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(x+5)/10 +5 = (7x-10)/6
(x+55)/10 =(7x -10)/6
6x + 330 =70x - 100
6x-70x =-100 -330
-64x =-430
x =-430/-64
x =215/32

I need:

an= n2-1

an= 2n3-2n

an= n2+n/2

Urgent!!!!!!!!

What do you want done with these? They appear to be formulas for the nth term of a sequence:

Example: Give the 10th term for each formula:

a(10) =10^2 -1 =99

a(10) =2*10^3 - 2*10 =1,980

a(10) =10^2 + 10/2=105

196 units squared

The sides have to be  14,14 and 14sqrt2

We do not know what you are asking for. The question is incomplete.

no solution to this system sorry

Thanks :)

Sorry Melody: I did forget to give the link. Here it is:

https://www.youtube.com/watch?v=xnE_sO7PbBs&list=PLDZcGqoKA84E2a0L6IS68hswD4iiUN2Cv

Thanks guest but did you forget to include the link? 

Mmm seems to be some disagreement here ://

Bill, Phil, and Jenny are siblings.

Bill is twice as old as Phil.                         2P=B    

Jenny is two years younger than Bill.        B-2=J        B=J+2

so     

B=2P=J+2 

2P=J+2

J=2P-2 

Currently, their dad is twice as old as the sum of their ages.  D=2(P+B+J)

In nine years, Dad's new age will be equal to the sum of his three kids' new ages.

D+9= P+B+J+27

D= P+B+J+18

But    D=2(P+B+J)

therefore Dad is 36

P+B+J=18

P+2P+(2P-2)=18

5P-2=18

P=4     Phil is 4

Bill is 8

Jenny is  6

check:

Bill 8,  Phil 4 , and Jenny 6  are siblings.    Dad is 36

Bill is twice as old as Phil.    yes

Jenny is two years younger than Bill.    yes

Currently, their dad is twice as old as the sum of their ages.   2*(8+4+6)=36    yes

In nine years, Dad's new age will be equal to the sum of his three kids' new ages. What is Jenny's current age?       45=18+27    yes

The codes consist of 5 numbers

1. There are 200 people who each have a code. How big is the possibility that one of them has the code 06479? 

1/10^5

2. One of the people gets a new code. How big is the possibility that this code begins with 55?

10^3 / 10^5 = 1/100

8 keys are no good there are only 6 possible keys left so the chance that the next one is good is 1/6

The next is 1 in 4 which means there are 4 keys left, 

tries 4th last, no good :(

tries 3rd last, no good :(

tries second last, good great, no good doesn't matter, it has to be the last one.

So the worst scenario is that 3 more keys must be tried :)

Suppose the cards a labelled ABCD and there are 4 of each so 20 cards in total.

 The cards are shook well, and then a card will get picked from the stack, which will be noted and then put back in the stack again. This happens three times. 

1. How many different sequences are there possible?   4^3 = 64

2. How big is the chance that there'll be chosen three of the same kind?  4/64 = 1/16

3. Now they pick five cards instead of three, how many different sequences are there possible now?

             4^5 = 1024

Here is another presentation for what our guest did. Thanks guest :)

\(\frac{(x+5)}{9} +5 =\frac{ (7x-10)}{6}\)

The lowest common denominator is 18 so multiply every term by 18

That way you will get rid of all the fractions and it will be much easier.

\(\frac{18(x+5)}{9} +18*5 =\frac{18 (7x-10)}{6}\\ \frac{2(x+5)}{1} +90 =\frac{3 (7x-10)}{1}\\ 2(x+5)+90 =3 (7x-10)\\ 2x+10+90 =21x-30\\ 2x+100 =21x-30\\ 130 =19x\\ x=\frac{130}{19}=6\frac{16}{19}\)

The target is cut into 12 sectors. One sector is black. Since the entire are is 18dm^2

It follows that the black are is 18 / 12 = 1.5dm^2

Expectations:

She'll hit the black part 1/12 of the time.     72/12= about 6 times

She'll hit the white part 3/12  of the time.     3*6= about 18 times

She'll hit the grey part 8/12 of the time.     8*6 = about 48 times

3)

If he hits the white part 5 times he will hit the grey part  \(\frac{5}{3}*8= 13.\dot3 \approx 13\;times\)

Here is the first one:

