questions on attachment!!!!
That do not make no sense to me.Use sigma notation to write the sum. -2.4-.4+1.6+3.6
By the formula's of arithmetical sequences:Find a formula for an for the arithmetic sequence. a1=7, a11=19
$\displaystyle a_{11}=a_1+10k$
Substitute know values,
$\displaystyle 19=7+10k$
Thus,
$\displaystyle 12=10k$
Thus,
$\displaystyle k=1.2$
"An arithemtica sequnce with initial term of 7 and constant diffrence of 1.2"
Same idea.Find a formula for an for the arithmetic sequence. a1=11, a11=15
This is a geometric series because the situation is,A rubber ball on a hard surface takes a sequence of vertical bounces. Each bounce is 1/6 as high as the proceeding one. If this ball is dropped from a height of 12 feet, find the total distance it has traveled when it hits the surface the fifth time.
$\displaystyle 12+12(1/6)+12(1/6)(1/6)+12(1/6)(1/6)(1/6)+...$
Thus,
$\displaystyle 12+12(1/6)+12(1/6)^2+...+12(1/6)^4$---> Because you only got 5 bounces (do not think the exponent is 4 a common mistake it should be 4)
Thus, factor
$\displaystyle 12(1+(1/6)+...+(1/6)^4)$
Use sum for geometric series,
$\displaystyle 12\cdot \left( \frac{1-(1/6)^6}{1-1/6} \right)$
-2,4,0,-16,326. Write the first five terms of the sequence.
a0=-2
a1=4
an=-2a n -1-4an-2
Which term is that? The third thus,7. Find the coefficient of b6a2 in the expansion of (b+2a)8
$\displaystyle {8 \choose 3}=56$
Tooooooooooooooooooooooooo long.8. Use the binomial theorem to expand and simplify the expression.
(s-u)5
Order no important thus,
9. A college has seven instructors qualified to teach a special computer lab course which requires two instructors to be present. How many different pairs of teachers could teach the class.
$\displaystyle {7 \choose 2}=21$
Same as before,10. Eight cards are drawn, without replacement, from a standard deck of 52 cards. How many sets of eight cards are possible?
$\displaystyle {52 \choose 8}=752538150$
Not doing this one you promised only 10.11.
Hello, Lane!
Use sigma notation to write the sum: $\displaystyle -2.4 - 0.4+1.6+3.6$
Here are two possible sigma-representations.
It starts with $\displaystyle -2.4$ ... and 2 is added three times: .$\displaystyle \sum^3_{n=0}(-2.4 + 2n)$
It starts with $\displaystyle -4.4$ and 2 is added four times: .$\displaystyle \sum^4_{n=1}(-4.4 + 2n)$
2. Find a formula for $\displaystyle a_n$ for the arithemtic sequence: $\displaystyle a_1 = 7,\;a_{11}=19$
Recall the formula for the $\displaystyle n^{th}$term: .$\displaystyle a_n\:=\:a_1 + d(n-1)$
. . where $\displaystyle a_1$ is the first term and $\displaystyle d$ is the common difference.
We are given $\displaystyle a_1=7$, so the formula becomes: .$\displaystyle a_n\:=\:7 + d(n-1)$
We are given $\displaystyle a_{11}=19$ ... the $\displaystyle 11^{th}$ term is $\displaystyle 19.$
So we have: .$\displaystyle 19\:=\:7 + d(11-1)\quad\Rightarrow\quad d = 1.2$
Therefore: .$\displaystyle a_n\:=\:7 + 1.2(n-1)$
5. A rubber ball on a hard surface takes a sequence of vertical bounces.
Each bounce is 1/6 as high as the preceeding one.
If the ball is dropped from a height of 12 feet,
find the total distance it has traveled when it hits the surface for the fifth time.
Warning: "The Bouncing Ball" is a classic trick question.
I'll baby-step through the explanation . . .
The ball falls $\displaystyle 12$ feet. .(Contact #1)
It bounces up .$\displaystyle \frac{1}{6}(12)$ feet . . . and falls .$\displaystyle \frac{1}{6}(12)$ feet. .(Contact #2)
It bounces up $\displaystyle \left(\frac{1}{6}\right)^2(12)$ feet . . . and falls $\displaystyle \left(\frac{1}{6}\right)^2(12)$ feet. .(Contact #3)
. . You get the idea . . .
By the 5th contact, the total distance is:
. . $\displaystyle D \:=\:12 + 2\cdot12\left(\frac{1}{6}\right) + 2\cdot12\left(\frac{1}{6}\right)^2 +$$\displaystyle 2\cdot12\left(\frac{1}{6}\right)^3 + 2\cdot12\left(\frac{1}{6}\right)^4$
. . $\displaystyle D\:=\:12 + 4\bigg[1 + \frac{1}{6} + \left(\frac{1}{6}\right)^2 +$$\displaystyle \left(\frac{1}{6}\right)^3\bigg]$ $\displaystyle \:=\:12 + 4\left(\frac{259}{216}\right)\:=\:\boxed{\frac{907 }{54}\text{ feet}}$
Hello again, Lane!
7) Find the coefficient of $\displaystyle b^6a^2$ in the expansion of $\displaystyle (b+2a)^8$
The expansion begins: .$\displaystyle \binom{8}{0}b^8 + \binom{8}{1}b^7(2a) + \underbrace{\binom{8}{2}b^6(2a)^2} + \hdots$
The third term is: .$\displaystyle \frac{8!}{2!\,6!}\,b^6(4a^2) \;=\;\boxed{112}\,b^6a^2$
8. Use the binomial theorem to expand and simplify: $\displaystyle (s - u)^5$
If you don't know the Binomial Theorem, we can't be much help.
$\displaystyle (s - u)^5\;=\;\binom{5}{0}s^5 + \binom{5}{1}s^4(-u) + \binom{5}{2}s^3(-u)^2 + \binom{5}{3}s^2(-u)^3$$\displaystyle + \binom{5}{4}s(-u)^4 + \binom{5}{5}(-u)^5$
. . I expanded it . . . you can simplify it.