  /11  = 4 R 5

4 /11  = 0 R 4

450122

31 * 32¹  +  8 * 32⁰   = 1000  =  v8

25₆  =  2*6 + 5  =   17  base 10

54₆ =   5*6 + 4  =    34 base 10

17 + 34  =  51  base 10

To convert back to base 6

51  /  6    =   8 R  3

8 / 6  =         1 R   2

1/6  =            0 R 1

123₆  

Convert 12121₃  to base 10

1 * 3^4  + 2 * 3^3  + 1 * 3^2 + 2 * 3  + 1    = 151

Convert 1222₃ to base 10

1* 3^3  + 2 * 3^2 + 2 * 3^1  + 2   = 53

151  + 53   =  204

Convert  204₁₀  to base 3

204 / 3  =    68  R 0

68 / 3  =    22  R 2

22 / 3  =  7 R 1

7 / 3  =    2 R 1

2 / 3  =    0 R 2

So

12121₃ + 1221₃  = 21120₃

We an represent this problem as \(\frac{1}{2}ab = 5(a+b)\), where a and b are the two legs. We can now simplify this equation further.

\(\frac{1}{2}ab = 5a+5b\)

\(ab=10a+10b\)

\(ab-10a-10b=0\)

We can now use Simon's Favourite Factoring Trick to simplify this.

\(ab-10a-10b+100=100\)

\((a-10)(b-10)=100\)

The factor pairs of 100 are (1, 100), (2,50), (4, 25), (5, 20), and (10, 10). We now choose a and b to make thepart in the parentheses equal to one of these pairs. Therefore, all possibilities for the legs (a and b) are (11, 110), (12, 60), (14, 35), (15, 30), and (20, 20). We can calculate the area for each of these.

\(\frac{1}{2}(11)(110)=605\)

\(\frac{1}{2}(12)(60)=360\)

\(\frac{1}{2}(14)(35)=245\)

\(\frac{1}{2}(15)(30)=225\)

\(\frac{1}{2}(20)(20)=200\)

Therefore, the sum of all areas is 605 + 360 + 245 + 225 + 200 = 1635

(x + 1) ( x -2p)

x^2 + ( 1 - 2p)x - 2p

The x cootdinate of the  vertex is  - [ 1 -2p ] / 2  = [ 2p - 1]  / 2

We   want   this point on the graph  ( [ 2p -1 ] /2 , 0 )

So

( [2p-1]/2 )^2 + (1 - 2p) ([2p-1] /2) -2p =  0

p^2 - p + 1/4  + p - 2p^2 -1/2 + p - 2p =  0

-p^2 - p -1/4 =  0

p^2 + p + 1/4  = 0

(p + 1/2)^2  = 0

p + 1/2   = 0

p  = -1/2

First, we must set some variables to complete this problem. Let's set HB as x. 

Note that \(AC^2 = AB^2 - BC^2 = (2 +x)^2 - 1^2 = x^2 + 4x + 3\)

We can write a system of equations to help us solve this question. We have

\(CH^2 = BC^2 - BH^2 = 1 - x^2 \\ CH^2 = AC^2 - AH^2 = (x^2 + 4x + 3) - 2^2 = x^2 + 4x -1 \)

Since both equations are equal to CH^2, we set both of them to equal each other, and we get

\(1-x^2 = x^2 + 4x -1 \\ 2x^2 + 4x -2 = 0 \\ x^2 + 2x - 1 = 0 \\ x^2 + 2x = 1\)

\(x^2 + 2x + 1 = 1 + 1 \\ (x + 1)^2 = 2\\ x + 1 =\sqrt{2} \\ x =\sqrt{2} - 1 \)

Now, we move on to the final step, which is to find CH. Plugging x back into the original formula, we get

\(CH^2 = BC^2 - BH^2 = 1^2 - (\sqrt{ 2} -1)^2 = 1 - (2 - 2\sqrt{ 2} + 1) = 2\sqrt {2} - 2 \\ CH = \sqrt {2\sqrt {2} - 2} = \sqrt{\sqrt {8} - 2} ≈ .91\)

So our final answer is about .91. 

Thanks! :)

We can use a really handy formula to help solve this question. 

First, we have to calculate the semiperimeter of the triangle, which is half the sum of the three side.

We get \(\frac{3+4+6}{2}=\frac{13}{2}\)

That may seem useless, but it will come in handy very soon. 

Now, we apply Heron's Formula, which states that the area of a triangle is equal to \(\sqrt{s(s-a)(s-b)(s-c)}\). Subbing everything in, We get

\(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\)

\(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)

So our final answer is \(\frac{\sqrt{455}}{4}\)

Thanks! :)

First, let's note that if AD = AB, then \(\angle ADB = \angle ABD\)

So, this means we have \(ABD = (180 - 17) / 2 = 81.5° \)

Our final answer is thus 81.5

Thanks! :)

Let's define some variables to help us solve the problem. Let's set the smallest number as n. 

We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

Since the sum of the integers add up to 28/5 times the largest number, which is n+6, we can write the equation

\(7n+21=28/5(n+6)\)

Now, we simply solve for n. We get

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

Multiplying both sides by 5/7, we get \(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

So our final answer is 9. 

Thanks! :)

We can find the diameter of the circle with the 3 x intercepts. 

The diameter of a circle is the distance between any two points that crosses the center. Since the x intercepts defintely cross the center, the diamter is

\(10-(-3)=13\)

Now, the same goes for the y intercept. Since the other y intercept must be 13 units away from 5, then we have \(5-13=-8\)

So the other y intercept is -8. 

Now, for the area, we already know the diameter, and the radius is half the diameter, so we have \(r=\frac{1}{2}13 = 6.5\)

The area of the circle is equal to \(r^2 \pi\), so plugging in 6.5, we get \(42.25\pi\) as the area. 

Thanks! :)

First, we need to simplify the equation, so let's combine all the like terms. 

We get \(-15x^2-kx+45\)

Now, in order for the solutions to be nonreal, the descriminant must be less than 0.

The descriminant is \(b^2-4ac\), so plugging in our values, we get

\((-15)^2 - 4(-k)(45) < 0 \\ 225+180k<0\\ k<-1.25\)

Therefore, the biggest interger k can be is -2. 

So -2 is our answer!

Thanks! :)

Let's first find out how many cubes have how many colors. All cubes must have 0-3 black sides. 8 of the cubes have 3 black spots (all of the corners), 48 have 2 black spots (12 edges each with 4 cubes), and 96 have one black spot (4 by 4 center for each side). If we roll a cube with 3 black sides, it will have a 1/2 chance of landing on a black side, if we roll a cube with 2 black sides, it will have a 1/3 chance of landing on a black side, and if we roll a cube with 1 black side, it will have a 1/6 chance of landing on a black side. Therefore, the total probability is:

\(\frac{8 \times \frac{1}{2} + 48 \times \frac{1}{3} + 96 \times \frac{1}{6}}{216}\)

\( \frac{4+16+16}{216}\)

\(\frac{36}{216}\)

\(\mathbf{\frac{1}{6}}\)

Therefore, you will have a 1/6 chance of rolling a black face.

Expanding the desired expression, we get \((2a-3)(4b-6)=8ab-12a-12b+18=8ab-12(a+b)+18\). This implies that we need the sum and product of the roots of the given equation, which are \(10/2=5\) and \(5/2\), respectively. Thus, the desired expression equals \(\)\(\left(8\cdot \frac{5}{2}\right) - (12 \cdot 5) + 18 = \boxed{-22}\).