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3 years ago

7

If a bank offers an annual interest rate for the money in your checking account but pays interest monthly that is the annual int

erest divided by what

1 answer:

Art [367]3 years ago

8 0

There are 12 months in a year.... so divide apr by 12....

You might be interested in

Is 1/3, 2/3, 4/3, 8/3, 16/3 arithmetic sequence

11111nata11111 [884]

The given sequence is not arithmetic sequence

<em><u>Solution:</u></em>

Given sequence is:

$\frac{1}{3} , \frac{2}{3} , \frac{4}{3} , \frac{8}{3} , \frac{16}{3}$

We have to find if the above sequence is arithmetic sequence or not

An arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant

<em><u>Here in the given sequence</u></em>

$\text{ first term } = a_1 = \frac{1}{3}\\\\\text{ second term } = a_2 = \frac{2}{3}\\\\\text{ third term } = a_3 = \frac{4}{3}\\\\\text{ fourth term } = a_4 = \frac{8}{3}\\\\\text{ fifth term } = a_5 = \frac{16}{3}$

<em><u>Let us find the difference between terms</u></em>

$a_2 - a_1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$

$a_3 - a_2 = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}$

$a_4 - a_3 = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}$

$a_5 - a_4 = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$

Thus the difference between terms is not constant

So the given sequence is not arithmetic sequence

5 0

3 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

Write the decomposition that helps you, and then round to the given place value. Draw number lines to explain your thinking. Cir

egoroff_w [7]

Further explanation

There are general rules used in rounding numbers

1. Determine the rounded digit

2. Pay attention to the digits afterward (right). If the digit shows the numbers 0, 1, 2, 3 or 4, then the digit to be rounded is fixed, and the digit on the right is 0 all

3. If the digit on the right shows the numbers 5, 6, 7, 8 or 9, then the digit to be rounded is added to one, and the digit on the right side is made all 0

In rounding the nearest hundreds number, then we see the digit in the number tens (right of the digit hundreds), if it is worth 0, 1, 2, 3 or 4, then it remains, but if it shows numbers 5, 6, 7, 8 or 9, then the digit to be rounded is added one, and the right-hand side is 0

Calculated from the decimal point, the place on the right represents tents, hundredths, thousandths, etc.

a. 32.697

Decomposition

ten : 3

one : 2

tenth : 6

hundredth : 9

thousandths : 7

For the tenth number is 6, the value on the right is 9 then rounded become 32.7

For the hundredth number is 9, the value on the right is 7 then rounded become 32.70

for the one number is 2 , the value on the right is 6 then rounded become 33

b. 141.999

Decomposition

hundred : 1

ten : 4

one : 1

tenth : 9

hundredth : 9

thousandths : 9

a. For the tenth number is 9, the value on the right is 9 then rounded become 142.0

For the hundredth number is 9, the value on the right is 9 then rounded become 142.00

For the ten number is 4, the value on the right is 1 then rounded become

14

For the ten number is 1, the value on the right is 4 then rounded become

142

<h3>Learn more </h3>

five numbers that are round up and round down to 600

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all the numbers that round to 50

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round 4,398,202 to the nearest 100

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Keywords: round up, round down, the rounded digit

7 0

3 years ago

What is the value of f(3)?

Anni [7]

Answer:

-12

Step-by-step explanation:

Note that when x < 6, you will use the expression -4x

f(x) = -4x

Plug in 3 for x

f(3) = -4(3)

Simplify. Multiply

f(3) = -4(3)

f(3) = -12

-12 is your answer

5 0

3 years ago

In 4th grade :) help plss

Soloha48 [4]

Answer:

1. 570

2. 5,700

3. 57,000

Step-by-step explanation:

Hope it helped

4 0

3 years ago

Read 2 more answers