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3 years ago

Jennifer is 20 miles north of her house, and she is driving north on the highway at a rate of 55 miles per hour whats the slpoe

Mathematics

1 answer:

LenaWriter [7]3 years ago

3 0

Answer:

The slope is 55.

Step-by-step explanation:

Slope determines the direction and the rate of increase or decrease between the two points on a surface or line.

In the question, it is given initially she is 20 miles north of her house

⇒ When x = 0 the y-coordinate is 20

So, the y-intercept is 20

Now, it is given that the rate of change is 55 miles per hour

Hence, the slope is same as 55 because it is the actual rate of change between the positions of Jennifer with time.

Hence, The slope is 55.

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A student has 2 pounds of cereal. If the student eats 1/6 pound of cereal each day, how many days can the student eat cereal

harina [27]

Answer:

12 days

Step-by-step explanation:

1/6 per day

in 6 days- 1 pound

6x2=12 days=2 pounds

hope this helped :)

7 0

2 years ago

Read 2 more answers

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: