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Simplify as

x + y + 3z   =10

6x - 6y - 13z = -7

( k-1)x - 2y + z = 3

Multiply the first equation by  6....add to the second equation

6x + 6y + 18z   = 60

6x - 6y  -13z  =  -7       

12x + 5z  = 53    (1)

Multiply the first equation by 2 ....add to the second equation

2x + 2y + 6z  = 20

(k-1)x - 2y + z  = 3         

( 1 + k)x + 7z  = 23      ( 2)

Multiply (1) by 7    and (2)   by 5

84x + 35z  = 371

5(1 + k)x  +  35z = -115

This will have no solutions when

5 (1 +k) =  84

5 + 5k = 84

5k = 79

k = 79/5

Let the time to complete the job with 5 workers = T days

So......the total man-days for the job  is  5T

If one additional worker is  hired, the man-days for the job is 6 (T - 8)

So....since the man-days will be the same for the job.....

5T  = 6(T - 8)

5T = 6T - 48

48 = T 

So   the number of man-days to complete the job is   5(48)  = 240

So.....to complete the job 28 days earlier

(number of  workers needed) * ( 48 - 28)  =  240

(number of workers needed)  * (20)   = 240

number of workers needed  = 240 / 20     =    12

So.....(12  - 5) =  7  additional workers should be hired to  complete the job 28 days earlier

Using the square of the distance  formula

(2 - a)^2  + (2a - 7 - -  1 )^2  = 5

(a - 2)^2  + (2a - 6)^2   = 5

a^2 - 4a + 4 + 4a^2 - 24a + 36  =  5

5a^2 - 28a + 35  =  0

The sum of the values for a  =   - (-28) / 5   =  28 / 5

How many additional workers should be hired?

Hello Lilliam0216!

I assume the 5 workers can complete the job in around 120 hours.

\(\psi =\frac{J}{5\cdot 120h}\\ J=5\cdot 120h\cdot \psi =(5+1)\cdot (120-8)\cdot \psi=(5+x)\cdot (120-28)\cdot \psi\\ 6\cdot 112=(5+x)\cdot 92\\ 672=460+92x\\ x=\frac{672-460}{92}=2.3\color{blue}\to 3\\ 6\cdot t=(5+\color{blue}3)\cdot 92\)

\(t=\frac{8\cdot 92}{6}=\color{blue}122\frac{2}{3}h\)

If the 5 workers completed the work in 122 2/3 hours, 3 additional workers should be hired to save 28 hours.

An additional time specification is absolutely necessary for your question.

  !

To divide a plane into \( 5 \) regions using straight lines, we need to determine the minimum number of lines required to achieve this configuration.

Let's break down the problem:

1. With no lines, the plane forms one region.
2. With one line, the plane is divided into two regions.
3. With two lines, the plane can form up to \( 4 \) regions.
4. With three lines, the plane can form up to \( 7 \) regions.

Now, we need to find the minimum number of lines required to form \( 5 \) regions.

If we draw \( 3 \) lines, we will have \( 7 \) regions. But if we add a fourth line, it will intersect with the existing regions, increasing the number of regions by \( 1 \). Thus, with \( 4 \) lines, we can have \( 8 \) regions, which is more than required.

Therefore, the smallest number of straight lines that will divide a plane into \( 5 \) regions is \( 3 \).

equation of a line in standard form is:  Ax + By + C = 0
x stands for x-intercept and y stands for y-intercept
the equation of a line in y-intercept form is y=mx+b

slope(m)= 8
y-intercept= (0,14)
Plugging this into the equation y=mx+b, we have:

y=8x+14   → change this into standard from by moving "y" to the other side
0=8x+y+14

substitute (0) for y to solve for x

0=8x+14

-14=8x

-14/8=x

-7/4=x

therefore, the x-intercept of the line is -7/4

We know \(\sin(2x)>0, \tan(2x)<0\), therefore x is in quadrant 3. So \(\cos(2x)\)must be negative.

Therefore, \({\cos}^{2}(2x)=1-{\sin}^{2}(2x)=\frac{49}{64}\), \(\cos(2x)=-\frac{7}{8}\).

Becuase cos(2x) is in the third quadrant, then we know x must be in the second quadrant, so cos must be negative

Using our half-angle formula for cos:

\(\cos(x)=-\sqrt{\frac{\frac{1}{8}}{2}}=-\frac{1}{4}\). Therefore cos(x) = -1/4

Yes, sorry the 14A3 is base 11.

Thanks!

I assume the 14A3  is also base 11  ???

