Answer:
15^3=15×15×15=3375
0^4=0×0×0×0=0
5^5=5×5×5×5×5=3125
3.5^2=3.5×3.5=12.25
6^5=6×6×6×6×6=7776
7^4=7×7×7×7=2401
1^8=1×1×1×1×1×1×1×1=1
2.5^3=2.5×2.5×2.5=15.625
Step-by-step explanation:
Answer: The best estimate for this is 52.
Answer:
987.76 Hz
Step-by-step explanation:
You have B4's frequency of 493.88 Hz and you want to know the frequency of the note one octave <u>above</u> B4.
When you go up an octave, you double the original frequency. When you go down an octave, you divide the original frequency by two.
So, since B4's frequency is about 493.88 Hz, an octave above would be 2 x 493.88 Hz or 987.76 Hz
If we were looking for an octave below B4, we would divide 493.88 Hz by 2 to get 246.94 Hz.
Answer:
All but A are repeating... did the question ask which is NOT a repeating decimal?
Step-by-step explanation:
Let
. Then
and
. By the intermediate value theorem, it follows that there is some
such that
![f(c)\in[f(-1),f(2)]=[-1,29]](https://tex.z-dn.net/?f=f%28c%29%5Cin%5Bf%28-1%29%2Cf%282%29%5D%3D%5B-1%2C29%5D)
.
This guarantees that there is some
between -1 and 2 such that
, i.e. there is some
such that
.
