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Aleks04 [339]
3 years ago
10

For each of the geometric sequences, determine the common ratio. Sequence A has common ratio

Mathematics
2 answers:
klemol [59]3 years ago
7 0

Answer:

4

Step-by-step explanation:

I did it

jolli1 [7]3 years ago
7 0

Sequence A has common ratio 4

Sequence C has common ratio 0.2

Sequence E has common ratio -1.5

EG. 2020

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PLEASE HELP WILL GIVE BRAINLIEST NOW
hoa [83]

Answer:

5. (2,4)  6. (-1,6)

Step-by-step explanation:

The solution to multiple graphs is where the two graphs meet.  

(-1,6) is the only point where both of the y values are the same.

7 0
2 years ago
Please help me out on this !!
ElenaW [278]
Would it be 43? 62-19=43? Sorry if this isn’t helpful I’m just trying to catch up on homewokr so I gotta answer some questions
6 0
2 years ago
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Can someone help me
erik [133]

Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

  (2x -12)/2 = arcsin(16/22×sin(24°))

  x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

  2x -12 = 0

  x -6 = 0 . . . . . divide by 2

  x = 6 . . . . . . . add 6

Then the range of values of x will be ...

  6 < x < 23.206

_____

* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

  16² = 2·AB²·(1 -cos(2x-12))

The value of AB drops out of the ratio of these equations, and the result for x is as above.

4 0
2 years ago
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A cooler contains eleven bottles of sports drink: five lemon-line flavored and six orange flavored. You randomly grab a bottle a
Elenna [48]

Answer:

The probability that you would choose lemon-lime and then orange is 3/11 =.273.  

Step-by-step explanation:

These are 'dependent events', which mean that your the event is affected by previous events.  So, because you have eleven total bottles (five lemon-lime and six orange) and you do not replace the first bottle, that would only leave you with ten bottles remaining.  The probability that you will pick the lemon-lime on the first choice is 5/11 because all of the bottles are there.  However, your second choice will only include ten total bottles since you already took one.  The probability that you would choose orange would be 6/10.  When you multiply these two fractions and reduce to simplest form, you get 3/11.  


5 0
2 years ago
Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6
Alla [95]

Given:

The expression is:

2x^3+mx^2+nx+c

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

m+n=-6

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

Let the given polynomial is:

P(x)=2x^3+mx^2+nx+c

It leaves the same remainder when divided by x -2 or by x+1. By using remainder theorem, we can say that

P(2)=P(-1)              ...(i)

Substituting x=-1 in the given polynomial.

P(-1)=2(-1)^3+m(-1)^2+n(-1)+c

P(-1)=-2+m-n+c

Substituting x=2 in the given polynomial.

P(2)=2(2)^3+m(2)^2+n(2)+c

P(2)=2(8)+m(4)+2n+c

P(2)=16+4m+2n+c

Now, substitute the values of P(2) and P(-1) in (i), we get

16+4m+2n+c=-2+m-n+c

16+4m+2n+c+2-m+n-c=0

18+3m+3n=0

3m+3n=-18

Divide both sides by 3.

\dfrac{3m+3n}{3}=\dfrac{-18}{3}

m+n=-6

Hence proved.

7 0
2 years ago
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