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r(x) can be 0, 2, 4, and 6.

So the sum of all possible values of s(r(x)) = s(0) + s(2) + s(4) + s(6)

FREEEZE! there is no s(0) or s(6) because the domain of s(x) is {1,2,3,4}, so get rid of those.

s(2) + s(4)

= (2+1) + (4+1)

= 3 + 5

= 8

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ 2*sqrt(6)*x + 4]

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [(2(*sqrt(6)*x + 2) ]   multiply both sides by 2

2sqrt (x) / [sqrt(3)*x + sqrt(2)]  = 1 / [ sqrt(6)*x + 2]  cross multiply

(2sqrt (x))  [ sqrt(6)*x + 2]  =  sqrt(3)*x + sqrt(2)

Factor  sqrt (2)  out of sqrt  sqrt(6)*x + 2

[2sqrt (x) ] ( [sqrt (2)] * [sqrt(3) *x + sqrt(2)] )  = [ sqrt (3)*x + sqrt(2) ]

2sqrt(2)* sqrt(x) [ sqrt (3)*x + sqrt(2)] = [sqrt(3)*x + sqrt(2)]

2sqrt(2)*sqrt(x) [ sqrt (3)*x + sqrt(2)] - [sqrt(3)*x + sqrt(2)]  = 0

[ sqrt (3)*x + sqrt(2)] [ 2sqrt(2)* sqrt(x) - 1 ]  = 0

Set both factors to 0  and either

sqrt (3) * x + sqrt (2) = 0  →   x  = -sqrt (2)/ sqrt(3)

But this gives a non-real result to the original problem

Or

2sqrt(2) * sqrt(x)   - 1  = 0

sqrt(8) * sqrt (x) = 1

sqrt (8x) = 1     square both sides

8x   = 1     divide both sides by 8

x = 1/8

21

You might want to learn how to use a calculator. 

21

You might want to learn how to use a calculator. 

idk

3y = 2x - 3

Divide everything by 3.

y = (2/3)x - 1

In this format, the coefficient of x is the slope. 2/3 is the slope.

Parallel lines have the same slope.

Plug in the point (-3, 4) using (2/3) as the slope and find the y-intercept, b.

4 = (2/3)(-3) + b

4 = -2 + b

6 = b

Parallel line: y = (2/3)x + 6

Perpendicular lines have a negative reciprocal slope.

Plug in the point (-3, 4) using (-3/2) as the slope and find the y-intercept, b.

4 = (-3/2)(-3) + b

4 = 9/2 + b

-1/2 = b

Perpendicular line: y = (-3/2)x - 1/2

Can I join??

On the restricted domain, the function will reach its highest point at y = 1 when x  = 0

[Note that this point will always be on the graph of y = a^x ]

As x gets larger and larger......y gets smaller and smaller, but always > 0

So....the range is  (0, 1 ]

Look at the graph : https://www.desmos.com/calculator/whogz6eb9o

993

We have that

E + B + J <  40

And....when all the trading of counters has ceased..we find that

{E - E/5) =  (B - B/3 + E/5 +  J/4)  = (J - J/4 + B/3)

(4/5)E =  [ (2/3)B + E/5 + J/4 ] = [ (3/4)J + B/3 ]

Using the first and third parts of the equality, we have

(4/5)E = B/3 + (3/4)J    multiply through by 60→   48E = 20B + 45J    (1)

Using the first two parts of this equality, we have

(4/5)E =  [ (2/3)B + E/5 + J/4 ]

(3/5)E = (2/3)B + J/4   multiply through by 60 →  36E = 40B + 15J

Then multiply through by -3 →   -108E = -120B - 45J     (2)
 

Add (1)  and (2)

-60E = -100B

B = (3/5)E

And using the first two parts of the equality with substitution for B we have that

(3/5)E  = (2/3)(3/5)E + J/4

(3/5)E = (2/5)E + J/4

(1/5)E = J/4

J = (4/5)E

So

E + B + J <  40

E + (3/5)E + (4/5)E < 40

(12/5)E < 40

E < 16+2/3

So......since E is divisible by 5, the greatest that Emma could have started with is 15

Which means that Ben started with (3/5)(15)  = 9

And Jack started with(4/5)(15) = 12

And

E + B + J   =

15 + 9 + 12 =

36 < 40

So....they could have started with 36 counters

BTW.....there are other possibilites

E = 10 , B  = 6   and J = 8   =  24 counters

E = 5, B = 3  and J = 4   =    12 counters

Multiply out the denominator and combine like terms:

(x - 1)(x - 1) + (x + 1)(x + 1)

(x² - 2x + 1) + (x² + 2x + 1)

x² - 2x + 1 + x² + 2x + 1

2x²  + 2

Now set this equal to 0 and solve for x.