A =10

If so

4392₁₁  = 4(11)^3 + 3(11)^2  + 9 (11) + 2  = 5788₁₀

14A3₁₁  =  (11)^3 + 4(11)^2 + 10(11) + 3  = 1928₁₀

13A₁₁  = 1(11)^2 + 3(11) + 10  = 164₁₀

5788 - 1928 + 164  =  4024₁₀

To  convert back to base 11

4024 / 11  = 365 R 9

365 / 11   =  33 R 2

33/11  = 3 R 0

3/11 = 0 R 3

So

4024₁₀  = 3029₁₁

Let the total number of games to be played = N

To have the same  number of wins as losses you need.50N wins

So far....you have played .75N

Number you have remaining = .25N

So far....you have won  .35 * .75N    = .2625N

So....the number of  remaining games  you must win to have the same number of wins as losses is .50N - .26.25N = .2375N

So

Number of  remaining wins you need  / Number of games left  =

.2375N / .25N  = 

.2375 / .25  = 

.95  = 

95% of your remaining games

DE = DF =    DA + AF  = 5

Law of Cosines

AE^2  = DA^2 + DE^2  - 2 (DA * DE)cos (60 degrees)

AE^2   = 3^2 + 5^2  - 2 ( 3 * 5) (1/2)

AE^2 = 19

AE = sqrt (19) ≈ 4.36

Side of the square  =  

(product of the leg lengths)  / ( sum of leg lengths)  = 

[ 1 * 3 ]  / [ 1 + 3] =  

3/4

Area  = (3/4)^2   = 9/16

See the graph below

The polygon is  bounded by  (0,2) , (0,4), (2,2)  and the point determined by the intersection of the lines

y= 2x  - 2 

y = (1/4)x + 4

To find the x intersection of these lines

2x -2 = (1/4)x + 4

(7/4)x = 6

x = 24/7

y = 2(24/7) - 2 = 34/7

The last boundary point is (24/7,34/7)

We can  split this into two areas

The first is a triangle with a base = 3 and a height =  (34/7 - 4) = 6/7

The area of this =  (1/2)(3)(6/7) =  9/7

The second area is a trapezoid with bases = 2,3  and a height = 2

Its area =  (1/2)(2)(2 + 3) = 5

The total area =  5 + 9/7  =  44/7

Are you sure it is a square and not a rectangle...

I don't have any answers, but day before yesterday I had 701 hearts, and today I have 674 hearts. 

They're a lot harder to earn nowadays; it's discouraging that they just somehow vanish into the night. 

_.

prime factorization of  2700  =  2^2 * 3^3 * 5^2

Possible Factorizations       Number  of  identifiable sequences

22333551                             8! / (2! * 3! * 2!)  = 3360

43335511                             8! /( 2! * 3! * 2! ) = 3360

62335511                             8! / ( 2! * 2! * 2!) =  5040

66355111                             8! / ( 2! * 2! * 3!) =  3360

Total possible sequences  =   3 (3360) + 5040  =  15120

a - b = 1            square both sides

a^2 - 2ab + b^2  =  1

a^2 + b^2  =  1 + 2ab

a^3 - b^3  =   (a - b) ( a^2  + ab +b^2)

1  =  (1) ( 1  + 3ab)

1 =   1  + 3ab

3ab = 0

ab = 0

ab          a           b                 a + b       

 0           1           0                    1

 0           0          -1                   -1

                Number of possible integers

11111                   1                     

1112                    4           

122                     3

113                     3

23                       2 

13 possible positive integers

Hello tomtom!

Under whom are the cone and cylinder located?

Let  the tangent line touch circle A at D   and  circle B  at E

Let the distance from  B to C  = x

We can set up similar triangles ADC  and BEC  such that

AD / AC  =  BE / BC

5 / ( 5 + 3 + x)  =  3  / x

5 ( 8 + x)  = 3   / x          cross-multiply

5x  = 3 (8 + x)

5x  = 24 + 3x

2x  = 24

x = 12  = BC

Impossible

To see why

ACB  = 90°

QCP  = 45°

If the cos ACP  = 2/3....then arcos  (2/3)  ≈ 48.2°

But this means that   QCP + ACP  =  45 + 48.2  =  93.2°   which would be >  ACB

Here :  https://web2.0calc.com/questions/pls-help_13322

Rewrite as

x =  2y^2  - 13y  -1

The y coordinate of the vertex  is    - (-13)  / ( 2*2) = 13/4

The x coordinate is

2(13/4)^2 -13(13/4) - 1  =   

-177/8

Vertex = ( -177/8 , 13/4)

Let the point we want  be    (x , (1/2)(x^2 - 8))

Using the square of the distance

D^2  =  x^2  +   [ (1/2)(x^2 -8) ]^2

Take the derivative and set to 0

2x + (x^2 - 8)(2x)  = 0

2x + 2x^3 - 16x  = 0

2x^3 - 14x  =  0

x^3 - 7x   = 0

x( x^2 - 7)  = 0

x = 0   (reject)

x^2 - 7  = 0

x = sqrt (7)

The point  is  ( sqrt 7 , -1/2)

The distance  is  sqrt  [ (sqrt 7)^2  + (-1/2)^2 ] =  sqrt [ 7 + 1/4 ] =  sqrt (29 ) / 2