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Andreyy89
3 years ago
10

The age of a father is equal to the square of his daughter’s age. If one year ago, the father’s age was 8 times the daughter’s a

ge. What would their current ages be?
A- 25 years and 5 years

B- 81 years and 9 years

C- 49 years and 7 years

D- 36 years and 6 years

E- None of these choices


Thank you for any help. I’m having a great deal of trouble with this problem. Any help would be much appreciated.
Mathematics
1 answer:
grin007 [14]3 years ago
6 0
I did process of elimination and for each option subtracted one from the current values to see if one was 8x the other.
A - 24 & 4 (6x)
B - 80 & 8 (10x)
C - 48 & 6 (8x)
D - 35 & 5 (7x)

I found the correct answer to be c
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PLEASE HELP!!!!! How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?
saul85 [17]

Answer:

I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

first = 1000 ,  last =  9990

we have a formula,   a_{n} = a + (n-1)d

here, a_{n} is the last 4-digit number divisible by 5.

n is the number of 4-digit even numbers divisible by 5

d is the common difference between the numbers, which is 10 in this case

a is the first 4-digit number divisible by 5

9990 = 1000 + (n-1)*10

899 = n-1

n = 900

Hence, there are 900 4-digit even numbers divisible by 5

7 0
2 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

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2 years ago
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Answer:

Step-by-step explanation:

x = 3y.......(1)

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Substituting x in (1) into (2)

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