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hammer [34]
3 years ago
10

Robert stands atop a 1,380-foot hill. He climbs down, reaching the bottom of the hill in 30 minutes. What is Robert’s average ra

te of elevation change in feet per minute?
A.

+460

B.

+46

C.

-46
or
D.

-460
Mathematics
1 answer:
Zigmanuir [339]3 years ago
3 0

Answer:

-46 feet/min

Step-by-step explanation:

Initial elevation = 1,380 ft

Final Elevation = 0 feet

Change in elevation

= Final elevation - Initial Elevation

= 0 - 1,380

= - 1,380 feet

Given that the journey took 30 min,

rate of descent = Change in elevation / time

= -1,380 / 30

= - 46 feet/min

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Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
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Which inequality represents all the solutions of -5x + 5 ≥ 160 − 10x? A. x ≤ -33 B. x ≥ 33 C. x ≥ 31 D. x ≤ 31
ikadub [295]

Answer:

The answer is B.

Step-by-step explanation:

Solve for x.

-5x + 5 \geq 160 - 10x

Add 10x to both sides, and subtract 5 from both sides.

5x \geq 155

Now, divide by 5.

x \geq 31          is the answer

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3 years ago
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The reasons as well as the answer
Radda [10]

Answer:

x = 46, y = 67

Step-by-step explanation:

x = ∠ AEC = 46 ( alternate angles )

Since Δ ACE is isosceles then the base angles are congruent, then

y = \frac{180-46}{2} = \frac{134}{2} = 67

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3 years ago
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For (-1,2) you go to the left one and up two.
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What is 12.87+3.905? Show how to solve please.
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