Age of Yusuf is 24 years more than half the age of Ajay. 5 years before sum of their ages was 41. Find their present ages.​

Yusuf = 37.5

Ajay = 13.5

Step-by-step explanation:

Lets say.. Yusuf = y and ajay = a

y = 24+a (since yusuf is 24 years older than ajay)

5 years before, the sum of their ages was 41 so subtract 5 from y and a and add the two.

(y-5)+(a-5)=41

We already have defined y in terms of a (y =24+a) so we substitute that in for y.

(24+a-5)+(a-5)= 41

24-5 = 19

19+a+a-5=41

a+a = 2a

2a+19-5 = 41

19-5 = 14

2a+14 = 41

subtract 14 from both sides

2a=27

a=13.5

substitute that in y=24+a

y=24+13.5

y=37.5

lets check these answers.

(37.5-5)+(13.5-5)= 41

37.5-5 = 32.5 and 13.5-5= 8.5

32.5+8.5=41??

Correct!

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Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}