Solve for x:
(x + 5)/9 + 5 = (7 x - 10)/6

Put each term in (x + 5)/9 + 5 over the common denominator 9: (x + 5)/9 + 5 = 45/9 + (x + 5)/9:
45/9 + (x + 5)/9 = (7 x - 10)/6

45/9 + (x + 5)/9 = ((x + 5) + 45)/9:
(x + 5 + 45)/9 = (7 x - 10)/6

Grouping like terms, x + 5 + 45 = x + (45 + 5):
(x + (45 + 5))/9 = (7 x - 10)/6

45 + 5 = 50:
(x + 50)/9 = (7 x - 10)/6

Multiply both sides by 18:
(18 (x + 50))/9 = (18 (7 x - 10))/6

18/9 = (9×2)/9 = 2:
2 (x + 50) = (18 (7 x - 10))/6

18/6 = (6×3)/6 = 3:
2 (x + 50) = 3 (7 x - 10)

Expand out terms of the left hand side:
2 x + 100 = 3 (7 x - 10)

Expand out terms of the right hand side:
2 x + 100 = 21 x - 30

Subtract 21 x from both sides:
(2 x - 21 x) + 100 = (21 x - 21 x) - 30

2 x - 21 x = -19 x:
-19 x + 100 = (21 x - 21 x) - 30

21 x - 21 x = 0:
100 - 19 x = -30

Subtract 100 from both sides:
(100 - 100) - 19 x = -100 - 30

100 - 100 = 0:
-19 x = -30 - 100

-30 - 100 = -130:
-19 x = -130

Divide both sides of -19 x = -130 by -19:
(-19 x)/(-19) = (-130)/(-19)

(-19)/(-19) = 1:
x = (-130)/(-19)

Multiply numerator and denominator of (-130)/(-19) by -1:
x = 130/19

Here is the second one:

Solve for x:
(x + 5)/10 + 5 = (7 x - 10)/6

Put each term in (x + 5)/10 + 5 over the common denominator 10: (x + 5)/10 + 5 = 50/10 + (x + 5)/10:
50/10 + (x + 5)/10 = (7 x - 10)/6

50/10 + (x + 5)/10 = ((x + 5) + 50)/10:
(x + 5 + 50)/10 = (7 x - 10)/6

Grouping like terms, x + 5 + 50 = x + (50 + 5):
(x + (50 + 5))/10 = (7 x - 10)/6

50 + 5 = 55:
(x + 55)/10 = (7 x - 10)/6

Multiply both sides by 30:
(30 (x + 55))/10 = (30 (7 x - 10))/6

30/10 = (10×3)/10 = 3:
3 (x + 55) = (30 (7 x - 10))/6

30/6 = (6×5)/6 = 5:
3 (x + 55) = 5 (7 x - 10)

Expand out terms of the left hand side:
3 x + 165 = 5 (7 x - 10)

Expand out terms of the right hand side:
3 x + 165 = 35 x - 50

Subtract 35 x from both sides:
(3 x - 35 x) + 165 = (35 x - 35 x) - 50

3 x - 35 x = -32 x:
-32 x + 165 = (35 x - 35 x) - 50

35 x - 35 x = 0:
165 - 32 x = -50

Subtract 165 from both sides:
(165 - 165) - 32 x = -165 - 50

165 - 165 = 0:
-32 x = -50 - 165

-50 - 165 = -215:
-32 x = -215

Divide both sides of -32 x = -215 by -32:
(-32 x)/(-32) = (-215)/(-32)

(-32)/(-32) = 1:
x = (-215)/(-32)

Multiply numerator and denominator of (-215)/(-32) by -1:
x = 215/32

r=sqrt(8^2 - 15^2)    Square both sides

r^2 =(8^2 - 15^2)

r^2 =(64 - 225)

r^2=-161 =-1 x 161 Take the sqrt of both sides

r =i sqrt(161)

r =12.688577 i

Step one is simplifying the equatation!

(x+5/9)+5 = (7x-10/6)

x + 5/9 + 5 = 7x − 10/6

x + 5/9 + 5 = 7x + −5/3

x + 50/9 = 7x + −5/3

Step two is subtracting 7x from both sides

−6x + 50/9 −7x = −5/3

Step three is subtracting 50/9 from both sides

−6x = −65/9

Step four is dividing both sides by −6

−6x/−6 =−65/9/−6

x = 65/54

Now the second one! First we simplify the equatation.

x + 1/2 + 5 = 7x + −5/3

x + 11/2 = 7x + −5/3

Step two is to subtract 7x from both sides

−6x + 11/2 = −5/3

Step three is subtracting 11/2 from both sides

(−6x + 11/2 − 11/2 = −5/3 − 11/2)

−6x = −43/6

Now we divide both sides by −6

−6x/−6 = −43/6/−6

x = 43/36

(a) A regular nonagon (9-sided polygon) is inscribed in a circle as shown below. Three of the 9 vertices are selected at random, and the triangle formed by those three points is drawn. What is the probability that the center of the circle lies inside the triangle? (A successful selection of 3 such points is shown below.)

