![\bf \textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x) = 6.4\implies \log_{10}(x)=6.4\implies 10^{6.4}=x\implies 2511886.43\approx x](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%20%3D%206.4%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D6.4%5Cimplies%2010%5E%7B6.4%7D%3Dx%5Cimplies%202511886.43%5Capprox%20x)
let's recall that when the base is omitted, "10" is implied.
Answer: y = 0
Note: of course you don't need to type in the "y=" part as its already taken care of
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The highest point is when y = 5
The lowest point is when y = -5
Find the midpoint of those y values. Add them up: 5 + (-5) = 0
Then divide that result by 2: 0/2 = 0
The midpoint of 5 and -5 is 0. So the midline is y = 0
This is the center horizontal line that runs through the middle of the sinusoidal curve
The vertical distance from the midline to either the peak or valley is the same (in this case, 5) which is known as the amplitude
Answer:
Jo is correct.
Step-by-step explanation:
No matter what number you multiply by, a fraction between 0 and 1 will give you less than the original whole number.
Answer:
25
Step-by-step explanation:
If you add 25 to both sides, you get
x^2-10x+25=26.
Now, you can simplify the left side into a perfect square
(x-5)^2=26
What exactly do you need to find? Does it need simplified?
