Answer:
255 m
Step-by-step explanation:
From the given question, the surface of the water body is the reference point. And this point is assumed to be 0, so that any distance above it is positive, and any distance below is negative. This is synonymous to the number line system.
Thus,
For the aircraft, its height = +200 m
For the submarine, its depth = -55 m
So that the difference between the submarine and aircraft can be determined as;
200 - (-55)
= 200 + 55
= 255 m
The distance between the submarine and aircraft is 255 m.
Answer: whats the question??
Step-by-step explanation:
Triangles RDO and RHC are similar, so you have
.. RO:RC = RD:RH
.. RO:(RD +CD) = RD:(RO +OH)
.. 6:(4 +CD) = 4:(6 +4) . . . . . . . fill in numbers
.. 6*10 = 4(4 +CD) . . . . . . . . . . product of means = product of extremes
.. 15 = 4 +CD
.. CD = 11
P(x) = (x^2)(x - 4)^2(x + 4) + some constant(b)
2025 = (1^2)(1 - 4)^2(1 + 4) + b
2025 = 45 + b
b = 1980
Complete Equation:
p(x) = (x^2)(x - 4)^2(x +4) + 1980
or expanded form
p(x) = x^5 - 4x^4 - 16x^3 + 64x^2 + 1980
Answer:
![x=\frac{6\ln \left(2\right)}{2\ln \left(5\right)-\ln \left(2\right)}\\x \approx 1.65](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B6%5Cln%20%5Cleft%282%5Cright%29%7D%7B2%5Cln%20%5Cleft%285%5Cright%29-%5Cln%20%5Cleft%282%5Cright%29%7D%5C%5Cx%20%5Capprox%201.65)
Step-by-step explanation:
Given
![f(x) = 2^{\left(x+6\right)}](https://tex.z-dn.net/?f=f%28x%29%20%3D%202%5E%7B%5Cleft%28x%2B6%5Cright%29%7D)
![g(x) = 5^{2x}](https://tex.z-dn.net/?f=g%28x%29%20%3D%205%5E%7B2x%7D)
We want to find ![f(x) = g(x)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20g%28x%29)
For this you need to:
if
, then ![\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28f%5Cleft%28x%5Cright%29%5Cright%29%3D%5Cln%20%5Cleft%28g%5Cleft%28x%5Cright%29%5Cright%29)
![\ln \left(2^{x+6}\right)=\ln \left(5^{2x}\right)](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%282%5E%7Bx%2B6%7D%5Cright%29%3D%5Cln%20%5Cleft%285%5E%7B2x%7D%5Cright%29)
Apply log rule: ![\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)](https://tex.z-dn.net/?f=%5Clog%20_a%5Cleft%28x%5Eb%5Cright%29%3Db%5Ccdot%20%5Clog%20_a%5Cleft%28x%5Cright%29)
![\left(x+6\right)\ln \left(2\right)=2x\ln \left(5\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2B6%5Cright%29%5Cln%20%5Cleft%282%5Cright%29%3D2x%5Cln%20%5Cleft%285%5Cright%29)
Solve for x
Expand ![\left(x+6\right)\ln \left(2\right) = \ln \left(2\right)x+6\ln \left(2\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2B6%5Cright%29%5Cln%20%5Cleft%282%5Cright%29%20%3D%20%5Cln%20%5Cleft%282%5Cright%29x%2B6%5Cln%20%5Cleft%282%5Cright%29)
![\ln \left(2\right)x+6\ln \left(2\right)=2x\ln \left(5\right)](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%282%5Cright%29x%2B6%5Cln%20%5Cleft%282%5Cright%29%3D2x%5Cln%20%5Cleft%285%5Cright%29)
![\ln \left(2\right)x=2x\ln \left(5\right)-6\ln \left(2\right)\\\ln \left(2\right)x-2x\ln \left(5\right)=-6\ln \left(2\right)\\\left(\ln \left(2\right)-2\ln \left(5\right)\right)x=-6\ln \left(2\right)\\\\x=\frac{6\ln \left(2\right)}{2\ln \left(5\right)-\ln \left(2\right)}\approx 1.65](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%282%5Cright%29x%3D2x%5Cln%20%5Cleft%285%5Cright%29-6%5Cln%20%5Cleft%282%5Cright%29%5C%5C%5Cln%20%5Cleft%282%5Cright%29x-2x%5Cln%20%5Cleft%285%5Cright%29%3D-6%5Cln%20%5Cleft%282%5Cright%29%5C%5C%5Cleft%28%5Cln%20%5Cleft%282%5Cright%29-2%5Cln%20%5Cleft%285%5Cright%29%5Cright%29x%3D-6%5Cln%20%5Cleft%282%5Cright%29%5C%5C%5C%5Cx%3D%5Cfrac%7B6%5Cln%20%5Cleft%282%5Cright%29%7D%7B2%5Cln%20%5Cleft%285%5Cright%29-%5Cln%20%5Cleft%282%5Cright%29%7D%5Capprox%201.65)