Find three consecutive odd integers whose sum is (-75)?

Find three consecutive odd integers whose sum is (-75)?
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Those three odd integers are 43, 45, and 47.
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Description : Find three consecutive even integers whose sum is -132?

Last Answer : 42, 44, 46.

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Description : What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?

Last Answer : Let the middle one be n, then:(n - 2)(n + 2) = 3n + 6→ n² - 4 = 3n + 6→ n² - 3n - 10 = 0→ (n + 2)(n - 5) = 0→ n = -2 or 5-2 can be discounted as n is odd→ n = 5→ largest of the three odd numbers is 5 + 2 = 7.

How do i solve this Find three consecutive odd integers so that three times the second is 9 more than twice the third. What is the smallest number?

Description : How do i solve this Find three consecutive odd integers so that three times the second is 9 more than twice the third. What is the smallest number?

Last Answer : Yeah, I hate these too. Identify the unknowns. Three consecutive odd integers will be x, x + 2 and x + 4 Set up the equation with the information you know. Three times the second is 3 times (x + 2) ... Since 39 is 9 more than 30, we got it right. The answer asked for (the smallest integer) is 11.

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Description : What are three consecutive integers the sum is 186?

Last Answer : 45, 46, 47, 48 = 186

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Description : What are three consecutive integers the sum is 186?

Last Answer : 45, 46, 47, 48 = 186

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Last Answer : The sum of three consecutive integers is 36. (x - 1) + x + (x + 1) = 36 13 is the largest of the three integers.

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Last Answer : 182/4 = 45.5so of the four numbers 45 will be the second44 + 45 + 46 + 47 = 89 + 93 = 182

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Description : What The sum of 3 consecutive integers that equals 252?

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In a pocket of A, the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Rs.1, 50p, 25p coins are reversed, find the new total value of coins in the pocket of A? A) Rs 68 B) Rs 43 C) Rs 75 D) Rs 82 E) NONE

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Last Answer : Answer: D  Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7. Let the number of ... =100 7k+50 5k+25 3k=1025k (In above we find the value of k). ⇒ 8200p = Rs 82.

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Description : What two integers whose sum is negative 9 and whose product is 20?

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Description : Two integers whose sum is 0 and product is -64?

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Last Answer : 6 x 8 = 4810 x 12 = 120120 - 48 = 72

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Description : When the smaller of the two consecutive integers is added to five times the larger the result is 47?

Last Answer : x + 5x + 5 = 476x = 42x = 7x + 1 = 8

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Description : When the smaller of the two consecutive integers is added to five times the larger the result is 47?

Last Answer : x + 5x + 5 = 476x = 42x = 7x + 1 = 8

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Last Answer : answer:The video lost me about 40 seconds in then he made this leap from 1+2+3+4… to 1+ x2 + 2x squared and so on. That may be obvious to a mathematician, but to me it was a giant leap of faith. And logically it doesn’t make sense either. So I am just as confused as you are.

When the celebrated German mathematician Karl Friedrich Gauss (1777-1855) was nine he was asked to add all the integers from 1 through 100. He quickly added 1 to 100, 2 to 99, and so on for 50 pairs of numbers each adding to 101. 50 X 101=5,050.Now find the sum of all the digits in integers from 1 through 1,000,000,000.That's all the digits in all the numbers, not all the numbers themselves. -Riddles

Description : When the celebrated German mathematician Karl Friedrich Gauss (1777-1855) was nine he was asked to add all the integers from 1 through 100. He quickly added 1 to 100, 2 to 99, and so on for 50 pairs ... 1,000,000,000.That's all the digits in all the numbers, not all the numbers themselves. -Riddles

Last Answer : The numbers can be grouped by pairs: 999,999,999 and 0; 999,999,998 and 1' 999,999,997 and 2; and so on.... There are half a billion pairs, and the sum of the digits in each pair is 81. The digits in the unpaired number, 1,000,000,000, add to 1. Then: (500,000,000 X 81) + 1= 40,500,000,001.

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Description : What three consecvtive odd numbers have the the sum of 45?

