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Put the terms with \(x\) on one side and the numbers on the other:

\(2x+x+7x \le 12+18+5\)

which simplifies to \(10x\le35\) 

So \(x= \dfrac{7}{2}\). 

I hope that helped!

The two circles below?

\(\overline{PQ}=\sqrt{(R+r)^2-(R-r)^2}=\sqrt{R^2+2Rr+r^2-R^2+2Rr-r^2}\\ \color{blue}\overline{PQ}=2 \sqrt{Rr}\)

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Let's assume there are 100 games :

.75 * 100 * .35  = games won so far 

.25 (100) x       games remaining  times the percentage needed to get to .5(100)  <==== wins = losses

.75 (100) .35   +   .25 (100) (x)  = .5 (100)

     26.25         +  25 x = 50 

          25 x = 23.75

             x = .95                     You need to win 95% of the remaining games                

Here is my attempt 

1. Members of the painting team can paint a 100 square foot wall in 50 minutes.

2. The work rate of the team is 2 square feet per minute.

3. It would take the 40 members 240 minutes to paint a 480 square foot wall using the same work rate

Let's denote each two-digit number as $10a + b,$ where $a$ is the tens digit and $b$ is the units digit. Let $n$ have $k$ two-digit numbers in its decomposition.

When we calculate the alternating sum of these two-digit numbers, we get $10a_1 + b_1 - 10a_2 - b_2 + 10a_3 + b_3 - \cdots \pm 10a_k + b_k,$ where $a_i$ is the tens digit of the $i$-th two-digit number and $b_i$ is the units digit.

This can be rewritten as $10(a_1 - a_2 + a_3 - \cdots \pm a_k) + (b_1 - b_2 + b_3 - \cdots \pm b_k).$ Let $A = a_1 - a_2 + a_3 - \cdots \pm a_k$ and $B = b_1 - b_2 + b_3 - \cdots \pm b_k.$ Then, the sum becomes $10A + B.$

Since $10A + B$ is divisible by $m,$ it implies $10A + B \equiv 0 \pmod{m},$ which gives $10A \equiv -B \pmod{m}.$

We know that $10 \equiv 1 \pmod{m},$ so $10A \equiv A \pmod{m}.$ Hence, we have $A \equiv -B \pmod{m}.$

Therefore, $n$ is divisible by $m$ if and only if $10a + b$ is divisible by $m,$ implying that the divisibility test works for $m = \boxed{11}.$

 2x - 5 <= -x + 12 - 7x + 18

Rearrange  as

10x <= 35

x <= 35  / 10

x <= 7 /  2

Rearrange L1  as

-3y = -11x - 14

Slope of L1   = 11/3

Slope of a perpendicular line  = -3/11

L2  equation L

y = (-3/11) (x - 5) - 6

y = (-3/11)x + 15/11 - 6

y = (-3/11)x + 15/11 - 66/11

y = (-3/11)x - 51/11

Rearrange the given equation as

18y = -5x - 8

The slope of  a parallel line is   -5/18

The equation of a parallel line  through  (-1,8)  is

y =(-5/18)(x + 1) + 8

y = (-5/18)x  - 5/18 + 8

y = (-5/18)x + 139/18

18y = -5x  + 139

5x + 18y  = 139

Let x be the days to complete the job in the first case

The amount of work to be done  is  number of workers * numer of days =   5x

So....in the second case

5x  = 6 ( x -15)

5x = 6x - 90

90 = x   =    the number of  days to complete the job in the frst scenario

So

5 * 90  =  n (90 - 40)  where n is the toal number of workers needed in the last case

450  = 50n

450 / 50 = n =  9

So....  9 - 5  =  4 additional workers should be hired

3600 / [ 50 *20 ] =  3.6  sq ft / min  painted by one worker

So

4000 / [ 25 * 3.6 ]  ≈  44.44  min

\( \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor \)

f(1)  = 0

f(2)  = 3

f(3) = 7

f(4)  = 14

f(5)  = 23

f(6)  = 34

f(7) = 47

f(8)  = 62

f(9) = 78

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.

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f (1000)  =  999,997

It appears that, from x = 3 to x = 8 inc;usive,  f(x)  =  x^2 - 2

And from x = 9  to x =1000 inclusive , f(x)  = x^2 - 3

So

So....the sum is

         8                      1000

3  +   Σ   x^2  - 2   +   Σ  x^2 - 3      =  333,830,510

       x = 3                 x = 9

x = 2x + 8                x = 13 -5x             x = 20 - 14x                     x = 40  + 5x

-x = 8                       6x = 13                 15x  = 20                         -4x  = 40

x  =  -8  NO             x = 13 / 6 OK          x = 20/15 = 4/3 NO         x = -40/4  = -10  NO

Only one value 

x = 13 / 6

floor (t)  =  -t^2 + 3t + 4

Here's a graphical solution  (subbing x for t )

https://web2.0calc.com/questions/help-stuck

After n steps, he has visited (n + 1)^2  lattice points (points with integer coordinates)

So

(2 + 1)^2  =  3^2  =  9

When Dirk finishes, Edith has run 14km  and Foley has run 9 km

So Foley runs  9/14  as fast as Edith

When Edith has run 24km, Foley has run  24 (9/14)  ≈  15.43  km

So Foley is   24 - 15.43  ≈  8.57 km behind

The system will have no solutions when the determinant =  0

1   1   3    1    1

1 -2    8    1  -2

k   0   5    k    0

Det  =  [ 1 * -2 * 5   + 1 * 8 * k ] - [ 3*-2 * k  + 5 * 1 * 1 ] =  0

[ -10 + 8k ] - [ -6k + 5 ] = 0

-15 + 14k =  0

14k  =  15

k = 15/14

The area of the triangle is (12 * 6)/2 = 36.

The area formula for a triangle is 1/2 * bh.

We already know the base; it is XY, which is 12.

So, 1/2 * 12 * h is less than or equal to 24.

This is only true when h is less than or equal to 4, in which the altitude from D to XY is less than 4.

So, the area of the triangle that does not satisfy this condition has height 2 and base 4, which works out to an area of 4.

So, the chance of faliure is 4/36 = 1/9, and the chance of success is 1 - 1/9 = 8/9.

The solution now has 200 + 150 + 250 = 600 mL of water.

As shown in the image, Franklin could be at (0, 1), (0, -1), (1, 0), or (-1, 0) after 1 step, as shown in the blue points in the image.

We add further possibilities to those on the 2nd step, giving the orange points in the image.

Note that we do not count the blue dots in our final answer. So, we count only the orange dots, in which there are 9. Therefore, there are 9 different points that Franklin could possibly be at after 2 steps.

\(What\ is\ the\ coefficient\ of\ x^2\ ? \\ .\\ (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 8x + 7x + 1)\\ =1 + 16 x + {\color{blue}16 x^2} + 19 x^3 + 19 x^4 + 4 x^5 + 4 x^6 + x^7\)

\(\color{blue }The\ coefficient\ of\ x^2\ is\ 16.\)

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If the 23rd term is 1/4, and the 33rd term is 1/5, then you would count down ten times to get - 0.05, therefore, (0.25=1/4) - 0.1 (20 terms) =

 3/20