Answer:
x=9
Step-by-step explanation:
2x+10=28
Subtract 10 from both sides
2x=18
Divide by 2 on both sides
x=9
![\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:
φ ≈ 1.19029 radians (≈ 68.2°)
Step-by-step explanation:
There are simple formulas for A and φ in this conversion, but it can be instructive to see how they are derived.
We want to compare ...
y(t) = Asin(ωt +φ)
to
y(t) = Psin(ωt) +Qcos(ωt)
Using trig identities to expand the first equation, we have ...
y(t) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)
Matching coefficients with the second equation, we have ...
P = Acos(φ)
Q = Asin(φ)
The ratio of these eliminates A and gives a relation for φ:
Q/P = sin(φ)/cos(φ)
Q/P = tan(φ)
φ = arctan(Q/P) . . . . taking quadrant into account
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We can also use our equations for P and Q to find A:
P² +Q² = (Acos(φ))² +(Asin(φ))² = A²(cos(φ)² +sin(φ)²) = A²
A = √(P² +Q²)
_____
Here, we want φ.
φ = arctan(Q/P) = arctan(5/2)
φ ≈ 1.19029 . . . radians
Answer:
Step-by-step explanation:
If 3⁄4 gallons of water come out of a hose in 15 minutes, to calculate the rate of water that comes out in an hour (60 minutes), we will use the expression;
3/4 gallons = 15 minutes
x gallons = 60minutes
cross multiply
15x = 3/4 * 60
15x = 180/4
15x = 45
x = 45/15
x = 3/1
x = 3:1
Hence the water rate in gallons per hour is 3 gallons per hour
The second one and the last one
