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You got this wrong:   8/12 = BD/DC

Should be:  6/12 = BD/DC        (But there's nothing you can do with this.)

The correct way is          6/12 = BD / (8 - BD)        correct!!!

so if S= (3.0) and the distance between the two is 5, if you add 5 to 0 you will get 5 for the y coordinate of R, (considering the fact the line is vertical). So the final answer would be 5-3, which is 2?

haha, sorry about the mistake from before, it was a pretty careless mistake, i forgot you can just subtract 5 from 8 inorder to get the X coordinate. So in this case, S would be (3,0)?

wait a second, that would be incorrect considering the x for S should be smaller than 8 

Yes!

so we can use the distance formula which is √ [ (x₂ - x₁)² + (y₂ - y₁)²], and plug S's coordinate in along with C, we get 

S= (13,0) correct? 

Let O be the origin.

Triangle(OAC) is similar to triangle(SRC).

OA = 8 and OC = 8 making OA = OC.

Because of similar triangles, SR = SC; call this length "x".

Area(triangle(SRC)) = ½·x·x = 12.5     --->     x² = 25     --->     x = 5

The distance from S to R is 5.

The distance from C to S is also 5.

Can you take it from here?

AD => x          BD => 17 - x

8² - x² = 16² - (17 - x)²

x = 97/34                 AD = 97/34

CD = sqrt(8² - x²)       CD = (15√287) / 34

[ACD] = 1/2(AD * CD) 

[ACD] = (1455√287) / 2312 = 10.66144601

y  =  (x + A) / ( Bx + C)

(4,0)     --->     0  =  (4 + A) / (B·4 + C)

                       0  =  (4 + A) / (4B + C)

                       0  =  4 + A

                       A  =  -4

(0,-4)     --->     -4  =  (0 + A) / (B·0 + C)

                         -4  =  A / C

                         -4  =  -4 / C

                          C  =  1

(-2,-2)     --->     -2  =  (-2 + A) / ( B·-2 + C)

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

                          -2  =  -6 / (-2B + 1)

                    4B - 2  =  -6

                         4B  =  -4

                           B  =  -1

y  =  (x + -4) / (-1X + 1)

You can calculate the final answer to the problem ...

We can use the distance formula, or s = d/t.

s = d/12 (since the distance remains constant we can use it for the second equation too)

s - 35 = d/27

s = d/27 + 35

d/12 = d/27 + 35

I believe you can solve for d, and then plug that back into one of the original equations to get the value of s. The speed in light traffic is "s", and the speed in heavy traffic is "s-35".

Hope this helps. Let me know if I did anything wrong!

It sure helped! I hadn't thought of the equation. Thanks InhumanCalculator!

~ PikachuLovesKetchup

Here is a list of prime numbers, Compare for yourself.

If Angle A is 27 degrees, and A, B, and C form an arithmatic sequence in that order, then we can say that:

A + (A + d) + (A + 2d) = 180, where d is the common difference.

= 27 + (27 + d) + (27 + 2d) = 180

You can then solve for d, and thus solve for C.

Hope this helped! Let me know if I did anything wrong! If you don't understand my method then you can reply and I will do my best to explain more thoroughly!

Sum( 423 , 843 , 1263 ==2,529)

Let her speed in heavy traffic==S

Her speed in light traffic ==S  +  35

[S  +  35] * 12 /60 ==S * 27/60, solve for S

S ==28 mph - her speed in heavy traffic.

28  +  35 ==63 mph - her speed in light traffic.

WOW! 2dtx.com & Coolmathgames9 are liars! did it feel like their staff mess up the script?

NO! None of them is prime. They are all composite numbers.

I misread the question and found all the ones between 1 and 1000  (not 10000)

I found 202 of them

The question did not aske how many, it simply ask if any existed.  The obvious answer is yes there are lots.

the smallest one is 31

31 is the smallest between 1 and 10000 that has a prime factor between 30 and 40 and no prime factors more than 140 or less then 30.

Here is the table I contructed to find how many there are between 1 and 1000

AX · XB  =  XY · XY

--->     15 · XB  =  8 · 8

--->            XB  =  64/15

AB  =  AX - XB

--->     AB  =  15 - 64/15

--->     AB  =  161/15

h(2)  =  f( g(2) )

g(2)  =  sqrt( f(2) ) - 3

f(2)  =  2(2) + 5  -  4 + 5  =  9

      --->     g(2)  =  sqrt( 9 ) - 3  =  3 - 3  =  0

      --->     h(2)  =  f( 0 )  =  2(0) + 5  =  0 + 5  =  5

I chose all except C.

If two triangles share the same size they are congruent, and if they are the same name they are the same triangle.

But if the triangle is the same shape, then it could be a similiar triangle, not a congruent triangle. Unless the triangle has the same name, then it is still congruent and is the same triangle.

We can use complementary counting, and see what are the odds of 0 heads to be up. Since there are 3 coins, and there is a 50% chance to get heads or tails, the odds of having 0 heads is (1/2)^3 = 1 in 8. 

Then, subtracting 1/8 from the total probability (8/8), we get the chance for at least one head to be up as 7/8.

pryap

Let x be the cost for one pound of almond and y be the cost for one pound of jelly bean

Making a system of equations:

\(3x + 5y = 24\)

\(9x+7y = 42\)

Multiplying the first equation by -3 and we get

\(-9x- 15y = -72\)

\(9x +7y = 42\)

Then we add both equations together

\(y = \frac{30}{8} = \boxed{3.75}\)

Then we plug 3.75 in for y in the first equation

\(3x + 5(3.75) = 24\)

\(3x + 18.75 = 24\)

\(3x = 5.25\)

\(x = \boxed{1.75}\)

So one pound of almond cost \($1.75\) and one pound of jelly bean cost \($3.75\).

Rewrite to find the vertex:

y  =  x² + 11x - 5 + 5x

y  =  x² + 16x - 5

y  =  (x² + 16x + .....) - 5

Complete the square:  (divide the coefficient of the x-term by 2 and square the answer):

y  =  (x² + 16x + 64) - 5 - 64

y  =  (x + 8)² - 69

The mininum will occur when x = -8; it is -69.

2x - 5  <=  x + 18

Subtract x from both sides:

  x - 5  <=  18

Add 5 to both sides:

       x  <=  23

Interval form:  (-infinity, 23]