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The sum of the first n odd natural number is

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Last Answer : If by "smaller than ten" you mean closer to zero than 10,thensum = -9 + -7 + ... + -1 + 1 + ... + 7 + 9 = 0If you mean the numbers less than 10, then:sum = 9 + 7 + ... + 1 + -1 + -3 + ... = -∞If you mean the ... numbers less than 10, then:sum = 1 + 3 + 5 + 7 + 9 = ½ 5 (1 + 9) = ½ 5 10 = 25

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How do you write If the sum of any two numbers is odd then the prodduct of those two numbers is even in logic statement?

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Write a program to calculate sum of all the odd numbers between 1 to 20.

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Write a program to sum all the odd numbers between 1 to 20.

Description : Write a program to sum all the odd numbers between 1 to 20. 

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The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Description : The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C ... 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)

In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to the sum of the other two digits. The number will decreased by 297, if its digit are reversed. approximately what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Description : In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to ... what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Last Answer : A) Maximum mark consist of a three digit number let's consider Unit digit place is Z,ten's place digit is Y and hundred's place digit is X. According to the question, Y=x+z---------1 X+Y ... Total number of Mahi, Nitesh and Ritesh =164/100 451=739.61 Required average =739.64/3=246.54~247.

A question (sum) is given to three boys whose chances of solving it are 1/3,1/4 and 1/5 respectively. What is the probability that the question will be solved? A) 4/5 B) 3/5 C) 3/4 D) 7/5

Description : A question (sum) is given to three boys whose chances of solving it are 1/3,1/4 and 1/5 respectively. What is the probability that the question will be solved? A) 4/5 B) 3/5 C) 3/4 D) 7/5

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Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \frac{3}{4} \)b) \( \frac{5}{8} \)c) \( \frac{9}{8} \)d) \( \frac{7}{8} \)47. The sum of perimeter of all triangle is (in \( cm \) )a) 144b) 625c) 400d) 16948. The area of all the triangle is (in sq cm)a) 576b) \( 144 \sqrt{3} \)c) \( 169 \sqrt{3} \)d) \( 192 \sqrt{3} \)49. The sum of perimeter of first 6 triangle is (in \( cm \) )a) 120b) \( \frac{567}{4} \)c) \( \frac{569}{4} \)d) 14450. The side of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3

Why Estimate the sum of 12.3 and 9.75?

Description : Why Estimate the sum of 12.3 and 9.75?

Last Answer : because it is hard to do the math

The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

The ratio of marks obtained by Vinod and Basu is 6:5. If the combined average of their percentage is 68.75 and their sum of the marks is 275, find the total marks for which exam was conducted. A) 150 B) 200 C) 400 D) 450 E) None

Description : The ratio of marks obtained by Vinod and Basu is 6:5. If the combined average of their percentage is 68.75 and their sum of the marks is 275, find the total marks for which exam was conducted. A) 150 B) 200 C) 400 D) 450 E) None

Last Answer : Answer: B The sum of the marks=6x+5x=11x But the sum of the marks is given as 275=11x. We get x=25x=25 therefore, Vinod's marks is 6x=1506x=150 and Basu's marks = 5x=125. ... Therefore, if their combined average of their percentage is 137.5 then the total marks would be 137.568.75 100=200

A, B and C are 3 partners in a business. Their investments are respectively Rs 2000, Rs 4,000 and Rs 3,000. A gets 30% of total profit for managing the business. The remaining profit is divided among them in the ratio of their investments. At the end of the year, the profit of Ais Rs 1100 less than the sum of the profit of B and C. What amount of income will C get? A) Rs 2100.75 B) Rs 2887.5 C) Rs 2705.75 D) Rs 2546.25

Description : A, B and C are 3 partners in a business. Their investments are respectively Rs 2000, Rs 4,000 and Rs 3,000. A gets 30% of total profit for managing the business. The remaining profit is divided among them in the ratio ... of income will C get? A) Rs 2100.75 B) Rs 2887.5 C) Rs 2705.75 D) Rs 2546.25

Last Answer : Answer: B) The ratio of profit of A, B and C is 2000:4000:3000=2:4:3. Let the annual profit be P. Then, A will get 0.3p for managing the business. And, remaining 0.7p will be distributed in the ratio of their ... ⇒ 7/9 0.7p−2/9 0.7p−0.3p=1100 ⇒ p=12,375 So, C's share=3/9 0.7p=Rs 2887.5

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