how do I solve this y^0 – 8 and y^1 – 8 when y = -3?
回答 (4)
2008-03-24 5:54 pm
✔ 最佳答案
y^0 = 1 (anything to the 0 power equals one)so, y^0-8 = 1-8 = -7
y^1 = y (anything to the first power, stays the same)
so, y^1 - 8 = -3 - 8 = -11
that's it! ;)
2008-03-25 12:54 am
y^0 - 8
= -3^0 - 8
= 1 - 8
= -7
y^1 - 8
= -3^1 - 8
= -3 - 8
= -11
= -3^0 - 8
= 1 - 8
= -7
y^1 - 8
= -3^1 - 8
= -3 - 8
= -11
2008-03-25 12:53 am
A number raised to the zeroeth power is always equal to 1.
y⁰ – 8 = 1 - 8
= -7
A number raised to the first power doesn't change.
y¹ – 8 = (-3)¹ - 8
= -3 - 8
= -11
y⁰ – 8 = 1 - 8
= -7
A number raised to the first power doesn't change.
y¹ – 8 = (-3)¹ - 8
= -3 - 8
= -11
2008-03-25 12:56 am
Y^0=1
so 1-8= -7
Y^1-8
=(-3)^1-8
=(-11)
so 1-8= -7
Y^1-8
=(-3)^1-8
=(-11)
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