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[Commex]

mewa hadanna dannavada bn ubala?

 

[Dr.Amdan]

Bump

 

[Commex]

Bump

thx bn

 

[PHBhagya]

@dilann

 

[helplesser]

Step 1.Image to text
Step 2. Chat gpt

 

[Commex]

Step 1.Image to text
Step 2. Chat gpt

hariyata enne nene machn oya warahan dala tiyena ewa ehema image to text karata

 

[kinkon]

 

[Solo Rider]

 

[jzx]

@NRTG

 

[NRTG]

 

[Any.key]

bump

 

[helayalk]

.

 

[Geshi95]

බම්ප්

 

[NRTG]

1.
The relation R is defined on Z by aRb if a+b is even. To show that R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive.

Reflexive: For all a in Z, a+a=2a is even. Therefore, aRa for all a in Z.
Symmetric: For all a,b in Z, if aRb then bRa. If a+b is even then b+a is also even. Therefore, if aRb then bRa.
Transitive: For all a,b,c in Z, if aRb and bRc then aRc. If a+b is even and b+c is even then (a+b)+(b+c) = (a+c)+2b is even. Therefore, if aRb and bRc then aRc.

2. (a)
The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $16000, n = 1 (compounded annually), and t = 10. We need to find r.
Substituting these values into the formula and solving for r, we get:
16000 = 6000(1 + r/1)^(1*10) 16000/6000 = (1 + r)^10 2.6667 = (1 + r)^10 log(2.6667) = log(1 + r)^10 log(2.6667) = 10log(1 + r) log(2.6667)/10 = log(1 + r) 0.0083 = log(1 + r) 10^0.0083 - 1 = r

Therefore the annual interest rate is approx 8.3 %

2.(b)
We can use the formula for compound interest to solve this problem. The formula is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $40000, n = 1 (compounded annually), and r = 8.3% (from previous question). We need to find t.
Substituting these values into the formula and solving for t, we get:
40000 = 6000(1 + 0.083/1)^(1t) 40000/6000 = (1 + 0.083)^t 6.6667 = (1.083)^t log(6.6667) = log(1.083)^t log(6.6667) = tlog(1.083) log(6.6667)/log(1.083) = t

Therefore, it will take approximately 21 years for the investment to be worth at least $40000 .

 

[jzx]

හයියෝ. Eyebrows උස්ස උස්ස ඉන්න නෙමේ මනුස්සයො, ගාන පුලුවන්නම් කියල දෙන්න කියල mention කලේ

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1.
The relation R is defined on Z by aRb if a+b is even. To show that R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive.

Reflexive: For all a in Z, a+a=2a is even. Therefore, aRa for all a in Z.
Symmetric: For all a,b in Z, if aRb then bRa. If a+b is even then b+a is also even. Therefore, if aRb then bRa.
Transitive: For all a,b,c in Z, if aRb and bRc then aRc. If a+b is even and b+c is even then (a+b)+(b+c) = (a+c)+2b is even. Therefore, if aRb and bRc then aRc.

2. (a)
The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $16000, n = 1 (compounded annually), and t = 10. We need to find r.
Substituting these values into the formula and solving for r, we get:
16000 = 6000(1 + r/1)^(1*10) 16000/6000 = (1 + r)^10 2.6667 = (1 + r)^10 log(2.6667) = log(1 + r)^10 log(2.6667) = 10log(1 + r) log(2.6667)/10 = log(1 + r) 0.0083 = log(1 + r) 10^0.0083 - 1 = r

Therefore the annual interest rate is approx 8.3 %

2.(b)
We can use the formula for compound interest to solve this problem. The formula is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $40000, n = 1 (compounded annually), and r = 8.3% (from previous question). We need to find t.
Substituting these values into the formula and solving for t, we get:
40000 = 6000(1 + 0.083/1)^(1t) 40000/6000 = (1 + 0.083)^t 6.6667 = (1.083)^t log(6.6667) = log(1.083)^t log(6.6667) = tlog(1.083) log(6.6667)/log(1.083) = t

Therefore, it will take approximately 21 years for the investment to be worth at least $40000 .

මේක දැක්කෙ දැං
------ Post added on Jun 6, 2023 at 10:02 PM

 

[NRTG]

හයියෝ. Eyebrows උස්ස උස්ස ඉන්න නෙමේ මනුස්සයො, ගාන පුලුවන්නම් කියල දෙන්න කියල mention කලේ

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මේක දැක්කෙ දැං
------ Post added on Jun 6, 2023 at 10:02 PM

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3. (a) - 12
(b) - 4/3

4. (a) - -71

(b) - -71 88 3
0 1 426
355 426 0

(c) - 71/71 - 88/71 - 3/71
0 -1/71 7/71
-355/71 -426/71 0
------ Post added on Jun 6, 2023 at 11:02 PM

 

[Solo Rider]

1.
The relation R is defined on Z by aRb if a+b is even. To show that R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive.

