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kondaur [170]
2 years ago
14

A land surveyor places two stakes 500 ft apart and creates a perpendicular to the line that connects these two stakes. He needs

to place a third stake 100 ft away that is equidistant to the two original stakes. To apply the Perpendicular Bisector Theorem, the land surveyor would need to identify
A. a line parallel to the line connecting the two stakes

B. a line that is parallel to the perpendicular line

C. the midpoint along the line connecting the two stakes

D. the midpoint along any line that is perpendicular to the line connecting the two stakes
Mathematics
2 answers:
cupoosta [38]2 years ago
8 0

Answer:
We will choose the last option is correct.
Step-by-step explanation:
A land surveyor places two stakes 500 ft apart and locates the midpoint between the stakes.
From the midpoint, he needs to place another stake 100 ft away that is equidistant to the two original stakes.  
Therefore, to apply the Perpendicular Bisector Theorem, the land surveyor would have to identify a line that is "perpendicular to the line connecting the two stakes and going through the midpoint of the two stakes".
Therefore we will choose the last option is correct. (Answer)

viktelen [127]2 years ago
8 0

Answer:

D

Step-by-step explanation:

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Answer:

47.25 pounds

Step-by-step explanation:

\dfrac{dA}{dt}=R_{in}-R_{out}

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}

Change In Volume of the tank, \frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

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Therefore:

\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3

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e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2

Multiplying the DE by the integrating factor, we have:

(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}

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A(t)=(50+t)-75000(50+t)^{-2}

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A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds

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1 year ago
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devlian [24]

Answer:

c- shifted 3 units right and 4 units up

Step-by-step explanation:

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X=his fortune
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Rudiy27

Answer:

Step-by-step explanation:

price of 1 pen= $ 2

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