Hi Mr Owl and guest :)

Here is my thinking.

Any point can be chosen as the first vertex. So P(correct first vertex) = 1

Now draw a line through the chosen vertex and through the centre thereby bisecting the nonogon.

Now the centre will be included so long as one vertex is in side A and the other is in Side B

There are 4 points on side A and there are 4 suitable matches for each of those on part B so that is 4*4=16 possible triangles that include the centre (this is for each of the 9 initial vertices)

Now the number of ways that 2 points can be chosen from the 8 without restriction is 8C2 = 28

P(the centre is included in the triangle is) = \(\frac{16}{28}=\frac{4}{7}\)

24+4n=350+3.5n

Subtract 4n-3.5n and 3.5-3.5

24+.5n=350

Subtract 24-24 and 350-24

.5n=326

n=652

Solve for n:  24 + 4n = 350 + 3.5n     

 24 + 4n = 350 + 3.5n     Isolate n to the LHS

4n - 3.5n =350 - 24

0.5n =326                          Divide both sides by 0.5

n =326 / 0.5

n =652

2/b=p

b-2=j

p+b+j=2d

b+p+j=18

3p+2p-2=18 or 5p=20

20/5=4

Penny is 4 years old.

Let Phil's age =P
Bill's age =2P
Jenny's age =2P - 2
[P + 2P + 2P - 2] x 2 =D
[P + 2P + 2P - 2] +27 =D+9, solve for D, P
P = 4 years - Phil's age
2*4 =8 years - Bills's age
8 - 2 =6 years- Jenny's current age.
D =36 years - Dad's age.
18+27 =36 + 9
45 Years = 45 yeas - nine years from now.

f(x)  is the number of kilowatts generated.

x  is the wind speed in meters per second.

f(10)  =  0.08(10)³ + (10)² + 10 + 0.26

f(10)  =  0.08(1000) + 100 + 10 + 0.26

f(10)  =  80 + 100 + 10 + 0.26

f(10)  =  190.26

So when the wind speed is 10 meters per second, the power generated by the wind turbine is 190.26 kilowatts.

1.   4x³ + 8x² - 96x

                                      Factor  4x  out of each term.

=    4x(x² + 2x - 24)

                                      Factor  x² + 2x - 24  as  (x - 4)(x + 6) .

=  4x(x - 4)(x + 6)

(The one you have selected is right.    )

3.     We want to solve this equation for  x :

0  =  x⁴ - 4x³ + 3x²

                                      Factor  x²  out of all the terms on the right side.

0  =  x²(x² - 4x + 3)

                                      Factor  x² - 4x + 3  as  (x - 1)(x - 3) .

0  =  x²(x - 1)(x - 3)

                                      And we can "factor"  x²  as  x * x .

0  =  (x)(x)(x - 1)(x - 3)

                                      Set each factor equal to zero and solve for  x .

x  =  0       or       x  =  0       or       x  =  1       or       x  =  3

x  =  0  occurs twice, so it has multiplicity  2 .

x  =  1  occurs once, so it has multiplicity  1 .

x  =  3  occurs once, so it has multiplicity  1 .

1) Show that the circumradius of a right triangle is equal to half the hypotenuse. 

Let   ABC  be a right triangle  and let B be the right angle

Let  B  = (0,0), A = (0, a)  and C =  (c ,0)

So......the midpoint of the hypotenuse is given by  (c/2, a/2)

And the distance ffrom this point to A  is    √ [ c^2 /4   + a^2/4]  = √ [ a^2 + c^2] / 2

And the distance from this point to C  is  √ [ c^2/4 + + a^2/4 ] =  √ [ a^2 + c^2] / 2

And the distance from this point to B  is √ [ a^2 + c^2] / 2

So.....a circle centered at the midpoint of the hypotenuse will pass through all three of the vertices since they are all equally distant from this midpoint