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sergiy2304 [10]
3 years ago
13

Radio signals travel at a rate of 3 × 108 meters per second. How many seconds will it take for a radio signal to travel from a s

atellite to the surface of the Earth if the satellite is orbiting at a height of 3.6 × 107 meters?
Mathematics
2 answers:
lys-0071 [83]3 years ago
5 0

Answer:

1.1888 (8 goes on forever) seconds to travel from the satellite to Earth.

Step-by-step explanation:

It is traveling at 3 x 108 m/s and needs to travel 3.6 x 107 m. 3 x 108=324 and 3.6 x 107=385.2 so 324 m/s for 385.2 m. 385.2/325=1.1888.

stellarik [79]3 years ago
3 0

Answer:

1.2 x 10^-1 seconds

Step-by-step explanation:

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Harry took a loan from the bank.
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Answer:

The size of Harry's loan is $9000.

Step-by-step explanation:

D(t) models Harry's remaining debt, in dollars, as a function of time t, in months  that is given by :

D(t) =-200t+ 9000

We can see 200 is in negative that means it is getting deducted from the function. So, Harry must be paying this each month against his loan.

Lets put t = 0, that shows no payments have been made.

This will get the amount of loan, before any payments.

D(t)=-200(0)+9000

So,D(t) =9000

Hence, the size of Harry's loan is $9000.

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2logx=3-2log(x+3) solve for x​
AVprozaik [17]

Answer:

\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}

Step-by-step explanation:

2\log x=3-2\log(x+3)\\\\Domain:\ x>0\ \wedge\ x+3>0\to x>-3\\\\D:x>0\\============================\\2\log x=3-2\log(x+3)\qquad\text{add}\ 2\log(x+3)\ \text{to both sides}\\\\2\log x+2\log(x+3)=3\qquad\text{divide both sides by 2}\\\\\log x+\log(x+3)=\dfrac{3}{2}\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log\bigg(x(x+3)\bigg)=\dfrac{3}{2}\qquad\text{use the de}\text{finition of a logarithm}\\\\x(x+3)=10^\frac{3}{2}\qquad\text{use the distributive property}

x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D

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3 years ago
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