# Math Help - 10 questions that I need HELP ON!!! Please i can't do them!!!

1. ## 10 questions that I need HELP ON!!! Please i can't do them!!!

questions on attachment!!!!

2. Use sigma notation to write the sum. -2.4-.4+1.6+3.6
That do not make no sense to me.

Find a formula for an for the arithmetic sequence. a1=7, a11=19
By the formula's of arithmetical sequences:
$a_{11}=a_1+10k$
Substitute know values,
$19=7+10k$
Thus,
$12=10k$
Thus,
$k=1.2$

"An arithemtica sequnce with initial term of 7 and constant diffrence of 1.2"

Find a formula for an for the arithmetic sequence. a1=11, a11=15
Same idea.

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.
This is a geometric series because the situation is,
$12+12(1/6)+12(1/6)(1/6)+12(1/6)(1/6)(1/6)+...$
Thus,
$12+12(1/6)+12(1/6)^2+...+12(1/6)^4$---> Because you only got 5 bounces (do not think the exponent is 4 a common mistake it should be 4)

Thus, factor
$12(1+(1/6)+...+(1/6)^4)$
Use sum for geometric series,
$12\cdot \left( \frac{1-(1/6)^6}{1-1/6} \right)$

6. Write the first five terms of the sequence.
a0=-2
a1=4
an=-2a n -1-4an-2
-2,4,0,-16,32

7. Find the coefficient of b6a2 in the expansion of (b+2a)8
Which term is that? The third thus,
${8 \choose 3}=56$

8. Use the binomial theorem to expand and simplify the expression.
(s-u)5
Tooooooooooooooooooooooooo long.

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.
Order no important thus,
${7 \choose 2}=21$

10. Eight cards are drawn, without replacement, from a standard deck of 52 cards. How many sets of eight cards are possible?
Same as before,
${52 \choose 8}=752538150$

11.
Not doing this one you promised only 10.

3. Hello, Lane!

Use sigma notation to write the sum: $-2.4 - 0.4+1.6+3.6$

Here are two possible sigma-representations.

It starts with $-2.4$ ... and 2 is added three times: . $\sum^3_{n=0}(-2.4 + 2n)$

It starts with $-4.4$ and 2 is added four times: . $\sum^4_{n=1}(-4.4 + 2n)$

2. Find a formula for $a_n$ for the arithemtic sequence: $a_1 = 7,\;a_{11}=19$

Recall the formula for the $n^{th}$term: . $a_n\:=\:a_1 + d(n-1)$
. . where $a_1$ is the first term and $d$ is the common difference.

We are given $a_1=7$, so the formula becomes: . $a_n\:=\:7 + d(n-1)$

We are given $a_{11}=19$ ... the $11^{th}$ term is $19.$
So we have: . $19\:=\:7 + d(11-1)\quad\Rightarrow\quad d = 1.2$

Therefore: . $a_n\:=\:7 + 1.2(n-1)$

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 $12$ feet. .(Contact #1)

It bounces up . $\frac{1}{6}(12)$ feet . . . and falls . $\frac{1}{6}(12)$ feet. .(Contact #2)

It bounces up $\left(\frac{1}{6}\right)^2(12)$ feet . . . and falls $\left(\frac{1}{6}\right)^2(12)$ feet. .(Contact #3)

. . You get the idea . . .

By the 5th contact, the total distance is:

. . $D \:=\:12 + 2\cdot12\left(\frac{1}{6}\right) + 2\cdot12\left(\frac{1}{6}\right)^2 +$ $2\cdot12\left(\frac{1}{6}\right)^3 + 2\cdot12\left(\frac{1}{6}\right)^4$

. . $D\:=\:12 + 4\bigg[1 + \frac{1}{6} + \left(\frac{1}{6}\right)^2 +$ $\left(\frac{1}{6}\right)^3\bigg]$ $\:=\:12 + 4\left(\frac{259}{216}\right)\:=\:\boxed{\frac{907 }{54}\text{ feet}}$

4. Hello again, Lane!

7) Find the coefficient of $b^6a^2$ in the expansion of $(b+2a)^8$

The expansion begins: . $\binom{8}{0}b^8 + \binom{8}{1}b^7(2a) + \underbrace{\binom{8}{2}b^6(2a)^2} + \hdots$

The third term is: . $\frac{8!}{2!\,6!}\,b^6(4a^2) \;=\;\boxed{112}\,b^6a^2$

8. Use the binomial theorem to expand and simplify: $(s - u)^5$

If you don't know the Binomial Theorem, we can't be much help.

$(s - u)^5\;=\;\binom{5}{0}s^5 + \binom{5}{1}s^4(-u) + \binom{5}{2}s^3(-u)^2 + \binom{5}{3}s^2(-u)^3$ $+ \binom{5}{4}s(-u)^4 + \binom{5}{5}(-u)^5$

. . I expanded it . . . you can simplify it.