All Answers 358773 Answers

For two absolute values to sum to zero, we must have 

\(|a|+|b|=0\), then \(\begin{cases} a = 0 \\ b = 0 \end{cases}\), because \(|x|\ge0\).

Following this logic, the only possible solution is \(\begin{cases} x + y - 7 = 0 \\ 4x - y + 12 = 0 \end{cases}\).

 \(5x+5=0\)

\(\begin{cases} x = -1\\y=8 \end{cases}\).

Therefore the ordered pair (-1, 8) is the only ordered pair that works.

\(\begin{cases} {n \choose 1}*a = -6 \\ {n \choose 2}*a^2 = 12 \end{cases}\)

\(\begin{cases} na=-6 \\ \frac{n(n-1)}{2}*a*a=12 \end{cases}\)

\(\begin{cases} na=-6 \\ na*a(n-1)=24 \end{cases}\)

\(-6*a(n-1)=24\)

\(a(n-1)=-4\)

\(an-a=-4\)

\(\begin{cases} a = -2\\n=3 \end{cases}\)

\({(1-2x)}^{3}=1-6x+12{x}^{2}-8{x}^{3}\)

\(\bf{c=-8}\).

\(\)

What is the unit conversion from 156.25 to 12.5 inches

weight of sugar = s

weight of flour = f

\(\begin{cases} 2s+3f \le 40 \\ f \le 6 + 2s \end{cases}\)

\(\begin{cases} (2s)+3f \le 40 \\ f - 6\le (2s) \end{cases}\)

\(f-6 + 3f \le 2s+3f\le40\)

\(4f-6\le 40\)

\(4f\le46\)

\(f\le\frac{23}{2}\)

23/2 pounds is greatest solution, this occurs when \(f-6=2s\), \(s=\frac{11}{4}\). (always make sure to check the solution works with inequalities).

\(\)

Alan has answered that question here: https://web2.0calc.com/questions/please-help_63707

Equations of the three medians

y  = x                                                       

y = (-2)x + 1      ( 1)

y = (-1/2)x + 1/2           (2)

x intersection of (1) , (2)

-2x + 1  = -(1/2)x + 1/2

(3/2)x = 1/2

x = 1/3

y = (-2)(1/3) + 1 =   -2/3 + 1  =  1/3

They pass through rhe point  (1/3 , 1/3)

Your team  can never win 60% of its games  because you only have a percentage of 1/12 ≈  8.33%

And if you  only win 1 game  per week  and lose 2, you can never raise your winning percentage to 60%

\((3-\sqrt{5})(4+\sqrt{5})(1+6\sqrt{5})\)

Product of first two

( 12 -sqrt 5) - 5)  =  (7 - sqrt (5) )

(7  -sqrt 5) ( 1+ 6sqrt 5)   =   7  + 41sqrt 5   - 30   =  -23 + 41sqrt (5)

Let the load that your truck  can  carry   =  x

Let the load that each of  your neighbors trucks can carry =  y

We have

x  -  y   ≥  500

x ≤ (3/8) (4y)      →  x ≤ (3/2)y

See the graph here :

Max your tuck can carry  =1500 lbs

\( 9^x - 2 \cdot 3^x + 1 \)

Take the derivative and  set to 0

9^x ( ln 9)  - 2 * 3^x (ln 3)  = 0

9^x (ln 9)  - ( 3^x) ( 2ln3)  = 0

9^x * (ln 9)  - (ln 3^2) * 3^x  = 0

(ln 9)  ( 9^x - 3^x)   = 0

9^x  - 3^x   = 0

9^x   = 3^x

(3^2)^(x)  = 3^(x)        take the log of both sides

log (3)^(2x) = log(3)^x

(2x) log 3  =  x log 3

(log 3) (2x - x)  = 0

2x - x  = 0

x = 0

The min is

9^(0)  - 2* 3^(0)  + 1 = 

1  -  2  + 1 =    0

Points on the graph =  (0,1)  and (1,0)

Slope of line =  [ 0 -1 ] / [ 1 - 0 ] = -1 / 1   = -1

Equation of line      y =  -x + 1     →   x + y =  1

Equation  of inequality is either

x + y   ≥ 1

or

x + y ≤  1

(1,1)  is in the solution area  and  makes  the first inequality true

Divide the given  equation by 3

x^2 - (4/3)x + 5  = 0

This has the same roots

(b, c)  = ( -4/3 , 5)

x^2 + 10x  + y^2 - 8y =  59           complete the square on x, y

x^2 + 10x + 25 + y^2 - 8y + 16  =  59 + 25 + 16

(x + 5)^2  + ( y - 4)^2  =  100

r^2  = 100

2x^2 - 3x - 35 =  -43 + jx

2x^2  - ( 3 + j)x + 8   =  0

To only have one solution,  the discriminant = 0

(3 + j)^2  - 4 * 2 * 8  =   0

(3 + j*2  -  64  = 0

(3 + j)^2  =   64               take both roots

3 + j  =  8                     3 + j =  -8

j =  5                              j =  -11

No real numbers make this true

See WolframAlpha's  solution here :  https://www.wolframalpha.com/input?i=x+%2B+y++%3D+10+%2C+x%5E3+%2B+y%5E3++%3D+x%5E2+%2B+y%5E2+%2B+162

Laverne says   1 , 6 , 11,  16,  21

Progressive sums   =  1 , 7  , 18 , 34 , 55

21

(a + 2)  /(a + 1)  =  b /  8

8(a + 2)  / (a + 1)  =  b

a               b

0               16

1               12

3               10

7                 9

-2                0

-3                4

-5                6

-9                7

Note that every second, the adults close the gap  by ( 4  -  3)  = 1 yd per second

So.....to  close the gap of  18 yds it  must have taken  18 seconds

The adults traveled = Time * Rate per second =    18 * 4    = 72 yds

6i * sqrt 3   +  6i * i  = 

6sqrt (3) i  + 6i^2             {i^2  = -1}

6sqrt (3) i   -  6

-6 + 6sqrt (3) i

First term  = 1  =  a

Second term  = 14

15th term =  1 + 14 * 13  = 183

S₁₅ =    [ first term + last term ]  (no.of terms / 2)  =

]1 + 183 ] / (15/2)  =

1380  = S₁₅

Here's a similar problem

https://web2.0calc.com/questions/help-with-geometry_22606