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4. Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0

Find     1 / [2a]  + 1 / [2b]    =  (1/2) (1/a + 1/b)  = (1/2) (a + b) / [ab] = 

 [ a + b ]  / [2ab]

This isn't as hard as it seems

The sum of the roots   =  - [-8 ]  / 2   =  4

So   a + b  =  4

The product of the roots  =  7/2

So   ab   =  7/2     ⇒     2ab  = 7

So

[ a + b ] / [ 2ab]    =       4 / 7

1. Expand the product   $$(t-2)(4t^2 + 16)(t+2)$$

Rewrite   as   ( t - 2) (t + 2)  (4t^2 + 16)    =

(t^2  - 4) * 4 * (t^2 + 4)  =

4* (t^2 - 4) (t^2 + 4)  =

4 (t^4 - 16)  =

4t^4  - 64

2. Find all values of x that satisfy the equation   $\frac {12x}{x^2 + 8} = 2$ 

Multiply  both sides by   x^2 + 8

12x  =  2 (x^2 + 8)       divide both sides by 2

6x  = x^2 + 8            subtract  6x from both sides and rearrange

x^2 - 6x + 8  = 0       factor

(x - 4) (x - 2)  = 0

Setting both factors to 0 and solve for x and we get that

x = 4   and x   = 2 

3. Compute the sum of all the solutions of  $$(3x+1)(x-7)+(x-3)(3x+1)=0$$ . Express your answer as a fraction.

We can factor this as    (3x + 1) ( x - 7 + x - 3)  =  0    simplify

(3x + 1)  (2x - 10)  = 0

(3x +1) * 2 * (x - 5)  = 0      divide both sides by 2

(3x + 1)(x - 5)  = 0

Setting each factor to 0 and solving for x gives that   x = -1/3   and x = 5

The sum of these  is   5 - 1/3   =   15/3 - 1/3  =    14 / 3    =  4 + 2/3

EDIT to correct a small typo....thanks to the guest for spotting my error  !!

Angle BCD and angle DAB are congruent, so the measure of BCD is also 115°.

m∠BCD  +  66   =   180
 

m∠BCD   =   180 - 66

m∠BCD   =   114

Here's my take on this....like Melody....not sure of the outcome !!!

Number of ways  of drawing all 5 cards from one suit  =  C(13,5)

And there are four ways of doing this =  C(4,1)

Number of ways of drawing 5 cards from two suits

We want to first choose any  one of four suits = C(4,1)  and from one of these we want to choose 3 cards  = C(13.3)

And then we want to choose one of the three remaining suits = C(3,1)  and from this suit we want to choose any 2 cards  =  C(13,2)

And the number of ways of choosing any 5 cards is C (52, 5)

So the probability of choosing 5 cards from two suits is

[ C(4,1) * C(13,5)  +  C(4,1) * C(13,3) * C(3,1) * C(13,2) ]  / C(52,5)  ⇒ 

1749 / 16660

So...the probability of drawing 5 cards from at  least 3 suits  = 

1 -

[ P(drawing 5 cards of all the same suit) + P( drawing 5 cards from two suits )] =

1  -  1749 / 16660  =    14911/ 16660   ≈  89.5%

The domain of the first function is   [ 0, 5].....i.e., from the time [ in seconds ] the ball is thrown until it hits the ground

For the second function.....the ball constantly rises......so the domain is  [ 0, infinity )

This is, of course, unrealistic.....however....no bounds are specified

1.   We are just multiplying the terms under the radical

⁴√ 110   

2.

³ √ [ 24a^10 b^6 ]   =

³ √  [  2^3  * 3  *  a^10 * b^6 ]   =

To determine how many  powers  of 2 to take out   ⇒  3/3  =  1

To determine how many powers of a to take out and leave in ⇒

10/3   =  3 + 1/3  ......3 powers come out and 1 stays in

To determine how many powers of b to take out  ⇒

6/3  =  2   .....2 powers come out 

2 * a^3 b^2    ³ √ [ 3a ]

(4x  - y ) ^9          4th term

Note the  pattern, Manuel

C(9,0) * (4x)^9 *y^0  -   C(9,1) * (4x)^8 * y^1  +  C(9,2) * (4x)^7 * (y)^2  - C(9,3) *(4x)^6 * y^3

The   exponents  always sum to the expansion power and   in the form (a - b)^n, the terms alternate signs from positive, negative, positive, etc. 

We are interested in the term in red...so we have

- C(9,3) * (4x)^6 * y^3   =

- 84  * 4096 x^6 y^3  =

- 344064 x^6 y^3

Here's the expansion  as a check :

262144 x^9 - 589824 x^8 y + 589824 x^7 y^2 - 344064 x^6 y^3 + 129024 x^5 y^4 - 32256 x^4 y^5 + 5376 x^3 y^6 - 576 x^2 y^7 + 36 x y^8 - y^9

3.

