find two consecutive even integers such that two fifth of the smaller exceeds two eleventh of the larger by 4
pl. answer it fast tomorrow is my test
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Answer:
20 and 22
Step-by-step explanation:
Let the smaller number be n.
Being consecutive even numbers, the other one is n+2.
The statement says
2/5 of smaller = (2/11 of larger) + 4
=> 2n / 5 = 2 ( n+2 ) / 11 + 4
=> 22n = 10 ( n + 2 ) + 220 (multiplied by 55 to get rid of fractions)
=> 12n = 240
=> n = 20
That's the smaller. The other one is n + 2 = 22.
Anonymous:
Hope that helps. Plzzz mark this the Brainliest!
hey one more question if u answer it i will mark u brainliest for sure :- two perpendicular sides of a right angle triangle are in ratio 3 ratio 4 and the perimeter is 96
find 3 sides of triangle
Since the two sides are in the ratio 3:4, one side is 3x and the other is 4x, for some x. By Pythagoras, the hypotenuse is then 5x. The perimeter is then 3x + 4x + 5x = 12x = 96, so x = 8. The three sides are then 3x = 24, 4x = 32 and 5x = 40
enjoy!
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