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2 years ago

7

How long does it take for the account to be worth about $750?

1 answer:

pochemuha2 years ago

7 0

Answer:

if look at the graph and find the point on the graph where the y-value is $750 then you will find your answer. I can give u any more than that cuz i cant really read the graph. look where the y-value is $750 and write the x-value that matches it.

Step-by-step explanation:

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Ill give brainliest I really need help but all of the answers are on chegg so does anyone have a chegg account that I can use?

Archy [21]

Answer:

No, but you could just ask the question here....

Step-by-step explanation:

Sorry, i don't.

What's da Q?

4 0

3 years ago

Martha has saved 2,700 to share with her family . She divides one -third of her saving equally between her two children , lola a

VikaD [51]

Answer:

150

Step-by-step explanation:

one third of 2700 = 900

900/2 = 450

450/3 = 150

6 0

3 years ago

A day can be evenly divided into 86,400 periods of 1 second; 43,200 periods of each 2 seconds; or in many other ways. In total,

Leni [432]

Answer:

96 ways

Step-by-step explanation:

Given

$Day = 86400\ seconds$

Required

Ways to divide it into period of seconds

What this question implies is to determine the total number of factors of 86400

To start with, we determine the prime factorization of 86400

To do this, we continually divide 86400 by 2; when it can not be further divided, we divide by 3, then 7, then 11...

$86400/2 = 43200$

$43200/2=21600$

$21600/2=10800$

$10800/2=5400$

$5400/2= 2700$

$2700/2 = 1350$

$1350/2=675$

$675/3=225$

$225/3=75$

$75/3=25$

$25/5=5$

$5/5 = 1$

This implies that:

$86400 = 2^7 * 3^3 * 5^2$

The number of factors d is the solved by:

$d = (a+1)*(b+1) *(c+1)$

Where

$n = 2^a * 3^b * 5^c$

By comparison:

$a = 7$

$b = 3$

$c=2$

So:

$d = (7+1)*(3+1) *(2+1)$

$d = 8*4 *3$

$d = 96$

<em>Hence, there are 96 total ways</em>

7 0

3 years ago

Evaluate the expression when m=4.<br> m^2 + 6m + 8

Vikki [24]

Answer:

According to me,

<h2>48 is the answers</h2>

3 0

3 years ago

Find a such that the solution of y'' + y' − 2y = 0, y(0) = a, y' (0) = 1 tends to zero as t → +∞.

Kay [80]

Answer:

Step-by-step explanation:

$y'' + y' − 2y = 0, y(0) = a, y' (0) = 1$

Auxialary equation is

$m^2+m-2=0\\m=-2,1$

General solution is

$y=Ae^{-2x} +Be^x$

$y(0) = A+B =a$

$y'(0) = -2Ae^{2x} +Be^x = -2A+B = 1$

Eliminate B to get

3A =a-1

We know that y tends to 0 when x tends to infinity for any finite A

i.e. a should be a finite real number.

8 0

3 years ago