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Q: What is the interval of 8 4 8 8 8 4 8 4 0 4 4 1 3 4 4 4 4 8 4 0 4 4 4 4 4 8 4 8 2 6 4 12 8 8 2 6 5 3 4 8 4 8?

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Any combination really. 1 + 1 + 10, or 12 + 0 + 0, or 1 + 2 + 9, etc.

12/6 = 2

The point (1, 13) lies on y = ax^2 + bx + 12 So the coordinates of the point satisfy the equation. 13 = a + b + 12 => a + b = 1 .. .. .. .. .. .. .. .. .. (1) The stationary point of the curve is reached when dy/dx = 0: that is 2ax + b = 0 at (1, 12) this gives 2a + b = 0 .. .. .. (2) Solving (1) and (2) gives a = -1, b = 2

1 1/2

They are 12, 11, 10, ..., 2, 1, 0, -1, -2, -3, ...

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13 23 12 1 2 0 123 13 23 12 1 2 0 23 12 1 2 0 12 1 2 0 1 2 0 23 12 1 2 0

0 13 12 1 0 12 2 0 2 12 0 1 12 13 0 0 12 0 0 0 0 12 0

There are no mixed numbers between 0 and 2 with an interval of 18.

13, 12, 2, 0, 13, 12, 2, 0, 1, 0, 2, 0

.12 = 0 + (1/10) + (2/100) 0.12 = (0 x 1) + (1/10) + (2/100)

8*12^0 + 6*12^1 + 0*12^2 + 2*12^3 8*1 + 6*12 + 0*144 + 2*1728 8 + 72 + 3456 3536 ■

-2x^2 - 8x + 1 = 0 x^2 + 4x - 1/2 = 0 x^2 + 4x = 1/2 x^2 + 4x + 4 = 9/2 (x + 2)^2 = 9/2 x + 2 = ±√(9/2) x = ±√[(9)(2)/(2)(2)] - 2 x = ±(3/2)√2 - 2 x = (3/2)√2 - 2 or x = -(3/2)√2 - 2 Check: -2x^2 - 8x + 1 = 0 -2[(3/2)√2 - 2]^2 - 8[(3/2)√2 - 2] + 1 =? 0 -2(9/2 - 6√2 + 4) - 12√2 + 16 + 1 =? 0 -9 + 12√2 - 8 - 12√2 + 16 + 1 =? 0 -17 + 17 =? 0 0 = 0 True -2x^2 - 8x + 1 = 0 -2[-(3/2)√2 - 2]^2 - 8[-(3/2)√2 - 2] + 1 =? 0 -2[-[(3/2)√2 + 2]]^2 - 8[-(3/2)√2 - 2] + 1 =? 0 -2(9/2 + 6√2 + 4) + 12√2 + 16 + 1 =? 0 -9 - 12√2 - 8 + 12√2 + 16 + 1 =? 0 -17 + 17 =? 0 0 = 0 True

It is 48/2= 24 r 0 ^ 24/2= 12 r 0 | 12/2 = 6 r 0 | 6/2 = 3 r 0 | 3/2 = 1 r 1 | 1/2 = 0 r 1 | So your read it from 1 to 48 :D Enjoy

1/5 is between 0 and 1/2

on 4 valve tuba F major scale- 4 12 2 0 4 12 2 0 12 2 0 1 0 2 0

If that's 0 and 12, halfway is 6/1 If that's 0 and 1/2, halfway is 1/4

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