Reflexive: For all a in Z, a+a=2a is even. Therefore, aRa for all a in Z.
Symmetric: For all a,b in Z, if aRb then bRa. If a+b is even then b+a is also even. Therefore, if aRb then bRa.
Transitive: For all a,b,c in Z, if aRb and bRc then aRc. If a+b is even and b+c is even then (a+b)+(b+c) = (a+c)+2b is even. Therefore, if aRb and bRc then aRc.

2. (a)
The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $16000, n = 1 (compounded annually), and t = 10. We need to find r.
Substituting these values into the formula and solving for r, we get:
16000 = 6000(1 + r/1)^(1*10) 16000/6000 = (1 + r)^10 2.6667 = (1 + r)^10 log(2.6667) = log(1 + r)^10 log(2.6667) = 10log(1 + r) log(2.6667)/10 = log(1 + r) 0.0083 = log(1 + r) 10^0.0083 - 1 = r

Therefore the annual interest rate is approx 8.3 %

2.(b)
We can use the formula for compound interest to solve this problem. The formula is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $40000, n = 1 (compounded annually), and r = 8.3% (from previous question). We need to find t.
Substituting these values into the formula and solving for t, we get:
40000 = 6000(1 + 0.083/1)^(1t) 40000/6000 = (1 + 0.083)^t 6.6667 = (1.083)^t log(6.6667) = log(1.083)^t log(6.6667) = tlog(1.083) log(6.6667)/log(1.083) = t

Therefore, it will take approximately 21 years for the investment to be worth at least $40000 .

මටත් පුලුවන්
එත් මෙච්චර ලේසි එව්වට මම කාලෙ වෙන් කරන්නෙ නෑ

 

[jzx]

මටත් පුලුවන්
එත් මෙච්චර ලේසි එව්වට මම කාලෙ වෙන් කරන්නෙ නෑ

 

[KissMass]

යාකො ඔවා ප්‍රශ්න ද? අපේ පුතාගෙන් අහපන්....

 

[Commex]

1.
The relation R is defined on Z by aRb if a+b is even. To show that R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive.

Reflexive: For all a in Z, a+a=2a is even. Therefore, aRa for all a in Z.
Symmetric: For all a,b in Z, if aRb then bRa. If a+b is even then b+a is also even. Therefore, if aRb then bRa.
Transitive: For all a,b,c in Z, if aRb and bRc then aRc. If a+b is even and b+c is even then (a+b)+(b+c) = (a+c)+2b is even. Therefore, if aRb and bRc then aRc.

2. (a)
The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $16000, n = 1 (compounded annually), and t = 10. We need to find r.
Substituting these values into the formula and solving for r, we get:
16000 = 6000(1 + r/1)^(1*10) 16000/6000 = (1 + r)^10 2.6667 = (1 + r)^10 log(2.6667) = log(1 + r)^10 log(2.6667) = 10log(1 + r) log(2.6667)/10 = log(1 + r) 0.0083 = log(1 + r) 10^0.0083 - 1 = r

Therefore the annual interest rate is approx 8.3 %

2.(b)
We can use the formula for compound interest to solve this problem. The formula is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, P is the principal amount invested, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
In this case, we have P = $6000, A = $40000, n = 1 (compounded annually), and r = 8.3% (from previous question). We need to find t.
Substituting these values into the formula and solving for t, we get:
40000 = 6000(1 + 0.083/1)^(1t) 40000/6000 = (1 + 0.083)^t 6.6667 = (1.083)^t log(6.6667) = log(1.083)^t log(6.6667) = tlog(1.083) log(6.6667)/log(1.083) = t

Therefore, it will take approximately 21 years for the investment to be worth at least $40000 .

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3. (a) - 12
(b) - 4/3

4. (a) - -71

(b) - -71 88 3
0 1 426
355 426 0

(c) - 71/71 - 88/71 - 3/71
0 -1/71 7/71
-355/71 -426/71 0
------ Post added on Jun 6, 2023 at 11:02 PM

Ado me udauwata nm ubata wadi wadiyen labenna one
thanx bokka..

යාකො ඔවා ප්‍රශ්න ද? අපේ පුතාගෙන් අහපන්....

mama melo deyak danne ne bn