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ladessa [460]
3 years ago
15

Divide (2m3+7m2−6)÷(m−7)

Mathematics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

2m^2 + 21m + 147 + 1023/(m - 7)

Step-by-step explanation:

m - 7 ) 2m^3 + 7m^2  +  0m -  6 ( 2m^2 + 21m + 147 <--- Quotient

          2m^3 - 14m^2

                        21m^2 + 0m

                        21m^2 - 147m

                                    + 147m - 6

                                    + 147m - 1029

                                                   1023  <--- remainder.

       

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55. (a) If alpha and beta are the roots of the equation xsquare+ px+q=0 and beta&gt;alpha find the square of the
densk [106]

Answer:

√(p²-4q)

Step-by-step explanation:

Using the Quadratic Formula, we can say that

x = ( -p ± √(p²-4(1)(q))) / 2(1)  with the 1 representing the coefficient of x². Simplifying, we get

x = ( -p ± √(p²-4q)) / 2

The roots of the function are therefore at

x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus

( -p + √(p²-4q)) / 2 - ( ( -p - √(p²-4q)) / 2)

= 0 + 2 √(p²-4q)/2

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Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
Evaluate each finite series for the specified number of terms. 1+2+4+...;n=5
zaharov [31]

Answer:

31

Step-by-step explanation:

The series are given as geometric series because these terms have common ratio and not common difference.

Our common ratio is 2 because:

1*2 = 2

2*2 = 4

The summation formula for geometric series (r ≠ 1) is:

\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}} or \displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}

You may use either one of these formulas but I’ll use the first formula.

We are also given that n = 5, meaning we are adding up 5 terms in the series, substitute n = 5 in along with r = 2 and first term = 1.

\displaystyle \large{S_5=\frac{1(2^5-1)}{2-1}}\\\displaystyle \large{S_5=\frac{2^5-1}{1}}\\\displaystyle \large{S_5=2^5-1}\\\displaystyle \large{S_5=32-1}\\\displaystyle \large{S_5=31}

Therefore, the solution is 31.

__________________________________________________________

Summary

If the sequence has common ratio then the sequence or series is classified as geometric sequence/series.

Common Ratio can be found by either multiplying terms with common ratio to get the exact next sequence which can be expressed as \displaystyle \large{a_{n-1} \cdot r = a_n} meaning “previous term times ratio = next term” or you can also get the next term to divide with previous term which can be expressed as:

\displaystyle \large{r=\frac{a_{n+1}}{a_n}}

Once knowing which sequence or series is it, apply an appropriate formula for the series. For geometric series, apply the following three formulas:

\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}\\\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}

Above should be applied for series that have common ratio not equal to 1.

\displaystyle \large{S_n=a_1 \cdot n}

Above should be applied for series that have common ratio exactly equal to 1.

__________________________________________________________

Topics

Sequence & Series — Geometric Series

__________________________________________________________

Others

Let me know if you have any doubts about my answer, explanation or this question through comment!

__________________________________________________________

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</span>
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