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Topic: **How to write a squared number****Question:**
Three times the square of a number decreased by eight times the number is three. What is the number?
How do you find the number?

June 16, 2019 / By Kelan

Write the information in the form of an equation: Let the number be x Three times the square of a number = 3x^2 8 times the number = 8x Three times the square of a number decreased by eight times the number is three => 3x^2-8x=3 Rearrange this into the form: 3x^2-8x-3=0 Now you can factorise this and solve it to get the two possibilities for x

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Did you like the answer? We found more questions related to the topic: **How to write a squared number**

this question is whack, but all you have to work with is multiples of 3 so if you meant TENTHS: 93.13 93.26 93.39 31.13 31.26 31.39 62.13 62.26 62.39 now if you meant tens after all- then 93.39 31.13 62.26 see the pattern? and finally, with your hundredths digit being 3 more than tens (which all other answerers seemed to have failed to double check), I think there is only one answer! 93.26

First, create equations: x = the tens digit, y = those digit. you recognize right here: x + y = 7 and x = 3y - a million Plug the 2d equation into the 1st: (3y - a million) + y = 7 3y - a million + y = 7 4y - a million = 7 4y = 8 y = 2 considering the fact that x + y = 7, you recognize that x = 7 - y = 7 - 2 = 5. So, the variety is fifty two

Let the number be x. Three times the square of a number decreased by eight times the number is three. Putting that in terms of x: 3x² - 8x = 3 or 3x² - 8x - 3 = 0 3x² - 9x + x - 3 = 0 3x.(x - 3) + 1.(x - 3) = 0 (3x + 1).(x - 3) = 0 x = 3 or -1/3

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locate the huge type such that? a. its sq. is 12 greater beneficial than the huge type.is 4 the place 4^2 = 4 + 12 b. its sq. decreased with the help of thrice the huge type is eighteen is 6 the place 6^2 -- 3(6) = 18. c. the made of the huge type and four below the huge type 32 is 8 the place 8*(8 -- 4) = 32. d. the sq. of one greater beneficial than the huge type is 4 greater beneficial than 4 instances the huge type is 4 the place (4 + one million)^2 = 4(4) + 4. e. its sq. greater beneficial with the help of 8 instances the huge type is -15 is --3 the place (--3)^2 + 8(--3) = -- 15..

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If you are lucky, the least number of times would be ONE (if by chance you select the non-gold coin to compare with a gold coin, you will know immediately which one is lighter). I think your question wants to know the MAXIMUM number of times you would need to use the scale; in that case it would be FOUR (if by chance the first four pairs of coins you pick up and measure happen to be the 8 gold coins, which would obviously balance out) - which means the leftover coin would HAVE to be the non-gold one, no weighing necessary :)

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