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2 years ago

Julio spent $40 and now has no money left. He had _before his purchase

Mathematics

1 answer:

ser-zykov [4K]2 years ago

7 0

Answer:

$40

Step-by-step explanation:

If he spent $40 and has none left that means he currently has $0.

If you subtract $40 - b = $0

b = $40

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Given a triangle with: a = 150, A = 75°, and C = 30° Using the law of sines gives: c = 0

koban [17]

Answer:

$c = 77.6$

Step-by-step explanation:

You may have entered the measure of a side as the measure of an angle.

$\dfrac{\sin A}{a} = \dfrac{\sin C}{c}$

$\dfrac{\sin 75^\circ}{150} = \dfrac{\sin 30^\circ}{c}$

$c\sin 75^\circ = 150 \sin 30^\circ$

$c = \dfrac{150 \sin 30^\circ}{\sin 75^\circ}$

$c = 77.6$

You are correct. Good job!

5 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

What does 0.03703703703 equal in fractions

Zinaida [17]

Step 1: Write down the decimal divided by 1, like this: decimal 1. Step 2: Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.) Step 3: Simplify (or reduce) the fraction.

6 0

3 years ago

Read 2 more answers

If a plane traveled 610 miles in 1 hour, how fast is it going?

GalinKa [24]

Answer:

very fast

Step-by-step explanation:

10.2 miles per minute :)

7 0

2 years ago

Read 2 more answers

Point A is at -4 and point B is at 6. Which describes one way to find the point that

blondinia [14]

Answer:

D. For a ratio of 3:2, divide AB into 5 equal parts. Each equal part is 2 units, so the point that divides AB into a 3:2 ratio is 2.

Step-by-step explanation:

The complete question is:

Point A is at –4 and point B is at 6. Which describes one way to find the point that divides AB into a 3:2 ratio?

In a 3:2 ratio there are 3 + 2 = 5 equal parts.

The distance between points A and B is: 6 - (-4) = 6 + 4 = 10 units.

Each equal part is 10/5 = 2 units.

Point 2 is at 2 - (-4) = 6 units from -4

Point 2 is at 6 - 2 = 4 units from 6

Ratio 6:4 is equivalent to ratio 3:2

6 0

3 years ago