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geniusboy [140]

3 years ago

5

The function below can be used to calculate the cost of making a long distance call, f(m), which is based on a $2.50 initial cha

rge and a fee of $.12 per minute, m, that the call lasts.
f(m)=2.5+0.12m

If Natalie paid $6.82 for one call, how many minutes long was it?

1 answer:

Tpy6a [65]3 years ago

6 0

Answer:

she was on a call for 36 minutes

Step-by-step explanation: 6.82-2.5= 4.32 4.32/.12= 36

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3 years ago

Suppose a laboratory has a 26-gram sample of polonium-210.The half-life of polonium-210 is about days. a.How many half-lives of

vovangra [49]

Answer:

a) Two half lives, b) $m(276) = 6.526\,g$

Step-by-step explanation:

a) The polonium-210 has a half life of 138.4 days. Therefore, 1.994 half lives have past.

b) Mass decay is described by the following exponential model:

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

The time constant for the isotope is:

$\tau = \frac{138.4\,days}{\ln 2}$

$\tau = 199.669\,days$

The mass of the isotope after 276 days is:

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3 years ago

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denis23 [38]

Answer:

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2 years ago

The nth term of an AP is given by 2n +9 find the first term

den301095 [7]

The first term of an AP can be defined as T1 i.e. n=1. According to the given equation,. T1=a=2(1)+9=11.

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3 years ago

y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the

vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

$y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},$

where $c_1, c_2 \in \mathbb{R}$ . We now want to find a solution $y$ such that $y(-1)=3$ and $y'(-1)=-3$ . Therefore, all we need to do is find the constants $c_1$ and $c_2$ that satisfy the initial conditions. For the first condition, we have: $y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.$

For the second condition, we need to find the derivative $y'$ first. In this case, we have:

$y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.$

Therefore:

$y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.$

This means that we must solve the following system of equations:

$\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.$

If we add the equations above, we get:

$\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.$

If we now substitute $c_1 = 0$ into either of the equations in the system, we get:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

$\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.$

Indeed, we can see that:

$y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3$

$y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,$

which do correspond to the desired initial conditions.

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3 years ago