All categories

3 years ago

B2+15b+56 please show work!!!!!!

Mathematics

2 answers:

noname [10]3 years ago

6 0

The answer for this one is c....kinda hard to show work for this one....just take the answer and do the FOIL method.

kenny6666 [7]3 years ago

3 0

What does B equal? This is a key factor to answering the question.

You might be interested in

Help me ASAP!!!!!!!!<br><br> See picture attached

inysia [295]

Answer:

47 very easy

Step-by-step explanation:

lalalalalalalalalaalalalallala

7 0

3 years ago

A small companys net income for the first six months of the year was $76,500 and for the last six months it was $100,000. What i

Alex_Xolod [135]

Divide until you get one moth alone then multiply then simplify if you don't know how to simplify please pm me.

6 0

4 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Y varies inversely to x. If y = 7 when x = -4, what is y when x =5 ?

djverab [1.8K]

Answer:

Step-by-step explanation:

y*1/x

y=k/x where k is the constant of variation

when y=7, x=-4

7=k/-4

k=-28

when x=5

y=-28/5

note: * is used as symbol of variation

7 0

4 years ago

Ok so these math questions are simple. Would you be able to do them?

alexandr402 [8]

Answer:

3) it would be the first one 3*(6/5) and it would be the second one 3/(5*6)

Step-by-step explanation:

when you put it in the calculator they come out to be 3.6

I have no idea for the question above it

3 0

3 years ago