2x² + 2 = 0

2x² = -2

x² = -1

x = i

There are no real numbers that make the y undefined. So the domain is all real numbers.

( - ∞, ∞)

Solve for x:
x/(3 x^2 + 2) = 1/(24 x^2 + 16)

Cross multiply:
x (24 x^2 + 16) = 3 x^2 + 2

Expand out terms of the left hand side:
24 x^3 + 16 x = 3 x^2 + 2

Subtract 3 x^2 + 2 from both sides:
24 x^3 - 3 x^2 + 16 x - 2 = 0

The left hand side factors into a product with two terms:
(8 x - 1) (3 x^2 + 2) = 0

Split into two equations:
8 x - 1 = 0 or 3 x^2 + 2 = 0

Add 1 to both sides:
8 x = 1 or 3 x^2 + 2 = 0

Divide both sides by 8:
x = 1/8 or 3 x^2 + 2 = 0

Subtract 2 from both sides:
x = 1/8 or 3 x^2 = -2

Divide both sides by 3:
x = 1/8 or x^2 = -2/3

Take the square root of both sides:
x = 1/8 or x = i sqrt(2/3) or x = -i sqrt(2/3)

x/(3 x^2 + 2) ⇒ 1/(8 (2 + 3 (1/8)^2)) = 8/131
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (1/8)^2) = 8/131:
So this solution is correct

x/(3 x^2 + 2) ⇒ -(i sqrt(2/3))/(2 + 3 (-i sqrt(2/3))^2) = ∞^~
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (-i sqrt(2/3))^2) = ∞^~:
So this solution is incorrect

x/(3 x^2 + 2) ⇒ (i sqrt(2/3))/(2 + 3 (i sqrt(2/3))^2) = ∞^~
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (i sqrt(2/3))^2) = ∞^~:
So this solution is incorrect

The solution is:
Answer: |x = 1/8

Anyone there! Help!!!!!!!!!!!!!!!!!!!!!!!!

I had no idea how to do this one, but I think I understand it now! 

Why in the world would you write it like that!!!???. It appears that you trying to find the Future Value of $558,000 @ 3.782% for 15 terms(years)?. If I'm reading it right, don't you know that there is a shortcut to this?? The formula is very common and simple:

FV = PV x [1 + R]^N

FV = 558,000 x [1 + 0.03782]^15

FV = 558,000 x 1.745141....

FV = $973,788.71

P.S. This formula compounds your $558,000 @ 3.782% for 15 years.

16V^2 + 98V - 2V128=0, solve for V

Solve for V:
16 V^2 - 158 V = 0

Factor V and constant terms from the left hand side:
2 V (8 V - 79) = 0

Divide both sides by 2:
V (8 V - 79) = 0

Split into two equations:
V = 0 or 8 V - 79 = 0

Add 79 to both sides:
V = 0 or 8 V = 79

Divide both sides by 8:
Answer: |V = 0            or                V = 79/8

Thanks so much! 

We have

x^2 - 14x + 3y + 70  + y^2 - 11y - 21  = 0

x^2 - 14x + y^2 - 8y + 49   = 0

x^2 - 14x + 49 + y^2 -8y + 16  = -49 + 49 + 16

(x - 7)^2 + (y - 4)^2  = 16

This is a circle with a radius of 4  centered at (7, 4)

Look at the graph, here : https://www.desmos.com/calculator/9aqppc3dkl

The line y = x - 3    passes through (7,4)  so the line segment bounded by the circle is the diameter of that circle......so....the region of interest is just (1/2) the area of this circle  = (pi/2)*(4)^2  =

8pi units^2  ≈ 25.13 units^2

pie = 31/7     .........?

Are you asking for the radius of a circle with a circumference of 264?

Simplify the expression, and you'll get 579933.06865659

The answer is [-1,1].

Oops....mis-read the denominator

x^2 +4x - 21    will have 0's  at

(x + 7) (x - 3)   = 0       set each factor to 0   and x = 3  or x = -7

So....the  domain  is  (-inf, -7) U (-7, 3) U (3, inf)

Thanks to hectictar and EP  for catching my error.....!!!!!

i know right,   who are the nerds saying it is NOT hard 

Thanks Electric Pavlov!