the sum of three numbers in an arithmetic progression is 15 and their product is 105. Find the numbers An exam has a weightage of 30%. If a student has a. We can put what Gauss discovered into an easy-to-use formula, which is: (n / 2) (first number + last number) = sum, where n is the number of integers. Let's use the example of adding the numbers 1.

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• Andy C.answered • 06/26/18 Tutor 4.9(27) Math/Physics Tutor See tutors like this See tutors like this X+Y = 21 ---> Y = 21-x X-y = 5 x - (21-x) = 5 2x - 21 = 5 2x = 26 x = 13 13 and 8 Upvote •8Downvote Comments •32 More Report Etfunyt L. thx Report 09/09/20 Andrea J.
• The sum of two variables x and y is 30. So expressing it in the form of an equation, x + y = 30 equation (i) The other fact that we know is that their difference is 10. Therefore, x - y = 10 equation (ii) When these two equations are solved, the required numbers are obtained. It is known that both the equations are true simultaneously so they ...
• Adding the above three numbers , we get. ⇒ a − d + a + a + d ⇒ 3 a. Since we know that the above sum equals to 21 . Taking the above sum equals to ford house events similar triangles desmos answers old dominion freight ...
• Solution: Given Product of two numbers = 44. Sum of two numbers = 24. Let's consider the numbers we need to find as x and y. To solve the problem x.y = 44. x+y = 24. y=24-x. Replace the value of y in the equation x.y=44. It's not necessarily y if you want you can interchange with the value of x too as x and y are interchangeable.
• Solution: Given sum of two numbers is 21 and difference of them is 5 Let the two numbers be x,y x + y = 21 --- [a] x - y = 5 --- [b] By elimination method, we get ⇒ 2x = 26 ⇒ x = 13 Put x = 13 in eq [a], we get 13 + y = 21 ⇒ y = 21 - 13 ⇒ y = 8 Therefore, the two numbers are 13, 8.