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4) The sequence is certainly arithmetic! Notice that \({\underbrace{13,}\underbrace{20,}\underbrace{27,}34,...}\\ +7+7+7\) . In other words, the common difference is 7.

5) This is another arithmetic sequence with a common difference of -7.

Remember that awesome formula that tells you the nth term of an arithmetic sequence? It's \(a_n=a_1+d(n-1)\) . Let's use it!

\(a_n=a_1+d(n-1)\\ a_{50}=5-7(50-1)\\ a_{50}=5-7*49\\ a_{50}=-338\)

6) The arithmetic mean is also the average. 

\(a_n=\frac{a_{n-1}+a_{n+1}}{2}\\ a_n=\frac{3.9+7.1}{2}\\ a_n=\frac{11}{2}\\ a_n=5.5\)

Thanks X²

I unintentionally answered this question already when I answered your previous question!

Thank You!!!

If it is difficult for you to comprehend this problem, then it may be best to think of an example. Don't overcomplicate matters, though! I think \(\frac{x^3+5x^2}{x}=0\) is an adequate example because this equation involved the elimination of a variable. 

If you were to solve it, it may look something like this!

\(\frac{x^3+5x^2}{x}=0\\ \frac{x^2(x+5)}{x}=0\\ x(x+5)=0\\ x=0\text{ and }x=-5\)

A) Confirming the validity of all roots is essential. Notice that \(x=0\) in the previous example is an extraneous solution because it results in a division-by-zero error.

B) Factoring was essential to recognize that there was a common factor in the first place!

C) Yes, this is a necessary precaution. If we set the common factor equal to zero, then we will get that \(x\neq0\) because this does not fit the domain.

D) This step is illogical; do not do this.

8)Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.

We have the secant-tangent theorem, first

AB^2  = CB (CB + CD)

12^2  = CB^2 + 18CB

CB^2  + 18CB - 144   = 0    factor as 

(CB - 6)  ( CB + 18)  = 0

The frist answer  gives the positive answer  for  CB - 6 =  0  ⇒  CB  = 6

Draw AC    ......and the tangent of  angle BAC  =   6/12   =  1/2   

So   arctan (1/2)  = BAC....and minor arc  AC  is twice this  = 2 arctan (1/2) 

And  AC  =  √   ( BA^2  + BC^2)  = √  (12^2  + 6^2)  = √ (144 + 36)  =  √160

Draw  OC  and OA    and angle  COA  = minor arc AC  =  2 arctan (1/2) 

Using the Law of Cosines to find the radius.....we have that

160  = 2r^2  - 2r^2 * cos (2 arctan (1/2) )  

Let   cos (2arctan(1/2) )  =   1  - 2sin^2 ( arctan(1/2) ) ⇒  sin (arctan (1/2) )  = 1/ √5  ....so we have

  1  -  2(1/√5)^2   =  1 - 2/5  = 3/5  = .6 

160  = 2r^2 - 2r^2 (.6)

160  = r^2  ( 2 - 1.2)

160 = r^2 (.8)

200  = r^2

r  = √200  =    10√2

Thank you X and thanks Clown Prince!! for the joke

Hi, ManuelBautista2019!
 

The discriminant of a quadratic equation is the expression of the radicand of the quadratic formula \(x ={-b \pm \sqrt{\textcolor{red}{b^2-4ac}} \over 2a}\). In the given quadratic equation x^2-5x+10, a=1, b=-5, and c=+10. The discriminant will allow you to answer every question here.

Let's see the value of the discriminant and interpret its meaning. 

\(x^2-5x+10\\ a=1,b=-5,c=10\\ b^2-4ac\\ (-5)^2-4*1*10\\ 25-40\\ -15 \) 

1) The discriminant of the original quadratic yields a negative value, as calculated above, so the roots are imaginary.

2) The roots of a quadratic can be equal if and only if the discriminant equals 0, so the roots are unequal.

3) We already determined that the roots are imaginary (look at #1), so the roots are neither of the options listed. 

Sorry about that!!.

I appreciate you being funny but I need a Real answer please

1.  We probably only need two terms, NSS

First term  = 12 (1)  - 11  = 1

Second term  = 12(2)  - 11  =  13

It appears that the third answer is good

2. The "formula"  is

a_n  = a_n-1  + 13

The next term  is

57 + 13  =

70

Second answer

3.  A litle tricky, NSS......it appears that the nth term is

10  - .5 ( n - 1)

We need the 8th term....so we have.....

10 - .5 ( 8 - 1)  =

10 - .5 (7)  =

10  - 3.5 =

6.5   ⇒    first answer

It will always be \(2^n-(2^{n-1}+2^{n-2} \cdots +2^1+2^0)=1\).Thus, the minimum value is \(\boxed{1}\)

log (2x + 4)  = 0

This will be true when       log (1)   =   0

So....we need to find this :

2x + 4  = 1           subtract 4 from  both sides

2x  = -3        divide both sides by  2

x  = -3/2

7. It's helpful if we can find a general  "formula"

One possibility  is :

a_n  = -7  + 10 ( n - 1)

So term 110  is  just

-7  + 10 (110 - 1)   =

-7  + 10 (109)  =

1083

8.  Note, NSS...looking at the last two terms, it appears that we are subtracting 5 from the previous term to get the next term

So.....second term  =  -7 - 5  = -12

And....third term =  -12 - 5  = -17

9.

Baskets left over  plus  baskets per day *  work days  =  total  baskets   =

13  + 12 ( 15)  =

13  +  180  =

193

Just a minor typo in the 3rd to the last step. The second x^3 should read y^3.

Thank You!!

x^2  + 7x  - 30   =  0

(x - 3)  ( x + 10)  = 0         set each factor to 0   and solve for x  and we get that the roots are   x = 3   and x  = - 10

So

1.Real

2. Unequal

3. Rational

10.

She does 4 more each day than on the previous day....so.....

The number of sit-ups on the nth day can  be modeled by :

17   + 4 ( n - 1)

So...on Day 12, she does

17  + 4(12 - 1)    =

17 + 4*11  = 

17   + 44 =

61  sit-ups

Here's the first, tertre

( x + y + z)^8  =   ( x  +  ( y + z) )^8  = 

C (8,5) * x^3 * (y + z)^5   .......we need to  find the coefficiient on y^3z^2  for the second part in red        

C (8,5) * x^3 * C(5,3)* y^3 * z^2  =

56* x^3  *  10 * y^3 * z^2   =

560   (x^3 * y^3 * z^2)

y_n  = -5n  - 5

First term.....n = 1   ⇒    -5(1)  - 5  = -10

Second term....n  = 2  ⇒  -5(2)  - 5   = -15

Note the pattern....NSS...we are  just subtracting 5 from the previous term to get the next one....so   -20, -25, -30    are the other terms

Answer  #  3

2. 

c_n  = 3n  - 1          replace n with  15    and we have

c₁₅  =  3(15)  - 1   =   45  - 1   =  44

3.  The recursive formula  is

a_n  = a_n-1 + 6,    a₁  = 7    { This  "formula" means that are taking the previouis term and adding 6 to it to get the next term  }

To get the next term we have

31  +  6   = 37    {The first answer }

4.  a_n  = a_n-1  - 3, where  a₁ = 7, -8

5. a_n  = -2n + 10

a₁₄  =   -2(14)  + 10  = - 28 + 10  =  -18

The last answer on this one, NSS

Math is not my strongest subject, english is but i think its answer #3 but im not positive.