0.15, 0.17, 0.19.<br><br> what is going up by ?

Find a 3 digit number with all these properties: all 3 digits are different, the 1st digit is the square of the second digit in

DanielleElmas [232]

Answer:

425 or 937

Step-by-step explanation:

First, I listed out all the possible numbers for the first digits, namely, the squares under 10.

1*1=1

2*2=4

3*3=9

Since all the digits have to be different, the first digit cannot be 1 because 1 squared is 1. So that leaves us 4 and 9 to work with, which I tried out one at a time.

Starting with 4:

2 squared is 4.

42

2 times 2 plus 1 equals 5.

425

Starting with 9:

3 squared is 9.

93

2 times 3 plus 1 equals 7.

937

So here we have two numbers that both work and meet the requirements (unless I understood the problem wrong at the part where it says "...the second digit in the 3rd digit on one more than twice the second digit")!

I hope this helped! :D

8 0

3 years ago

Show work please. thank you

Eduardwww [97]

Problem 17, part A

i = simple interest = 900 dolars
p = principle = 1500 dollars
r = interest rate = 0.03 (decimal form of 3%)
t = time in years = unknown (leave it as t for now)

Plug in those values mentioned above and solve for t.

Simple interest formula
i = p*r*t
900 = 1500*0.03*t
900 = 45*t
900/45 = 45*t/45
20 = t
t = 20

Answer: 20 years

=============================================

Problem 17, part B

p = 1500
r = 0.03
t = 5 years

A = p+p*r*t
A = 1500+1500*0.03*5
A = 1500+225
A = 1725

Answer: 1725 dollars

=============================================

Problem 17, part C

Factor out the GCF p, then divide both sides by 1+rt

A = p+p*r*t
A = p*1+p*r*t
A = p*(1+rt)
A/(1+rt) = p*(1+rt)/(1+rt)
A/(1+rt) = p
p = A/(1+rt)

Answer: p = A/(1+rt)

The right side is the fraction of A all over the quantity (1+rt) which may be more readable if written like so
$p = \frac{A}{1+rt}$
If you're going to write it in the way p = A/(1+rt) then don't forget to use parenthesis.

4 0

3 years ago

A spinner has 7 equal sections which are numbered from 1-7. Which of the following are complementary events? (check all that app

Yuliya22 [10]

Spinning an odd number.
spinning 4 or not 4

8 0

3 years ago

Read 2 more answers

Plz help me I've been dealing with this for 4 days.

Sloan [31]

Step-by-step explanation:

(qvp)5=4-5=54643434346

3 0

3 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago

0.15, 0.17, 0.19. what is going up by ?