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2 years ago

A man walks 15 blocks to work every morning at a rate of 2 miles per hour. If there are

1 answer:

6 0

7mins 30 secs because if you do speed,distance,time then it is distance/speed then you get 7.5 and .5 of a minute that is 30 secs

You might be interested in

A. Find 33 and a third % of 879

Anvisha [2.4K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Check all of the ordered pairs that satisfy the equation below.

stiv31 [10]

Answer:

Hey there!

y=3/4x shows that the y-value is always 3/4 of the x-value.

Thus, if x is 4, y is 3. It always fits the 4:3 ratio.

The ordered pairs that satisfy this equation are: 28:21 and 16:12

Let me know if this helps :)

6 0

3 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

Logan has 20 action figures. He is shipping them to a friend. He can fit 3 action figures in a box. How many boxes will he need

nasty-shy [4]

He will need seven boxes

5 0

3 years ago

Solve the exponential equation: 2^3x+1 = 128

dsp73

Answer:

So x=2.32956156226

Step-by-step explanation:

First subtract 1 on both sides

2^3x=127

Now you can rewrite this as

(2^3)^x

As it still equals the same thing

Now 2^3 is 8

So the equation is now

8^x=127

So we use log formula

log8 127=x

So how many times do we have to multiply 8 by itself to get 127?

The answer is: 2.32956156226

Hope this helps!

3 0

2 years ago