³√ [ 4x^2  *  ³ √ [ 8x^7 ]    =

³√ [  4* 8  x^2  x^7 ]  =

³√ [ 32  x^9 ]  =                            {note that 32  = 2^5 ]

³√  [ 2^5 x^9 ]

To determine how many powers of 2 to take out and how many to leave in, we have

5/3   =  1  + 2/3          1 power of 2 comes out and 2 stay in

To determine how many powers of x come out and how many stay in, we have

9 / 3   =  3  powers come out      .....    so we end up with

 2 x^3 ³√ [2^2]     =    2x^3 *  ³√ 4  

4.

⁴√ 810

______    =      ⁴ √ [ 810 / 2]    =  ⁴√405   = ⁴ √ [ 81 * 5 ]  = ⁴√81   * ⁴√5

⁴√2

Note that 81  =  3^4   ....so we have...

⁴√81  * ⁴√ 5      =

⁴√ [ 3^4] *  ⁴√5    =

To determine how many powers of 3 to take out    4/4  =  1 power of 3

3 ⁴√5

√ [ 3 x^12 y^10 ]

_____________

√ [5 x^6  y ^3 ]

We  are using the property that   a^m / a^n   =  a^(m - n)

√ [ 3 x^(12 - 6)  y ^ (10 - 3) ]

_______________________

√ 5

√ [ 3 x^6  y^7]

___________

√5

Now...to determine how many powers to take out and how many to leave in :

Take   6 / 2   =  3 powers of x come out

And  7/2  =   3 + 1/2       =    3 powers of y come out  and 1 stays in

So we have

x^3 y^3 √ [ 3y ]

____________                  rationalize the denominator by multiplying op/bottom by √5

         √5

x^3 y^3 √ [ 3y ] √5

_______________

     √ 5 √ 5

x^3 y^3  √ [ 15 y ]

_______________          the second answer is correct   

           5

How did you get the domain?

Sum of a geometric series is given by

First term [  1 -  common ratio^n ] / [ 1 - common ratio]    

Where n is the sum of the first n terms

So we have

65 = 5 [ 1 - 3^n ] / [ 1 - 3 ]       simplify

65 = 5 [ 1 - 3^n ] / -2       multiply by -1 on top/bottom of rthe right side

65 = 5 [ 3^n - 1] / 2            multiply both sides by 2/5

26  =  3^n - 1         add 1 to both sides

27  =  3^n    

Because  3^3  = 27......then   n  = 3 terms

See the graphs, here :

https://www.desmos.com/calculator/kwkh3bcwpk

First one

[ x^2 + 3x - 28 ]  / [ x + 7]        factor the numerator

[ (x + 7) (x - 4) ] / ( x + 7)  =

x - 4

The "hole" will occur  where    x + 7  = 0   ⇒  x   =  - 7

So     plugging in  x = -7   into   x - 4, we get   -7 - 4  = -11

So....the coordinates are  ( -7, - 11)

Second one.....the third graph is the correct one

See the graph here :    https://www.desmos.com/calculator/8ytbtiboqq

This graph should help:

.

Why don't you try graphing them yourself on Desmos and see what happens :))

I suggest you study the other one. 

also, I know a lot of teachers talk about cross multiply but iI think it is a confusing way to teach it.

It is easy I guess but the logic is not very obvious.

What you must ALWAYS do is the SAME thing to each side. The equation must always balance!

the pic is showing that if you add 10 to one side then you must add 10 to the other side too,

otherwise the scale would become unbalanced!

Now ot your p[roblem:

$y = \frac{1}{x}\\ \text{First make both sides a fraction}\\ \frac{y}{1} = \frac{1}{x}\\ \text{Now you want x on the top so turn BOTH sides upside down (take the reciprocals)}\\ \frac{1}{y}= \frac{x}{1}\\ \frac{1}{y}= x\\ \text{Now swap sides}\\ x=\frac{1}{y}$

It is NOT    y=x/z + 1      IT IS      t=x/(z+1)      there is a big difference  

$x = yz + y$

the aim is to make y the subject so the first thing you have to do it to facot out the y (so that you will only see one y)

$yz+y =y*z+y*1= y(z+1)$

so

$x = yz + y\\ x=y(z+1)$

Now you want the y bt itself. At the moment it is multipled by (z+1) the opposite of multiply is divide. 

SO you have to divide BOTH sides by (z+1)

$x = yz + y\\ x=y(z+1)\\ \frac{x}{z+1}=\frac{y(z+1)}{z+1}\\ now \;\;cancel\\ \frac{x}{z+1}=y\\ \text{Now swap sides}\\ y=\frac{x}{z+1}$

9.Corresponding parts of similar triangles are proportional 

10. Substitution 

11. Multiplication Property of Equality 

Your Welcome!!!!

Quadratic Trinomial  Since degree is 2 

Cubic Trinomial   since degree is 3

since the type of polynomail stresses upon the highest exponent or degree of the polynomial . This all determines the nature of polynomial

Thank you Guest

help me Please

Binomial Theorem

Find the indicated term of the expansion.

4th term of (4x-y)^9