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Why is it easier to count by 5s or 10s and 2s instead of counting by 3s or 4s?
It is only because we count in tens - and 5 10 and 2 are factors.
What is the sum of one-half of s one-fifth of x and 7?
(1/2s)+(1/5x)+7 5s+2x+7 5s=2x+7 s=2/5x+7/5 (1/2(2/5x+7/5))+(1/5x)+7 1/5x+3.5/5+1/5x+7 2/5x+7 7/10 2/5x=-7 7/10 x=-19.25 (1/2s)+(1/5x)+7 1/2s-3.85+7 1/2s+3.15 1/2s=-3.15 s=-6.3 (1/2s)+(1/5x)+7 1/2(-6.3)+1/5(-19.25)+7 -3.15-3.85+7 0 In the end, it equals 0 because there were no values for x and s and since I started with just an equation with nothing on the other side, I used (by default) 0 on the other side.
How can you make 2s equal 5?
If 2s = 5 then s = 2.5. So by letting s equal 2.5 (two and one half) you will allow 2s to equal 5.
-v plus 5 plus 6v equals 1 plus 5v plus 3?
-v + 5 + 6v = 1 + 5v + 3Clean up and combine each side:(-v + 6v) + 5 = (1 + 3) + 5v5v + 5 = 4 + 5vSubtract 5v from each side:5 = 4.There is no solution.No value of 'v' can make [ 5 = 4 ].
How do you factor -20v plus 4v2 - 5y plus vy?
-20v + 4v2 - 5y + vy = 4v2 - 20v + vy - 5y = 4v*(v - 5) + y*(v - 5) = (4v + y)*(v - 5)
